∫ x2csch2(a + bx)sech2(a + bx) dx [496]

3.5.96 \(\int x^2 \text {csch}^2(a+b x) \text {sech}^2(a+b x) \, dx\) [496]

Optimal. Leaf size=64 \[ -\frac {2 x^2}{b}-\frac {2 x^2 \coth (2 a+2 b x)}{b}+\frac {2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}+\frac {\text {PolyLog}\left (2,e^{4 (a+b x)}\right )}{2 b^3} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5569, 4269, 3797, 2221, 2317, 2438} \begin {gather*} \frac {\text {Li}_2\left (e^{4 (a+b x)}\right )}{2 b^3}+\frac {2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac {2 x^2 \coth (2 a+2 b x)}{b}-\frac {2 x^2}{b} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Csch[a + b*x]^2*Sech[a + b*x]^2,x]

[Out]

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

*b^3)

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3797

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))/(1 + E^(2*((-I)*e + f*

fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4269

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 5569

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis

t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rubi steps

\begin {align*} \int x^2 \text {csch}^2(a+b x) \text {sech}^2(a+b x) \, dx &=4 \int x^2 \text {csch}^2(2 a+2 b x) \, dx\\ &=-\frac {2 x^2 \coth (2 a+2 b x)}{b}+\frac {4 \int x \coth (2 a+2 b x) \, dx}{b}\\ &=-\frac {2 x^2}{b}-\frac {2 x^2 \coth (2 a+2 b x)}{b}-\frac {8 \int \frac {e^{2 (2 a+2 b x)} x}{1-e^{2 (2 a+2 b x)}} \, dx}{b}\\ &=-\frac {2 x^2}{b}-\frac {2 x^2 \coth (2 a+2 b x)}{b}+\frac {2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac {2 \int \log \left (1-e^{2 (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac {2 x^2}{b}-\frac {2 x^2 \coth (2 a+2 b x)}{b}+\frac {2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac {\text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 (2 a+2 b x)}\right )}{2 b^3}\\ &=-\frac {2 x^2}{b}-\frac {2 x^2 \coth (2 a+2 b x)}{b}+\frac {2 x \log \left (1-e^{4 (a+b x)}\right )}{b^2}+\frac {\text {Li}_2\left (e^{4 (a+b x)}\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]
time = 2.60, size = 76, normalized size = 1.19 \begin {gather*} \frac {\text {PolyLog}\left (2,e^{4 (a+b x)}\right )+4 b x \left (-\frac {2 b e^{4 a} x}{-1+e^{4 a}}+\log \left (1-e^{4 (a+b x)}\right )+b x \text {csch}(2 a) \text {csch}(2 (a+b x)) \sinh (2 b x)\right )}{2 b^3} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Csch[a + b*x]^2*Sech[a + b*x]^2,x]

[Out]

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

a]*Csch[2*(a + b*x)]*Sinh[2*b*x]))/(2*b^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(198\) vs. \(2(62)=124\).
time = 1.92, size = 199, normalized size = 3.11


method	result	size

risch	\(-\frac {4 x^{2}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right ) \left ({\mathrm e}^{2 b x +2 a}+1\right )}-\frac {4 x^{2}}{b}-\frac {8 a x}{b^{2}}-\frac {4 a^{2}}{b^{3}}+\frac {2 x \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b^{2}}+\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}+\frac {2 \ln \left ({\mathrm e}^{b x +a}+1\right ) x}{b^{2}}+\frac {2 \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {2 \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{3}}+\frac {2 \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}+\frac {8 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}\)	\(199\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*csch(b*x+a)^2*sech(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-4*x^2/b/(exp(2*b*x+2*a)-1)/(exp(2*b*x+2*a)+1)-4*x^2/b-8*a*x/b^2-4/b^3*a^2+2*x*ln(exp(2*b*x+2*a)+1)/b^2+polylo

g(2,-exp(2*b*x+2*a))/b^3+2/b^2*ln(exp(b*x+a)+1)*x+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^3-2/b^3*a*ln(exp(b*x+a)-1)+8/b^3*a*ln(exp(b*x+a))

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Maxima [A]
time = 0.29, size = 118, normalized size = 1.84 \begin {gather*} -\frac {4 \, x^{2}}{b e^{\left (4 \, b x + 4 \, a\right )} - b} - \frac {4 \, x^{2}}{b} + \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\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}} + \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 {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x^2/(b*e^(4*b*x + 4*a) - b) - 4*x^2/b + (2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2*a)))/b^3 + 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 [C] Result contains complex when optimal does not.
time = 0.38, size = 1327, normalized size = 20.73 \begin {gather*} \text {Too large to display} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*(b^2*x^2 - a^2)*cosh(b*x + a)^4 + 8*(b^2*x^2 - a^2)*cosh(b*x + a)^3*sinh(b*x + a) + 12*(b^2*x^2 - a^2)*c

osh(b*x + a)^2*sinh(b*x + a)^2 + 8*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + 2*(b^2*x^2 - a^2)*sinh(b*x

+ a)^4 + 2*a^2 - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*co

sh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) - (cosh(b*x + a)^4 + 4

*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*

x + a)^4 - 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) +

6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*dilog(-I*cosh(b*x +

a) - I*sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^

2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - (b*x*cosh(b

*x + a)^4 + 4*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b*x*cosh(b*x + a)*

sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - b*x)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^4 + 4*a

*cosh(b*x + a)^3*sinh(b*x + a) + 6*a*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*s

inh(b*x + a)^4 - a)*log(cosh(b*x + a) + sinh(b*x + a) + I) + (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)^3*sinh(b*x

+ a) + 6*a*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - a)*log(c

osh(b*x + a) + sinh(b*x + a) - I) + (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)^3*sinh(b*x + a) + 6*a*cosh(b*x + a)

^2*sinh(b*x + a)^2 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - a)*log(cosh(b*x + a) + sinh(b*x +

a) - 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)^3*sinh(b*x + a) + 6*(b*x + a)*cosh(b*x + a)^

2*sinh(b*x + a)^2 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*sinh(b*x + a)^4 - b*x - a)*log(I*cos

h(b*x + a) + I*sinh(b*x + a) + 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)^3*sinh(b*x + a) + 6

*(b*x + a)*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*sinh(b*x +

a)^4 - b*x - a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*

x + a)^3*sinh(b*x + a) + 6*(b*x + a)*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)

^3 + (b*x + a)*sinh(b*x + a)^4 - b*x - a)*log(-cosh(b*x + a) - sinh(b*x + a) + 1))/(b^3*cosh(b*x + a)^4 + 4*b^

3*cosh(b*x + a)^3*sinh(b*x + a) + 6*b^3*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b^3*cosh(b*x + a)*sinh(b*x + a)^3

+ b^3*sinh(b*x + a)^4 - b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {csch}^{2}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*csch(b*x+a)**2*sech(b*x+a)**2,x)

[Out]

Integral(x**2*csch(a + b*x)**2*sech(a + b*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(cosh(a + b*x)^2*sinh(a + b*x)^2),x)

[Out]

int(x^2/(cosh(a + b*x)^2*sinh(a + b*x)^2), x)

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