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

Optimal. Leaf size=176 \[ \frac {\left (a^2-b^2\right ) \sin ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac {b^2 \left (3 a^2-b^2\right ) \sin (x)}{\left (a^2+b^2\right )^3}-\frac {2 a b \cos ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac {2 a b \left (a^2-b^2\right ) \cos (x)}{\left (a^2+b^2\right )^3}-\frac {a b^2 \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac {a^2 b^3}{\left (a^2+b^2\right )^3 (a \cos (x)+b \sin (x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.70, antiderivative size = 238, normalized size of antiderivative = 1.35, number of steps used = 33, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3111, 3109, 2633, 2565, 30, 3100, 2637, 3074, 206, 2564, 2638, 3155} \[ -\frac {b^2 \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac {a^2 \sin ^3(x)}{3 \left (a^2+b^2\right )^2}+\frac {b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {4 a^2 b^2 \sin (x)}{\left (a^2+b^2\right )^3}-\frac {2 a b \cos ^3(x)}{3 \left (a^2+b^2\right )^2}-\frac {2 a b^3 \cos (x)}{\left (a^2+b^2\right )^3}+\frac {2 a^3 b \cos (x)}{\left (a^2+b^2\right )^3}-\frac {a^2 b^3}{\left (a^2+b^2\right )^3 (a \cos (x)+b \sin (x))}+\frac {2 a b^4 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac {3 a^3 b^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 3155

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 1.27, size = 198, normalized size = 1.12 \[ \frac {2 a b^2 \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}-\frac {-2 a^5 \sin (2 x)+a^5 \sin (4 x)-21 a^4 b+16 a^3 b^2 \sin (2 x)+2 a^3 b^2 \sin (4 x)+90 a^2 b^3+b \left (a^2+b^2\right )^2 \cos (4 x)-4 b \left (3 a^4+a^2 b^2-2 b^4\right ) \cos (2 x)+18 a b^4 \sin (2 x)+a b^4 \sin (4 x)-9 b^5}{24 \left (a^2+b^2\right )^3 (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 369, normalized size = 2.10 \[ \frac {2 \, a^{6} b - 22 \, a^{4} b^{3} - 20 \, a^{2} b^{5} + 4 \, b^{7} - 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{4} + 2 \, {\left (4 \, a^{6} b + 7 \, a^{4} b^{3} + 2 \, a^{2} b^{5} - b^{7}\right )} \cos \relax (x)^{2} - 3 \, \sqrt {a^{2} + b^{2}} {\left ({\left (3 \, a^{4} b^{2} - 2 \, a^{2} b^{4}\right )} \cos \relax (x) + {\left (3 \, a^{3} b^{3} - 2 \, a b^{5}\right )} \sin \relax (x)\right )} \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \relax (x)^{3} - {\left (a^{7} - 2 \, a^{5} b^{2} - 7 \, a^{3} b^{4} - 4 \, a b^{6}\right )} \cos \relax (x)\right )} \sin \relax (x)}{6 \, {\left ({\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} \cos \relax (x) + {\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^6*b - 22*a^4*b^3 - 20*a^2*b^5 + 4*b^7 - 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(x)^4 + 2*(4*a^6*b
+ 7*a^4*b^3 + 2*a^2*b^5 - b^7)*cos(x)^2 - 3*sqrt(a^2 + b^2)*((3*a^4*b^2 - 2*a^2*b^4)*cos(x) + (3*a^3*b^3 - 2*
a*b^5)*sin(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^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(x
)^3 - (a^7 - 2*a^5*b^2 - 7*a^3*b^4 - 4*a*b^6)*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))

