3.1464 \(\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=222 \[ -\frac {2 a^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}-\frac {a^4 \cos (c+d x)}{b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}-\frac {x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2897, 2648, 2664, 12, 2660, 618, 204} \[ -\frac {2 a^5 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}+\frac {4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}-\frac {a^4 \cos (c+d x)}{b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}-\frac {x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (-\frac {1}{b^2}-\frac {1}{2 (a+b)^2 (-1+\sin (c+d x))}+\frac {1}{2 (a-b)^2 (1+\sin (c+d x))}+\frac {a^4}{b^2 \left (-a^2+b^2\right ) (a+b \sin (c+d x))^2}+\frac {2 \left (a^5-2 a^3 b^2\right )}{b^2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac {x}{b^2}+\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{2 (a-b)^2}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^2}+\frac {\left (2 a^3 \left (a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac {a^4 \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac {x}{b^2}+\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 {a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {a^4 \int \frac {a}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}+\frac {\left (4 a^3 \left (a^2-2 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 )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac {x}{b^2}+\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 {a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {a^5 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac {\left (8 a^3 \left (a^2-2 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 )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac {x}{b^2}+\frac {4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} 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 {a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\left (2 a^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 )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac {x}{b^2}+\frac {4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} 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 {a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\left (4 a^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 )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac {x}{b^2}-\frac {2 a^5 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac {4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} 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 {a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.98, size = 236, normalized size = 1.06 \[ \frac {-\frac {a^4 \cos (c+d x)}{b (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-\frac {a^4 (c+d x)-2 a^2 b^2 (c+d x)+2 a b^3+b^4 (c+d x)}{\left (b^3-a^2 b\right )^2}+\frac {2 a^3 \left (a^2-4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2}}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 724, normalized size = 3.26 \[ \left [-\frac {2 \, a^{4} b^{3} - 4 \, a^{2} b^{5} + 2 \, b^{7} + 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{6} b - b^{7}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{5} b - 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{6} - 4 \, a^{4} b^{2}\right )} \cos \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 ) - 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6} - {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d x \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} d \cos \left (d x + c\right )\right )}}, -\frac {a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d x \cos \left (d x + c\right ) + {\left (a^{6} b - b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left ({\left (a^{5} b - 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{6} - 4 \, a^{4} b^{2}\right )} \cos \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 ) - {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6} - {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d x \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 264, normalized size = 1.19 \[ \frac {\frac {2 \, {\left (a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (2 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{4} - 2 \, a^{2} b^{2}\right )}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}} - \frac {d x + c}{b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

(2*(a^5 - 4*a^3*b^2)*(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^4*b^2 - 2*a^2*b^4 + b^6)*sqrt(a^2 - b^2)) - 2*(2*a^3*b*tan(1/2*d*x + 1/2*c)^3 + a*b^3*tan(1/2*d*x
+ 1/2*c)^3 + a^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^4*tan(1/2*d*x + 1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c) - a^4 -
2*a^2*b^2)/((a^4*b - 2*a^2*b^3 + b^5)*(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)) - (d*x + c)/b^2)/d

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maple [A]  time = 0.55, size = 303, normalized size = 1.36 \[ -\frac {1}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {1}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2 a^{4}}{d b \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 a^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{2} \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {8 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x)

[Out]

-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-2/d/b^2*arctan(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^3/(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^4/b/(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)+2/d*a^5/b^2/(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))-8/d*a^3/(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(sec(d*x+c)^2*sin(d*x+c)^4/(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 = 19.77, size = 4797, normalized size = 21.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4/(cos(c + d*x)^2*(a + b*sin(c + d*x))^2),x)

[Out]

