### 3.446 $$\int x^3 \coth (a+b x) \text{csch}^2(a+b x) \, dx$$

Optimal. Leaf size=83 $\frac{3 \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^4}-\frac{3 x^2 \coth (a+b x)}{2 b^2}+\frac{3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}-\frac{3 x^2}{2 b^2}$

[Out]

(-3*x^2)/(2*b^2) - (3*x^2*Coth[a + b*x])/(2*b^2) - (x^3*Csch[a + b*x]^2)/(2*b) + (3*x*Log[1 - E^(2*(a + b*x))]
)/b^3 + (3*PolyLog[2, E^(2*(a + b*x))])/(2*b^4)

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Rubi [A]  time = 0.163233, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {5419, 4184, 3716, 2190, 2279, 2391} $\frac{3 \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^4}-\frac{3 x^2 \coth (a+b x)}{2 b^2}+\frac{3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}-\frac{3 x^2}{2 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Coth[a + b*x]*Csch[a + b*x]^2,x]

[Out]

(-3*x^2)/(2*b^2) - (3*x^2*Coth[a + b*x])/(2*b^2) - (x^3*Csch[a + b*x]^2)/(2*b) + (3*x*Log[1 - E^(2*(a + b*x))]
)/b^3 + (3*PolyLog[2, E^(2*(a + b*x))])/(2*b^4)

Rule 5419

Int[Coth[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Csch[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csch[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 4184

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

Rule 3716

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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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]

Rubi steps

\begin{align*} \int x^3 \coth (a+b x) \text{csch}^2(a+b x) \, dx &=-\frac{x^3 \text{csch}^2(a+b x)}{2 b}+\frac{3 \int x^2 \text{csch}^2(a+b x) \, dx}{2 b}\\ &=-\frac{3 x^2 \coth (a+b x)}{2 b^2}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}+\frac{3 \int x \coth (a+b x) \, dx}{b^2}\\ &=-\frac{3 x^2}{2 b^2}-\frac{3 x^2 \coth (a+b x)}{2 b^2}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}-\frac{6 \int \frac{e^{2 (a+b x)} x}{1-e^{2 (a+b x)}} \, dx}{b^2}\\ &=-\frac{3 x^2}{2 b^2}-\frac{3 x^2 \coth (a+b x)}{2 b^2}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}+\frac{3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}-\frac{3 \int \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac{3 x^2}{2 b^2}-\frac{3 x^2 \coth (a+b x)}{2 b^2}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}+\frac{3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}-\frac{3 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=-\frac{3 x^2}{2 b^2}-\frac{3 x^2 \coth (a+b x)}{2 b^2}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}+\frac{3 x \log \left (1-e^{2 (a+b x)}\right )}{b^3}+\frac{3 \text{Li}_2\left (e^{2 (a+b x)}\right )}{2 b^4}\\ \end{align*}

Mathematica [C]  time = 6.1293, size = 228, normalized size = 2.75 $\frac{3 \text{csch}(a) \text{sech}(a) \left (-b^2 x^2 e^{-\tanh ^{-1}(\tanh (a))}+\frac{i \tanh (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (i \tanh ^{-1}(\tanh (a))+i b x\right )}\right )-b x \left (-\pi +2 i \tanh ^{-1}(\tanh (a))\right )-2 \left (i \tanh ^{-1}(\tanh (a))+i b x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\tanh (a))+i b x\right )}\right )+2 i \tanh ^{-1}(\tanh (a)) \log \left (i \sinh \left (\tanh ^{-1}(\tanh (a))+b x\right )\right )-\pi \log \left (e^{2 b x}+1\right )+\pi \log (\cosh (b x))\right )}{\sqrt{1-\tanh ^2(a)}}\right )}{2 b^4 \sqrt{\text{sech}^2(a) \left (\cosh ^2(a)-\sinh ^2(a)\right )}}+\frac{3 x^2 \text{csch}(a) \sinh (b x) \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \text{csch}^2(a+b x)}{2 b}$

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Coth[a + b*x]*Csch[a + b*x]^2,x]

[Out]

-(x^3*Csch[a + b*x]^2)/(2*b) + (3*x^2*Csch[a]*Csch[a + b*x]*Sinh[b*x])/(2*b^2) + (3*Csch[a]*Sech[a]*(-((b^2*x^
2)/E^ArcTanh[Tanh[a]]) + (I*(-(b*x*(-Pi + (2*I)*ArcTanh[Tanh[a]])) - Pi*Log[1 + E^(2*b*x)] - 2*(I*b*x + I*ArcT
anh[Tanh[a]])*Log[1 - E^((2*I)*(I*b*x + I*ArcTanh[Tanh[a]]))] + Pi*Log[Cosh[b*x]] + (2*I)*ArcTanh[Tanh[a]]*Log
[I*Sinh[b*x + ArcTanh[Tanh[a]]]] + I*PolyLog[2, E^((2*I)*(I*b*x + I*ArcTanh[Tanh[a]]))])*Tanh[a])/Sqrt[1 - Tan
h[a]^2]))/(2*b^4*Sqrt[Sech[a]^2*(Cosh[a]^2 - Sinh[a]^2)])

