3.1259 \(\int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=266 \[ \frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d}-\frac {2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^5 d}+\frac {2 a x \left (2 a^2-3 b^2\right )}{b^5}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {a \sin (c+d x) \cos (c+d x)}{b^3 d}+\frac {a x}{b^3}-\frac {\cos ^3(c+d x)}{3 b^2 d}+\frac {\cos (c+d x)}{b^2 d} \]

[Out]

a*x/b^3+2*a*(2*a^2-3*b^2)*x/b^5+2*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^5/d-2*(a^
2-b^2)^(3/2)*(5*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/b^5/d-arctanh(cos(d*x+c))/a^2/d+
cos(d*x+c)/b^2/d+3*(a^2-b^2)*cos(d*x+c)/b^4/d-1/3*cos(d*x+c)^3/b^2/d-a*cos(d*x+c)*sin(d*x+c)/b^3/d+(a^2-b^2)^2
*cos(d*x+c)/a/b^4/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.35, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {2897, 3770, 2638, 2635, 8, 2633, 2664, 12, 2660, 618, 204} \[ \frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}+\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d}-\frac {2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^5 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {2 a x \left (2 a^2-3 b^2\right )}{b^5}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {a \sin (c+d x) \cos (c+d x)}{b^3 d}+\frac {a x}{b^3}-\frac {\cos ^3(c+d x)}{3 b^2 d}+\frac {\cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

(a*x)/b^3 + (2*a*(2*a^2 - 3*b^2)*x)/b^5 + (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]
])/(b^5*d) - (2*(a^2 - b^2)^(3/2)*(5*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*b^5*d)
- ArcTanh[Cos[c + d*x]]/(a^2*d) + Cos[c + d*x]/(b^2*d) + (3*(a^2 - b^2)*Cos[c + d*x])/(b^4*d) - Cos[c + d*x]^3
/(3*b^2*d) - (a*Cos[c + d*x]*Sin[c + d*x])/(b^3*d) + ((a^2 - b^2)^2*Cos[c + d*x])/(a*b^4*d*(a + b*Sin[c + d*x]
))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (-\frac {2 \left (-2 a^3+3 a b^2\right )}{b^5}+\frac {\csc (c+d x)}{a^2}+\frac {3 \left (-a^2+b^2\right ) \sin (c+d x)}{b^4}+\frac {2 a \sin ^2(c+d x)}{b^3}-\frac {\sin ^3(c+d x)}{b^2}+\frac {\left (a^2-b^2\right )^3}{a b^5 (a+b \sin (c+d x))^2}-\frac {\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )}{a^2 b^5 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {2 a \left (2 a^2-3 b^2\right ) x}{b^5}+\frac {\int \csc (c+d x) \, dx}{a^2}+\frac {(2 a) \int \sin ^2(c+d x) \, dx}{b^3}-\frac {\int \sin ^3(c+d x) \, dx}{b^2}-\frac {\left (3 \left (a^2-b^2\right )\right ) \int \sin (c+d x) \, dx}{b^4}+\frac {\left (a^2-b^2\right )^3 \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{a b^5}-\frac {\left (\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^2 b^5}\\ &=\frac {2 a \left (2 a^2-3 b^2\right ) x}{b^5}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}-\frac {a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {a \int 1 \, dx}{b^3}+\frac {\left (a^2-b^2\right )^2 \int \frac {a}{a+b \sin (c+d x)} \, dx}{a b^5}+\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{b^2 d}-\frac {\left (2 \left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 b^5 d}\\ &=\frac {a x}{b^3}+\frac {2 a \left (2 a^2-3 b^2\right ) x}{b^5}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{b^2 d}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}-\frac {\cos ^3(c+d x)}{3 b^2 d}-\frac {a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^5}+\frac {\left (4 \left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 b^5 d}\\ &=\frac {a x}{b^3}+\frac {2 a \left (2 a^2-3 b^2\right ) x}{b^5}-\frac {2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^5 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{b^2 d}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}-\frac {\cos ^3(c+d x)}{3 b^2 d}-\frac {a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=\frac {a x}{b^3}+\frac {2 a \left (2 a^2-3 b^2\right ) x}{b^5}-\frac {2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^5 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{b^2 d}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}-\frac {\cos ^3(c+d x)}{3 b^2 d}-\frac {a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}-\frac {\left (4 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=\frac {a x}{b^3}+\frac {2 a \left (2 a^2-3 b^2\right ) x}{b^5}+\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 d}-\frac {2 \left (a^2-b^2\right )^{3/2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^5 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{b^2 d}+\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{b^4 d}-\frac {\cos ^3(c+d x)}{3 b^2 d}-\frac {a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.74, size = 207, normalized size = 0.78 \[ \frac {\frac {12 a \left (4 a^2-5 b^2\right ) (c+d x)}{b^5}-\frac {24 \left (4 a^2+b^2\right ) \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^5}+\frac {9 \left (4 a^2-3 b^2\right ) \cos (c+d x)}{b^4}+\frac {12 \left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 (a+b \sin (c+d x))}+\frac {12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}-\frac {12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}-\frac {6 a \sin (2 (c+d x))}{b^3}-\frac {\cos (3 (c+d x))}{b^2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((12*a*(4*a^2 - 5*b^2)*(c + d*x))/b^5 - (24*(a^2 - b^2)^(3/2)*(4*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sq
rt[a^2 - b^2]])/(a^2*b^5) + (9*(4*a^2 - 3*b^2)*Cos[c + d*x])/b^4 - Cos[3*(c + d*x)]/b^2 - (12*Log[Cos[(c + d*x
)/2]])/a^2 + (12*Log[Sin[(c + d*x)/2]])/a^2 + (12*(a^2 - b^2)^2*Cos[c + d*x])/(a*b^4*(a + b*Sin[c + d*x])) - (
6*a*Sin[2*(c + d*x)])/b^3)/(12*d)

