3.29 \(\int (c+d x)^2 \text{csch}^2(a+b x) \, dx\)

Optimal. Leaf size=74 \[ \frac{d^2 \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}+\frac{2 d (c+d x) \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{(c+d x)^2 \coth (a+b x)}{b}-\frac{(c+d x)^2}{b} \]

[Out]

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

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Rubi [A]  time = 0.147543, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4184, 3716, 2190, 2279, 2391} \[ \frac{d^2 \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}+\frac{2 d (c+d x) \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{(c+d x)^2 \coth (a+b x)}{b}-\frac{(c+d x)^2}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csch[a + b*x]^2,x]

[Out]

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

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

Mathematica [C]  time = 5.39859, size = 198, normalized size = 2.68 \[ \frac{\text{csch}(a) \left (d^2 \left (-\sinh (a) \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}(\tanh (a))+b x\right )}\right )-b^2 x^2 \cosh (a) e^{-\tanh ^{-1}(\tanh (a))} \sqrt{\text{sech}^2(a)}+i \pi b x \sinh (a)-i \pi \sinh (a) \log \left (e^{2 b x}+1\right )+2 b x \sinh (a) \log \left (1-e^{-2 \left (\tanh ^{-1}(\tanh (a))+b x\right )}\right )+2 \sinh (a) \tanh ^{-1}(\tanh (a)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\tanh (a))+b x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\tanh (a))+b x\right )\right )+b x\right )+i \pi \sinh (a) \log (\cosh (b x))\right )+b^2 \sinh (b x) (c+d x)^2 \text{csch}(a+b x)-2 b c d (b x \cosh (a)-\sinh (a) \log (\sinh (a+b x)))\right )}{b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Csch[a + b*x]^2,x]

[Out]

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

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Maple [B]  time = 0.03, size = 240, normalized size = 3.2 \begin{align*} -2\,{\frac{{d}^{2}{x}^{2}+2\,cdx+{c}^{2}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-4\,{\frac{cd\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}}-2\,{\frac{{d}^{2}{x}^{2}}{b}}-4\,{\frac{a{d}^{2}x}{{b}^{2}}}-2\,{\frac{{a}^{2}{d}^{2}}{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+4\,{\frac{a{d}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{a{d}^{2}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*csch(b*x+a)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -2 \, d^{2}{\left (\frac{x^{2}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} + 2 \, \int \frac{x}{2 \,{\left (b e^{\left (b x + a\right )} + b\right )}}\,{d x} - 2 \, \int \frac{x}{2 \,{\left (b e^{\left (b x + a\right )} - b\right )}}\,{d x}\right )} - 2 \, c d{\left (\frac{2 \, x e^{\left (2 \, b x + 2 \, a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac{\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{2}} - \frac{\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{2}}\right )} + \frac{2 \, c^{2}}{b{\left (e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

-2*d^2*(x^2/(b*e^(2*b*x + 2*a) - b) + 2*integrate(1/2*x/(b*e^(b*x + a) + b), x) - 2*integrate(1/2*x/(b*e^(b*x
+ a) - b), x)) - 2*c*d*(2*x*e^(2*b*x + 2*a)/(b*e^(2*b*x + 2*a) - b) - log((e^(b*x + a) + 1)*e^(-a))/b^2 - log(
(e^(b*x + a) - 1)*e^(-a))/b^2) + 2*c^2/(b*(e^(-2*b*x - 2*a) - 1))

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Fricas [B]  time = 2.78976, size = 1521, normalized size = 20.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

-2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2 + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)^2 + 2*(b^2
*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)*sinh(b*x + a) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b
*c*d - a^2*d^2)*sinh(b*x + a)^2 - (d^2*cosh(b*x + a)^2 + 2*d^2*cosh(b*x + a)*sinh(b*x + a) + d^2*sinh(b*x + a)
^2 - d^2)*dilog(cosh(b*x + a) + sinh(b*x + a)) - (d^2*cosh(b*x + a)^2 + 2*d^2*cosh(b*x + a)*sinh(b*x + a) + d^
2*sinh(b*x + a)^2 - d^2)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + (b*d^2*x + b*c*d - (b*d^2*x + b*c*d)*cosh(b*x
+ a)^2 - 2*(b*d^2*x + b*c*d)*cosh(b*x + a)*sinh(b*x + a) - (b*d^2*x + b*c*d)*sinh(b*x + a)^2)*log(cosh(b*x +
a) + sinh(b*x + a) + 1) + (b*c*d - a*d^2 - (b*c*d - a*d^2)*cosh(b*x + a)^2 - 2*(b*c*d - a*d^2)*cosh(b*x + a)*s
inh(b*x + a) - (b*c*d - a*d^2)*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + (b*d^2*x + a*d^2 - (b
*d^2*x + a*d^2)*cosh(b*x + a)^2 - 2*(b*d^2*x + a*d^2)*cosh(b*x + a)*sinh(b*x + a) - (b*d^2*x + a*d^2)*sinh(b*x
+ a)^2)*log(-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., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} \operatorname{csch}^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*csch(b*x+a)**2,x)

[Out]

Integral((c + d*x)**2*csch(a + b*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*csch(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*csch(b*x + a)^2, x)