3.1.1 \(\int x^5 (a+b \text {sech}(c+d x^2)) \, dx\) [1]

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

[Out]

1/6*a*x^6+b*x^4*arctan(exp(d*x^2+c))/d-I*b*x^2*polylog(2,-I*exp(d*x^2+c))/d^2+I*b*x^2*polylog(2,I*exp(d*x^2+c)
)/d^2+I*b*polylog(3,-I*exp(d*x^2+c))/d^3-I*b*polylog(3,I*exp(d*x^2+c))/d^3

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Rubi [A]
time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, 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, 5544, 4265, 2611, 2320, 6724} \begin {gather*} \frac {a x^6}{6}+\frac {b x^4 \text {ArcTan}\left (e^{c+d x^2}\right )}{d}+\frac {i b \text {Li}_3\left (-i e^{d x^2+c}\right )}{d^3}-\frac {i b \text {Li}_3\left (i e^{d x^2+c}\right )}{d^3}-\frac {i b x^2 \text {Li}_2\left (-i e^{d x^2+c}\right )}{d^2}+\frac {i b x^2 \text {Li}_2\left (i e^{d x^2+c}\right )}{d^2} \end {gather*}

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 2320

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 2611

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 4265

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 5544

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 6724

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 \text {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) \text {Subst}\left (\int x \log \left (1-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}+\frac {(i b) \text {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) \text {Subst}\left (\int \text {Li}_2\left (-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d^2}-\frac {(i b) \text {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) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}-\frac {(i b) \text {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*}

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Mathematica [A]
time = 1.02, size = 143, normalized size = 1.14 \begin {gather*} \frac {a x^6}{6}+\frac {i b \left (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 )-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 )\right )}{2 d^3} \end {gather*}

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 = 1.11, size = 0, normalized size = 0.00 \[\int x^{5} \left (a +b \,\mathrm {sech}\left (d \,x^{2}+c \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (100) = 200\).
time = 0.39, size = 256, normalized size = 2.05 \begin {gather*} \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 ) - 3 \, {\left (i \, b d^{2} x^{4} - 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 ) - 3 \, {\left (-i \, b d^{2} x^{4} + 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 {gather*}

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 - 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 + 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*sin
h(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \left (a + b \operatorname {sech}{\left (c + d x^{2} \right )}\right )\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^5\,\left (a+\frac {b}{\mathrm {cosh}\left (d\,x^2+c\right )}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b/cosh(c + d*x^2)),x)

[Out]

int(x^5*(a + b/cosh(c + d*x^2)), x)

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