3.1 \(\int x^5 (a+b \text{sech}(c+d x^2)) \, dx\)

Optimal. Leaf size=125 \[ -\frac{i b x^2 \text{PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac{i b x^2 \text{PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac{i b \text{PolyLog}\left (3,-i e^{c+d x^2}\right )}{d^3}-\frac{i b \text{PolyLog}\left (3,i e^{c+d x^2}\right )}{d^3}+\frac{a x^6}{6}+\frac{b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d} \]

[Out]

(a*x^6)/6 + (b*x^4*ArcTan[E^(c + d*x^2)])/d - (I*b*x^2*PolyLog[2, (-I)*E^(c + d*x^2)])/d^2 + (I*b*x^2*PolyLog[
2, I*E^(c + d*x^2)])/d^2 + (I*b*PolyLog[3, (-I)*E^(c + d*x^2)])/d^3 - (I*b*PolyLog[3, I*E^(c + d*x^2)])/d^3

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Rubi [A]  time = 0.137792, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {14, 5436, 4180, 2531, 2282, 6589} \[ -\frac{i b x^2 \text{PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac{i b x^2 \text{PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac{i b \text{PolyLog}\left (3,-i e^{c+d x^2}\right )}{d^3}-\frac{i b \text{PolyLog}\left (3,i e^{c+d x^2}\right )}{d^3}+\frac{a x^6}{6}+\frac{b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sech[c + d*x^2]),x]

[Out]

(a*x^6)/6 + (b*x^4*ArcTan[E^(c + d*x^2)])/d - (I*b*x^2*PolyLog[2, (-I)*E^(c + d*x^2)])/d^2 + (I*b*x^2*PolyLog[
2, I*E^(c + d*x^2)])/d^2 + (I*b*PolyLog[3, (-I)*E^(c + d*x^2)])/d^3 - (I*b*PolyLog[3, I*E^(c + d*x^2)])/d^3

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x^5 \left (a+b \text{sech}\left (c+d x^2\right )\right ) \, dx &=\int \left (a x^5+b x^5 \text{sech}\left (c+d x^2\right )\right ) \, dx\\ &=\frac{a x^6}{6}+b \int x^5 \text{sech}\left (c+d x^2\right ) \, dx\\ &=\frac{a x^6}{6}+\frac{1}{2} b \operatorname{Subst}\left (\int x^2 \text{sech}(c+d x) \, dx,x,x^2\right )\\ &=\frac{a x^6}{6}+\frac{b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{(i b) \operatorname{Subst}\left (\int x \log \left (1-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}+\frac{(i b) \operatorname{Subst}\left (\int x \log \left (1+i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}\\ &=\frac{a x^6}{6}+\frac{b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{i b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{i b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac{(i b) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d^2}-\frac{(i b) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{c+d x}\right ) \, dx,x,x^2\right )}{d^2}\\ &=\frac{a x^6}{6}+\frac{b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{i b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{i b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}\\ &=\frac{a x^6}{6}+\frac{b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{i b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{i b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac{i b \text{Li}_3\left (-i e^{c+d x^2}\right )}{d^3}-\frac{i b \text{Li}_3\left (i e^{c+d x^2}\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 1.31328, size = 143, normalized size = 1.14 \[ \frac{a x^6}{6}+\frac{i b \left (-2 d x^2 \text{PolyLog}\left (2,-i e^{c+d x^2}\right )+2 d x^2 \text{PolyLog}\left (2,i e^{c+d x^2}\right )+2 \text{PolyLog}\left (3,-i e^{c+d x^2}\right )-2 \text{PolyLog}\left (3,i e^{c+d x^2}\right )+d^2 x^4 \log \left (1-i e^{c+d x^2}\right )-d^2 x^4 \log \left (1+i e^{c+d x^2}\right )\right )}{2 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*Sech[c + d*x^2]),x]

[Out]

(a*x^6)/6 + ((I/2)*b*(d^2*x^4*Log[1 - I*E^(c + d*x^2)] - d^2*x^4*Log[1 + I*E^(c + d*x^2)] - 2*d*x^2*PolyLog[2,
(-I)*E^(c + d*x^2)] + 2*d*x^2*PolyLog[2, I*E^(c + d*x^2)] + 2*PolyLog[3, (-I)*E^(c + d*x^2)] - 2*PolyLog[3, I
*E^(c + d*x^2)]))/d^3

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Maple [F]  time = 0.15, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \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^5*(a+b*sech(d*x^2+c)),x)

[Out]

int(x^5*(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}{6} \, a x^{6} + 2 \, b \int \frac{x^{5}}{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^5*(a+b*sech(d*x^2+c)),x, algorithm="maxima")

[Out]

1/6*a*x^6 + 2*b*integrate(x^5/(e^(d*x^2 + c) + e^(-d*x^2 - c)), x)

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Fricas [C]  time = 2.33502, size = 706, normalized size = 5.65 \begin{align*} \frac{a d^{3} x^{6} + 6 i \, b d x^{2}{\rm Li}_2\left (i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right )\right ) - 6 i \, b d x^{2}{\rm Li}_2\left (-i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right )\right ) + 3 i \, b c^{2} \log \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right ) + i\right ) - 3 i \, b c^{2} \log \left (\cosh \left (d x^{2} + c\right ) + \sinh \left (d x^{2} + c\right ) - i\right ) +{\left (-3 i \, b d^{2} x^{4} + 3 i \, b c^{2}\right )} \log \left (i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right ) + 1\right ) +{\left (3 i \, b d^{2} x^{4} - 3 i \, b c^{2}\right )} \log \left (-i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right ) + 1\right ) - 6 i \, b{\rm polylog}\left (3, i \, \cosh \left (d x^{2} + c\right ) + i \, \sinh \left (d x^{2} + c\right )\right ) + 6 i \, b{\rm polylog}\left (3, -i \, \cosh \left (d x^{2} + c\right ) - i \, \sinh \left (d x^{2} + c\right )\right )}{6 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sech(d*x^2+c)),x, algorithm="fricas")

[Out]

1/6*(a*d^3*x^6 + 6*I*b*d*x^2*dilog(I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c)) - 6*I*b*d*x^2*dilog(-I*cosh(d*x^2 +
c) - I*sinh(d*x^2 + c)) + 3*I*b*c^2*log(cosh(d*x^2 + c) + sinh(d*x^2 + c) + I) - 3*I*b*c^2*log(cosh(d*x^2 + c)
+ sinh(d*x^2 + c) - I) + (-3*I*b*d^2*x^4 + 3*I*b*c^2)*log(I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c) + 1) + (3*I*b
*d^2*x^4 - 3*I*b*c^2)*log(-I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c) + 1) - 6*I*b*polylog(3, I*cosh(d*x^2 + c) + I
*sinh(d*x^2 + c)) + 6*I*b*polylog(3, -I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c)))/d^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \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**5*(a+b*sech(d*x**2+c)),x)

[Out]

Integral(x**5*(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^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*sech(d*x^2+c)),x, algorithm="giac")

[Out]

integrate((b*sech(d*x^2 + c) + a)*x^5, x)