((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (2*tan(c/2 + (d*x)/2)^2*(a^4 + 2*b^4))/(b*
(a^4 + b^4 - 2*a^2*b^2)) - (2*a^2*(a^2 + 2*b^2))/(b*(a^2 - b^2)^2) - (6*a*b^2*tan(c/2 + (d*x)/2))/(a^2 - b^2)^
2)/(d*(a + 2*b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^4 - 2*b*tan(c/2 + (d*x)/2)^3)) - (2*atan((64*a*b^23*t
an(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^1
3 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) - (704*a^3*b^21*tan(c/2 + (
d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a
^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) + (3520*a^5*b^19*tan(c/2 + (d*x)/2))/
(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11
+ 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) - (9536*a^7*b^17*tan(c/2 + (d*x)/2))/(64*a*b^2
3 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^
15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) + (14464*a^9*b^15*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*
a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 -
4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) - (11072*a^11*b^13*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^2
1 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a
^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) + (1024*a^13*b^11*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520
*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7
+ 1536*a^19*b^5 - 192*a^21*b^3) + (5440*a^15*b^9*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19
- 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^
19*b^5 - 192*a^21*b^3) - (4544*a^17*b^7*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a
^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 -
192*a^21*b^3) + (1536*a^19*b^5*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 +
14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*
b^3) - (192*a^21*b^3*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9
*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3)))/(b^
2*d) + (a^3*atan(((a^3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(64*a*b^25 - 672*a
^3*b^23 + 3200*a^5*b^21 - 9632*a^7*b^19 + 20608*a^9*b^17 - 32096*a^11*b^15 + 35776*a^13*b^13 - 27680*a^15*b^11
+ 14272*a^17*b^9 - 4608*a^19*b^7 + 832*a^21*b^5 - 64*a^23*b^3) + 32*a^2*b^24 - 320*a^4*b^22 + 1440*a^6*b^20 -
3840*a^8*b^18 + 6720*a^10*b^16 - 8064*a^12*b^14 + 6720*a^14*b^12 - 3840*a^16*b^10 + 1440*a^18*b^8 - 320*a^20*
b^6 + 32*a^22*b^4 + (a^3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(256*a^4*b^24 -
2112*a^6*b^22 + 7680*a^8*b^20 - 16128*a^10*b^18 + 21504*a^12*b^16 - 18816*a^14*b^14 + 10752*a^16*b^12 - 3840*a
^18*b^10 + 768*a^20*b^8 - 64*a^22*b^6) - 32*a*b^27 + 352*a^3*b^25 - 1632*a^5*b^23 + 4224*a^7*b^21 - 6720*a^9*b
^19 + 6720*a^11*b^17 - 4032*a^13*b^15 + 1152*a^15*b^13 + 96*a^17*b^11 - 160*a^19*b^9 + 32*a^21*b^7 + (a^3*(a -
2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(96*a*b^29 - 1024*a^3*b^27 + 4960*a^5*b^25 -
