3.1.7 \(\int x^2 \coth ^2(a+b x) \, dx\) [7]

Optimal. Leaf size=65 \[ -\frac {x^2}{b}+\frac {x^3}{3}-\frac {x^2 \coth (a+b x)}{b}+\frac {2 x \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac {\text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 65, 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 = {3801, 3797, 2221, 2317, 2438, 30} \begin {gather*} \frac {\text {Li}_2\left (e^{2 (a+b x)}\right )}{b^3}+\frac {2 x \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac {x^2 \coth (a+b x)}{b}-\frac {x^2}{b}+\frac {x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Coth[a + b*x]^2,x]

[Out]

-(x^2/b) + x^3/3 - (x^2*Coth[a + b*x])/b + (2*x*Log[1 - E^(2*(a + b*x))])/b^2 + PolyLog[2, E^(2*(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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3797

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

Rule 3801

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

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Mathematica [C] Result contains complex when optimal does not.
time = 3.74, size = 163, normalized size = 2.51 \begin {gather*} \frac {x^3}{3}+\frac {i b \pi x-i \pi \log \left (1+e^{2 b x}\right )+2 b x \log \left (1-e^{-2 \left (b x+\tanh ^{-1}(\tanh (a))\right )}\right )+i \pi \log (\cosh (b x))+2 \tanh ^{-1}(\tanh (a)) \left (b x+\log \left (1-e^{-2 \left (b x+\tanh ^{-1}(\tanh (a))\right )}\right )-\log \left (i \sinh \left (b x+\tanh ^{-1}(\tanh (a))\right )\right )\right )-\text {PolyLog}\left (2,e^{-2 \left (b x+\tanh ^{-1}(\tanh (a))\right )}\right )-b^2 e^{-\tanh ^{-1}(\tanh (a))} x^2 \coth (a) \sqrt {\text {sech}^2(a)}}{b^3}+\frac {x^2 \text {csch}(a) \text {csch}(a+b x) \sinh (b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Coth[a + b*x]^2,x]

[Out]

x^3/3 + (I*b*Pi*x - I*Pi*Log[1 + E^(2*b*x)] + 2*b*x*Log[1 - E^(-2*(b*x + ArcTanh[Tanh[a]]))] + I*Pi*Log[Cosh[b
*x]] + 2*ArcTanh[Tanh[a]]*(b*x + Log[1 - E^(-2*(b*x + ArcTanh[Tanh[a]]))] - Log[I*Sinh[b*x + ArcTanh[Tanh[a]]]
]) - PolyLog[2, E^(-2*(b*x + ArcTanh[Tanh[a]]))] - (b^2*x^2*Coth[a]*Sqrt[Sech[a]^2])/E^ArcTanh[Tanh[a]])/b^3 +
(x^2*Csch[a]*Csch[a + b*x]*Sinh[b*x])/b

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(155\) vs. \(2(63)=126\).
time = 0.97, size = 156, normalized size = 2.40

method	result	size
risch	\(\frac {x^{3}}{3}-\frac {2 x^{2}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {2 x^{2}}{b}-\frac {4 a x}{b^{2}}-\frac {2 a^{2}}{b^{3}}+\frac {2 \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {2 \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{3}}+\frac {2 \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 \ln \left ({\mathrm e}^{b x +a}+1\right ) x}{b^{2}}+\frac {2 \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {4 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*coth(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.33, size = 108, normalized size = 1.66 \begin {gather*} -\frac {2 \, x^{2}}{b} + \frac {b x^{3} e^{\left (2 \, b x + 2 \, a\right )} - b x^{3} - 6 \, x^{2}}{3 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} - b\right )}} + \frac {2 \, {\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{3}} + \frac {2 \, {\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^2/b + 1/3*(b*x^3*e^(2*b*x + 2*a) - b*x^3 - 6*x^2)/(b*e^(2*b*x + 2*a) - b) + 2*(b*x*log(e^(b*x + a) + 1) +
dilog(-e^(b*x + a)))/b^3 + 2*(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 453 vs. \(2 (62) = 124\).
time = 0.35, size = 453, normalized size = 6.97 \begin {gather*} -\frac {b^{3} x^{3} - {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 6 \, a^{2} - 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 6 \, {\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} - a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, {\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )^{2} - b x - a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )}{3 \, {\left (b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} - b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 - (b^3*x^3 - 6*b^2*x^2 + 6*a^2)*cosh(b*x + a)^2 - 2*(b^3*x^3 - 6*b^2*x^2 + 6*a^2)*cosh(b*x + a)*
sinh(b*x + a) - (b^3*x^3 - 6*b^2*x^2 + 6*a^2)*sinh(b*x + a)^2 + 6*a^2 - 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*s
inh(b*x + a) + sinh(b*x + a)^2 - 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a
)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 6*(b*x*cosh(b*x + a)^2 + 2*b*x*
cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 - b*x)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 6*(a*cosh(b*
x + a)^2 + 2*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2 - a)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 6
*((b*x + a)*cosh(b*x + a)^2 + 2*(b*x + a)*cosh(b*x + a)*sinh(b*x + a) + (b*x + a)*sinh(b*x + a)^2 - b*x - a)*l
og(-cosh(b*x + a) - sinh(b*x + a) + 1))/(b^3*cosh(b*x + a)^2 + 2*b^3*cosh(b*x + a)*sinh(b*x + a) + b^3*sinh(b*
x + a)^2 - b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \coth ^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*coth(a + b*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*coth(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,{\mathrm {coth}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*coth(a + b*x)^2,x)

[Out]

int(x^2*coth(a + b*x)^2, x)

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