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

Optimal. Leaf size=107 \[ \frac{-b \left (a^2+b^2\right ) \sin (2 x)+a \left (a^2+b^2\right ) \cos (2 x)+3 a \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}+\frac{6 a^2 b \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [B]  time = 1.16922, antiderivative size = 283, normalized size of antiderivative = 2.64, number of steps used = 19, number of rules used = 11, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.688, Rules used = {4401, 2637, 2638, 6742, 639, 203, 638, 618, 206, 3100, 3074} \[ \frac{3 a^3 \sin (x)}{b^3 \left (a^2+b^2\right )}+\frac{3 a^2 \cos (x)}{b^2 \left (a^2+b^2\right )}+\frac{2 a^2 \left (a+b \tan \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac{x}{2}\right )+a+2 b \tan \left (\frac{x}{2}\right )\right )}-\frac{2 a^3 \cos ^2\left (\frac{x}{2}\right ) \left (\left (a^2-b^2\right ) \tan \left (\frac{x}{2}\right )+2 a b\right )}{b^3 \left (a^2+b^2\right )^2}+\frac{2 a^2 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b \left (a^2+b^2\right )^{5/2}}-\frac{2 a^2 b \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{3 a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{b \left (a^2+b^2\right )^{3/2}}-\frac{2 a \sin (x)}{b^3}-\frac{\cos (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 203

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

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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 206

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

Rule 3100

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

Rule 3074

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

Rubi steps

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

Mathematica [A]  time = 0.425441, size = 107, normalized size = 1. \[ \frac{-b \left (a^2+b^2\right ) \sin (2 x)+a \left (a^2+b^2\right ) \cos (2 x)+3 a \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}+\frac{6 a^2 b \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.107, size = 141, normalized size = 1.3 \begin{align*} 4\,{\frac{-ab\tan \left ( x/2 \right ) +1/2\,{a}^{2}-1/2\,{b}^{2}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-4\,{\frac{{a}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}} \left ({\frac{1/2\,b\tan \left ( x/2 \right ) +a/2}{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a}}-3/2\,{\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/(a^4+2*a^2*b^2+b^4)*(-a*b*tan(1/2*x)+1/2*a^2-1/2*b^2)/(tan(1/2*x)^2+1)-4*a^2/(a^4+2*a^2*b^2+b^4)*((1/2*b*tan
(1/2*x)+1/2*a)/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-a)-3/2*b/(a^2+b^2)^(1/2)*arctanh(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., 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)^3/(a*cos(x)+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.546829, size = 570, normalized size = 5.33 \begin{align*} \frac{2 \, a^{5} - 2 \, a^{3} b^{2} - 4 \, a b^{4} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{2} - 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \,{\left (a^{3} b \cos \left (x\right ) + a^{2} b^{2} \sin \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (-\frac{2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right )}{2 \,{\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right ) +{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^5 - 2*a^3*b^2 - 4*a*b^4 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(x)^2 - 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(x)*s
in(x) + 3*(a^3*b*cos(x) + a^2*b^2*sin(x))*sqrt(a^2 + b^2)*log(-(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 - 2
*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(x) - a*sin(x)))/(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2)))/((a
^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(x) + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*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)**3/(a*cos(x)+b*sin(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20356, size = 251, normalized size = 2.35 \begin{align*} -\frac{3 \, a^{2} b \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{3} - 3 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 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 \, b \tan \left (\frac{1}{2} \, x\right ) - a\right )}{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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