3.185 \(\int \frac{\sin ^4(x)}{(a+b \sin (x))^2} \, dx\)

Optimal. Leaf size=169 \[ \frac{x \left (6 a^2+b^2\right )}{2 b^4}+\frac{a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac{2 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\left (3 a^2-b^2\right ) \sin (x) \cos (x)}{2 b^2 \left (a^2-b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.382792, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2792, 3049, 3023, 2735, 2660, 618, 204} \[ \frac{x \left (6 a^2+b^2\right )}{2 b^4}+\frac{a \left (3 a^2-2 b^2\right ) \cos (x)}{b^3 \left (a^2-b^2\right )}-\frac{2 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2}}+\frac{a^2 \sin ^2(x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\left (3 a^2-b^2\right ) \sin (x) \cos (x)}{2 b^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Sin[x])^2,x]

[Out]

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

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.528877, size = 115, normalized size = 0.68 \[ \frac{-\frac{8 a^3 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+4 a b \cos (x) \left (\frac{a^3}{(a-b) (a+b) (a+b \sin (x))}+2\right )+12 a^2 x+2 b^2 x-b^2 \sin (2 x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Sin[x])^2,x]

[Out]

(12*a^2*x + 2*b^2*x - (8*a^3*(3*a^2 - 4*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 4*a
*b*Cos[x]*(2 + a^3/((a - b)*(a + b)*(a + b*Sin[x]))) - b^2*Sin[2*x])/(4*b^4)

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Maple [A]  time = 0.051, size = 266, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+4\,{\frac{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}a}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{1}{{b}^{2}}\tan \left ({\frac{x}{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+4\,{\frac{a}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+6\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ){a}^{2}}{{b}^{4}}}+2\,{\frac{{a}^{3}\tan \left ( x/2 \right ) }{{b}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{{a}^{4}}{{b}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-6\,{\frac{{a}^{5}}{{b}^{4} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+8\,{\frac{{a}^{3}}{{b}^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{x}{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2/(tan(1/2*x)^2+1)^2*tan(1/2*x)^3+4/b^3/(tan(1/2*x)^2+1)^2*tan(1/2*x)^2*a-1/b^2/(tan(1/2*x)^2+1)^2*tan(1/2
*x)+4/b^3/(tan(1/2*x)^2+1)^2*a+6/b^4*arctan(tan(1/2*x))*a^2+2*a^3/b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b
^2)*tan(1/2*x)+2*a^4/b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)-6*a^5/b^4/(a^2-b^2)^(3/2)*arctan(1/2*(2*a
*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+8*a^3/b^2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+1
/2*x/b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01208, size = 1277, normalized size = 7.56 \begin{align*} \left [\frac{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} -{\left (3 \, a^{6} - 4 \, a^{4} b^{2} +{\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\right )} \sin \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) +{\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} x +{\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \cos \left (x\right ) +{\left ({\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} x + 3 \,{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \,{\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8} +{\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}}, \frac{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} + 2 \,{\left (3 \, a^{6} - 4 \, a^{4} b^{2} +{\left (3 \, a^{5} b - 4 \, a^{3} b^{3}\right )} \sin \left (x\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) +{\left (6 \, a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + a b^{6}\right )} x +{\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \cos \left (x\right ) +{\left ({\left (6 \, a^{6} b - 11 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}\right )} x + 3 \,{\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{2 \,{\left (a^{5} b^{4} - 2 \, a^{3} b^{6} + a b^{8} +{\left (a^{4} b^{5} - 2 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.76866, size = 248, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (a^{3} b \tan \left (\frac{1}{2} \, x\right ) + a^{4}\right )}}{{\left (a^{2} b^{3} - b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}} + \frac{{\left (6 \, a^{2} + b^{2}\right )} x}{2 \, b^{4}} + \frac{b \tan \left (\frac{1}{2} \, x\right )^{3} + 4 \, a \tan \left (\frac{1}{2} \, x\right )^{2} - b \tan \left (\frac{1}{2} \, x\right ) + 4 \, a}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*a^5 - 4*a^3*b^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^2*b^
4 - b^6)*sqrt(a^2 - b^2)) + 2*(a^3*b*tan(1/2*x) + a^4)/((a^2*b^3 - b^5)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a))
+ 1/2*(6*a^2 + b^2)*x/b^4 + (b*tan(1/2*x)^3 + 4*a*tan(1/2*x)^2 - b*tan(1/2*x) + 4*a)/((tan(1/2*x)^2 + 1)^2*b^
3)