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

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

[Out]

a^3*b^3*arctanh((b*cos(x)-a*sin(x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(7/2)-a^3*b^2*cos(x)/(a^2+b^2)^3+1/3*a*b^2*cos(
x)^3/(a^2+b^2)^2-1/3*a*cos(x)^3/(a^2+b^2)+1/5*a*cos(x)^5/(a^2+b^2)+a^2*b^3*sin(x)/(a^2+b^2)^3-1/3*a^2*b*sin(x)
^3/(a^2+b^2)^2+1/3*b*sin(x)^3/(a^2+b^2)-1/5*b*sin(x)^5/(a^2+b^2)

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Rubi [A]  time = 0.36, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, 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 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 30

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

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 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 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 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 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]

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*}

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Mathematica [A]  time = 1.48, size = 223, normalized size = 1.16 \[ \frac {3 a^5 \cos (5 x)-30 a^4 b \sin (x)+15 a^4 b \sin (3 x)-3 a^4 b \sin (5 x)+6 a^3 b^2 \cos (5 x)+240 a^2 b^3 \sin (x)+10 a^2 b^3 \sin (3 x)-6 a^2 b^3 \sin (5 x)-30 a \left (a^4+8 a^2 b^2-b^4\right ) \cos (x)-5 a \left (a^4-2 a^2 b^2-3 b^4\right ) \cos (3 x)+3 a b^4 \cos (5 x)+30 b^5 \sin (x)-5 b^5 \sin (3 x)-3 b^5 \sin (5 x)}{240 \left (a^2+b^2\right )^3}-\frac {2 a^3 b^3 \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

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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fricas [A]  time = 0.60, size = 307, normalized size = 1.59 \[ \frac {15 \, \sqrt {a^{2} + b^{2}} a^{3} b^{3} \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) + 6 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \relax (x)^{5} - 10 \, {\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \relax (x)^{3} - 30 \, {\left (a^{5} b^{2} + a^{3} b^{4}\right )} \cos \relax (x) - 2 \, {\left (3 \, a^{6} b - 11 \, a^{4} b^{3} - 16 \, a^{2} b^{5} - 2 \, b^{7} + 3 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{4} - {\left (6 \, a^{6} b + 13 \, a^{4} b^{3} + 8 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{30 \, {\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )}} \]

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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giac [B]  time = 1.69, size = 361, normalized size = 1.87 \[ \frac {a^{3} b^{3} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (15 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{9} + 15 \, a b^{4} \tan \left (\frac {1}{2} \, x\right )^{8} + 80 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{7} + 20 \, b^{5} \tan \left (\frac {1}{2} \, x\right )^{7} - 30 \, a^{5} \tan \left (\frac {1}{2} \, x\right )^{6} - 90 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right )^{6} - 48 \, a^{4} b \tan \left (\frac {1}{2} \, x\right )^{5} + 34 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 8 \, b^{5} \tan \left (\frac {1}{2} \, x\right )^{5} + 10 \, a^{5} \tan \left (\frac {1}{2} \, x\right )^{4} - 50 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right )^{4} + 30 \, a b^{4} \tan \left (\frac {1}{2} \, x\right )^{4} + 80 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 20 \, b^{5} \tan \left (\frac {1}{2} \, x\right )^{3} - 10 \, a^{5} \tan \left (\frac {1}{2} \, x\right )^{2} - 70 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right ) - 2 \, a^{5} - 14 \, a^{3} b^{2} + 3 \, a b^{4}\right )}}{15 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{5}} \]

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)

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maple [A]  time = 0.11, size = 305, normalized size = 1.58 \[ -\frac {16 a^{3} b^{3} \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (8 a^{6}+24 a^{4} b^{2}+24 a^{2} b^{4}+8 b^{6}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (-a^{2} b^{3} \left (\tan ^{9}\left (\frac {x}{2}\right )\right )-a \,b^{4} \left (\tan ^{8}\left (\frac {x}{2}\right )\right )+\left (-\frac {16}{3} a^{2} b^{3}-\frac {4}{3} b^{5}\right ) \left (\tan ^{7}\left (\frac {x}{2}\right )\right )+\left (2 a^{5}+6 a^{3} b^{2}\right ) \left (\tan ^{6}\left (\frac {x}{2}\right )\right )+\left (\frac {16}{5} a^{4} b -\frac {34}{15} a^{2} b^{3}+\frac {8}{15} b^{5}\right ) \left (\tan ^{5}\left (\frac {x}{2}\right )\right )+\left (-\frac {2}{3} a^{5}+\frac {10}{3} a^{3} b^{2}-2 a \,b^{4}\right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+\left (-\frac {16}{3} a^{2} b^{3}-\frac {4}{3} b^{5}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\left (\frac {2}{3} a^{5}+\frac {14}{3} a^{3} b^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-a^{2} b^{3} \tan \left (\frac {x}{2}\right )+\frac {2 a^{5}}{15}+\frac {14 a^{3} b^{2}}{15}-\frac {a \,b^{4}}{5}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{5}} \]

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]

-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))-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

