### 3.349 $$\int x^3 \text{sech}^2(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=83 $-\frac{3 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}-\frac{3 x \log \left (e^{2 (a+b x)}+1\right )}{b^3}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2}{2 b^2}$

[Out]

(3*x^2)/(2*b^2) - (3*x*Log[1 + E^(2*(a + b*x))])/b^3 - (3*PolyLog[2, -E^(2*(a + b*x))])/(2*b^4) - (x^3*Sech[a
+ b*x]^2)/(2*b) + (3*x^2*Tanh[a + b*x])/(2*b^2)

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Rubi [A]  time = 0.182871, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {5418, 4184, 3718, 2190, 2279, 2391} $-\frac{3 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}-\frac{3 x \log \left (e^{2 (a+b x)}+1\right )}{b^3}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2}{2 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Sech[a + b*x]^2*Tanh[a + b*x],x]

[Out]

(3*x^2)/(2*b^2) - (3*x*Log[1 + E^(2*(a + b*x))])/b^3 - (3*PolyLog[2, -E^(2*(a + b*x))])/(2*b^4) - (x^3*Sech[a
+ b*x]^2)/(2*b) + (3*x^2*Tanh[a + b*x])/(2*b^2)

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 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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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^3 \text{sech}^2(a+b x) \tanh (a+b x) \, dx &=-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 \int x^2 \text{sech}^2(a+b x) \, dx}{2 b}\\ &=-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}-\frac{3 \int x \tanh (a+b x) \, dx}{b^2}\\ &=\frac{3 x^2}{2 b^2}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}-\frac{6 \int \frac{e^{2 (a+b x)} x}{1+e^{2 (a+b x)}} \, dx}{b^2}\\ &=\frac{3 x^2}{2 b^2}-\frac{3 x \log \left (1+e^{2 (a+b x)}\right )}{b^3}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}+\frac{3 \int \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=\frac{3 x^2}{2 b^2}-\frac{3 x \log \left (1+e^{2 (a+b x)}\right )}{b^3}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=\frac{3 x^2}{2 b^2}-\frac{3 x \log \left (1+e^{2 (a+b x)}\right )}{b^3}-\frac{3 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}+\frac{3 x^2 \tanh (a+b x)}{2 b^2}\\ \end{align*}

Mathematica [C]  time = 6.12298, size = 228, normalized size = 2.75 $\frac{3 \text{csch}(a) \text{sech}(a) \left (-b^2 x^2 e^{-\tanh ^{-1}(\coth (a))}+\frac{i \coth (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )-b x \left (-\pi +2 i \tanh ^{-1}(\coth (a))\right )-2 \left (i \tanh ^{-1}(\coth (a))+i b x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )+2 i \tanh ^{-1}(\coth (a)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )-\pi \log \left (e^{2 b x}+1\right )+\pi \log (\cosh (b x))\right )}{\sqrt{1-\coth ^2(a)}}\right )}{2 b^4 \sqrt{\text{csch}^2(a) \left (\sinh ^2(a)-\cosh ^2(a)\right )}}+\frac{3 x^2 \text{sech}(a) \sinh (b x) \text{sech}(a+b x)}{2 b^2}-\frac{x^3 \text{sech}^2(a+b x)}{2 b}$

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Sech[a + b*x]^2*Tanh[a + b*x],x]

[Out]

-(x^3*Sech[a + b*x]^2)/(2*b) + (3*Csch[a]*(-((b^2*x^2)/E^ArcTanh[Coth[a]]) + (I*Coth[a]*(-(b*x*(-Pi + (2*I)*Ar
cTanh[Coth[a]])) - Pi*Log[1 + E^(2*b*x)] - 2*(I*b*x + I*ArcTanh[Coth[a]])*Log[1 - E^((2*I)*(I*b*x + I*ArcTanh[
Coth[a]]))] + Pi*Log[Cosh[b*x]] + (2*I)*ArcTanh[Coth[a]]*Log[I*Sinh[b*x + ArcTanh[Coth[a]]]] + I*PolyLog[2, E^
((2*I)*(I*b*x + I*ArcTanh[Coth[a]]))]))/Sqrt[1 - Coth[a]^2])*Sech[a])/(2*b^4*Sqrt[Csch[a]^2*(-Cosh[a]^2 + Sinh
[a]^2)]) + (3*x^2*Sech[a]*Sech[a + b*x]*Sinh[b*x])/(2*b^2)