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fricas [A]  time = 1.27, size = 698, normalized size = 2.62 \[ \left [\frac {4 \, a^{3} b^{3} \cos \left (d x + c\right )^{3} + 6 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2}\right )} d x - 3 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} - a b^{4} + {\left (4 \, a^{4} b - 3 \, a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 6 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - 3 \, {\left (b^{6} \sin \left (d x + c\right ) + a b^{5}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (b^{6} \sin \left (d x + c\right ) + a b^{5}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (a^{2} b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3}\right )} d x - 6 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{6} d \sin \left (d x + c\right ) + a^{3} b^{5} d\right )}}, \frac {4 \, a^{3} b^{3} \cos \left (d x + c\right )^{3} + 6 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2}\right )} d x + 6 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} - a b^{4} + {\left (4 \, a^{4} b - 3 \, a^{2} b^{3} - b^{5}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 6 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - 3 \, {\left (b^{6} \sin \left (d x + c\right ) + a b^{5}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (b^{6} \sin \left (d x + c\right ) + a b^{5}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (a^{2} b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3}\right )} d x - 6 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{6} d \sin \left (d x + c\right ) + a^{3} b^{5} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/6*(4*a^3*b^3*cos(d*x + c)^3 + 6*(4*a^6 - 5*a^4*b^2)*d*x - 3*(4*a^5 - 3*a^3*b^2 - a*b^4 + (4*a^4*b - 3*a^2*b
^3 - b^5)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 -
2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) -
a^2 - b^2)) + 6*(4*a^5*b - 5*a^3*b^3 + a*b^5)*cos(d*x + c) - 3*(b^6*sin(d*x + c) + a*b^5)*log(1/2*cos(d*x + c
) + 1/2) + 3*(b^6*sin(d*x + c) + a*b^5)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^2*b^4*cos(d*x + c)^3 - 3*(4*a^5*b
- 5*a^3*b^3)*d*x - 6*(a^4*b^2 - a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(a^2*b^6*d*sin(d*x + c) + a^3*b^5*d), 1/6
*(4*a^3*b^3*cos(d*x + c)^3 + 6*(4*a^6 - 5*a^4*b^2)*d*x + 6*(4*a^5 - 3*a^3*b^2 - a*b^4 + (4*a^4*b - 3*a^2*b^3 -
b^5)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 6*(4*a^5*b
- 5*a^3*b^3 + a*b^5)*cos(d*x + c) - 3*(b^6*sin(d*x + c) + a*b^5)*log(1/2*cos(d*x + c) + 1/2) + 3*(b^6*sin(d*x
+ c) + a*b^5)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^2*b^4*cos(d*x + c)^3 - 3*(4*a^5*b - 5*a^3*b^3)*d*x - 6*(a^4*
b^2 - a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(a^2*b^6*d*sin(d*x + c) + a^3*b^5*d)]

