3.186 \(\int \frac {\sin ^3(x)}{(a+b \sin (x))^2} \, dx\)

Optimal. Leaf size=124 \[ -\frac {\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}-\frac {2 a x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2792, 3023, 2735, 2660, 618, 204} \[ -\frac {\left (2 a^2-b^2\right ) \cos (x)}{b^2 \left (a^2-b^2\right )}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}+\frac {a^2 \sin (x) \cos (x)}{b \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {2 a x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + b*Sin[x])^2,x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.43, size = 94, normalized size = 0.76 \[ \frac {b \cos (x) \left (-\frac {a^3}{(a-b) (a+b) (a+b \sin (x))}-1\right )+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-2 a x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*x + (2*a^2*(2*a^2 - 3*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + b*Cos[x]*(-1 -
a^3/((a - b)*(a + b)*(a + b*Sin[x]))))/b^3

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fricas [A]  time = 0.56, size = 483, normalized size = 3.90 \[ \left [-\frac {{\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} + 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) + 4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} x + 2 \, {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \relax (x) + 2 \, {\left (2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} x + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \relax (x)\right )} \sin \relax (x)}{2 \, {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sin \relax (x)\right )}}, -\frac {{\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \relax (x)\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} x + {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5}\right )} \cos \relax (x) + {\left (2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} x + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \relax (x)\right )} \sin \relax (x)}{a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7} + {\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sin \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 204, normalized size = 1.65 \[ \frac {2 \, {\left (2 \, a^{4} - 3 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {2 \, a x}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 196, normalized size = 1.58 \[ -\frac {2}{b^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}-\frac {4 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{3}}-\frac {2 a^{2} \tan \left (\frac {x}{2}\right )}{b \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}-\frac {2 a^{3}}{b^{2} \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}+\frac {4 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{3} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}-\frac {6 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a+b*sin(x))^2,x)

[Out]

-2/b^2/(tan(1/2*x)^2+1)-4/b^3*a*arctan(tan(1/2*x))-2*a^2/b/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)*tan(1/2
*x)-2*a^3/b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)+4*a^4/b^3/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)
+2*b)/(a^2-b^2)^(1/2))-6*a^2/b/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+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(sin(x)^3/(a+b*sin(x))^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 = 9.98, size = 2578, normalized size = 20.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a + b*sin(x))^2,x)

[Out]

