3.385 \(\int x^2 \sinh (a+b x) \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=104 \[ \frac {2 i \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}-\frac {2 i \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {2 \cosh (a+b x)}{b^3}-\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x \sinh (a+b x)}{b^2}+\frac {x^2 \cosh (a+b x)}{b}+\frac {x^2 \text {sech}(a+b x)}{b} \]

[Out]

-4*x*arctan(exp(b*x+a))/b^2+2*cosh(b*x+a)/b^3+x^2*cosh(b*x+a)/b+2*I*polylog(2,-I*exp(b*x+a))/b^3-2*I*polylog(2
,I*exp(b*x+a))/b^3+x^2*sech(b*x+a)/b-2*x*sinh(b*x+a)/b^2

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Rubi [A]  time = 0.12, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 = {5449, 3296, 2638, 5418, 4180, 2279, 2391} \[ \frac {2 i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac {2 i \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}-\frac {4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {2 x \sinh (a+b x)}{b^2}+\frac {2 \cosh (a+b x)}{b^3}+\frac {x^2 \cosh (a+b x)}{b}+\frac {x^2 \text {sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

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

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 2638

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

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 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 5449

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

Rubi steps

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

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Mathematica [A]  time = 0.51, size = 172, normalized size = 1.65 \[ \frac {b^2 x^2 \text {sech}(a+b x)+\cosh (b x) \left (\cosh (a) \left (b^2 x^2+2\right )-2 b x \sinh (a)\right )+\sinh (b x) \left (\sinh (a) \left (b^2 x^2+2\right )-2 b x \cosh (a)\right )+2 i \left (\text {Li}_2\left (-i e^{a+b x}\right )-\text {Li}_2\left (i e^{a+b x}\right )\right )+(-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-(\pi -2 i a) \log \left (\cot \left (\frac {1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

(((-2*I)*a + Pi - (2*I)*b*x)*(Log[1 - I*E^(a + b*x)] - Log[1 + I*E^(a + b*x)]) - ((-2*I)*a + Pi)*Log[Cot[((2*I
)*a + Pi + (2*I)*b*x)/4]] + (2*I)*(PolyLog[2, (-I)*E^(a + b*x)] - PolyLog[2, I*E^(a + b*x)]) + b^2*x^2*Sech[a
+ b*x] + Cosh[b*x]*((2 + b^2*x^2)*Cosh[a] - 2*b*x*Sinh[a]) + (-2*b*x*Cosh[a] + (2 + b^2*x^2)*Sinh[a])*Sinh[b*x
])/b^3

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fricas [B]  time = 0.73, size = 869, normalized size = 8.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sech(b*x+a)^2*sinh(b*x+a)^3,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*I*cosh(b*x + a)^3 - 12*I*cosh(b*x + a)*sinh(b*x + a)^
2 - 4*I*sinh(b*x + a)^3 + (-12*I*cosh(b*x + a)^2 - 4*I)*sinh(b*x + a) - 4*I*cosh(b*x + a))*dilog(I*cosh(b*x +
a) + I*sinh(b*x + a)) + (4*I*cosh(b*x + a)^3 + 12*I*cosh(b*x + a)*sinh(b*x + a)^2 + 4*I*sinh(b*x + a)^3 + (12*
I*cosh(b*x + a)^2 + 4*I)*sinh(b*x + a) + 4*I*cosh(b*x + a))*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (4*I*a
*cosh(b*x + a)^3 + 12*I*a*cosh(b*x + a)*sinh(b*x + a)^2 + 4*I*a*sinh(b*x + a)^3 + 4*I*a*cosh(b*x + a) + (12*I*
a*cosh(b*x + a)^2 + 4*I*a)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + I) + (-4*I*a*cosh(b*x + a)^3 - 1
2*I*a*cosh(b*x + a)*sinh(b*x + a)^2 - 4*I*a*sinh(b*x + a)^3 - 4*I*a*cosh(b*x + a) + (-12*I*a*cosh(b*x + a)^2 -
4*I*a)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - I) + ((4*I*b*x + 4*I*a)*cosh(b*x + a)^3 + (12*I*b*x
+ 12*I*a)*cosh(b*x + a)*sinh(b*x + a)^2 + (4*I*b*x + 4*I*a)*sinh(b*x + a)^3 + (4*I*b*x + 4*I*a)*cosh(b*x + a)
+ ((12*I*b*x + 12*I*a)*cosh(b*x + a)^2 + 4*I*b*x + 4*I*a)*sinh(b*x + a))*log(I*cosh(b*x + a) + I*sinh(b*x + a
) + 1) + ((-4*I*b*x - 4*I*a)*cosh(b*x + a)^3 + (-12*I*b*x - 12*I*a)*cosh(b*x + a)*sinh(b*x + a)^2 + (-4*I*b*x
- 4*I*a)*sinh(b*x + a)^3 + (-4*I*b*x - 4*I*a)*cosh(b*x + a) + ((-12*I*b*x - 12*I*a)*cosh(b*x + a)^2 - 4*I*b*x
- 4*I*a)*sinh(b*x + a))*log(-I*cosh(b*x + a) - I*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*sech(b*x + a)^2*sinh(b*x + a)^3, x)

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maple [B]  time = 0.48, size = 205, normalized size = 1.97 \[ \frac {\left (x^{2} b^{2}-2 b x +2\right ) {\mathrm e}^{b x +a}}{2 b^{3}}+\frac {\left (x^{2} b^{2}+2 b x +2\right ) {\mathrm e}^{-b x -a}}{2 b^{3}}+\frac {2 x^{2} {\mathrm e}^{b x +a}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}+\frac {2 i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {2 i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{b^{3}}-\frac {2 i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {2 i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{b^{3}}+\frac {2 i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {4 a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sech(b*x+a)^2*sinh(b*x+a)^3,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/(1+exp(2*b*x+2*a
))+2*I/b^2*ln(1+I*exp(b*x+a))*x+2*I/b^3*ln(1+I*exp(b*x+a))*a-2*I/b^2*ln(1-I*exp(b*x+a))*x-2*I/b^3*ln(1-I*exp(b
*x+a))*a+2*I/b^3*dilog(1+I*exp(b*x+a))-2*I/b^3*dilog(1-I*exp(b*x+a))+4/b^3*a*arctan(exp(b*x+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + 2 \, e^{\left (4 \, a\right )}\right )} e^{\left (3 \, b x\right )} + 2 \, {\left (3 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 2 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} + {\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x\right )}}{2 \, {\left (b^{3} e^{\left (2 \, b x + 3 \, a\right )} + b^{3} e^{a}\right )}} - 4 \, \int \frac {x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sech(b*x+a)^2*sinh(b*x+a)^3,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) - 4*integrate(x*e^(b*x + a)/(b*e^(2*b*x + 2*a) +
b), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sech(b*x+a)**2*sinh(b*x+a)**3,x)

[Out]

Integral(x**2*sinh(a + b*x)**3*sech(a + b*x)**2, x)

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