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

Optimal. Leaf size=99 \[ -\frac{2 d (c+d x) \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{2 d (c+d x) \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{2 d^2 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{2 d^2 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

(-2*(c + d*x)^2*ArcTanh[E^(a + b*x)])/b - (2*d*(c + d*x)*PolyLog[2, -E^(a + b*x)])/b^2 + (2*d*(c + d*x)*PolyLo
g[2, E^(a + b*x)])/b^2 + (2*d^2*PolyLog[3, -E^(a + b*x)])/b^3 - (2*d^2*PolyLog[3, E^(a + b*x)])/b^3

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Rubi [A]  time = 0.0903547, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4182, 2531, 2282, 6589} \[ -\frac{2 d (c+d x) \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{2 d (c+d x) \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{2 d^2 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{2 d^2 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csch[a + b*x],x]

[Out]

(-2*(c + d*x)^2*ArcTanh[E^(a + b*x)])/b - (2*d*(c + d*x)*PolyLog[2, -E^(a + b*x)])/b^2 + (2*d*(c + d*x)*PolyLo
g[2, E^(a + b*x)])/b^2 + (2*d^2*PolyLog[3, -E^(a + b*x)])/b^3 - (2*d^2*PolyLog[3, E^(a + b*x)])/b^3

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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)^2 \text{csch}(a+b x) \, dx &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{(2 d) \int (c+d x) \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac{(2 d) \int (c+d x) \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{2 d (c+d x) \text{Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac{2 d (c+d x) \text{Li}_2\left (e^{a+b x}\right )}{b^2}+\frac{\left (2 d^2\right ) \int \text{Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}-\frac{\left (2 d^2\right ) \int \text{Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{2 d (c+d x) \text{Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac{2 d (c+d x) \text{Li}_2\left (e^{a+b x}\right )}{b^2}+\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{2 d (c+d x) \text{Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac{2 d (c+d x) \text{Li}_2\left (e^{a+b x}\right )}{b^2}+\frac{2 d^2 \text{Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac{2 d^2 \text{Li}_3\left (e^{a+b x}\right )}{b^3}\\ \end{align*}

Mathematica [A]  time = 2.29854, size = 118, normalized size = 1.19 \[ \frac{-\frac{2 d \left (b (c+d x) \text{PolyLog}\left (2,-e^{a+b x}\right )-d \text{PolyLog}\left (3,-e^{a+b x}\right )\right )}{b^2}+\frac{2 d \left (b (c+d x) \text{PolyLog}\left (2,e^{a+b x}\right )-d \text{PolyLog}\left (3,e^{a+b x}\right )\right )}{b^2}+(c+d x)^2 \log \left (1-e^{a+b x}\right )-(c+d x)^2 \log \left (e^{a+b x}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.03, size = 306, normalized size = 3.1 \begin{align*} -2\,{\frac{{c}^{2}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{b}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{d}^{2}{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{cd\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-2\,{\frac{cd\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}-2\,{\frac{cd\ln \left ( 1+{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+2\,{\frac{cd\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}+4\,{\frac{cda{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{b}}-{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-{\frac{{d}^{2}\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{b}}+{\frac{{d}^{2}\ln \left ( 1+{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{3}}}-2\,{\frac{{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-2\,{\frac{{a}^{2}{d}^{2}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{cd{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{cd{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.54884, size = 263, normalized size = 2.66 \begin{align*} -c^{2}{\left (\frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac{\log \left (e^{\left (-b x - a\right )} - 1\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 )} c d}{b^{2}} + \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 )} c d}{b^{2}} - \frac{{\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^{2}}{b^{3}} + \frac{{\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^{2}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.64391, size = 635, normalized size = 6.41 \begin{align*} -\frac{2 \, d^{2}{\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 2 \, d^{2}{\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 2 \,{\left (b d^{2} x + b c d\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 2 \,{\left (b d^{2} x + b c d\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) +{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) -{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) -{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*d^2*polylog(3, cosh(b*x + a) + sinh(b*x + a)) - 2*d^2*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) - 2*(b*d^
2*x + b*c*d)*dilog(cosh(b*x + a) + sinh(b*x + a)) + 2*(b*d^2*x + b*c*d)*dilog(-cosh(b*x + a) - sinh(b*x + a))
+ (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^
2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-cosh(b*x +
a) - sinh(b*x + a) + 1))/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}{\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),x)

[Out]

Integral((c + d*x)**2*csch(a + b*x), 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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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