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maxima [B]  time = 0.45, size = 521, normalized size = 2.70 \[ \frac {a^{3} b^{3} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\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 (2 \, a^{5} + 14 \, a^{3} b^{2} - 3 \, a b^{4} - \frac {15 \, a^{2} b^{3} \sin \relax (x)}{\cos \relax (x) + 1} - \frac {15 \, a b^{4} \sin \relax (x)^{8}}{{\left (\cos \relax (x) + 1\right )}^{8}} - \frac {15 \, a^{2} b^{3} \sin \relax (x)^{9}}{{\left (\cos \relax (x) + 1\right )}^{9}} + \frac {10 \, {\left (a^{5} + 7 \, a^{3} b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {20 \, {\left (4 \, a^{2} b^{3} + b^{5}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {10 \, {\left (a^{5} - 5 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {2 \, {\left (24 \, a^{4} b - 17 \, a^{2} b^{3} + 4 \, b^{5}\right )} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {30 \, {\left (a^{5} + 3 \, a^{3} b^{2}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} - \frac {20 \, {\left (4 \, a^{2} b^{3} + b^{5}\right )} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}}{15 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6} + \frac {5 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {10 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {10 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {5 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \relax (x)^{8}}{{\left (\cos \relax (x) + 1\right )}^{8}} + \frac {{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sin \relax (x)^{10}}{{\left (\cos \relax (x) + 1\right )}^{10}}\right )}} \]

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]

a^3*b^3*log((b - a*sin(x)/(cos(x) + 1) + sqrt(a^2 + b^2))/(b - a*sin(x)/(cos(x) + 1) - 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*(2*a^5 + 14*a^3*b^2 - 3*a*b^4 - 15*a^2*b^3*sin(x)/(cos
(x) + 1) - 15*a*b^4*sin(x)^8/(cos(x) + 1)^8 - 15*a^2*b^3*sin(x)^9/(cos(x) + 1)^9 + 10*(a^5 + 7*a^3*b^2)*sin(x)
^2/(cos(x) + 1)^2 - 20*(4*a^2*b^3 + b^5)*sin(x)^3/(cos(x) + 1)^3 - 10*(a^5 - 5*a^3*b^2 + 3*a*b^4)*sin(x)^4/(co
s(x) + 1)^4 + 2*(24*a^4*b - 17*a^2*b^3 + 4*b^5)*sin(x)^5/(cos(x) + 1)^5 + 30*(a^5 + 3*a^3*b^2)*sin(x)^6/(cos(x
) + 1)^6 - 20*(4*a^2*b^3 + b^5)*sin(x)^7/(cos(x) + 1)^7)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + 5*(a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6)*sin(x)^2/(cos(x) + 1)^2 + 10*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(x)^4/(cos(x) + 1)^4
+ 10*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(x)^6/(cos(x) + 1)^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(
x)^8/(cos(x) + 1)^8 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sin(x)^10/(cos(x) + 1)^10)

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mupad [B]  time = 1.64, size = 600, normalized size = 3.11 \[ \frac {\frac {8\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (4\,a^2\,b^3+b^5\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^5+7\,a^3\,b^2\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (a^5+3\,a^3\,b^2\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}-\frac {2\,\left (2\,a^5+14\,a^3\,b^2-3\,a\,b^4\right )}{15\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (a^5-5\,a^3\,b^2+3\,a\,b^4\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {8\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (4\,a^2+b^2\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {2\,a^2\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {2\,a\,b^4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^8}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {2\,a^2\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^9}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}-\frac {4\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (24\,a^4-17\,a^2\,b^2+4\,b^4\right )}{15\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+\frac {2\,a^3\,b^3\,\mathrm {atanh}\left (\frac {2\,a^6\,b+2\,b^7+6\,a^2\,b^5+6\,a^4\,b^3-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}{2\,{\left (a^2+b^2\right )}^{7/2}}\right )}{{\left (a^2+b^2\right )}^{7/2}} \]

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]

((8*tan(x/2)^3*(b^5 + 4*a^2*b^3))/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (4*tan(x/2)^2*(a^5 + 7*a^3*b^2))/(
3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (4*tan(x/2)^6*(a^5 + 3*a^3*b^2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)
- (2*(2*a^5 - 3*a*b^4 + 14*a^3*b^2))/(15*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (4*tan(x/2)^4*(3*a*b^4 + a^5 -
5*a^3*b^2))/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (8*b^3*tan(x/2)^7*(4*a^2 + b^2))/(3*(a^6 + b^6 + 3*a^2*
b^4 + 3*a^4*b^2)) + (2*a^2*b^3*tan(x/2))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (2*a*b^4*tan(x/2)^8)/(a^6 + b^6
+ 3*a^2*b^4 + 3*a^4*b^2) + (2*a^2*b^3*tan(x/2)^9)/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) - (4*b*tan(x/2)^5*(24*a
^4 + 4*b^4 - 17*a^2*b^2))/(15*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))/(5*tan(x/2)^2 + 10*tan(x/2)^4 + 10*tan(x/2
)^6 + 5*tan(x/2)^8 + tan(x/2)^10 + 1) + (2*a^3*b^3*atanh((2*a^6*b + 2*b^7 + 6*a^2*b^5 + 6*a^4*b^3 - 2*a*tan(x/
2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))/(2*(a^2 + b^2)^(7/2))))/(a^2 + b^2)^(7/2)

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

[Out]

Timed out

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