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giac [A]  time = 0.26, size = 353, normalized size = 1.33 \[ \frac {\frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {3 \, {\left (4 \, a^{3} - 5 \, a b^{2}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {6 \, {\left (4 \, a^{6} - 7 \, a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2} b^{5}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 9 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} - 7 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}} + \frac {6 \, {\left (a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{2} b^{4}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(3*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*(4*a^3 - 5*a*b^2)*(d*x + c)/b^5 - 6*(4*a^6 - 7*a^4*b^2 + 2*a^2*b
^4 + b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sq
rt(a^2 - b^2)*a^2*b^5) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^4 - 9*b^2*tan(1/2*d*x +
1/2*c)^4 + 18*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 9*a^2
- 7*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4) + 6*(a^4*b*tan(1/2*d*x + 1/2*c) - 2*a^2*b^3*tan(1/2*d*x + 1/2*c)
+ b^5*tan(1/2*d*x + 1/2*c) + a^5 - 2*a^3*b^2 + a*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) +
a)*a^2*b^4))/d

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maple [B]  time = 0.70, size = 778, normalized size = 2.92 \[ \frac {2 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {12 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 a^{2}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {14}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{5}}-\frac {10 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 a^{3}}{d \,b^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {4 a}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2}{d a \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {8 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{5} \sqrt {a^{2}-b^{2}}}+\frac {14 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{3} \sqrt {a^{2}-b^{2}}}-\frac {4 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d b \sqrt {a^{2}-b^{2}}}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)/(a+b*sin(d*x+c))^2,x)

[Out]

2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c
)^4*a^2-6/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+12/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*
x+1/2*c)^2*a^2-8/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a*ta
n(1/2*d*x+1/2*c)+6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2-14/3/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3+8/d/b^5*arctan(t
an(1/2*d*x+1/2*c))*a^3-10/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a+1/d/a^2*ln(tan(1/2*d*x+1/2*c))+2/d*a^2/b^3/(tan(1
/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)-4/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2
*c)*b+a)*tan(1/2*d*x+1/2*c)+2/d/a^2*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)+2/d
*a^3/b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-4/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c
)*b+a)*a+2/d/a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)-8/d*a^4/b^5/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*t
an(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+14/d*a^2/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(
a^2-b^2)^(1/2))-4/d/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-2/d/a^2*b/(a^2-
b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 13.60, size = 5197, normalized size = 19.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)*(a + b*sin(c + d*x))^2),x)

[Out]

