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

Optimal. Leaf size=193 \[ -\frac{b \sin ^5(x)}{5 \left (a^2+b^2\right )}+\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^2 b \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac{a^2 b^3 \sin (x)}{\left (a^2+b^2\right )^3}+\frac{a \cos ^5(x)}{5 \left (a^2+b^2\right )}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac{a b^2 \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{a^3 b^2 \cos (x)}{\left (a^2+b^2\right )^3}+\frac{a^3 b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

[Out]

(a^3*b^3*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) - (a^3*b^2*Cos[x])/(a^2 + b^2)^3 +
(a*b^2*Cos[x]^3)/(3*(a^2 + b^2)^2) - (a*Cos[x]^3)/(3*(a^2 + b^2)) + (a*Cos[x]^5)/(5*(a^2 + b^2)) + (a^2*b^3*Si
n[x])/(a^2 + b^2)^3 - (a^2*b*Sin[x]^3)/(3*(a^2 + b^2)^2) + (b*Sin[x]^3)/(3*(a^2 + b^2)) - (b*Sin[x]^5)/(5*(a^2
 + b^2))

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Rubi [A]  time = 0.358671, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {3109, 2564, 14, 2565, 30, 2637, 2638, 3074, 206} \[ -\frac{b \sin ^5(x)}{5 \left (a^2+b^2\right )}+\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^2 b \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac{a^2 b^3 \sin (x)}{\left (a^2+b^2\right )^3}+\frac{a \cos ^5(x)}{5 \left (a^2+b^2\right )}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac{a b^2 \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{a^3 b^2 \cos (x)}{\left (a^2+b^2\right )^3}+\frac{a^3 b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*b^3*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) - (a^3*b^2*Cos[x])/(a^2 + b^2)^3 +
(a*b^2*Cos[x]^3)/(3*(a^2 + b^2)^2) - (a*Cos[x]^3)/(3*(a^2 + b^2)) + (a*Cos[x]^5)/(5*(a^2 + b^2)) + (a^2*b^3*Si
n[x])/(a^2 + b^2)^3 - (a^2*b*Sin[x]^3)/(3*(a^2 + b^2)^2) + (b*Sin[x]^3)/(3*(a^2 + b^2)) - (b*Sin[x]^5)/(5*(a^2
 + b^2))

Rule 3109

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
 !(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
 &&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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]

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])

