3.1470 \(\int \frac {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=295 \[ \frac {2 b \cot (c+d x)}{a^3 d}+\frac {2 b^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {2 b^5 \left (5 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}+\frac {b^6 \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

[Out]

2*b^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(5/2)/d+2*b^5*(5*a^2-3*b^2)*arctan((b+a*t
an(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(5/2)/d-1/2*arctanh(cos(d*x+c))/a^2/d-(a^2+3*b^2)*arctanh(co
s(d*x+c))/a^4/d+2*b*cot(d*x+c)/a^3/d-1/2*cot(d*x+c)*csc(d*x+c)/a^2/d+1/2*cos(d*x+c)/(a+b)^2/d/(1-sin(d*x+c))+1
/2*cos(d*x+c)/(a-b)^2/d/(1+sin(d*x+c))+b^6*cos(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.39, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2897, 3770, 3767, 8, 3768, 2648, 2664, 12, 2660, 618, 204} \[ \frac {2 b^5 \left (5 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}+\frac {2 b^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac {b^6 \cos (c+d x)}{a^3 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(5/2)*d) + (2*b^5*(5*a^2 - 3*b^2)*Ar
cTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(5/2)*d) - ArcTanh[Cos[c + d*x]]/(2*a^2*d) -
((a^2 + 3*b^2)*ArcTanh[Cos[c + d*x]])/(a^4*d) + (2*b*Cot[c + d*x])/(a^3*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^
2*d) + Cos[c + d*x]/(2*(a + b)^2*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) + (b^
6*Cos[c + d*x])/(a^3*(a^2 - b^2)^2*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 2648

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

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 3767

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

Rule 3768

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

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 {\csc ^3(c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (\frac {\left (a^2+3 b^2\right ) \csc (c+d x)}{a^4}-\frac {2 b \csc ^2(c+d x)}{a^3}+\frac {\csc ^3(c+d x)}{a^2}-\frac {1}{2 (a+b)^2 (-1+\sin (c+d x))}-\frac {1}{2 (a-b)^2 (1+\sin (c+d x))}+\frac {b^5}{a^3 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}+\frac {b^5 \left (5 a^2-3 b^2\right )}{a^4 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {\int \csc ^3(c+d x) \, dx}{a^2}-\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{2 (a-b)^2}-\frac {(2 b) \int \csc ^2(c+d x) \, dx}{a^3}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^2}+\frac {\left (b^5 \left (5 a^2-3 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )^2}+\frac {b^5 \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{a^3 \left (a^2-b^2\right )}+\frac {\left (a^2+3 b^2\right ) \int \csc (c+d x) \, dx}{a^4}\\ &=-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^6 \cos (c+d x)}{a^3 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {b^5 \int \frac {a}{a+b \sin (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^2}+\frac {(2 b) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}+\frac {\left (2 b^5 \left (5 a^2-3 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^4 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^6 \cos (c+d x)}{a^3 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {b^5 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^2}-\frac {\left (4 b^5 \left (5 a^2-3 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^4 \left (a^2-b^2\right )^2 d}\\ &=\frac {2 b^5 \left (5 a^2-3 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^4 \left (a^2-b^2\right )^{5/2} d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^6 \cos (c+d x)}{a^3 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\left (2 b^5\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 \left (a^2-b^2\right )^2 d}\\ &=\frac {2 b^5 \left (5 a^2-3 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^4 \left (a^2-b^2\right )^{5/2} d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^6 \cos (c+d x)}{a^3 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\left (4 b^5\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 \left (a^2-b^2\right )^2 d}\\ &=\frac {2 b^5 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}+\frac {2 b^5 \left (5 a^2-3 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^4 \left (a^2-b^2\right )^{5/2} d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {\left (a^2+3 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {2 b \cot (c+d x)}{a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {b^6 \cos (c+d x)}{a^3 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.48, size = 356, normalized size = 1.21 \[ \frac {b^6 \cos (c+d x)}{a^3 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-\frac {b \tan \left (\frac {1}{2} (c+d x)\right )}{a^3 d}+\frac {b \cot \left (\frac {1}{2} (c+d x)\right )}{a^3 d}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {3 \left (a^2+2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}-\frac {3 \left (a^2+2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {6 b^5 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a \sin \left (\frac {1}{2} (c+d x)\right )+b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 d \left (a^2-b^2\right )^{5/2}}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{d (a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{d (a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(6*b^5*(2*a^2 - b^2)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^
4*(a^2 - b^2)^(5/2)*d) + (b*Cot[(c + d*x)/2])/(a^3*d) - Csc[(c + d*x)/2]^2/(8*a^2*d) - (3*(a^2 + 2*b^2)*Log[Co
s[(c + d*x)/2]])/(2*a^4*d) + (3*(a^2 + 2*b^2)*Log[Sin[(c + d*x)/2]])/(2*a^4*d) + Sec[(c + d*x)/2]^2/(8*a^2*d)
+ Sin[(c + d*x)/2]/((a + b)^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - Sin[(c + d*x)/2]/((a - b)^2*d*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])) + (b^6*Cos[c + d*x])/(a^3*(a - b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])) - (b*Tan
[(c + d*x)/2])/(a^3*d)