log(tan(c/2 + (d*x)/2))/(a^2*d) + ((2*(12*a^4 + 3*b^4 - 13*a^2*b^2))/(3*a*b^4) + (2*tan(c/2 + (d*x)/2)*(18*a^4
+ 3*b^4 - 20*a^2*b^2))/(3*a^2*b^3) + (2*tan(c/2 + (d*x)/2)^7*(2*a^4 + b^4 - 2*a^2*b^2))/(a^2*b^3) + (2*tan(c/
2 + (d*x)/2)^6*(4*a^4 + b^4 - 3*a^2*b^2))/(a*b^4) + (2*tan(c/2 + (d*x)/2)^5*(10*a^4 + 3*b^4 - 12*a^2*b^2))/(a^
2*b^3) + (2*tan(c/2 + (d*x)/2)^4*(12*a^4 + 3*b^4 - 13*a^2*b^2))/(a*b^4) + (2*tan(c/2 + (d*x)/2)^3*(14*a^4 + 3*
b^4 - 14*a^2*b^2))/(a^2*b^3) + (2*tan(c/2 + (d*x)/2)^2*(36*a^4 + 9*b^4 - 43*a^2*b^2))/(3*a*b^4))/(d*(a + 2*b*t
an(c/2 + (d*x)/2) + 4*a*tan(c/2 + (d*x)/2)^2 + 6*a*tan(c/2 + (d*x)/2)^4 + 4*a*tan(c/2 + (d*x)/2)^6 + a*tan(c/2
+ (d*x)/2)^8 + 6*b*tan(c/2 + (d*x)/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 + 2*b*tan(c/2 + (d*x)/2)^7)) + (2*a*atan((
(a*(4*a^2 - 5*b^2)*((a*(4*a^2 - 5*b^2)*((32*(4*b^16 + 7*a^2*b^14 - 25*a^4*b^12 + 114*a^6*b^10 - 235*a^8*b^8 +
184*a^10*b^6 - 48*a^12*b^4))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(20*a^4*b^18 - 19*a^2*b^20 + 610*a^6*b^16 - 1
596*a^8*b^14 + 1434*a^10*b^12 - 480*a^12*b^10 + 32*a^14*b^8))/(a^3*b^16) + (a*(4*a^2 - 5*b^2)*((32*(8*a^2*b^16
+ 4*a^4*b^14 - 25*a^6*b^12 + 14*a^8*b^10))/(a^2*b^11) + (a*((32*(4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*t
an(c/2 + (d*x)/2)*(16*a^4*b^22 - 17*a^6*b^20 + 2*a^8*b^18))/(a^3*b^16))*(4*a^2 - 5*b^2)*1i)/b^5 + (32*tan(c/2
+ (d*x)/2)*(16*a^2*b^22 + 3*a^4*b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^10*b^14))/(a^3*b^16))*1i)/b^5)*1i)/b^5
- (32*(5*a^2*b^12 - 224*a^14 - 184*a^4*b^10 + 154*a^6*b^8 + 722*a^8*b^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2
*b^11) + (32*tan(c/2 + (d*x)/2)*(b^20 + 4*a^2*b^18 + 390*a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b
^10 - 3792*a^12*b^8 + 1600*a^14*b^6 - 256*a^16*b^4))/(a^3*b^16)))/b^5 - (a*(4*a^2 - 5*b^2)*((32*(5*a^2*b^12 -
224*a^14 - 184*a^4*b^10 + 154*a^6*b^8 + 722*a^8*b^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) + (a*(4*a^2 -
5*b^2)*((32*(4*b^16 + 7*a^2*b^14 - 25*a^4*b^12 + 114*a^6*b^10 - 235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*b^4))/(a^
2*b^11) + (32*tan(c/2 + (d*x)/2)*(20*a^4*b^18 - 19*a^2*b^20 + 610*a^6*b^16 - 1596*a^8*b^14 + 1434*a^10*b^12 -
480*a^12*b^10 + 32*a^14*b^8))/(a^3*b^16) - (a*(4*a^2 - 5*b^2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*b^12 + 14
*a^8*b^10))/(a^2*b^11) - (a*((32*(4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^22 -
17*a^6*b^20 + 2*a^8*b^18))/(a^3*b^16))*(4*a^2 - 5*b^2)*1i)/b^5 + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^22 + 3*a^4*
b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^10*b^14))/(a^3*b^16))*1i)/b^5)*1i)/b^5 - (32*tan(c/2 + (d*x)/2)*(b^20
+ 4*a^2*b^18 + 390*a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1600*a^14*b^6 -
256*a^16*b^4))/(a^3*b^16)))/b^5)/((64*(224*a^12 - 5*b^12 + 84*a^2*b^10 + 81*a^4*b^8 - 906*a^6*b^6 + 1482*a^8*b
^4 - 960*a^10*b^2))/(a^2*b^11) - (64*tan(c/2 + (d*x)/2)*(2048*a^18 + 500*a^4*b^14 - 1200*a^6*b^12 - 4540*a^8*b
^10 + 21944*a^10*b^8 - 36160*a^12*b^6 + 29696*a^14*b^4 - 12288*a^16*b^2))/(a^3*b^16) + (a*(4*a^2 - 5*b^2)*((a*
(4*a^2 - 5*b^2)*((32*(4*b^16 + 7*a^2*b^14 - 25*a^4*b^12 + 114*a^6*b^10 - 235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*
b^4))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(20*a^4*b^18 - 19*a^2*b^20 + 610*a^6*b^16 - 1596*a^8*b^14 + 1434*a^1