14400*a^7*b^23 + 27840*a^9*b^21 - 37632*a^11*b^19 + 36288*a^13*b^17 - 24960*a^15*b^15 + 12000*a^17*b^13 - 3840
*a^19*b^11 + 736*a^21*b^9 - 64*a^23*b^7) + 32*a^2*b^28 - 320*a^4*b^26 + 1440*a^6*b^24 - 3840*a^8*b^22 + 6720*a
^10*b^20 - 8064*a^12*b^18 + 6720*a^14*b^16 - 3840*a^16*b^14 + 1440*a^18*b^12 - 320*a^20*b^10 + 32*a^22*b^8))/(
b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2)))/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6
*b^6 + 5*a^8*b^4 - a^10*b^2))*1i)/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2) + (a^3*
(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(64*a*b^25 - 672*a^3*b^23 + 3200*a^5*b^21
- 9632*a^7*b^19 + 20608*a^9*b^17 - 32096*a^11*b^15 + 35776*a^13*b^13 - 27680*a^15*b^11 + 14272*a^17*b^9 - 460
8*a^19*b^7 + 832*a^21*b^5 - 64*a^23*b^3) + 32*a^2*b^24 - 320*a^4*b^22 + 1440*a^6*b^20 - 3840*a^8*b^18 + 6720*a
^10*b^16 - 8064*a^12*b^14 + 6720*a^14*b^12 - 3840*a^16*b^10 + 1440*a^18*b^8 - 320*a^20*b^6 + 32*a^22*b^4 - (a^
3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(256*a^4*b^24 - 2112*a^6*b^22 + 7680*a^
8*b^20 - 16128*a^10*b^18 + 21504*a^12*b^16 - 18816*a^14*b^14 + 10752*a^16*b^12 - 3840*a^18*b^10 + 768*a^20*b^8
- 64*a^22*b^6) - 32*a*b^27 + 352*a^3*b^25 - 1632*a^5*b^23 + 4224*a^7*b^21 - 6720*a^9*b^19 + 6720*a^11*b^17 -
4032*a^13*b^15 + 1152*a^15*b^13 + 96*a^17*b^11 - 160*a^19*b^9 + 32*a^21*b^7 - (a^3*(a - 2*b)*(a + 2*b)*(-(a +
b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(96*a*b^29 - 1024*a^3*b^27 + 4960*a^5*b^25 - 14400*a^7*b^23 + 27840*
a^9*b^21 - 37632*a^11*b^19 + 36288*a^13*b^17 - 24960*a^15*b^15 + 12000*a^17*b^13 - 3840*a^19*b^11 + 736*a^21*b
^9 - 64*a^23*b^7) + 32*a^2*b^28 - 320*a^4*b^26 + 1440*a^6*b^24 - 3840*a^8*b^22 + 6720*a^10*b^20 - 8064*a^12*b^
18 + 6720*a^14*b^16 - 3840*a^16*b^14 + 1440*a^18*b^12 - 320*a^20*b^10 + 32*a^22*b^8))/(b^12 - 5*a^2*b^10 + 10*
a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2)))/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10
*b^2))*1i)/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2))/(2*tan(c/2 + (d*x)/2)*(256*a^
4*b^20 - 2112*a^6*b^18 + 7680*a^8*b^16 - 16128*a^10*b^14 + 21504*a^12*b^12 - 18816*a^14*b^10 + 10752*a^16*b^8
- 3840*a^18*b^6 + 768*a^20*b^4 - 64*a^22*b^2) + 256*a^5*b^19 - 1088*a^7*b^17 + 1024*a^9*b^15 + 2368*a^11*b^13
- 7040*a^13*b^11 + 7744*a^15*b^9 - 4352*a^17*b^7 + 1216*a^19*b^5 - 128*a^21*b^3 + (a^3*(a - 2*b)*(a + 2*b)*(-(
a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(64*a*b^25 - 672*a^3*b^23 + 3200*a^5*b^21 - 9632*a^7*b^19 + 2060
8*a^9*b^17 - 32096*a^11*b^15 + 35776*a^13*b^13 - 27680*a^15*b^11 + 14272*a^17*b^9 - 4608*a^19*b^7 + 832*a^21*b
^5 - 64*a^23*b^3) + 32*a^2*b^24 - 320*a^4*b^22 + 1440*a^6*b^20 - 3840*a^8*b^18 + 6720*a^10*b^16 - 8064*a^12*b^
14 + 6720*a^14*b^12 - 3840*a^16*b^10 + 1440*a^18*b^8 - 320*a^20*b^6 + 32*a^22*b^4 + (a^3*(a - 2*b)*(a + 2*b)*(