________________________________________________________________________________________

giac [A]  time = 0.26, size = 335, normalized size = 1.90 \[ -\frac {{\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (a b^{4} \tan \left (\frac {1}{2} \, x\right ) + a^{2} b^{3}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )}} - \frac {2 \, {\left (9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} - 3 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{5} + 12 \, a b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} - 4 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} - 12 \, a^{3} b \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, a b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right ) - 3 \, b^{4} \tan \left (\frac {1}{2} \, x\right ) - 4 \, a^{3} b + 8 \, a b^{3}\right )}}{3 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a^3*b^2 - 2*a*b^4)*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*b^4*tan(1/2*x) + a^2*b^3)/((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*(9*a^2*b^2*tan(1/2*x)^5 - 3*b^4*tan(1/2*x
)^5 + 12*a*b^3*tan(1/2*x)^4 - 4*a^4*tan(1/2*x)^3 + 18*a^2*b^2*tan(1/2*x)^3 - 2*b^4*tan(1/2*x)^3 - 12*a^3*b*tan
(1/2*x)^2 + 12*a*b^3*tan(1/2*x)^2 + 9*a^2*b^2*tan(1/2*x) - 3*b^4*tan(1/2*x) - 4*a^3*b + 8*a*b^3)/((a^6 + 3*a^4
*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*x)^2 + 1)^3)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 261, normalized size = 1.48 \[ -\frac {2 a \,b^{2} \left (\frac {-\tan \left (\frac {x}{2}\right ) b^{2}-a b}{\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a -2 b \tan \left (\frac {x}{2}\right )-a}-\frac {\left (3 a^{2}-2 b^{2}\right ) \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (a^{2}+b^{2}\right )}+\frac {2 \left (-3 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{5}\left (\frac {x}{2}\right )\right )-8 b^{3} a \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+2 \left (\frac {4}{3} a^{4}-6 a^{2} b^{2}+\frac {2}{3} b^{4}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+2 \left (4 a^{3} b -4 b^{3} a \right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 \left (-3 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {x}{2}\right )+\frac {8 a^{3} b}{3}-\frac {16 b^{3} a}{3}}{\left (a^{2}+b^{2}\right ) \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b^2/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*((-tan(1/2*x)*b^2-a*b)/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-a)-(3*a^2-2*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)))+2/(a^2+b^2)/(a^4+2*a^2*b^2+b^4)*((-3*a^2*b^
2+b^4)*tan(1/2*x)^5-4*b^3*a*tan(1/2*x)^4+(4/3*a^4-6*a^2*b^2+2/3*b^4)*tan(1/2*x)^3+(4*a^3*b-4*a*b^3)*tan(1/2*x)
^2+(-3*a^2*b^2+b^4)*tan(1/2*x)+4/3*a^3*b-8/3*b^3*a)/(tan(1/2*x)^2+1)^3

________________________________________________________________________________________

maxima [B]  time = 0.48, size = 606, normalized size = 3.44 \[ -\frac {{\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} a \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 (4 \, a^{4} b - 11 \, a^{2} b^{3} - \frac {{\left (a^{3} b^{2} + 16 \, a b^{4}\right )} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {{\left (8 \, a^{4} b - 31 \, a^{2} b^{3} + 6 \, b^{5}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {{\left (4 \, a^{5} + 15 \, a^{3} b^{2} - 34 \, a b^{4}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {{\left (4 \, a^{4} b + 45 \, a^{2} b^{3} - 4 \, b^{5}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {{\left (4 \, a^{5} - 9 \, a^{3} b^{2} + 32 \, a b^{4}\right )} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {3 \, {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {3 \, {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}}\right )}}{3 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6} + \frac {2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {6 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {6 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sin \relax (x)^{7}}{{\left (\cos \relax (x) + 1\right )}^{7}} - \frac {{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \sin \relax (x)^{8}}{{\left (\cos \relax (x) + 1\right )}^{8}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a^2*b^2 - 2*b^4)*a*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)) + 2/3*(4*a^4*b - 11*a^2*b^3 - (a^3*b^2 + 16*a*
b^4)*sin(x)/(cos(x) + 1) + (8*a^4*b - 31*a^2*b^3 + 6*b^5)*sin(x)^2/(cos(x) + 1)^2 + (4*a^5 + 15*a^3*b^2 - 34*a
*b^4)*sin(x)^3/(cos(x) + 1)^3 - (4*a^4*b + 45*a^2*b^3 - 4*b^5)*sin(x)^4/(cos(x) + 1)^4 - (4*a^5 - 9*a^3*b^2 +
32*a*b^4)*sin(x)^5/(cos(x) + 1)^5 - 3*(3*a^2*b^3 - 2*b^5)*sin(x)^6/(cos(x) + 1)^6 + 3*(3*a^3*b^2 - 2*a*b^4)*si
n(x)^7/(cos(x) + 1)^7)/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sin(x)/(
cos(x) + 1) + 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*sin(x)^2/(cos(x) + 1)^2 + 6*(a^6*b + 3*a^4*b^3 + 3*a^2*b
^5 + b^7)*sin(x)^3/(cos(x) + 1)^3 + 6*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sin(x)^5/(cos(x) + 1)^5 - 2*(a^7 +
3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*sin(x)^6/(cos(x) + 1)^6 + 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sin(x)^7/(co
s(x) + 1)^7 - (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*sin(x)^8/(cos(x) + 1)^8)