0*b^12 - 480*a^12*b^10 + 32*a^14*b^8))/(a^3*b^16) + (a*(4*a^2 - 5*b^2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*
b^12 + 14*a^8*b^10))/(a^2*b^11) + (a*((32*(4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a
^4*b^22 - 17*a^6*b^20 + 2*a^8*b^18))/(a^3*b^16))*(4*a^2 - 5*b^2)*1i)/b^5 + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^22
+ 3*a^4*b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^10*b^14))/(a^3*b^16))*1i)/b^5)*1i)/b^5 - (32*(5*a^2*b^12 - 22
4*a^14 - 184*a^4*b^10 + 154*a^6*b^8 + 722*a^8*b^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) + (32*tan(c/2 +
(d*x)/2)*(b^20 + 4*a^2*b^18 + 390*a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1
600*a^14*b^6 - 256*a^16*b^4))/(a^3*b^16))*1i)/b^5 + (a*(4*a^2 - 5*b^2)*((32*(5*a^2*b^12 - 224*a^14 - 184*a^4*b
^10 + 154*a^6*b^8 + 722*a^8*b^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) + (a*(4*a^2 - 5*b^2)*((32*(4*b^16
+ 7*a^2*b^14 - 25*a^4*b^12 + 114*a^6*b^10 - 235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*b^4))/(a^2*b^11) + (32*tan(c/
2 + (d*x)/2)*(20*a^4*b^18 - 19*a^2*b^20 + 610*a^6*b^16 - 1596*a^8*b^14 + 1434*a^10*b^12 - 480*a^12*b^10 + 32*a
^14*b^8))/(a^3*b^16) - (a*(4*a^2 - 5*b^2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*b^12 + 14*a^8*b^10))/(a^2*b^1
1) - (a*((32*(4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^22 - 17*a^6*b^20 + 2*a^8
*b^18))/(a^3*b^16))*(4*a^2 - 5*b^2)*1i)/b^5 + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^22 + 3*a^4*b^20 - 62*a^6*b^18 +
60*a^8*b^16 - 16*a^10*b^14))/(a^3*b^16))*1i)/b^5)*1i)/b^5 - (32*tan(c/2 + (d*x)/2)*(b^20 + 4*a^2*b^18 + 390*a
^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1600*a^14*b^6 - 256*a^16*b^4))/(a^3*
b^16))*1i)/b^5))*(4*a^2 - 5*b^2))/(b^5*d) + (atan((((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 +
(d*x)/2)*(b^20 + 4*a^2*b^18 + 390*a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1
600*a^14*b^6 - 256*a^16*b^4))/(a^3*b^16) - (32*(5*a^2*b^12 - 224*a^14 - 184*a^4*b^10 + 154*a^6*b^8 + 722*a^8*b
^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*b^16 + 7*
a^2*b^14 - 25*a^4*b^12 + 114*a^6*b^10 - 235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*b^4))/(a^2*b^11) + (32*tan(c/2 +
(d*x)/2)*(20*a^4*b^18 - 19*a^2*b^20 + 610*a^6*b^16 - 1596*a^8*b^14 + 1434*a^10*b^12 - 480*a^12*b^10 + 32*a^14*
b^8))/(a^3*b^16) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*b^12 + 1
4*a^8*b^10))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^22 + 3*a^4*b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^
10*b^14))/(a^3*b^16) + (((32*(4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^22 - 17*
a^6*b^20 + 2*a^8*b^18))/(a^3*b^16))*(4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^2*b^5)))/(a^2*b^5)))/(a^2*b
^5))*1i)/(a^2*b^5) - ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(5*a^2*b^12 - 224*a^14 - 184*a^4*b^10 +
154*a^6*b^8 + 722*a^8*b^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) - (32*tan(c/2 + (d*x)/2)*(b^20 + 4*a^2*b
^18 + 390*a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1600*a^14*b^6 - 256*a^16*
b^4))/(a^3*b^16) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*b^16 + 7*a^2*b^14 - 25*a^4*b^12 + 114*a
