3.28 \(\int (c+d x)^3 \text{csch}^2(a+b x) \, dx\)

Optimal. Leaf size=103 \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}-\frac{3 d^3 \text{PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 d (c+d x)^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{(c+d x)^3 \coth (a+b x)}{b}-\frac{(c+d x)^3}{b} \]

[Out]

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

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Rubi [A]  time = 0.227086, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4184, 3716, 2190, 2531, 2282, 6589} \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}-\frac{3 d^3 \text{PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 d (c+d x)^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{(c+d x)^3 \coth (a+b x)}{b}-\frac{(c+d x)^3}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Csch[a + b*x]^2,x]

[Out]

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

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int (c+d x)^3 \text{csch}^2(a+b x) \, dx &=-\frac{(c+d x)^3 \coth (a+b x)}{b}+\frac{(3 d) \int (c+d x)^2 \coth (a+b x) \, dx}{b}\\ &=-\frac{(c+d x)^3}{b}-\frac{(c+d x)^3 \coth (a+b x)}{b}-\frac{(6 d) \int \frac{e^{2 (a+b x)} (c+d x)^2}{1-e^{2 (a+b x)}} \, dx}{b}\\ &=-\frac{(c+d x)^3}{b}-\frac{(c+d x)^3 \coth (a+b x)}{b}+\frac{3 d (c+d x)^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac{\left (6 d^2\right ) \int (c+d x) \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{(c+d x)^3}{b}-\frac{(c+d x)^3 \coth (a+b x)}{b}+\frac{3 d (c+d x)^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac{3 d^2 (c+d x) \text{Li}_2\left (e^{2 (a+b x)}\right )}{b^3}-\frac{\left (3 d^3\right ) \int \text{Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac{(c+d x)^3}{b}-\frac{(c+d x)^3 \coth (a+b x)}{b}+\frac{3 d (c+d x)^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac{3 d^2 (c+d x) \text{Li}_2\left (e^{2 (a+b x)}\right )}{b^3}-\frac{\left (3 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=-\frac{(c+d x)^3}{b}-\frac{(c+d x)^3 \coth (a+b x)}{b}+\frac{3 d (c+d x)^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}+\frac{3 d^2 (c+d x) \text{Li}_2\left (e^{2 (a+b x)}\right )}{b^3}-\frac{3 d^3 \text{Li}_3\left (e^{2 (a+b x)}\right )}{2 b^4}\\ \end{align*}

Mathematica [A]  time = 2.29771, size = 185, normalized size = 1.8 \[ \frac{-6 d^2 \left (b (c+d x) \text{PolyLog}\left (2,-e^{-a-b x}\right )+d \text{PolyLog}\left (3,-e^{-a-b x}\right )\right )-6 d^2 \left (b (c+d x) \text{PolyLog}\left (2,e^{-a-b x}\right )+d \text{PolyLog}\left (3,e^{-a-b x}\right )\right )-\frac{2 b^3 (c+d x)^3}{e^{2 a}-1}+3 b^2 d (c+d x)^2 \log \left (1-e^{-a-b x}\right )+3 b^2 d (c+d x)^2 \log \left (e^{-a-b x}+1\right )+b^3 \text{csch}(a) \sinh (b x) (c+d x)^3 \text{csch}(a+b x)}{b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Csch[a + b*x]^2,x]

[Out]

((-2*b^3*(c + d*x)^3)/(-1 + E^(2*a)) + 3*b^2*d*(c + d*x)^2*Log[1 - E^(-a - b*x)] + 3*b^2*d*(c + d*x)^2*Log[1 +
E^(-a - b*x)] - 6*d^2*(b*(c + d*x)*PolyLog[2, -E^(-a - b*x)] + d*PolyLog[3, -E^(-a - b*x)]) - 6*d^2*(b*(c + d
*x)*PolyLog[2, E^(-a - b*x)] + d*PolyLog[3, E^(-a - b*x)]) + b^3*(c + d*x)^3*Csch[a]*Csch[a + b*x]*Sinh[b*x])/
b^4

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Maple [B]  time = 0.059, size = 473, normalized size = 4.6 \begin{align*} 12\,{\frac{c{d}^{2}a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+6\,{\frac{c{d}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+6\,{\frac{c{d}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+6\,{\frac{c{d}^{2}\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-6\,{\frac{c{d}^{2}a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{3}}}-12\,{\frac{c{d}^{2}ax}{{b}^{2}}}+6\,{\frac{{d}^{3}{a}^{2}x}{{b}^{3}}}-6\,{\frac{c{d}^{2}{x}^{2}}{b}}-6\,{\frac{c{d}^{2}{a}^{2}}{{b}^{3}}}-6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+3\,{\frac{{c}^{2}d\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+3\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}}+3\,{\frac{{d}^{3}\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}+6\,{\frac{{d}^{3}{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+3\,{\frac{{d}^{3}\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}+6\,{\frac{{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}-3\,{\frac{{d}^{3}{a}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-6\,{\frac{{d}^{3}{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{{d}^{3}{a}^{2}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{4}}}+6\,{\frac{c{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+6\,{\frac{c{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+4\,{\frac{{a}^{3}{d}^{3}}{{b}^{4}}}-2\,{\frac{{d}^{3}{x}^{3}}{b}}-6\,{\frac{{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-6\,{\frac{{d}^{3}{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-2\,{\frac{{d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*csch(b*x+a)^2,x)

[Out]