((2*(a*b^2 - 2*a^3))/(b^2*(a^2 - b^2)) - (2*a^2*tan(x/2)^3)/(b*(a^2 - b^2)) + (2*tan(x/2)^2*(a*b^2 - 2*a^3))/(
b^2*(a^2 - b^2)) - (2*tan(x/2)*(3*a^2 - 2*b^2))/(b*(a^2 - b^2)))/(a + 2*b*tan(x/2) + 2*a*tan(x/2)^2 + a*tan(x/
2)^4 + 2*b*tan(x/2)^3) - (4*a*atan((512*a^4*b^5*tan(x/2))/((512*a^4*b^14)/(b^9 - 2*a^2*b^7 + a^4*b^5) - (1408*
a^6*b^12)/(b^9 - 2*a^2*b^7 + a^4*b^5) + (1280*a^8*b^10)/(b^9 - 2*a^2*b^7 + a^4*b^5) - (384*a^10*b^8)/(b^9 - 2*
a^2*b^7 + a^4*b^5)) - (384*a^6*b^3*tan(x/2))/((512*a^4*b^14)/(b^9 - 2*a^2*b^7 + a^4*b^5) - (1408*a^6*b^12)/(b^
9 - 2*a^2*b^7 + a^4*b^5) + (1280*a^8*b^10)/(b^9 - 2*a^2*b^7 + a^4*b^5) - (384*a^10*b^8)/(b^9 - 2*a^2*b^7 + a^4
*b^5))))/b^3 - (a^2*atan(((a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*a^4*b^6 - 8*a^6*b^4 + 4*a^
8*b^2))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(8*a^3*b^8 - 29*a^5*b^6 + 28*a^7*b^4 - 8*a^9*b^2))/(b^10 -
2*a^2*b^8 + a^4*b^6) + (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^2*b^10 - 3*a^4*b^8 + a^6*b^
6))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(6*a^3*b^10 - 10*a^5*b^8 + 4*a^7*b^6))/(b^10 - 2*a^2*b^8 + a^4*
b^6) + (a^2*((32*(a^2*b^12 - 2*a^4*b^10 + a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(3*a*b^14 - 8*a
^3*b^12 + 7*a^5*b^10 - 2*a^7*b^8))/(b^10 - 2*a^2*b^8 + a^4*b^6))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))
/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*1i)/(b^9 - 3*a^2*b^7 + 3*a
^4*b^5 - a^6*b^3) + (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*a^4*b^6 - 8*a^6*b^4 + 4*a^8*b^2)
)/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(8*a^3*b^8 - 29*a^5*b^6 + 28*a^7*b^4 - 8*a^9*b^2))/(b^10 - 2*a^2*
b^8 + a^4*b^6) - (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^2*b^10 - 3*a^4*b^8 + a^6*b^6))/(b
^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(6*a^3*b^10 - 10*a^5*b^8 + 4*a^7*b^6))/(b^10 - 2*a^2*b^8 + a^4*b^6) -
(a^2*((32*(a^2*b^12 - 2*a^4*b^10 + a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(3*a*b^14 - 8*a^3*b^1
2 + 7*a^5*b^10 - 2*a^7*b^8))/(b^10 - 2*a^2*b^8 + a^4*b^6))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2))/(b^9
- 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*1i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5
- a^6*b^3))/((64*(4*a^8 - 6*a^6*b^2))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (64*tan(x/2)*(16*a^9 + 24*a^5*b^4 - 40*a^
7*b^2))/(b^10 - 2*a^2*b^8 + a^4*b^6) + (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*a^4*b^6 - 8*a
^6*b^4 + 4*a^8*b^2))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(8*a^3*b^8 - 29*a^5*b^6 + 28*a^7*b^4 - 8*a^9*b
^2))/(b^10 - 2*a^2*b^8 + a^4*b^6) + (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^2*b^10 - 3*a^4
*b^8 + a^6*b^6))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(6*a^3*b^10 - 10*a^5*b^8 + 4*a^7*b^6))/(b^10 - 2*a
^2*b^8 + a^4*b^6) + (a^2*((32*(a^2*b^12 - 2*a^4*b^10 + a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(3
*a*b^14 - 8*a^3*b^12 + 7*a^5*b^10 - 2*a^7*b^8))/(b^10 - 2*a^2*b^8 + a^4*b^6))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a -
b)^3)^(1/2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2
*b^7 + 3*a^4*b^5 - a^6*b^3) - (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(4*a^4*b^6 - 8*a^6*b^4 +
4*a^8*b^2))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(8*a^3*b^8 - 29*a^5*b^6 + 28*a^7*b^4 - 8*a^9*b^2))/(b^1
0 - 2*a^2*b^8 + a^4*b^6) - (a^2*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((32*(2*a^2*b^10 - 3*a^4*b^8 + a^
6*b^6))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(6*a^3*b^10 - 10*a^5*b^8 + 4*a^7*b^6))/(b^10 - 2*a^2*b^8 +
a^4*b^6) - (a^2*((32*(a^2*b^12 - 2*a^4*b^10 + a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(x/2)*(3*a*b^14 -
8*a^3*b^12 + 7*a^5*b^10 - 2*a^7*b^8))/(b^10 - 2*a^2*b^8 + a^4*b^6))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1
/2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))/(b^9 - 3*a^2*b^7 + 3*
a^4*b^5 - a^6*b^3)))*(2*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*2i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)

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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(sin(x)**3/(a+b*sin(x))**2,x)

[Out]

Timed out

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