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Maple [A]  time = 0.028, size = 121, normalized size = 1.5 \begin{align*} -{\frac{{x}^{2} \left ( 2\,bx{{\rm e}^{2\,bx+2\,a}}+3\,{{\rm e}^{2\,bx+2\,a}}+3 \right ) }{{b}^{2} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}+3\,{\frac{{x}^{2}}{{b}^{2}}}+6\,{\frac{ax}{{b}^{3}}}+3\,{\frac{{a}^{2}}{{b}^{4}}}-3\,{\frac{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{3}}}-{\frac{3\,{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{4}}}-6\,{\frac{a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*(2*b*x*exp(2*b*x+2*a)+3*exp(2*b*x+2*a)+3)/b^2/(1+exp(2*b*x+2*a))^2+3*x^2/b^2+6/b^3*a*x+3/b^4*a^2-3*x*ln(1
+exp(2*b*x+2*a))/b^3-3/2*polylog(2,-exp(2*b*x+2*a))/b^4-6/b^4*a*ln(exp(b*x+a))

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Maxima [A]  time = 1.5117, size = 149, normalized size = 1.8 \begin{align*} -\frac{3 \, x^{2} +{\left (2 \, b x^{3} e^{\left (2 \, a\right )} + 3 \, x^{2} e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac{3 \, x^{2}}{b^{2}} - \frac{3 \,{\left (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 )\right )}}{2 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.05874, size = 2966, normalized size = 35.73 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(b^2*x^2 - a^2)*cosh(b*x + a)^4 + 12*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + 3*(b^2*x^2 - a^2)*sinh
(b*x + a)^4 - (2*b^3*x^3 - 3*b^2*x^2 + 6*a^2)*cosh(b*x + a)^2 - (2*b^3*x^3 - 3*b^2*x^2 - 18*(b^2*x^2 - a^2)*co
sh(b*x + a)^2 + 6*a^2)*sinh(b*x + a)^2 - 3*a^2 - 3*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b
*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a)
)*sinh(b*x + a) + 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 3*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x
+ a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3
+ cosh(b*x + a))*sinh(b*x + a) + 1)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 3*(a*cosh(b*x + a)^4 + 4*a*cos
h(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 + 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh(b*x + a)^2 + a)*sinh(b*x +
a)^2 + 4*(a*cosh(b*x + a)^3 + a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a) + sinh(b*x + a) + I) + 3*(
a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 + 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh(
b*x + a)^2 + a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 + a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a)
+ sinh(b*x + a) - I) - 3*((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*s
inh(b*x + a)^4 + 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 + b*x + a)*sinh(b*x + a)^2 + b*x
+ 4*((b*x + a)*cosh(b*x + a)^3 + (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(I*cosh(b*x + a) + I*sinh(b*x
+ a) + 1) - 3*((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*sinh(b*x + a
)^4 + 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 + b*x + a)*sinh(b*x + a)^2 + b*x + 4*((b*x
+ a)*cosh(b*x + a)^3 + (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1)
+ 2*(6*(b^2*x^2 - a^2)*cosh(b*x + a)^3 - (2*b^3*x^3 - 3*b^2*x^2 + 6*a^2)*cosh(b*x + a))*sinh(b*x + a))/(b^4*c
osh(b*x + a)^4 + 4*b^4*cosh(b*x + a)*sinh(b*x + a)^3 + b^4*sinh(b*x + a)^4 + 2*b^4*cosh(b*x + a)^2 + b^4 + 2*(
3*b^4*cosh(b*x + a)^2 + b^4)*sinh(b*x + a)^2 + 4*(b^4*cosh(b*x + a)^3 + b^4*cosh(b*x + a))*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**3*sech(b*x+a)**3*sinh(b*x+a),x)

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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