3.7 \(\int x^2 \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=74 \[ -\frac {i \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {x^2 \cot (a+b x)}{b}-\frac {i x^2}{b}-\frac {x^3}{3} \]

[Out]

-I*x^2/b-1/3*x^3-x^2*cot(b*x+a)/b+2*x*ln(1-exp(2*I*(b*x+a)))/b^2-I*polylog(2,exp(2*I*(b*x+a)))/b^3

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Rubi [A]  time = 0.12, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3720, 3717, 2190, 2279, 2391, 30} \[ -\frac {i \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac {2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {x^2 \cot (a+b x)}{b}-\frac {i x^2}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cot[a + b*x]^2,x]

[Out]

((-I)*x^2)/b - x^3/3 - (x^2*Cot[a + b*x])/b + (2*x*Log[1 - E^((2*I)*(a + b*x))])/b^2 - (I*PolyLog[2, E^((2*I)*
(a + b*x))])/b^3

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3720

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

Rubi steps

\begin {align*} \int x^2 \cot ^2(a+b x) \, dx &=-\frac {x^2 \cot (a+b x)}{b}+\frac {2 \int x \cot (a+b x) \, dx}{b}-\int x^2 \, dx\\ &=-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {x^2 \cot (a+b x)}{b}-\frac {(4 i) \int \frac {e^{2 i (a+b x)} x}{1-e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {x^2 \cot (a+b x)}{b}+\frac {2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {2 \int \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {x^2 \cot (a+b x)}{b}+\frac {2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}+\frac {i \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{b^3}\\ &=-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {x^2 \cot (a+b x)}{b}+\frac {2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {i \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}\\ \end {align*}

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Mathematica [B]  time = 5.35, size = 153, normalized size = 2.07 \[ \frac {-b^2 x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt {\sec ^2(a)}-i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+i b x \left (\pi -2 \tan ^{-1}(\tan (a))\right )+2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+\pi \log \left (1+e^{-2 i b x}\right )-\pi \log (\cos (b x))}{b^3}+\frac {x^2 \csc (a) \sin (b x) \csc (a+b x)}{b}-\frac {x^3}{3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Cot[a + b*x]^2,x]

[Out]

-1/3*x^3 + (I*b*x*(Pi - 2*ArcTan[Tan[a]]) + Pi*Log[1 + E^((-2*I)*b*x)] + 2*(b*x + ArcTan[Tan[a]])*Log[1 - E^((
2*I)*(b*x + ArcTan[Tan[a]]))] - Pi*Log[Cos[b*x]] - 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] - I*PolyLog
[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))] - b^2*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a]^2])/b^3 + (x^2*Csc[a]*
Csc[a + b*x]*Sin[b*x])/b

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fricas [B]  time = 1.16, size = 281, normalized size = 3.80 \[ -\frac {2 \, b^{3} x^{3} \sin \left (2 \, b x + 2 \, a\right ) + 6 \, b^{2} x^{2} \cos \left (2 \, b x + 2 \, a\right ) + 6 \, b^{2} x^{2} + 6 \, a \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) + 6 \, a \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b x + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b x + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 i \, {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right )}{6 \, b^{3} \sin \left (2 \, b x + 2 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^3*x^3*sin(2*b*x + 2*a) + 6*b^2*x^2*cos(2*b*x + 2*a) + 6*b^2*x^2 + 6*a*log(-1/2*cos(2*b*x + 2*a) + 1/
2*I*sin(2*b*x + 2*a) + 1/2)*sin(2*b*x + 2*a) + 6*a*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*a) + 1/2)*s
in(2*b*x + 2*a) - 6*(b*x + a)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1)*sin(2*b*x + 2*a) - 6*(b*x + a)*l
og(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1)*sin(2*b*x + 2*a) + 3*I*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2
*a))*sin(2*b*x + 2*a) - 3*I*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a))*sin(2*b*x + 2*a))/(b^3*sin(2*b*x + 2*
a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cot \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*cot(b*x + a)^2, x)

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maple [B]  time = 0.90, size = 183, normalized size = 2.47 \[ -\frac {x^{3}}{3}-\frac {2 i x^{2}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {2 i x^{2}}{b}-\frac {4 i a x}{b^{2}}-\frac {2 i a^{2}}{b^{3}}+\frac {2 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 i \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3-2*I*x^2/b/(exp(2*I*(b*x+a))-1)-2*I/b*x^2-4*I/b^2*a*x-2*I/b^3*a^2+2/b^2*ln(exp(I*(b*x+a))+1)*x-2*I/b^3
*polylog(2,-exp(I*(b*x+a)))+2/b^2*ln(1-exp(I*(b*x+a)))*x+2/b^3*ln(1-exp(I*(b*x+a)))*a-2*I/b^3*polylog(2,exp(I*
(b*x+a)))+4/b^3*a*ln(exp(I*(b*x+a)))-2/b^3*a*ln(exp(I*(b*x+a))-1)

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maxima [B]  time = 0.62, size = 384, normalized size = 5.19 \[ \frac {-i \, b^{3} x^{3} + 6 \, {\left (b x \cos \left (2 \, b x + 2 \, a\right ) + i \, b x \sin \left (2 \, b x + 2 \, a\right ) - b x\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 6 \, {\left (b x \cos \left (2 \, b x + 2 \, a\right ) + i \, b x \sin \left (2 \, b x + 2 \, a\right ) - b x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + {\left (i \, b^{3} x^{3} - 6 \, b^{2} x^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) - 1\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) - 1\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left (-3 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b x \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + {\left (-3 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b x \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (b^{3} x^{3} + 6 i \, b^{2} x^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{-3 i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b^{3} \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-I*b^3*x^3 + 6*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) - b*x)*arctan2(sin(b*x + a), cos(b*x + a) + 1)
- 6*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) - b*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (I*b^3*x^
3 - 6*b^2*x^2)*cos(2*b*x + 2*a) - 6*(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) - 1)*dilog(-e^(I*b*x + I*a)) - 6*(c
os(2*b*x + 2*a) + I*sin(2*b*x + 2*a) - 1)*dilog(e^(I*b*x + I*a)) + (-3*I*b*x*cos(2*b*x + 2*a) + 3*b*x*sin(2*b*
x + 2*a) + 3*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (-3*I*b*x*cos(2*b*x + 2*a) + 3
*b*x*sin(2*b*x + 2*a) + 3*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - (b^3*x^3 + 6*I*b^
2*x^2)*sin(2*b*x + 2*a))/(-3*I*b^3*cos(2*b*x + 2*a) + 3*b^3*sin(2*b*x + 2*a) + 3*I*b^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {cot}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cot(a + b*x)^2,x)

[Out]

int(x^2*cot(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cot ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*cot(a + b*x)**2, x)

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