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

Optimal. Leaf size=98 \[ -\frac{b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3}+\frac{2 a b}{\left (a^2+b^2\right )^2 (a \cot (x)+b)}+\frac{a}{2 \left (a^2+b^2\right ) (a \cot (x)+b)^2}+\frac{a \left (a^2-3 b^2\right ) \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^3} \]

[Out]

-((b*(3*a^2 - b^2)*x)/(a^2 + b^2)^3) + a/(2*(a^2 + b^2)*(b + a*Cot[x])^2) + (2*a*b)/((a^2 + b^2)^2*(b + a*Cot[
x])) + (a*(a^2 - 3*b^2)*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^3

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Rubi [A]  time = 0.198367, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3085, 3483, 3529, 3531, 3530} \[ -\frac{b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3}+\frac{2 a b}{\left (a^2+b^2\right )^2 (a \cot (x)+b)}+\frac{a}{2 \left (a^2+b^2\right ) (a \cot (x)+b)^2}+\frac{a \left (a^2-3 b^2\right ) \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(3*a^2 - b^2)*x)/(a^2 + b^2)^3) + a/(2*(a^2 + b^2)*(b + a*Cot[x])^2) + (2*a*b)/((a^2 + b^2)^2*(b + a*Cot[
x])) + (a*(a^2 - 3*b^2)*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^3

Rule 3085

Int[sin[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(b + a*Cot[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [C]  time = 0.771634, size = 114, normalized size = 1.16 \[ \frac{b x \left (b^2-3 a^2\right )}{\left (a^2+b^2\right )^3}+\frac{3 a b \sin (x)}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))}+\frac{a \left (a^2-3 b^2\right ) \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^3}+\frac{a^3}{2 (a-i b)^2 (a+i b)^2 (a \cos (x)+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(-3*a^2 + b^2)*x)/(a^2 + b^2)^3 + (a*(a^2 - 3*b^2)*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^3 + a^3/(2*(a - I*
b)^2*(a + I*b)^2*(a*Cos[x] + b*Sin[x])^2) + (3*a*b*Sin[x])/((a^2 + b^2)^2*(a*Cos[x] + b*Sin[x]))

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Maple [A]  time = 0.108, size = 193, normalized size = 2. \begin{align*}{\frac{{a}^{3}\ln \left ( a+b\tan \left ( x \right ) \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{a\ln \left ( a+b\tan \left ( x \right ) \right ){b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{{a}^{4}}{{b}^{2} \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( x \right ) \right ) }}-3\,{\frac{{a}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( x \right ) \right ) }}+{\frac{{a}^{3}}{2\,{b}^{2} \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( x \right ) \right ) ^{2}}}-{\frac{\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ){a}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{3\,\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) a{b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( x \right ) \right ){a}^{2}b}{ \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( x \right ) \right ){b}^{3}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3/(a^2+b^2)^3*ln(a+b*tan(x))-3*a/(a^2+b^2)^3*ln(a+b*tan(x))*b^2-a^4/(a^2+b^2)^2/b^2/(a+b*tan(x))-3*a^2/(a^2+
b^2)^2/(a+b*tan(x))+1/2*a^3/b^2/(a^2+b^2)/(a+b*tan(x))^2-1/2/(a^2+b^2)^3*ln(tan(x)^2+1)*a^3+3/2/(a^2+b^2)^3*ln
(tan(x)^2+1)*a*b^2-3/(a^2+b^2)^3*arctan(tan(x))*a^2*b+1/(a^2+b^2)^3*arctan(tan(x))*b^3

