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

Optimal. Leaf size=83 \[ -\frac {3 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac {3 x \log \left (e^{2 (a+b x)}+1\right )}{b^3}+\frac {3 x^2 \tanh (a+b x)}{2 b^2}-\frac {x^3 \text {sech}^2(a+b x)}{2 b}+\frac {3 x^2}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 6.14, size = 227, normalized size = 2.73 \[ \frac {3 x^2 \text {sech}(a) \sinh (b x) \text {sech}(a+b x)}{2 b^2}-\frac {3 \text {csch}(a) \text {sech}(a) \left (b^2 x^2 e^{-\tanh ^{-1}(\coth (a))}-\frac {i \coth (a) \left (i \text {Li}_2\left (e^{2 i \left (i b x+i \tanh ^{-1}(\coth (a))\right )}\right )-b x \left (-\pi +2 i \tanh ^{-1}(\coth (a))\right )-2 \left (i \tanh ^{-1}(\coth (a))+i b x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )+2 i \tanh ^{-1}(\coth (a)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )-\pi \log \left (e^{2 b x}+1\right )+\pi \log (\cosh (b x))\right )}{\sqrt {1-\coth ^2(a)}}\right )}{2 b^4 \sqrt {\text {csch}^2(a) \left (\sinh ^2(a)-\cosh ^2(a)\right )}}-\frac {x^3 \text {sech}^2(a+b x)}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(x^3*Sech[a + b*x]^2)/b - (3*Csch[a]*((b^2*x^2)/E^ArcTanh[Coth[a]] - (I*Coth[a]*(-(b*x*(-Pi + (2*I)*ArcTa
nh[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[Cot
h[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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fricas [C]  time = 0.93, size = 1113, normalized size = 13.41 \[ \text {result too large to display} \]

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

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)

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maple [A]  time = 0.23, size = 121, normalized size = 1.46 \[ -\frac {x^{2} \left (2 b x \,{\mathrm e}^{2 b x +2 a}+3 \,{\mathrm e}^{2 b x +2 a}+3\right )}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}+\frac {3 x^{2}}{b^{2}}+\frac {6 a x}{b^{3}}+\frac {3 a^{2}}{b^{4}}-\frac {3 x \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}-\frac {3 \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}}-\frac {6 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}} \]

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*a*x/b^3+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 = 0.43, size = 110, normalized size = 1.33 \[ -\frac {3 \, x^{2} + {\left (2 \, b x^{3} e^{\left (2 \, a\right )} + 3 \, x^{2} e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac {3 \, x^{2}}{b^{2}} - \frac {3 \, {\left (2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )\right )}}{2 \, b^{4}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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