3.158 \(\int \frac{x \csc ^2(x)}{(a+b \cot (x))^2} \, dx\)

Optimal. Leaf size=50 \[ -\frac{a x}{b \left (a^2+b^2\right )}+\frac{\log (a \sin (x)+b \cos (x))}{a^2+b^2}+\frac{x}{b (a+b \cot (x))} \]

[Out]

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

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Rubi [A]  time = 0.0817342, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4425, 3484, 3530} \[ -\frac{a x}{b \left (a^2+b^2\right )}+\frac{\log (a \sin (x)+b \cos (x))}{a^2+b^2}+\frac{x}{b (a+b \cot (x))} \]

Antiderivative was successfully verified.

[In]

Int[(x*Csc[x]^2)/(a + b*Cot[x])^2,x]

[Out]

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

Rule 4425

Int[Csc[(c_.) + (d_.)*(x_)]^2*(Cot[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbo
l] :> -Simp[((e + f*x)^m*(a + b*Cot[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*Cot[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -
1]

Rule 3484

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2),
 Int[(b - a*Tan[c + d*x])/(a + b*Tan[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 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{x \csc ^2(x)}{(a+b \cot (x))^2} \, dx &=\frac{x}{b (a+b \cot (x))}-\frac{\int \frac{1}{a+b \cot (x)} \, dx}{b}\\ &=-\frac{a x}{b \left (a^2+b^2\right )}+\frac{x}{b (a+b \cot (x))}+\frac{\int \frac{-b+a \cot (x)}{a+b \cot (x)} \, dx}{a^2+b^2}\\ &=-\frac{a x}{b \left (a^2+b^2\right )}+\frac{x}{b (a+b \cot (x))}+\frac{\log (b \cos (x)+a \sin (x))}{a^2+b^2}\\ \end{align*}

Mathematica [A]  time = 0.182409, size = 48, normalized size = 0.96 \[ \frac{b \log (a \sin (x)+b \cos (x))-a x}{a^2 b+b^3}+\frac{x \sin (x)}{a b \sin (x)+b^2 \cos (x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Csc[x]^2)/(a + b*Cot[x])^2,x]

[Out]

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

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Maple [C]  time = 0.147, size = 87, normalized size = 1.7 \begin{align*}{\frac{-2\,ix}{{a}^{2}+{b}^{2}}}-{\frac{2\,ix}{ \left ( ib{{\rm e}^{2\,ix}}+a{{\rm e}^{2\,ix}}+ib-a \right ) \left ( ib+a \right ) }}+{\frac{1}{{a}^{2}+{b}^{2}}\ln \left ({{\rm e}^{2\,ix}}+{\frac{ib-a}{ib+a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*csc(x)^2/(a+b*cot(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.05116, size = 338, normalized size = 6.76 \begin{align*} -\frac{8 \, a b x \cos \left (2 \, x\right ) + 4 \,{\left (a^{2} - b^{2}\right )} x \sin \left (2 \, x\right ) -{\left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, x\right )^{2} + 4 \, a b \sin \left (2 \, x\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, x\right )^{2} + a^{2} + b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, x\right )\right )} \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, x\right )^{2} + 4 \, a b \sin \left (2 \, x\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, x\right )^{2} + a^{2} + b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, x\right )}{a^{2} + b^{2}}\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (2 \, x\right )^{2} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (2 \, x\right )^{2} - 2 \,{\left (a^{4} - b^{4}\right )} \cos \left (2 \, x\right ) + 4 \,{\left (a^{3} b + a b^{3}\right )} \sin \left (2 \, x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(x)^2/(a+b*cot(x))^2,x, algorithm="maxima")

[Out]

-1/2*(8*a*b*x*cos(2*x) + 4*(a^2 - b^2)*x*sin(2*x) - ((a^2 + b^2)*cos(2*x)^2 + 4*a*b*sin(2*x) + (a^2 + b^2)*sin
(2*x)^2 + a^2 + b^2 - 2*(a^2 - b^2)*cos(2*x))*log(((a^2 + b^2)*cos(2*x)^2 + 4*a*b*sin(2*x) + (a^2 + b^2)*sin(2
*x)^2 + a^2 + b^2 - 2*(a^2 - b^2)*cos(2*x))/(a^2 + b^2)))/(a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cos
(2*x)^2 + (a^4 + 2*a^2*b^2 + b^4)*sin(2*x)^2 - 2*(a^4 - b^4)*cos(2*x) + 4*(a^3*b + a*b^3)*sin(2*x))

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Fricas [A]  time = 2.64612, size = 216, normalized size = 4.32 \begin{align*} -\frac{2 \, a x \cos \left (x\right ) - 2 \, b x \sin \left (x\right ) -{\left (b \cos \left (x\right ) + a \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} + a^{2}\right )}{2 \,{\left ({\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) +{\left (a^{3} + a b^{2}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \csc ^{2}{\left (x \right )}}{\left (a + b \cot{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*csc(x)**2/(a+b*cot(x))**2,x)

[Out]

Integral(x*csc(x)**2/(a + b*cot(x))**2, x)

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Giac [B]  time = 1.30335, size = 435, normalized size = 8.7 \begin{align*} -\frac{2 \, a x \tan \left (\frac{1}{2} \, x\right )^{2} - b \log \left (\frac{4 \,{\left (b^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} + 4 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + b^{2}\right )}}{\tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, b x \tan \left (\frac{1}{2} \, x\right ) + 2 \, a \log \left (\frac{4 \,{\left (b^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} + 4 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + b^{2}\right )}}{\tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, x\right ) - 2 \, a x + b \log \left (\frac{4 \,{\left (b^{2} \tan \left (\frac{1}{2} \, x\right )^{4} - 4 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} + 4 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + b^{2}\right )}}{\tan \left (\frac{1}{2} \, x\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, x\right )^{2} + 1}\right )}{2 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, x\right )^{2} + b^{3} \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, a^{3} \tan \left (\frac{1}{2} \, x\right ) - 2 \, a b^{2} \tan \left (\frac{1}{2} \, x\right ) - a^{2} b - b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*x*tan(1/2*x)^2 - b*log(4*(b^2*tan(1/2*x)^4 - 4*a*b*tan(1/2*x)^3 + 4*a^2*tan(1/2*x)^2 - 2*b^2*tan(1/2
*x)^2 + 4*a*b*tan(1/2*x) + b^2)/(tan(1/2*x)^4 + 2*tan(1/2*x)^2 + 1))*tan(1/2*x)^2 + 4*b*x*tan(1/2*x) + 2*a*log
(4*(b^2*tan(1/2*x)^4 - 4*a*b*tan(1/2*x)^3 + 4*a^2*tan(1/2*x)^2 - 2*b^2*tan(1/2*x)^2 + 4*a*b*tan(1/2*x) + b^2)/
(tan(1/2*x)^4 + 2*tan(1/2*x)^2 + 1))*tan(1/2*x) - 2*a*x + b*log(4*(b^2*tan(1/2*x)^4 - 4*a*b*tan(1/2*x)^3 + 4*a
^2*tan(1/2*x)^2 - 2*b^2*tan(1/2*x)^2 + 4*a*b*tan(1/2*x) + b^2)/(tan(1/2*x)^4 + 2*tan(1/2*x)^2 + 1)))/(a^2*b*ta
n(1/2*x)^2 + b^3*tan(1/2*x)^2 - 2*a^3*tan(1/2*x) - 2*a*b^2*tan(1/2*x) - a^2*b - b^3)