12*d^2/b^3*c*a*ln(exp(b*x+a))+6*d^2/b^2*c*ln(1-exp(b*x+a))*x+6*d^2/b^3*c*ln(1-exp(b*x+a))*a+6*d^2/b^2*c*ln(1+e
xp(b*x+a))*x-6*d^2/b^3*c*a*ln(exp(b*x+a)-1)-12*d^2/b^2*c*a*x+6*d^3/b^3*a^2*x-6*d^2/b*c*x^2-6*d^2/b^3*c*a^2-6*d
/b^2*c^2*ln(exp(b*x+a))+3*d/b^2*c^2*ln(1+exp(b*x+a))+3*d/b^2*c^2*ln(exp(b*x+a)-1)+3*d^3/b^2*ln(1-exp(b*x+a))*x
^2+6*d^3/b^3*polylog(2,exp(b*x+a))*x+3*d^3/b^2*ln(1+exp(b*x+a))*x^2+6*d^3/b^3*polylog(2,-exp(b*x+a))*x-3*d^3/b
^4*a^2*ln(1-exp(b*x+a))-6*d^3/b^4*a^2*ln(exp(b*x+a))+3*d^3/b^4*a^2*ln(exp(b*x+a)-1)+6*d^2/b^3*c*polylog(2,-exp
(b*x+a))+6*d^2/b^3*c*polylog(2,exp(b*x+a))+4*d^3/b^4*a^3-2*d^3/b*x^3-6*d^3/b^4*polylog(3,-exp(b*x+a))-6*d^3/b^
4*polylog(3,exp(b*x+a))-2/b*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/(exp(2*b*x+2*a)-1)

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Maxima [B]  time = 1.99967, size = 432, normalized size = 4.19 \begin{align*} -3 \, c^{2} 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{6 \,{\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 )} c d^{2}}{b^{3}} + \frac{6 \,{\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 )} c d^{2}}{b^{3}} + \frac{2 \, c^{3}}{b{\left (e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )}} - \frac{2 \,{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2}\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} + \frac{3 \,{\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (b x + a\right )})\right )} d^{3}}{b^{4}} + \frac{3 \,{\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (b x + a\right )})\right )} d^{3}}{b^{4}} - \frac{2 \,{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2}\right )}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

-3*c^2*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) + 6*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))*c*d^2/b^3 + 6*(b*x*log(-e^(b*x + a) + 1)
+ dilog(e^(b*x + a)))*c*d^2/b^3 + 2*c^3/(b*(e^(-2*b*x - 2*a) - 1)) - 2*(d^3*x^3 + 3*c*d^2*x^2)/(b*e^(2*b*x +
2*a) - b) + 3*(b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*polylog(3, -e^(b*x + a)))*d^3/b^4
+ 3*(b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilog(e^(b*x + a)) - 2*polylog(3, e^(b*x + a)))*d^3/b^4 - 2*(b^3*d^
3*x^3 + 3*b^3*c*d^2*x^2)/b^4

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Fricas [C]  time = 2.85253, size = 2689, normalized size = 26.11 \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)^3*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

-(2*b^3*c^3 - 6*a*b^2*c^2*d + 6*a^2*b*c*d^2 - 2*a^3*d^3 + 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3
*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*cosh(b*x + a)^2 + 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3
*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*cosh(b*x + a)*sinh(b*x + a) + 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3
*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*sinh(b*x + a)^2 + 6*(b*d^3*x + b*c*d^2 - (b*d^3*x + b*c*d^
2)*cosh(b*x + a)^2 - 2*(b*d^3*x + b*c*d^2)*cosh(b*x + a)*sinh(b*x + a) - (b*d^3*x + b*c*d^2)*sinh(b*x + a)^2)*
dilog(cosh(b*x + a) + sinh(b*x + a)) + 6*(b*d^3*x + b*c*d^2 - (b*d^3*x + b*c*d^2)*cosh(b*x + a)^2 - 2*(b*d^3*x
+ b*c*d^2)*cosh(b*x + a)*sinh(b*x + a) - (b*d^3*x + b*c*d^2)*sinh(b*x + a)^2)*dilog(-cosh(b*x + a) - sinh(b*x
+ a)) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - (b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cosh(b*x + a)^
2 - 2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cosh(b*x + a)*sinh(b*x + a) - (b^2*d^3*x^2 + 2*b^2*c*d^2*x + b
^2*c^2*d)*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3 - (b^
2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cosh(b*x + a)^2 - 2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cosh(b*x + a)*sinh(b*
x + a) - (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sinh(b*x + a)^2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 3*(b^2*
d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3 - (b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cosh(b
*x + a)^2 - 2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cosh(b*x + a)*sinh(b*x + a) - (b^2*d^3*x^2
+ 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*sinh(b*x + a)^2)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*(d^3*co
sh(b*x + a)^2 + 2*d^3*cosh(b*x + a)*sinh(b*x + a) + d^3*sinh(b*x + a)^2 - d^3)*polylog(3, cosh(b*x + a) + sinh
(b*x + a)) + 6*(d^3*cosh(b*x + a)^2 + 2*d^3*cosh(b*x + a)*sinh(b*x + a) + d^3*sinh(b*x + a)^2 - d^3)*polylog(3
, -cosh(b*x + a) - sinh(b*x + a)))/(b^4*cosh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b*x + a) + b^4*sinh(b*x + a
)^2 - b^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} \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)**3*csch(b*x+a)**2,x)

[Out]

Integral((c + d*x)**3*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 )}^{3} \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)^3*csch(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^3*csch(b*x + a)^2, x)