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Maxima [B]  time = 1.67194, size = 485, normalized size = 4.95 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (-a - \frac{2 \, b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (\frac{2 \, a^{2} b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac{2 \, a^{2} b \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{{\left (a^{3} + 5 \, a b^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + \frac{4 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac{2 \,{\left (a^{6} - 3 \, a^{2} b^{4} - 2 \, b^{6}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac{4 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*a^2*b - b^3)*arctan(sin(x)/(cos(x) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(-a -
2*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (a^3 - 3*a*b^2)*log
(sin(x)^2/(cos(x) + 1)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(2*a^2*b*sin(x)/(cos(x) + 1) - 2*a^2*b*s
in(x)^3/(cos(x) + 1)^3 + (a^3 + 5*a*b^2)*sin(x)^2/(cos(x) + 1)^2)/(a^6 + 2*a^4*b^2 + a^2*b^4 + 4*(a^5*b + 2*a^
3*b^3 + a*b^5)*sin(x)/(cos(x) + 1) - 2*(a^6 - 3*a^2*b^4 - 2*b^6)*sin(x)^2/(cos(x) + 1)^2 - 4*(a^5*b + 2*a^3*b^
3 + a*b^5)*sin(x)^3/(cos(x) + 1)^3 + (a^6 + 2*a^4*b^2 + a^2*b^4)*sin(x)^4/(cos(x) + 1)^4)

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Fricas [B]  time = 0.535953, size = 632, normalized size = 6.45 \begin{align*} \frac{a^{5} + 7 \, a^{3} b^{2} - 2 \,{\left (6 \, a^{3} b^{2} +{\left (3 \, a^{4} b - 4 \, a^{2} b^{3} + b^{5}\right )} x\right )} \cos \left (x\right )^{2} + 2 \,{\left (3 \, a^{4} b - 3 \, a^{2} b^{3} - 2 \,{\left (3 \, a^{3} b^{2} - a b^{4}\right )} x\right )} \cos \left (x\right ) \sin \left (x\right ) - 2 \,{\left (3 \, a^{2} b^{3} - b^{5}\right )} x +{\left (a^{3} b^{2} - 3 \, a b^{4} +{\left (a^{5} - 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (x\right ) \sin \left (x\right )\right )} \log \left (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^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8} +{\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \cos \left (x\right ) \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^5 + 7*a^3*b^2 - 2*(6*a^3*b^2 + (3*a^4*b - 4*a^2*b^3 + b^5)*x)*cos(x)^2 + 2*(3*a^4*b - 3*a^2*b^3 - 2*(3*
a^3*b^2 - a*b^4)*x)*cos(x)*sin(x) - 2*(3*a^2*b^3 - b^5)*x + (a^3*b^2 - 3*a*b^4 + (a^5 - 4*a^3*b^2 + 3*a*b^4)*c
os(x)^2 + 2*(a^4*b - 3*a^2*b^3)*cos(x)*sin(x))*log(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2))/(a^6*b^2
+ 3*a^4*b^4 + 3*a^2*b^6 + b^8 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b
^5 + a*b^7)*cos(x)*sin(x))

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.18225, size = 327, normalized size = 3.34 \begin{align*} -\frac{{\left (3 \, a^{2} b - b^{3}\right )} x}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac{{\left (a^{3} b - 3 \, a b^{3}\right )} \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{3 \, a^{3} b^{4} \tan \left (x\right )^{2} - 9 \, a b^{6} \tan \left (x\right )^{2} + 2 \, a^{6} b \tan \left (x\right ) + 14 \, a^{4} b^{3} \tan \left (x\right ) - 12 \, a^{2} b^{5} \tan \left (x\right ) + a^{7} + 9 \, a^{5} b^{2} - 4 \, a^{3} b^{4}}{2 \,{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )}{\left (b \tan \left (x\right ) + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a^2*b - b^3)*x/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 1/2*(a^3 - 3*a*b^2)*log(tan(x)^2 + 1)/(a^6 + 3*a^4*b^
2 + 3*a^2*b^4 + b^6) + (a^3*b - 3*a*b^3)*log(abs(b*tan(x) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 1/2*(3
*a^3*b^4*tan(x)^2 - 9*a*b^6*tan(x)^2 + 2*a^6*b*tan(x) + 14*a^4*b^3*tan(x) - 12*a^2*b^5*tan(x) + a^7 + 9*a^5*b^
2 - 4*a^3*b^4)/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*(b*tan(x) + a)^2)