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fricas [B]  time = 2.30, size = 1844, normalized size = 6.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/4*(4*a^9 - 8*a^7*b^2 + 4*a^5*b^4 - 4*(4*a^7*b^2 - 8*a^5*b^4 + 7*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^4 - 6*(a^9
- 5*a^7*b^2 + 7*a^5*b^4 - 5*a^3*b^6 + 2*a*b^8)*cos(d*x + c)^2 - 6*((2*a^3*b^5 - a*b^7)*cos(d*x + c)^3 - (2*a^
3*b^5 - a*b^7)*cos(d*x + c) + ((2*a^2*b^6 - b^8)*cos(d*x + c)^3 - (2*a^2*b^6 - b^8)*cos(d*x + c))*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)) + 3*((a^9
- a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^
8)*cos(d*x + c) + ((a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4
*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((a^9 - a^7*b^2 - 3*a^5*
b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c) + (
(a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 -
2*b^9)*cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^8*b - 4*a^6*b^3 + 2*a^4*b^5 - (5*a^8*
b - 13*a^6*b^3 + 11*a^4*b^5 - 3*a^2*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^
6)*d*cos(d*x + c)^3 - (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c) + ((a^10*b - 3*a^8*b^3 + 3*a^6*b
^5 - a^4*b^7)*d*cos(d*x + c)^3 - (a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c))*sin(d*x + c)), -1/
4*(4*a^9 - 8*a^7*b^2 + 4*a^5*b^4 - 4*(4*a^7*b^2 - 8*a^5*b^4 + 7*a^3*b^6 - 3*a*b^8)*cos(d*x + c)^4 - 6*(a^9 - 5
*a^7*b^2 + 7*a^5*b^4 - 5*a^3*b^6 + 2*a*b^8)*cos(d*x + c)^2 + 12*((2*a^3*b^5 - a*b^7)*cos(d*x + c)^3 - (2*a^3*b
^5 - a*b^7)*cos(d*x + c) + ((2*a^2*b^6 - b^8)*cos(d*x + c)^3 - (2*a^2*b^6 - b^8)*cos(d*x + c))*sin(d*x + c))*s
qrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 3*((a^9 - a^7*b^2 - 3*a^5*b^4 +
5*a^3*b^6 - 2*a*b^8)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c) + ((a^8*b
- a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)
*cos(d*x + c))*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8
)*cos(d*x + c)^3 - (a^9 - a^7*b^2 - 3*a^5*b^4 + 5*a^3*b^6 - 2*a*b^8)*cos(d*x + c) + ((a^8*b - a^6*b^3 - 3*a^4*
b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c)^3 - (a^8*b - a^6*b^3 - 3*a^4*b^5 + 5*a^2*b^7 - 2*b^9)*cos(d*x + c))*sin(
d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^8*b - 4*a^6*b^3 + 2*a^4*b^5 - (5*a^8*b - 13*a^6*b^3 + 11*a^4*b
^5 - 3*a^2*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^3 - (a^
11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c) + ((a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x +
c)^3 - (a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cos(d*x + c))*sin(d*x + c))]

