### 3.440 $$\int x^2 \cosh (a+b x) \coth ^2(a+b x) \, dx$$

Optimal. Leaf size=95 $-\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}+\frac{2 \sinh (a+b x)}{b^3}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{x^2 \sinh (a+b x)}{b}-\frac{x^2 \text{csch}(a+b x)}{b}$

[Out]

(-4*x*ArcTanh[E^(a + b*x)])/b^2 - (2*x*Cosh[a + b*x])/b^2 - (x^2*Csch[a + b*x])/b - (2*PolyLog[2, -E^(a + b*x)
])/b^3 + (2*PolyLog[2, E^(a + b*x)])/b^3 + (2*Sinh[a + b*x])/b^3 + (x^2*Sinh[a + b*x])/b

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Rubi [A]  time = 0.117816, antiderivative size = 95, 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 = {5450, 3296, 2637, 5419, 4182, 2279, 2391} $-\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}+\frac{2 \sinh (a+b x)}{b^3}-\frac{2 x \cosh (a+b x)}{b^2}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{x^2 \sinh (a+b x)}{b}-\frac{x^2 \text{csch}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-4*x*ArcTanh[E^(a + b*x)])/b^2 - (2*x*Cosh[a + b*x])/b^2 - (x^2*Csch[a + b*x])/b - (2*PolyLog[2, -E^(a + b*x)
])/b^3 + (2*PolyLog[2, E^(a + b*x)])/b^3 + (2*Sinh[a + b*x])/b^3 + (x^2*Sinh[a + b*x])/b

Rule 5450

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*Coth[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int
[(c + d*x)^m*Cosh[a + b*x]^n*Coth[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cosh[a + b*x]^(n - 2)*Coth[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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]

Rubi steps

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

Mathematica [B]  time = 0.438756, size = 230, normalized size = 2.42 $\frac{\text{csch}\left (\frac{1}{2} (a+b x)\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right ) \left (4 \sinh (a+b x) \text{PolyLog}\left (2,-e^{-a-b x}\right )-4 \sinh (a+b x) \text{PolyLog}\left (2,e^{-a-b x}\right )+b^2 x^2 \cosh (2 (a+b x))-2 b x \sinh (2 (a+b x))+2 \cosh (2 (a+b x))+4 b x \log \left (1-e^{-a-b x}\right ) \sinh (a+b x)-4 b x \log \left (e^{-a-b x}+1\right ) \sinh (a+b x)+4 a \log \left (1-e^{-a-b x}\right ) \sinh (a+b x)-4 a \log \left (e^{-a-b x}+1\right ) \sinh (a+b x)-4 a \sinh (a+b x) \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )-3 b^2 x^2-2\right )}{4 b^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Csch[(a + b*x)/2]*Sech[(a + b*x)/2]*(-2 - 3*b^2*x^2 + 2*Cosh[2*(a + b*x)] + b^2*x^2*Cosh[2*(a + b*x)] + 4*a*L
og[1 - E^(-a - b*x)]*Sinh[a + b*x] + 4*b*x*Log[1 - E^(-a - b*x)]*Sinh[a + b*x] - 4*a*Log[1 + E^(-a - b*x)]*Sin
h[a + b*x] - 4*b*x*Log[1 + E^(-a - b*x)]*Sinh[a + b*x] - 4*a*Log[Tanh[(a + b*x)/2]]*Sinh[a + b*x] + 4*PolyLog[
2, -E^(-a - b*x)]*Sinh[a + b*x] - 4*PolyLog[2, E^(-a - b*x)]*Sinh[a + b*x] - 2*b*x*Sinh[2*(a + b*x)]))/(4*b^3)

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Maple [A]  time = 0.075, size = 185, normalized size = 2. \begin{align*}{\frac{ \left ({x}^{2}{b}^{2}-2\,bx+2 \right ){{\rm e}^{bx+a}}}{2\,{b}^{3}}}-{\frac{ \left ({x}^{2}{b}^{2}+2\,bx+2 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{3}}}-2\,{\frac{{{\rm e}^{bx+a}}{x}^{2}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-2\,{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+2\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+4\,{\frac{a{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44217, size = 212, normalized size = 2.23 \begin{align*} \frac{{\left (b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + 2 \, e^{\left (4 \, a\right )}\right )} e^{\left (3 \, b x\right )} - 2 \,{\left (3 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 2 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} +{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x\right )}}{2 \,{\left (b^{3} e^{\left (2 \, b x + 3 \, a\right )} - b^{3} e^{a}\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 )}}{b^{3}} + \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 )}}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.41477, size = 1952, 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(x^2*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

1/2*((b^2*x^2 - 2*b*x + 2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - 2*b*x + 2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2
- 2*b*x + 2)*sinh(b*x + a)^4 + b^2*x^2 - 2*(3*b^2*x^2 + 2)*cosh(b*x + a)^2 - 2*(3*b^2*x^2 - 3*(b^2*x^2 - 2*b*x
+ 2)*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^2 + 2*b*x + 4*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + si
nh(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - cosh(b*x + a))*dilog(cosh(b*x + a) + sinh(b*x + a)) -
4*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)
- cosh(b*x + a))*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 4*(b*x*cosh(b*x + a)^3 + 3*b*x*cosh(b*x + a)*sinh(b*
x + a)^2 + b*x*sinh(b*x + a)^3 - b*x*cosh(b*x + a) + (3*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a))*log(cosh(b*x
+ a) + sinh(b*x + a) + 1) - 4*(a*cosh(b*x + a)^3 + 3*a*cosh(b*x + a)*sinh(b*x + a)^2 + a*sinh(b*x + a)^3 - a*
cosh(b*x + a) + (3*a*cosh(b*x + a)^2 - a)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 4*((b*x + a)
*cosh(b*x + a)^3 + 3*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^2 + (b*x + a)*sinh(b*x + a)^3 - (b*x + a)*cosh(b*x
+ a) + (3*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 4*((b^
2*x^2 - 2*b*x + 2)*cosh(b*x + a)^3 - (3*b^2*x^2 + 2)*cosh(b*x + a))*sinh(b*x + a) + 2)/(b^3*cosh(b*x + a)^3 +
3*b^3*cosh(b*x + a)*sinh(b*x + a)^2 + b^3*sinh(b*x + a)^3 - b^3*cosh(b*x + a) + (3*b^3*cosh(b*x + a)^2 - b^3)*
sinh(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**2*cosh(b*x+a)**3*csch(b*x+a)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cosh \left (b x + a\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(x^2*cosh(b*x+a)^3*csch(b*x+a)^2,x, algorithm="giac")

[Out]

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