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

Optimal. Leaf size=172 \[ \frac{2 a b \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac{2 a b \left (a^2-b^2\right ) \sin (x)}{\left (a^2+b^2\right )^3}+\frac{\left (a^2-b^2\right ) \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{a^2 \left (a^2-3 b^2\right ) \cos (x)}{\left (a^2+b^2\right )^3}+\frac{a^3 b^2}{\left (a^2+b^2\right )^3 (a \cos (x)+b \sin (x))}+\frac{a^2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.677315, antiderivative size = 238, normalized size of antiderivative = 1.38, number of steps used = 33, number of rules used = 12, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3111, 3109, 2565, 30, 2564, 2637, 2638, 3074, 206, 2633, 3099, 3154} \[ \frac{2 a b \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac{2 a^3 b \sin (x)}{\left (a^2+b^2\right )^3}-\frac{2 a b^3 \sin (x)}{\left (a^2+b^2\right )^3}+\frac{a^2 \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{b^2 \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac{a^2 \cos (x)}{\left (a^2+b^2\right )^2}+\frac{4 a^2 b^2 \cos (x)}{\left (a^2+b^2\right )^3}+\frac{a^3 b^2}{\left (a^2+b^2\right )^3 (a \cos (x)+b \sin (x))}+\frac{2 a^4 b \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac{3 a^2 b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*b*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) - (3*a^2*b^3*ArcTanh[(b*Cos[x] - a*
Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) + (4*a^2*b^2*Cos[x])/(a^2 + b^2)^3 - (a^2*Cos[x])/(a^2 + b^2)^2 +
(a^2*Cos[x]^3)/(3*(a^2 + b^2)^2) - (b^2*Cos[x]^3)/(3*(a^2 + b^2)^2) + (2*a^3*b*Sin[x])/(a^2 + b^2)^3 - (2*a*b^
3*Sin[x])/(a^2 + b^2)^3 + (2*a*b*Sin[x]^3)/(3*(a^2 + b^2)^2) + (a^3*b^2)/((a^2 + b^2)^3*(a*Cos[x] + b*Sin[x]))

Rule 3111

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

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 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 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 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])

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3099

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

Rule 3154

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> -Simp[(b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Co
s[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - c*C)/(a^2 - b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d +
e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - c*C, 0]

Rubi steps

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

Mathematica [A]  time = 1.21688, size = 200, normalized size = 1.16 \[ \frac{16 a^2 b^3 \sin (2 x)-2 a^2 b^3 \sin (4 x)+a \left (a^2+b^2\right )^2 \cos (4 x)+\left (4 a^3 b^2-8 a^5+12 a b^4\right ) \cos (2 x)+90 a^3 b^2+18 a^4 b \sin (2 x)-a^4 b \sin (4 x)-9 a^5-21 a b^4-2 b^5 \sin (2 x)-b^5 \sin (4 x)}{24 \left (a^2+b^2\right )^3 (a \cos (x)+b \sin (x))}-\frac{2 a^2 b \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*b*(2*a^2 - 3*b^2)*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) + (-9*a^5 + 90*a^3*b^2
- 21*a*b^4 + (-8*a^5 + 4*a^3*b^2 + 12*a*b^4)*Cos[2*x] + a*(a^2 + b^2)^2*Cos[4*x] + 18*a^4*b*Sin[2*x] + 16*a^2
*b^3*Sin[2*x] - 2*b^5*Sin[2*x] - a^4*b*Sin[4*x] - 2*a^2*b^3*Sin[4*x] - b^5*Sin[4*x])/(24*(a^2 + b^2)^3*(a*Cos[
x] + b*Sin[x]))

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Maple [A]  time = 0.133, size = 269, normalized size = 1.6 \begin{align*} -4\,{\frac{ \left ( -{a}^{3}b+a{b}^{3} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{5}+ \left ( -3/2\,{a}^{2}{b}^{2}+1/2\,{b}^{4} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{4}+ \left ( -10/3\,{a}^{3}b+2/3\,a{b}^{3} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{3}+ \left ({a}^{4}-3\,{a}^{2}{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}+ \left ( -{a}^{3}b+a{b}^{3} \right ) \tan \left ( x/2 \right ) +1/3\,{a}^{4}-3/2\,{a}^{2}{b}^{2}+1/6\,{b}^{4}}{ \left ({a}^{2}+{b}^{2} \right ) \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+4\,{\frac{{a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) } \left ({\frac{-1/2\,\tan \left ( x/2 \right ){b}^{2}-1/2\,ab}{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a}}-1/2\,{\frac{2\,{a}^{2}-3\,{b}^{2}}{\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(cos(x)^2*sin(x)^3/(a*cos(x)+b*sin(x))^2,x)