-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(256*a^4*b^24 - 2112*a^6*b^22 + 7680*a^8*b^20 - 16128*a^10*b^1
8 + 21504*a^12*b^16 - 18816*a^14*b^14 + 10752*a^16*b^12 - 3840*a^18*b^10 + 768*a^20*b^8 - 64*a^22*b^6) - 32*a*
b^27 + 352*a^3*b^25 - 1632*a^5*b^23 + 4224*a^7*b^21 - 6720*a^9*b^19 + 6720*a^11*b^17 - 4032*a^13*b^15 + 1152*a
^15*b^13 + 96*a^17*b^11 - 160*a^19*b^9 + 32*a^21*b^7 + (a^3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(
tan(c/2 + (d*x)/2)*(96*a*b^29 - 1024*a^3*b^27 + 4960*a^5*b^25 - 14400*a^7*b^23 + 27840*a^9*b^21 - 37632*a^11*b
^19 + 36288*a^13*b^17 - 24960*a^15*b^15 + 12000*a^17*b^13 - 3840*a^19*b^11 + 736*a^21*b^9 - 64*a^23*b^7) + 32*
a^2*b^28 - 320*a^4*b^26 + 1440*a^6*b^24 - 3840*a^8*b^22 + 6720*a^10*b^20 - 8064*a^12*b^18 + 6720*a^14*b^16 - 3
840*a^16*b^14 + 1440*a^18*b^12 - 320*a^20*b^10 + 32*a^22*b^8))/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 +
5*a^8*b^4 - a^10*b^2)))/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2)))/(b^12 - 5*a^2*b
^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2) - (a^3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*
(tan(c/2 + (d*x)/2)*(64*a*b^25 - 672*a^3*b^23 + 3200*a^5*b^21 - 9632*a^7*b^19 + 20608*a^9*b^17 - 32096*a^11*b^
15 + 35776*a^13*b^13 - 27680*a^15*b^11 + 14272*a^17*b^9 - 4608*a^19*b^7 + 832*a^21*b^5 - 64*a^23*b^3) + 32*a^2
*b^24 - 320*a^4*b^22 + 1440*a^6*b^20 - 3840*a^8*b^18 + 6720*a^10*b^16 - 8064*a^12*b^14 + 6720*a^14*b^12 - 3840
*a^16*b^10 + 1440*a^18*b^8 - 320*a^20*b^6 + 32*a^22*b^4 - (a^3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2
)*(tan(c/2 + (d*x)/2)*(256*a^4*b^24 - 2112*a^6*b^22 + 7680*a^8*b^20 - 16128*a^10*b^18 + 21504*a^12*b^16 - 1881
6*a^14*b^14 + 10752*a^16*b^12 - 3840*a^18*b^10 + 768*a^20*b^8 - 64*a^22*b^6) - 32*a*b^27 + 352*a^3*b^25 - 1632
*a^5*b^23 + 4224*a^7*b^21 - 6720*a^9*b^19 + 6720*a^11*b^17 - 4032*a^13*b^15 + 1152*a^15*b^13 + 96*a^17*b^11 -
160*a^19*b^9 + 32*a^21*b^7 - (a^3*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*(tan(c/2 + (d*x)/2)*(96*a*b
^29 - 1024*a^3*b^27 + 4960*a^5*b^25 - 14400*a^7*b^23 + 27840*a^9*b^21 - 37632*a^11*b^19 + 36288*a^13*b^17 - 24
960*a^15*b^15 + 12000*a^17*b^13 - 3840*a^19*b^11 + 736*a^21*b^9 - 64*a^23*b^7) + 32*a^2*b^28 - 320*a^4*b^26 +
1440*a^6*b^24 - 3840*a^8*b^22 + 6720*a^10*b^20 - 8064*a^12*b^18 + 6720*a^14*b^16 - 3840*a^16*b^14 + 1440*a^18*
b^12 - 320*a^20*b^10 + 32*a^22*b^8))/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2)))/(b
^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2)))/(b^12 - 5*a^2*b^10 + 10*a^4*b^8 - 10*a^6*
b^6 + 5*a^8*b^4 - a^10*b^2)))*(a - 2*b)*(a + 2*b)*(-(a + b)^5*(a - b)^5)^(1/2)*2i)/(d*(b^12 - 5*a^2*b^10 + 10*
a^4*b^8 - 10*a^6*b^6 + 5*a^8*b^4 - a^10*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**4/(a+b*sin(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)**4*sec(c + d*x)**2/(a + b*sin(c + d*x))**2, x)

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