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

Optimal. Leaf size=123 \[ -\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac{b \sin (x)}{a^2+b^2}-\frac{a^2 b \sin (x)}{\left (a^2+b^2\right )^2}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \cos (x)}{\left (a^2+b^2\right )^2}+\frac{a b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.157631, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3109, 2633, 2565, 30, 3100, 2637, 3074, 206} \[ -\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac{b \sin (x)}{a^2+b^2}-\frac{a^2 b \sin (x)}{\left (a^2+b^2\right )^2}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a b^2 \cos (x)}{\left (a^2+b^2\right )^2}+\frac{a b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^3*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a*b^2*Cos[x])/(a^2 + b^2)^2 - (a*C
os[x]^3)/(3*(a^2 + b^2)) - (a^2*b*Sin[x])/(a^2 + b^2)^2 + (b*Sin[x])/(a^2 + b^2) - (b*Sin[x]^3)/(3*(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 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 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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 (x)}{a \cos (x)+b \sin (x)} \, dx &=\frac{a \int \cos ^2(x) \sin (x) \, dx}{a^2+b^2}+\frac{b \int \cos ^3(x) \, dx}{a^2+b^2}-\frac{(a b) \int \frac{\cos ^2(x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}\\ &=-\frac{a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac{\left (a^2 b\right ) \int \cos (x) \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (a b^3\right ) \int \frac{1}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int x^2 \, dx,x,\cos (x)\right )}{a^2+b^2}-\frac{b \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (x)\right )}{a^2+b^2}\\ &=-\frac{a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^2 b \sin (x)}{\left (a^2+b^2\right )^2}+\frac{b \sin (x)}{a^2+b^2}-\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac{\left (a 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 )^2}\\ &=\frac{a b^3 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac{a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac{a^2 b \sin (x)}{\left (a^2+b^2\right )^2}+\frac{b \sin (x)}{a^2+b^2}-\frac{b \sin ^3(x)}{3 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [A]  time = 1.06909, size = 112, normalized size = 0.91 \[ -\frac{3 a \left (a^2+5 b^2\right ) \cos (x)+a \left (a^2+b^2\right ) \cos (3 x)-2 b \sin (x) \left (\left (a^2+b^2\right ) \cos (2 x)-a^2+5 b^2\right )}{12 \left (a^2+b^2\right )^2}-\frac{2 a b^3 \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b^3*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (3*a*(a^2 + 5*b^2)*Cos[x] + a*(a^2 +
b^2)*Cos[3*x] - 2*b*(-a^2 + 5*b^2 + (a^2 + b^2)*Cos[2*x])*Sin[x])/(12*(a^2 + b^2)^2)

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Maple [A]  time = 0.091, size = 170, normalized size = 1.4 \begin{align*} -2\,{\frac{-{b}^{3} \left ( \tan \left ( x/2 \right ) \right ) ^{5}+ \left ({a}^{3}+2\,a{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{4}+ \left ( 4/3\,{a}^{2}b-2/3\,{b}^{3} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{3}+2\,a{b}^{2} \left ( \tan \left ( x/2 \right ) \right ) ^{2}-{b}^{3}\tan \left ( x/2 \right ) +1/3\,{a}^{3}+4/3\,a{b}^{2}}{ \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{b}^{3}}{ \left ( 2\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \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)/(a*cos(x)+b*sin(x)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.53742, size = 494, normalized size = 4.02 \begin{align*} \frac{3 \, \sqrt{a^{2} + b^{2}} a 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 ) - 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} - 6 \,{\left (a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right ) - 2 \,{\left (a^{4} b - a^{2} b^{3} - 2 \, b^{5} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.21764, size = 271, normalized size = 2.2 \begin{align*} \frac{a 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^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (3 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{5} - 3 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{4} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 3 \, b^{3} \tan \left (\frac{1}{2} \, x\right ) - a^{3} - 4 \, a b^{2}\right )}}{3 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\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)^3*sin(x)/(a*cos(x)+b*sin(x)),x, algorithm="giac")

[Out]

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