Rubi steps

\begin{align*} \int \frac{\cos ^3(x) \sin ^3(x)}{a \cos (x)+b \sin (x)} \, dx &=\frac{a \int \cos ^2(x) \sin ^3(x) \, dx}{a^2+b^2}+\frac{b \int \cos ^3(x) \sin ^2(x) \, dx}{a^2+b^2}-\frac{(a b) \int \frac{\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}\\ &=-\frac{\left (a^2 b\right ) \int \cos (x) \sin ^2(x) \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (a b^2\right ) \int \cos ^2(x) \sin (x) \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (a^2 b^2\right ) \int \frac{\cos (x) \sin (x)}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (x)\right )}{a^2+b^2}+\frac{b \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (x)\right )}{a^2+b^2}\\ &=\frac{\left (a^3 b^2\right ) \int \sin (x) \, dx}{\left (a^2+b^2\right )^3}+\frac{\left (a^2 b^3\right ) \int \cos (x) \, dx}{\left (a^2+b^2\right )^3}-\frac{\left (a^3 b^3\right ) \int \frac{1}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^3}-\frac{\left (a^2 b\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\sin (x)\right )}{\left (a^2+b^2\right )^2}+\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\cos (x)\right )}{\left (a^2+b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (x)\right )}{a^2+b^2}+\frac{b \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (x)\right )}{a^2+b^2}\\ &=-\frac{a^3 b^2 \cos (x)}{\left (a^2+b^2\right )^3}+\frac{a b^2 \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac{a \cos ^5(x)}{5 \left (a^2+b^2\right )}+\frac{a^2 b^3 \sin (x)}{\left (a^2+b^2\right )^3}-\frac{a^2 b \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac{b \sin ^5(x)}{5 \left (a^2+b^2\right )}+\frac{\left (a^3 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{\left (a^2+b^2\right )^3}\\ &=\frac{a^3 b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac{a^3 b^2 \cos (x)}{\left (a^2+b^2\right )^3}+\frac{a b^2 \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac{a \cos ^5(x)}{5 \left (a^2+b^2\right )}+\frac{a^2 b^3 \sin (x)}{\left (a^2+b^2\right )^3}-\frac{a^2 b \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac{b \sin ^5(x)}{5 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [A]  time = 1.55243, size = 223, normalized size = 1.16 \[ \frac{240 a^2 b^3 \sin (x)+10 a^2 b^3 \sin (3 x)-6 a^2 b^3 \sin (5 x)+6 a^3 b^2 \cos (5 x)-30 a \left (8 a^2 b^2+a^4-b^4\right ) \cos (x)-5 a \left (-2 a^2 b^2+a^4-3 b^4\right ) \cos (3 x)-30 a^4 b \sin (x)+15 a^4 b \sin (3 x)-3 a^4 b \sin (5 x)+3 a^5 \cos (5 x)+3 a b^4 \cos (5 x)+30 b^5 \sin (x)-5 b^5 \sin (3 x)-3 b^5 \sin (5 x)}{240 \left (a^2+b^2\right )^3}-\frac{2 a^3 b^3 \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*b^3*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) + (-30*a*(a^4 + 8*a^2*b^2 - b^4)*Cos
[x] - 5*a*(a^4 - 2*a^2*b^2 - 3*b^4)*Cos[3*x] + 3*a^5*Cos[5*x] + 6*a^3*b^2*Cos[5*x] + 3*a*b^4*Cos[5*x] - 30*a^4
*b*Sin[x] + 240*a^2*b^3*Sin[x] + 30*b^5*Sin[x] + 15*a^4*b*Sin[3*x] + 10*a^2*b^3*Sin[3*x] - 5*b^5*Sin[3*x] - 3*
a^4*b*Sin[5*x] - 6*a^2*b^3*Sin[5*x] - 3*b^5*Sin[5*x])/(240*(a^2 + b^2)^3)

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Maple [A]  time = 0.111, size = 305, normalized size = 1.6 \begin{align*} -2\,{\frac{1}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{5}} \left ( -{a}^{2}{b}^{3} \left ( \tan \left ( x/2 \right ) \right ) ^{9}-a{b}^{4} \left ( \tan \left ( x/2 \right ) \right ) ^{8}+ \left ( -16/3\,{a}^{2}{b}^{3}-4/3\,{b}^{5} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{7}+ \left ( 2\,{a}^{5}+6\,{a}^{3}{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{6}+ \left ({\frac{16\,{a}^{4}b}{5}}-{\frac{34\,{a}^{2}{b}^{3}}{15}}+{\frac{8\,{b}^{5}}{15}} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{5}+ \left ( -2/3\,{a}^{5}+10/3\,{a}^{3}{b}^{2}-2\,a{b}^{4} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{4}+ \left ( -16/3\,{a}^{2}{b}^{3}-4/3\,{b}^{5} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{3}+ \left ( 2/3\,{a}^{5}+14/3\,{a}^{3}{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}-{a}^{2}{b}^{3}\tan \left ( x/2 \right ) +2/15\,{a}^{5}+{\frac{14\,{a}^{3}{b}^{2}}{15}}-1/5\,a{b}^{4} \right ) }-16\,{\frac{{a}^{3}{b}^{3}}{ \left ( 8\,{a}^{6}+24\,{a}^{4}{b}^{2}+24\,{a}^{2}{b}^{4}+8\,{b}^{6} \right ) \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*(-a^2*b^3*tan(1/2*x)^9-a*b^4*tan(1/2*x)^8+(-16/3*a^2*b^3-4/3*b^5)*tan(1/2*x)^
7+(2*a^5+6*a^3*b^2)*tan(1/2*x)^6+(16/5*a^4*b-34/15*a^2*b^3+8/15*b^5)*tan(1/2*x)^5+(-2/3*a^5+10/3*a^3*b^2-2*a*b
^4)*tan(1/2*x)^4+(-16/3*a^2*b^3-4/3*b^5)*tan(1/2*x)^3+(2/3*a^5+14/3*a^3*b^2)*tan(1/2*x)^2-a^2*b^3*tan(1/2*x)+2
/15*a^5+14/15*a^3*b^2-1/5*a*b^4)/(tan(1/2*x)^2+1)^5-16*a^3*b^3/(8*a^6+24*a^4*b^2+24*a^2*b^4+8*b^6)/(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(cos(x)^3*sin(x)^3/(a*cos(x)+b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.583998, size = 706, normalized size = 3.66 \begin{align*} \frac{15 \, \sqrt{a^{2} + b^{2}} a^{3} b^{3} \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 ) + 6 \,{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )^{5} - 10 \,{\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (x\right )^{3} - 30 \,{\left (a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (x\right ) - 2 \,{\left (3 \, a^{6} b - 11 \, a^{4} b^{3} - 16 \, a^{2} b^{5} - 2 \, b^{7} + 3 \,{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{4} -{\left (6 \, a^{6} b + 13 \, a^{4} b^{3} + 8 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{30 \,{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*sqrt(a^2 + b^2)*a^3*b^3*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)) + 6*(a^7 + 3*a^5*b^2 + 3*a^3*b
^4 + a*b^6)*cos(x)^5 - 10*(a^7 + 2*a^5*b^2 + a^3*b^4)*cos(x)^3 - 30*(a^5*b^2 + a^3*b^4)*cos(x) - 2*(3*a^6*b -
11*a^4*b^3 - 16*a^2*b^5 - 2*b^7 + 3*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(x)^4 - (6*a^6*b + 13*a^4*b^3 + 8
*a^2*b^5 + b^7)*cos(x)^2)*sin(x))/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)

