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

Optimal. Leaf size=108 \[ -\frac{a b \text{PolyLog}\left (2,-e^{c+d x^2}\right )}{d^2}+\frac{a b \text{PolyLog}\left (2,e^{c+d x^2}\right )}{d^2}+\frac{a^2 x^4}{4}-\frac{2 a b x^2 \tanh ^{-1}\left (e^{c+d x^2}\right )}{d}+\frac{b^2 \log \left (\sinh \left (c+d x^2\right )\right )}{2 d^2}-\frac{b^2 x^2 \coth \left (c+d x^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.164728, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {5437, 4190, 4182, 2279, 2391, 4184, 3475} \[ -\frac{a b \text{PolyLog}\left (2,-e^{c+d x^2}\right )}{d^2}+\frac{a b \text{PolyLog}\left (2,e^{c+d x^2}\right )}{d^2}+\frac{a^2 x^4}{4}-\frac{2 a b x^2 \tanh ^{-1}\left (e^{c+d x^2}\right )}{d}+\frac{b^2 \log \left (\sinh \left (c+d x^2\right )\right )}{2 d^2}-\frac{b^2 x^2 \coth \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Csch[c + d*x^2])^2,x]

[Out]

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

Rule 5437

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

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 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]

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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 4.40151, size = 260, normalized size = 2.41 \[ \frac{8 a b \left (\frac{\text{sech}(c) \left (\text{PolyLog}\left (2,-e^{-\tanh ^{-1}(\tanh (c))-d x^2}\right )-\text{PolyLog}\left (2,e^{-\tanh ^{-1}(\tanh (c))-d x^2}\right )+\left (\tanh ^{-1}(\tanh (c))+d x^2\right ) \left (\log \left (1-e^{-\tanh ^{-1}(\tanh (c))-d x^2}\right )-\log \left (e^{-\tanh ^{-1}(\tanh (c))-d x^2}+1\right )\right )\right )}{\sqrt{\text{sech}^2(c)}}+2 \tanh ^{-1}(\tanh (c)) \tanh ^{-1}\left (\sinh (c) \tanh \left (\frac{d x^2}{2}\right )+\cosh (c)\right )\right )+2 d x^2 \left (a^2 d x^2-2 b^2 \coth (c)\right )+4 b^2 d x^2 \coth (c)+2 b^2 d x^2 \text{csch}\left (\frac{c}{2}\right ) \sinh \left (\frac{d x^2}{2}\right ) \text{csch}\left (\frac{1}{2} \left (c+d x^2\right )\right )-2 b^2 d x^2 \text{sech}\left (\frac{c}{2}\right ) \sinh \left (\frac{d x^2}{2}\right ) \text{sech}\left (\frac{1}{2} \left (c+d x^2\right )\right )-4 b^2 \left (d x^2 \coth (c)-\log \left (\sinh \left (c+d x^2\right )\right )\right )}{8 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*b^2*d*x^2*Coth[c] + 2*d*x^2*(a^2*d*x^2 - 2*b^2*Coth[c]) - 4*b^2*(d*x^2*Coth[c] - Log[Sinh[c + d*x^2]]) + 8*
a*b*(2*ArcTanh[Tanh[c]]*ArcTanh[Cosh[c] + Sinh[c]*Tanh[(d*x^2)/2]] + (((d*x^2 + ArcTanh[Tanh[c]])*(Log[1 - E^(
-(d*x^2) - ArcTanh[Tanh[c]])] - Log[1 + E^(-(d*x^2) - ArcTanh[Tanh[c]])]) + PolyLog[2, -E^(-(d*x^2) - ArcTanh[
Tanh[c]])] - PolyLog[2, E^(-(d*x^2) - ArcTanh[Tanh[c]])])*Sech[c])/Sqrt[Sech[c]^2]) + 2*b^2*d*x^2*Csch[c/2]*Cs
ch[(c + d*x^2)/2]*Sinh[(d*x^2)/2] - 2*b^2*d*x^2*Sech[c/2]*Sech[(c + d*x^2)/2]*Sinh[(d*x^2)/2])/(8*d^2)