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Maple [B]  time = 0.032, size = 177, normalized size = 2.1 \begin{align*} -{\frac{{x}^{2} \left ( 2\,bx{{\rm e}^{2\,bx+2\,a}}+3\,{{\rm e}^{2\,bx+2\,a}}-3 \right ) }{{b}^{2} \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-3\,{\frac{{x}^{2}}{{b}^{2}}}-6\,{\frac{ax}{{b}^{3}}}-3\,{\frac{{a}^{2}}{{b}^{4}}}+3\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+3\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{4}}}+3\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-3\,{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{4}}}+6\,{\frac{a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(b*x+a)*csch(b*x+a)^3,x)

[Out]

-x^2*(2*b*x*exp(2*b*x+2*a)+3*exp(2*b*x+2*a)-3)/b^2/(exp(2*b*x+2*a)-1)^2-3*x^2/b^2-6/b^3*a*x-3/b^4*a^2+3/b^3*ln
(1+exp(b*x+a))*x+3/b^4*polylog(2,-exp(b*x+a))+3/b^3*ln(1-exp(b*x+a))*x+3/b^4*ln(1-exp(b*x+a))*a+3/b^4*polylog(
2,exp(b*x+a))-3/b^4*a*ln(exp(b*x+a)-1)+6/b^4*a*ln(exp(b*x+a))

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Maxima [A]  time = 1.53725, size = 176, normalized size = 2.12 \begin{align*} \frac{3 \, x^{2} -{\left (2 \, b x^{3} e^{\left (2 \, a\right )} + 3 \, x^{2} e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} - \frac{3 \, x^{2}}{b^{2}} + \frac{3 \,{\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^{4}} + \frac{3 \,{\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^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

(3*x^2 - (2*b*x^3*e^(2*a) + 3*x^2*e^(2*a))*e^(2*b*x))/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) + b^2) - 3*
x^2/b^2 + 3*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^4 + 3*(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x
+ a)))/b^4

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Fricas [B]  time = 2.35122, size = 2545, normalized size = 30.66 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-(3*(b^2*x^2 - a^2)*cosh(b*x + a)^4 + 12*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + 3*(b^2*x^2 - a^2)*sin
h(b*x + a)^4 + (2*b^3*x^3 - 3*b^2*x^2 + 6*a^2)*cosh(b*x + a)^2 + (2*b^3*x^3 - 3*b^2*x^2 + 18*(b^2*x^2 - a^2)*c
osh(b*x + a)^2 + 6*a^2)*sinh(b*x + a)^2 - 3*a^2 - 3*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(
b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a
))*sinh(b*x + a) + 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 3*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a
)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - c
osh(b*x + a))*sinh(b*x + a) + 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 3*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b
*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 - b*x)*sinh(b
*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a)
+ 1) + 3*(a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - 2*a*cosh(b*x + a)^2 + 2
*(3*a*cosh(b*x + a)^2 - a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 - a*cosh(b*x + a))*sinh(b*x + a) + a)*log(co
sh(b*x + a) + sinh(b*x + a) - 1) - 3*((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 +
(b*x + a)*sinh(b*x + a)^4 - 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x +
a)^2 + b*x + 4*((b*x + a)*cosh(b*x + a)^3 - (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(-cosh(b*x + a) -
sinh(b*x + a) + 1) + 2*(6*(b^2*x^2 - a^2)*cosh(b*x + a)^3 + (2*b^3*x^3 - 3*b^2*x^2 + 6*a^2)*cosh(b*x + a))*sin
h(b*x + a))/(b^4*cosh(b*x + a)^4 + 4*b^4*cosh(b*x + a)*sinh(b*x + a)^3 + b^4*sinh(b*x + a)^4 - 2*b^4*cosh(b*x
+ a)^2 + b^4 + 2*(3*b^4*cosh(b*x + a)^2 - b^4)*sinh(b*x + a)^2 + 4*(b^4*cosh(b*x + a)^3 - b^4*cosh(b*x + a))*s
inh(b*x + a))

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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**3*cosh(b*x+a)*csch(b*x+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^3*cosh(b*x + a)*csch(b*x + a)^3, x)