________________________________________________________________________________________

mupad [B]  time = 2.87, size = 586, normalized size = 3.33 \[ -\frac {\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (4\,a^4\,b+45\,a^2\,b^3-4\,b^5\right )}{3\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}-\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (2\,b^5-3\,a^2\,b^3\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (4\,a^5-9\,a^3\,b^2+32\,a\,b^4\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (4\,a^5+15\,a^3\,b^2-34\,a\,b^4\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (8\,a^4\,b-31\,a^2\,b^3+6\,b^5\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,\left (11\,a\,b^3-4\,a^3\,b\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7\,\left (2\,a\,b^3-3\,a^3\,b\right )}{a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^3\,b+16\,a\,b^3\right )}{3\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^8+2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7-2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+6\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+6\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}-\frac {a\,b^2\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^7-a^6\,b\,1{}\mathrm {i}+3{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^5\,b^2-a^4\,b^3\,3{}\mathrm {i}+3{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^4-a^2\,b^5\,3{}\mathrm {i}+1{}\mathrm {i}\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^6-b^7\,1{}\mathrm {i}}{{\left (a^2+b^2\right )}^{7/2}}\right )\,\left (3\,a^2-2\,b^2\right )\,2{}\mathrm {i}}{{\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)^2)/(a*cos(x) + b*sin(x))^2,x)

[Out]

- ((2*tan(x/2)^4*(4*a^4*b - 4*b^5 + 45*a^2*b^3))/(3*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (2*tan(x/2)^6*(2*b^
5 - 3*a^2*b^3))/(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2) + (2*tan(x/2)^5*(32*a*b^4 + 4*a^5 - 9*a^3*b^2))/(3*(a^2 +
b^2)*(a^4 + b^4 + 2*a^2*b^2)) - (2*tan(x/2)^3*(4*a^5 - 34*a*b^4 + 15*a^3*b^2))/(3*(a^2 + b^2)*(a^4 + b^4 + 2*a
^2*b^2)) - (2*tan(x/2)^2*(8*a^4*b + 6*b^5 - 31*a^2*b^3))/(3*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) + (2*a*(11*a*
b^3 - 4*a^3*b))/(3*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) + (2*b*tan(x/2)^7*(2*a*b^3 - 3*a^3*b))/(a^6 + b^6 + 3*
a^2*b^4 + 3*a^4*b^2) + (2*b*tan(x/2)*(16*a*b^3 + a^3*b))/(3*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)))/(a + 2*b*tan
(x/2) + 2*a*tan(x/2)^2 - 2*a*tan(x/2)^6 - a*tan(x/2)^8 + 6*b*tan(x/2)^3 + 6*b*tan(x/2)^5 + 2*b*tan(x/2)^7) - (
a*b^2*atan((a^7*tan(x/2)*1i - a^6*b*1i - b^7*1i - a^2*b^5*3i - a^4*b^3*3i + a^3*b^4*tan(x/2)*3i + a^5*b^2*tan(
x/2)*3i + a*b^6*tan(x/2)*1i)/(a^2 + b^2)^(7/2))*(3*a^2 - 2*b^2)*2i)/(a^2 + b^2)^(7/2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________