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Maple [F]  time = 0.118, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b{\rm csch} \left (d{x}^{2}+c\right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a^{2} x^{4} - \frac{1}{2} \,{\left (\frac{2 \, x^{2} e^{\left (2 \, d x^{2} + 2 \, c\right )}}{d e^{\left (2 \, d x^{2} + 2 \, c\right )} - d} - \frac{\log \left ({\left (e^{\left (d x^{2} + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{d^{2}} - \frac{\log \left ({\left (e^{\left (d x^{2} + c\right )} - 1\right )} e^{\left (-c\right )}\right )}{d^{2}}\right )} b^{2} + 4 \, a b{\left (\int \frac{x^{3}}{2 \,{\left (e^{\left (d x^{2} + c\right )} + 1\right )}}\,{d x} + \int \frac{x^{3}}{2 \,{\left (e^{\left (d x^{2} + c\right )} - 1\right )}}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 - 1/2*(2*x^2*e^(2*d*x^2 + 2*c)/(d*e^(2*d*x^2 + 2*c) - d) - log((e^(d*x^2 + c) + 1)*e^(-c))/d^2 - l
og((e^(d*x^2 + c) - 1)*e^(-c))/d^2)*b^2 + 4*a*b*(integrate(1/2*x^3/(e^(d*x^2 + c) + 1), x) + integrate(1/2*x^3
/(e^(d*x^2 + c) - 1), x))

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Fricas [B]  time = 1.77467, size = 1602, normalized size = 14.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*d^2*x^4 - 4*b^2*c - (a^2*d^2*x^4 - 4*b^2*d*x^2 - 4*b^2*c)*cosh(d*x^2 + c)^2 - 2*(a^2*d^2*x^4 - 4*b^2
*d*x^2 - 4*b^2*c)*cosh(d*x^2 + c)*sinh(d*x^2 + c) - (a^2*d^2*x^4 - 4*b^2*d*x^2 - 4*b^2*c)*sinh(d*x^2 + c)^2 -
4*(a*b*cosh(d*x^2 + c)^2 + 2*a*b*cosh(d*x^2 + c)*sinh(d*x^2 + c) + a*b*sinh(d*x^2 + c)^2 - a*b)*dilog(cosh(d*x
^2 + c) + sinh(d*x^2 + c)) + 4*(a*b*cosh(d*x^2 + c)^2 + 2*a*b*cosh(d*x^2 + c)*sinh(d*x^2 + c) + a*b*sinh(d*x^2
+ c)^2 - a*b)*dilog(-cosh(d*x^2 + c) - sinh(d*x^2 + c)) - 2*(2*a*b*d*x^2 - (2*a*b*d*x^2 - b^2)*cosh(d*x^2 + c
)^2 - 2*(2*a*b*d*x^2 - b^2)*cosh(d*x^2 + c)*sinh(d*x^2 + c) - (2*a*b*d*x^2 - b^2)*sinh(d*x^2 + c)^2 - b^2)*log
(cosh(d*x^2 + c) + sinh(d*x^2 + c) + 1) - 2*(2*a*b*c - (2*a*b*c - b^2)*cosh(d*x^2 + c)^2 - 2*(2*a*b*c - b^2)*c
osh(d*x^2 + c)*sinh(d*x^2 + c) - (2*a*b*c - b^2)*sinh(d*x^2 + c)^2 - b^2)*log(cosh(d*x^2 + c) + sinh(d*x^2 + c
) - 1) + 4*(a*b*d*x^2 + a*b*c - (a*b*d*x^2 + a*b*c)*cosh(d*x^2 + c)^2 - 2*(a*b*d*x^2 + a*b*c)*cosh(d*x^2 + c)*
sinh(d*x^2 + c) - (a*b*d*x^2 + a*b*c)*sinh(d*x^2 + c)^2)*log(-cosh(d*x^2 + c) - sinh(d*x^2 + c) + 1))/(d^2*cos
h(d*x^2 + c)^2 + 2*d^2*cosh(d*x^2 + c)*sinh(d*x^2 + c) + d^2*sinh(d*x^2 + c)^2 - d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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