3.3 \(\int x^3 (a+b \text{sech}(c+d x^2)) \, dx\)

Optimal. Leaf size=77 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{c+d x^2}\right )}{2 d^2}+\frac{i b \text{PolyLog}\left (2,i e^{c+d x^2}\right )}{2 d^2}+\frac{a x^4}{4}+\frac{b x^2 \tan ^{-1}\left (e^{c+d x^2}\right )}{d} \]

[Out]

(a*x^4)/4 + (b*x^2*ArcTan[E^(c + d*x^2)])/d - ((I/2)*b*PolyLog[2, (-I)*E^(c + d*x^2)])/d^2 + ((I/2)*b*PolyLog[
2, I*E^(c + d*x^2)])/d^2

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Rubi [A]  time = 0.0765736, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {14, 5436, 4180, 2279, 2391} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{c+d x^2}\right )}{2 d^2}+\frac{i b \text{PolyLog}\left (2,i e^{c+d x^2}\right )}{2 d^2}+\frac{a x^4}{4}+\frac{b x^2 \tan ^{-1}\left (e^{c+d x^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sech[c + d*x^2]),x]

[Out]

(a*x^4)/4 + (b*x^2*ArcTan[E^(c + d*x^2)])/d - ((I/2)*b*PolyLog[2, (-I)*E^(c + d*x^2)])/d^2 + ((I/2)*b*PolyLog[
2, I*E^(c + d*x^2)])/d^2

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5436

Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sech[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.178493, size = 135, normalized size = 1.75 \[ \frac{1}{4} \left (a x^4-\frac{b \left (2 i \left (\text{PolyLog}\left (2,-i e^{c+d x^2}\right )-\text{PolyLog}\left (2,i e^{c+d x^2}\right )\right )+\left (-2 i c-2 i d x^2+\pi \right ) \left (\log \left (1-i e^{c+d x^2}\right )-\log \left (1+i e^{c+d x^2}\right )\right )-(\pi -2 i c) \log \left (\cot \left (\frac{1}{4} \left (2 i c+2 i d x^2+\pi \right )\right )\right )\right )}{d^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Sech[c + d*x^2]),x]

[Out]

(a*x^4 - (b*(((-2*I)*c + Pi - (2*I)*d*x^2)*(Log[1 - I*E^(c + d*x^2)] - Log[1 + I*E^(c + d*x^2)]) - ((-2*I)*c +
Pi)*Log[Cot[((2*I)*c + Pi + (2*I)*d*x^2)/4]] + (2*I)*(PolyLog[2, (-I)*E^(c + d*x^2)] - PolyLog[2, I*E^(c + d*
x^2)])))/d^2)/4

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b{\rm sech} \left (d{x}^{2}+c\right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*sech(d*x^2+c)),x)

[Out]

int(x^3*(a+b*sech(d*x^2+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a x^{4} + 2 \, b \int \frac{x^{3}}{e^{\left (d x^{2} + c\right )} + e^{\left (-d x^{2} - c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sech(d*x^2+c)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 2*b*integrate(x^3/(e^(d*x^2 + c) + e^(-d*x^2 - c)), x)

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Fricas [B]  time = 2.24856, size = 516, normalized size = 6.7 \begin{align*} \frac{a d^{2} x^{4} - 2 i \, b c \log \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right ) + i\right ) + 2 i \, b c \log \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right ) - i\right ) + 2 i \, b{\rm Li}_2\left (i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right )\right ) - 2 i \, b{\rm Li}_2\left (-i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right )\right ) +{\left (-2 i \, b d x^{2} - 2 i \, b c\right )} \log \left (i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right ) + 1\right ) +{\left (2 i \, b d x^{2} + 2 i \, b c\right )} \log \left (-i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right ) + 1\right )}{4 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sech(d*x^2+c)),x, algorithm="fricas")

[Out]

1/4*(a*d^2*x^4 - 2*I*b*c*log(cosh(d*x^2 + c) + sinh(d*x^2 + c) + I) + 2*I*b*c*log(cosh(d*x^2 + c) + sinh(d*x^2
+ c) - I) + 2*I*b*dilog(I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c)) - 2*I*b*dilog(-I*cosh(d*x^2 + c) - I*sinh(d*x^
2 + c)) + (-2*I*b*d*x^2 - 2*I*b*c)*log(I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c) + 1) + (2*I*b*d*x^2 + 2*I*b*c)*lo
g(-I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c) + 1))/d^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \operatorname{sech}{\left (c + d x^{2} \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{sech}\left (d x^{2} + c\right ) + a\right )} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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