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

Optimal. Leaf size=59 \[ \frac{x}{b (a+b \cos (x))}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b \sqrt{a-b} \sqrt{a+b}} \]

[Out]

(-2*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]) + x/(b*(a + b*Cos[x]))

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Rubi [A]  time = 0.0592501, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4423, 2659, 205} \[ \frac{x}{b (a+b \cos (x))}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b \sqrt{a-b} \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sin[x])/(a + b*Cos[x])^2,x]

[Out]

(-2*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]) + x/(b*(a + b*Cos[x]))

Rule 4423

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.103584, size = 58, normalized size = 0.98 \[ \frac{2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{b \sqrt{b^2-a^2}}+\frac{x}{b (a+b \cos (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.125, size = 154, normalized size = 2.6 \begin{align*} 2\,{\frac{x{{\rm e}^{ix}}}{b \left ( b{{\rm e}^{2\,ix}}+2\,a{{\rm e}^{ix}}+b \right ) }}-{\frac{i}{b}\ln \left ({{\rm e}^{ix}}+{\frac{1}{b} \left ( a\sqrt{{a}^{2}-{b}^{2}}+{a}^{2}-{b}^{2} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}+{\frac{i}{b}\ln \left ({{\rm e}^{ix}}+{\frac{1}{b} \left ( a\sqrt{{a}^{2}-{b}^{2}}-{a}^{2}+{b}^{2} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*exp(I*x)/b/(b*exp(2*I*x)+2*a*exp(I*x)+b)-I/(a^2-b^2)^(1/2)/b*ln(exp(I*x)+(a*(a^2-b^2)^(1/2)+a^2-b^2)/(a^2-
b^2)^(1/2)/b)+I/(a^2-b^2)^(1/2)/b*ln(exp(I*x)+(a*(a^2-b^2)^(1/2)-a^2+b^2)/(a^2-b^2)^(1/2)/b)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.52556, size = 516, normalized size = 8.75 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (x\right ) + a\right )} \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) - 2 \,{\left (a^{2} - b^{2}\right )} x}{2 \,{\left (a^{3} b - a b^{3} +{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right )\right )}}, -\frac{\sqrt{a^{2} - b^{2}}{\left (b \cos \left (x\right ) + a\right )} \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) -{\left (a^{2} - b^{2}\right )} x}{a^{3} b - a b^{3} +{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*(b*cos(x) + a)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(
x) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) - 2*(a^2 - b^2)*x)/(a^3*b - a*b^3 + (a^2*b^
2 - b^4)*cos(x)), -(sqrt(a^2 - b^2)*(b*cos(x) + a)*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x))) - (a^2 - b
^2)*x)/(a^3*b - a*b^3 + (a^2*b^2 - b^4)*cos(x))]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin \left (x\right )}{{\left (b \cos \left (x\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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