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

Optimal. Leaf size=15 \[ \frac {1}{2 a (a \cot (x)+b)^2} \]

[Out]

1/2/a/(b+a*cot(x))^2

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Rubi [A]  time = 0.03, antiderivative size = 19, normalized size of antiderivative = 1.27, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3087, 37} \[ \frac {\tan ^2(x)}{2 a (a+b \tan (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

Tan[x]^2/(2*a*(a + b*Tan[x])^2)

Rule 37

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

Rule 3087

Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[1/d, Subst[Int[(x^m*(a + b*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Tan[c + d*x]], x] /; FreeQ[{a,
b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

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

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Mathematica [B]  time = 0.10, size = 47, normalized size = 3.13 \[ \frac {a (a+b \sin (2 x))+2 b^2 \sin ^2(x)}{2 a \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.65, size = 116, normalized size = 7.73 \[ -\frac {4 \, a b^{2} \cos \relax (x)^{2} - a^{3} - 3 \, a b^{2} - 2 \, {\left (a^{2} b - b^{3}\right )} \cos \relax (x) \sin \relax (x)}{2 \, {\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6} + {\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \relax (x) \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 20, normalized size = 1.33 \[ -\frac {2 \, b \tan \relax (x) + a}{2 \, {\left (b \tan \relax (x) + a\right )}^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*tan(x) + a)/((b*tan(x) + a)^2*b^2)

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maple [B]  time = 0.69, size = 29, normalized size = 1.93 \[ \frac {a}{2 b^{2} \left (a +b \tan \relax (x )\right )^{2}}-\frac {1}{b^{2} \left (a +b \tan \relax (x )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a/b^2/(a+b*tan(x))^2-1/b^2/(a+b*tan(x))

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maxima [B]  time = 0.34, size = 84, normalized size = 5.60 \[ \frac {2 \, \sin \relax (x)^{2}}{{\left (a^{3} + \frac {4 \, a^{2} b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {4 \, a^{2} b \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {a^{3} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {2 \, {\left (a^{3} - 2 \, a b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )} {\left (\cos \relax (x) + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.49, size = 48, normalized size = 3.20 \[ \frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a-\frac {2\,a^2-4\,b^2}{2\,a}\right )}{b^2\,{\left (-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a*cos(x) + b*sin(x))^3,x)

[Out]

(tan(x/2)^2*(a - (2*a^2 - 4*b^2)/(2*a)))/(b^2*(a + 2*b*tan(x/2) - a*tan(x/2)^2)^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(sin(x)/(a*cos(x)+b*sin(x))**3,x)

[Out]

Timed out

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