^6*b^10 - 235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*b^4))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(20*a^4*b^18 - 19*a^2
*b^20 + 610*a^6*b^16 - 1596*a^8*b^14 + 1434*a^10*b^12 - 480*a^12*b^10 + 32*a^14*b^8))/(a^3*b^16) - ((4*a^2 + b
^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*b^12 + 14*a^8*b^10))/(a^2*b^11) + (32*
tan(c/2 + (d*x)/2)*(16*a^2*b^22 + 3*a^4*b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^10*b^14))/(a^3*b^16) - (((32*(
4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^22 - 17*a^6*b^20 + 2*a^8*b^18))/(a^3*b
^16))*(4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^2*b^5)))/(a^2*b^5)))/(a^2*b^5))*1i)/(a^2*b^5))/((64*(224*
a^12 - 5*b^12 + 84*a^2*b^10 + 81*a^4*b^8 - 906*a^6*b^6 + 1482*a^8*b^4 - 960*a^10*b^2))/(a^2*b^11) - (64*tan(c/
2 + (d*x)/2)*(2048*a^18 + 500*a^4*b^14 - 1200*a^6*b^12 - 4540*a^8*b^10 + 21944*a^10*b^8 - 36160*a^12*b^6 + 296
96*a^14*b^4 - 12288*a^16*b^2))/(a^3*b^16) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2
)*(b^20 + 4*a^2*b^18 + 390*a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1600*a^1
4*b^6 - 256*a^16*b^4))/(a^3*b^16) - (32*(5*a^2*b^12 - 224*a^14 - 184*a^4*b^10 + 154*a^6*b^8 + 722*a^8*b^6 - 14
34*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*b^16 + 7*a^2*b^1
4 - 25*a^4*b^12 + 114*a^6*b^10 - 235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*b^4))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2
)*(20*a^4*b^18 - 19*a^2*b^20 + 610*a^6*b^16 - 1596*a^8*b^14 + 1434*a^10*b^12 - 480*a^12*b^10 + 32*a^14*b^8))/(
a^3*b^16) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*b^12 + 14*a^8*b
^10))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^2*b^22 + 3*a^4*b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^10*b^14
))/(a^3*b^16) + (((32*(4*a^4*b^16 - 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^22 - 17*a^6*b^2
0 + 2*a^8*b^18))/(a^3*b^16))*(4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^2*b^5)))/(a^2*b^5)))/(a^2*b^5)))/(
a^2*b^5) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(5*a^2*b^12 - 224*a^14 - 184*a^4*b^10 + 154*a^6*b^
8 + 722*a^8*b^6 - 1434*a^10*b^4 + 960*a^12*b^2))/(a^2*b^11) - (32*tan(c/2 + (d*x)/2)*(b^20 + 4*a^2*b^18 + 390*
a^4*b^16 - 560*a^6*b^14 - 1414*a^8*b^12 + 4028*a^10*b^10 - 3792*a^12*b^8 + 1600*a^14*b^6 - 256*a^16*b^4))/(a^3
*b^16) + ((4*a^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*b^16 + 7*a^2*b^14 - 25*a^4*b^12 + 114*a^6*b^10 -
235*a^8*b^8 + 184*a^10*b^6 - 48*a^12*b^4))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(20*a^4*b^18 - 19*a^2*b^20 + 61
0*a^6*b^16 - 1596*a^8*b^14 + 1434*a^10*b^12 - 480*a^12*b^10 + 32*a^14*b^8))/(a^3*b^16) - ((4*a^2 + b^2)*(-(a +
b)^3*(a - b)^3)^(1/2)*((32*(8*a^2*b^16 + 4*a^4*b^14 - 25*a^6*b^12 + 14*a^8*b^10))/(a^2*b^11) + (32*tan(c/2 +
(d*x)/2)*(16*a^2*b^22 + 3*a^4*b^20 - 62*a^6*b^18 + 60*a^8*b^16 - 16*a^10*b^14))/(a^3*b^16) - (((32*(4*a^4*b^16
- 3*a^6*b^14))/(a^2*b^11) + (32*tan(c/2 + (d*x)/2)*(16*a^4*b^22 - 17*a^6*b^20 + 2*a^8*b^18))/(a^3*b^16))*(4*a
^2 + b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(a^2*b^5)))/(a^2*b^5)))/(a^2*b^5)))/(a^2*b^5)))*(4*a^2 + b^2)*(-(a + b
)^3*(a - b)^3)^(1/2)*2i)/(a^2*b^5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{6}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**6*csc(c + d*x)/(a + b*sin(c + d*x))**2, x)

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