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giac [A]  time = 0.26, size = 423, normalized size = 1.43 \[ \frac {\frac {48 \, {\left (2 \, a^{2} b^{5} - b^{7}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {16 \, {\left (2 \, a^{6} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{7} - a^{5} b^{2} - a b^{6}\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}} + \frac {12 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(48*(2*a^2*b^5 - b^7)*(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)))/((a^8 - 2*a^6*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + 16*(2*a^6*b*tan(1/2*d*x + 1/2*c)^3 + b^7*tan(1/2*
d*x + 1/2*c)^3 - a^7*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*b^2*tan(1/2*d*x + 1/2*c)^2 + a*b^6*tan(1/2*d*x + 1/2*c)^2
- 2*a^4*b^3*tan(1/2*d*x + 1/2*c) - b^7*tan(1/2*d*x + 1/2*c) - a^7 - a^5*b^2 - a*b^6)/((a^8 - 2*a^6*b^2 + a^4*b
^4)*(a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c) - a)) + 12*(a^2 + 2*b^2)
*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 + (a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c))/a^4 - (18*a^2*
tan(1/2*d*x + 1/2*c)^2 + 36*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c) + a^2)/(a^4*tan(1/2*d*x +
1/2*c)^2))/d

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maple [A]  time = 0.66, size = 404, normalized size = 1.37 \[ -\frac {1}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \,a^{3}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{4}}+\frac {b}{d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {1}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{7} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 b^{6}}{d \,a^{3} \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {12 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} a^{2} \sqrt {a^{2}-b^{2}}}-\frac {6 b^{7} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{4} \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c))^2,x)

[Out]