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

[Out]

Timed out

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Giac [B]  time = 1.23626, size = 487, normalized size = 2.52 \begin{align*} \frac{a^{3} b^{3} \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^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (15 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right )^{9} + 15 \, a b^{4} \tan \left (\frac{1}{2} \, x\right )^{8} + 80 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right )^{7} + 20 \, b^{5} \tan \left (\frac{1}{2} \, x\right )^{7} - 30 \, a^{5} \tan \left (\frac{1}{2} \, x\right )^{6} - 90 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, x\right )^{6} - 48 \, a^{4} b \tan \left (\frac{1}{2} \, x\right )^{5} + 34 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right )^{5} - 8 \, b^{5} \tan \left (\frac{1}{2} \, x\right )^{5} + 10 \, a^{5} \tan \left (\frac{1}{2} \, x\right )^{4} - 50 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, x\right )^{4} + 30 \, a b^{4} \tan \left (\frac{1}{2} \, x\right )^{4} + 80 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 20 \, b^{5} \tan \left (\frac{1}{2} \, x\right )^{3} - 10 \, a^{5} \tan \left (\frac{1}{2} \, x\right )^{2} - 70 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 15 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right ) - 2 \, a^{5} - 14 \, a^{3} b^{2} + 3 \, a b^{4}\right )}}{15 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*b^3*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^6
 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) + 2/15*(15*a^2*b^3*tan(1/2*x)^9 + 15*a*b^4*tan(1/2*x)^8 + 80*
a^2*b^3*tan(1/2*x)^7 + 20*b^5*tan(1/2*x)^7 - 30*a^5*tan(1/2*x)^6 - 90*a^3*b^2*tan(1/2*x)^6 - 48*a^4*b*tan(1/2*
x)^5 + 34*a^2*b^3*tan(1/2*x)^5 - 8*b^5*tan(1/2*x)^5 + 10*a^5*tan(1/2*x)^4 - 50*a^3*b^2*tan(1/2*x)^4 + 30*a*b^4
*tan(1/2*x)^4 + 80*a^2*b^3*tan(1/2*x)^3 + 20*b^5*tan(1/2*x)^3 - 10*a^5*tan(1/2*x)^2 - 70*a^3*b^2*tan(1/2*x)^2
+ 15*a^2*b^3*tan(1/2*x) - 2*a^5 - 14*a^3*b^2 + 3*a*b^4)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*x)^2 + 1
)^5)