[Out]

-4/(a^2+b^2)/(a^4+2*a^2*b^2+b^4)*((-a^3*b+a*b^3)*tan(1/2*x)^5+(-3/2*a^2*b^2+1/2*b^4)*tan(1/2*x)^4+(-10/3*a^3*b
+2/3*a*b^3)*tan(1/2*x)^3+(a^4-3*a^2*b^2)*tan(1/2*x)^2+(-a^3*b+a*b^3)*tan(1/2*x)+1/3*a^4-3/2*a^2*b^2+1/6*b^4)/(
tan(1/2*x)^2+1)^3+4*a^2*b/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*((-1/2*tan(1/2*x)*b^2-1/2*a*b)/(tan(1/2*x)^2*a-2*b*tan
(1/2*x)-a)-1/2*(2*a^2-3*b^2)/(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)^2*sin(x)^3/(a*cos(x)+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.600247, size = 818, normalized size = 4.76 \begin{align*} \frac{22 \, a^{5} b^{2} + 14 \, a^{3} b^{4} - 8 \, a b^{6} + 2 \,{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (x\right )^{4} - 2 \,{\left (3 \, a^{7} + 4 \, a^{5} b^{2} - a^{3} b^{4} - 2 \, a b^{6}\right )} \cos \left (x\right )^{2} - 3 \, \sqrt{a^{2} + b^{2}}{\left ({\left (2 \, a^{5} b - 3 \, a^{3} b^{3}\right )} \cos \left (x\right ) +{\left (2 \, a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \sin \left (x\right )\right )} \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^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{3} - 5 \,{\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{6 \,{\left ({\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} \cos \left (x\right ) +{\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(22*a^5*b^2 + 14*a^3*b^4 - 8*a*b^6 + 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(x)^4 - 2*(3*a^7 + 4*a^5*b
^2 - a^3*b^4 - 2*a*b^6)*cos(x)^2 - 3*sqrt(a^2 + b^2)*((2*a^5*b - 3*a^3*b^3)*cos(x) + (2*a^4*b^2 - 3*a^2*b^4)*s
in(x))*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)) - 2*((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(x)^3 - 5
*(a^6*b + 2*a^4*b^3 + a^2*b^5)*cos(x))*sin(x))/((a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 + a*b^8)*cos(x) + (a^
8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*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(cos(x)**2*sin(x)**3/(a*cos(x)+b*sin(x))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.2139, size = 462, normalized size = 2.69 \begin{align*} \frac{{\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \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 (a^{2} b^{3} \tan \left (\frac{1}{2} \, x\right ) + a^{3} b^{2}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, x\right ) - a\right )}} + \frac{2 \,{\left (6 \, a^{3} b \tan \left (\frac{1}{2} \, x\right )^{5} - 6 \, a b^{3} \tan \left (\frac{1}{2} \, x\right )^{5} + 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 3 \, b^{4} \tan \left (\frac{1}{2} \, x\right )^{4} + 20 \, a^{3} b \tan \left (\frac{1}{2} \, x\right )^{3} - 4 \, a b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} - 6 \, a^{4} \tan \left (\frac{1}{2} \, x\right )^{2} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 6 \, a^{3} b \tan \left (\frac{1}{2} \, x\right ) - 6 \, a b^{3} \tan \left (\frac{1}{2} \, x\right ) - 2 \, a^{4} + 9 \, a^{2} b^{2} - b^{4}\right )}}{3 \,{\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 )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^4*b - 3*a^2*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*(a^2*b^3*tan(1/2*x) + a^3*b^2)/((a^6 + 3*a^
4*b^2 + 3*a^2*b^4 + b^6)*(a*tan(1/2*x)^2 - 2*b*tan(1/2*x) - a)) + 2/3*(6*a^3*b*tan(1/2*x)^5 - 6*a*b^3*tan(1/2*
x)^5 + 9*a^2*b^2*tan(1/2*x)^4 - 3*b^4*tan(1/2*x)^4 + 20*a^3*b*tan(1/2*x)^3 - 4*a*b^3*tan(1/2*x)^3 - 6*a^4*tan(
1/2*x)^2 + 18*a^2*b^2*tan(1/2*x)^2 + 6*a^3*b*tan(1/2*x) - 6*a*b^3*tan(1/2*x) - 2*a^4 + 9*a^2*b^2 - b^4)/((a^6
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*x)^2 + 1)^3)