-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)+1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^3*tan(1/2*d*x+1/2*c)*b-1/8/a^2/d/tan(
1/2*d*x+1/2*c)^2+3/2/d/a^2*ln(tan(1/2*d*x+1/2*c))+3/d/a^4*ln(tan(1/2*d*x+1/2*c))*b^2+1/d*b/a^3/tan(1/2*d*x+1/2
*c)+1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)+2/d/a^4*b^7/(a-b)^2/(a+b)^2/(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^6/(a-b)^2/(a+b)^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+12/d*b
^5/(a-b)^2/(a+b)^2/a^2/(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))-6/d/a^4*b^7/(a
-b)^2/(a+b)^2/(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(csc(d*x+c)^3*sec(d*x+c)^2/(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 = 14.09, size = 2302, normalized size = 7.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*sin(c + d*x)^3*(a + b*sin(c + d*x))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) + ((tan(c/2 + (d*x)/2)^4*(17*a^6 - 32*b^6 + 33*a^2*b^4 - 66*a^4*b^2))/(2*(a^4 +
b^4 - 2*a^2*b^2)) - a^2/2 + (8*tan(c/2 + (d*x)/2)^2*(a^6 + 2*b^6 - 2*a^2*b^4 + 2*a^4*b^2))/(a^4 + b^4 - 2*a^2
*b^2) + 3*a*b*tan(c/2 + (d*x)/2) - (4*tan(c/2 + (d*x)/2)^5*(5*a^6*b + 2*b^7 + a^2*b^5 - 2*a^4*b^3))/(a*(a^4 +
b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^3*(a^6*b + 8*b^7 + a^2*b^5 + 14*a^4*b^3))/(a*(a^2 - b^2)^2))/(d*(4*a^4
*tan(c/2 + (d*x)/2)^2 - 4*a^4*tan(c/2 + (d*x)/2)^6 + 8*a^3*b*tan(c/2 + (d*x)/2)^3 - 8*a^3*b*tan(c/2 + (d*x)/2)
^5)) - (b*tan(c/2 + (d*x)/2))/(a^3*d) + (log(tan(c/2 + (d*x)/2))*(3*a^2 + 6*b^2))/(2*a^4*d) - (b^5*atan(((b^5*
(2*a^2 - b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(24*a^22 - 192*a^6*b^16 + 1152*a^8*b^14 - 2760*
a^10*b^12 + 3312*a^12*b^10 - 1944*a^14*b^8 + 288*a^16*b^6 + 264*a^18*b^4 - 144*a^20*b^2) - 24*a^21*b - 96*a^7*
b^15 + 552*a^9*b^13 - 1248*a^11*b^11 + 1368*a^13*b^9 - 672*a^15*b^7 + 24*a^17*b^5 + 96*a^19*b^3 + (3*b^5*(2*a^
2 - b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(16*a^23*b - tan(c/2 + (d*x)/2)*(48*a^24 - 64*a^10*b^14 + 432*a^12*b^12
- 1248*a^14*b^10 + 2000*a^16*b^8 - 1920*a^18*b^6 + 1104*a^20*b^4 - 352*a^22*b^2) + 16*a^11*b^13 - 96*a^13*b^11
+ 240*a^15*b^9 - 320*a^17*b^7 + 240*a^19*b^5 - 96*a^21*b^3))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a
^10*b^4 - 5*a^12*b^2))*3i)/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2) - (b^5*(2*a^2
- b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(24*a^21*b - tan(c/2 + (d*x)/2)*(24*a^22 - 192*a^6*b^16 + 1152*a^8*b^14 -
2760*a^10*b^12 + 3312*a^12*b^10 - 1944*a^14*b^8 + 288*a^16*b^6 + 264*a^18*b^4 - 144*a^20*b^2) + 96*a^7*b^15 -
552*a^9*b^13 + 1248*a^11*b^11 - 1368*a^13*b^9 + 672*a^15*b^7 - 24*a^17*b^5 - 96*a^19*b^3 + (3*b^5*(2*a^2 - b^
2)*(-(a + b)^5*(a - b)^5)^(1/2)*(16*a^23*b - tan(c/2 + (d*x)/2)*(48*a^24 - 64*a^10*b^14 + 432*a^12*b^12 - 1248
*a^14*b^10 + 2000*a^16*b^8 - 1920*a^18*b^6 + 1104*a^20*b^4 - 352*a^22*b^2) + 16*a^11*b^13 - 96*a^13*b^11 + 240
*a^15*b^9 - 320*a^17*b^7 + 240*a^19*b^5 - 96*a^21*b^3))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^
4 - 5*a^12*b^2))*3i)/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2))/(2*tan(c/2 + (d*x)
/2)*(144*a^4*b^16 - 576*a^6*b^14 + 864*a^8*b^12 - 864*a^10*b^10 + 720*a^12*b^8 - 288*a^14*b^6) + 288*a^3*b^17
- 1584*a^5*b^15 + 3168*a^7*b^13 - 2592*a^9*b^11 + 288*a^11*b^9 + 720*a^13*b^7 - 288*a^15*b^5 + (3*b^5*(2*a^2 -
b^2)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(24*a^22 - 192*a^6*b^16 + 1152*a^8*b^14 - 2760*a^10*b^1
2 + 3312*a^12*b^10 - 1944*a^14*b^8 + 288*a^16*b^6 + 264*a^18*b^4 - 144*a^20*b^2) - 24*a^21*b - 96*a^7*b^15 + 5
52*a^9*b^13 - 1248*a^11*b^11 + 1368*a^13*b^9 - 672*a^15*b^7 + 24*a^17*b^5 + 96*a^19*b^3 + (3*b^5*(2*a^2 - b^2)
*(-(a + b)^5*(a - b)^5)^(1/2)*(16*a^23*b - tan(c/2 + (d*x)/2)*(48*a^24 - 64*a^10*b^14 + 432*a^12*b^12 - 1248*a
^14*b^10 + 2000*a^16*b^8 - 1920*a^18*b^6 + 1104*a^20*b^4 - 352*a^22*b^2) + 16*a^11*b^13 - 96*a^13*b^11 + 240*a
^15*b^9 - 320*a^17*b^7 + 240*a^19*b^5 - 96*a^21*b^3))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4
- 5*a^12*b^2)))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2) + (3*b^5*(2*a^2 - b^2)*(
-(a + b)^5*(a - b)^5)^(1/2)*(24*a^21*b - tan(c/2 + (d*x)/2)*(24*a^22 - 192*a^6*b^16 + 1152*a^8*b^14 - 2760*a^1
0*b^12 + 3312*a^12*b^10 - 1944*a^14*b^8 + 288*a^16*b^6 + 264*a^18*b^4 - 144*a^20*b^2) + 96*a^7*b^15 - 552*a^9*
b^13 + 1248*a^11*b^11 - 1368*a^13*b^9 + 672*a^15*b^7 - 24*a^17*b^5 - 96*a^19*b^3 + (3*b^5*(2*a^2 - b^2)*(-(a +
b)^5*(a - b)^5)^(1/2)*(16*a^23*b - tan(c/2 + (d*x)/2)*(48*a^24 - 64*a^10*b^14 + 432*a^12*b^12 - 1248*a^14*b^1
0 + 2000*a^16*b^8 - 1920*a^18*b^6 + 1104*a^20*b^4 - 352*a^22*b^2) + 16*a^11*b^13 - 96*a^13*b^11 + 240*a^15*b^9
- 320*a^17*b^7 + 240*a^19*b^5 - 96*a^21*b^3))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^1
2*b^2)))/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2)))*(2*a^2 - b^2)*(-(a + b)^5*(a
- b)^5)^(1/2)*6i)/(d*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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