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

Optimal. Leaf size=349 \[ \frac {b \text {Li}_3\left (-\frac {a e^{d x^2+c}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {b \text {Li}_3\left (-\frac {a e^{d x^2+c}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {b x^2 \text {Li}_2\left (-\frac {a e^{d x^2+c}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {b x^2 \text {Li}_2\left (-\frac {a e^{d x^2+c}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^4 \log \left (\frac {a e^{c+d x^2}}{b-\sqrt {b^2-a^2}}+1\right )}{2 a d \sqrt {b^2-a^2}}+\frac {b x^4 \log \left (\frac {a e^{c+d x^2}}{\sqrt {b^2-a^2}+b}+1\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^6}{6 a} \]

[Out]

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

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Rubi [A]  time = 0.86, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5436, 4191, 3320, 2264, 2190, 2531, 2282, 6589} \[ -\frac {b x^2 \text {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {b x^2 \text {PolyLog}\left (2,-\frac {a e^{c+d x^2}}{\sqrt {b^2-a^2}+b}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {b \text {PolyLog}\left (3,-\frac {a e^{c+d x^2}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {b \text {PolyLog}\left (3,-\frac {a e^{c+d x^2}}{\sqrt {b^2-a^2}+b}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {b x^4 \log \left (\frac {a e^{c+d x^2}}{b-\sqrt {b^2-a^2}}+1\right )}{2 a d \sqrt {b^2-a^2}}+\frac {b x^4 \log \left (\frac {a e^{c+d x^2}}{\sqrt {b^2-a^2}+b}+1\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^6}{6 a} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*Sech[c + d*x^2]),x]

[Out]

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

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[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 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 3320

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol]
:> Dist[2, Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(E^(I*Pi*(k - 1/2))*(b + (2*a*E^(-(I*e) + f*fz*x))/E^(I*Pi*(k
- 1/2)) - (b*E^(2*(-(I*e) + f*fz*x)))/E^(2*I*k*Pi))), x], x] /; FreeQ[{a, b, c, d, e, f, fz}, x] && IntegerQ[
2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

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

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Mathematica [A]  time = 1.57, size = 376, normalized size = 1.08 \[ \frac {d^3 x^6 \sqrt {e^{2 c} \left (b^2-a^2\right )}-3 b e^c d^2 x^4 \log \left (\frac {a e^{2 c+d x^2}}{b e^c-\sqrt {e^{2 c} \left (b^2-a^2\right )}}+1\right )+3 b e^c d^2 x^4 \log \left (\frac {a e^{2 c+d x^2}}{\sqrt {e^{2 c} \left (b^2-a^2\right )}+b e^c}+1\right )-6 b e^c d x^2 \text {Li}_2\left (-\frac {a e^{d x^2+2 c}}{b e^c-\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )+6 b e^c d x^2 \text {Li}_2\left (-\frac {a e^{d x^2+2 c}}{e^c b+\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )+6 b e^c \text {Li}_3\left (-\frac {a e^{d x^2+2 c}}{b e^c-\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )-6 b e^c \text {Li}_3\left (-\frac {a e^{d x^2+2 c}}{e^c b+\sqrt {\left (b^2-a^2\right ) e^{2 c}}}\right )}{6 a d^3 \sqrt {e^{2 c} \left (b^2-a^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.43, size = 731, normalized size = 2.09 \[ \frac {{\left (a^{2} - b^{2}\right )} d^{3} x^{6} + 6 \, a b d x^{2} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} {\rm Li}_2\left (-\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) + {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} + a}{a} + 1\right ) - 6 \, a b d x^{2} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} {\rm Li}_2\left (-\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) - {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} + a}{a} + 1\right ) - 3 \, a b c^{2} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} \log \left (2 \, a \cosh \left (d x^{2} + c\right ) + 2 \, a \sinh \left (d x^{2} + c\right ) + 2 \, a \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} + 2 \, b\right ) + 3 \, a b c^{2} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} \log \left (2 \, a \cosh \left (d x^{2} + c\right ) + 2 \, a \sinh \left (d x^{2} + c\right ) - 2 \, a \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} + 2 \, b\right ) - 6 \, a b \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} {\rm polylog}\left (3, -\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) + {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}}}{a}\right ) + 6 \, a b \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} {\rm polylog}\left (3, -\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) - {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}}}{a}\right ) + 3 \, {\left (a b d^{2} x^{4} - a b c^{2}\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} \log \left (\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) + {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} + a}{a}\right ) - 3 \, {\left (a b d^{2} x^{4} - a b c^{2}\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} \log \left (\frac {b \cosh \left (d x^{2} + c\right ) + b \sinh \left (d x^{2} + c\right ) - {\left (a \cosh \left (d x^{2} + c\right ) + a \sinh \left (d x^{2} + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{a^{2}}} + a}{a}\right )}{6 \, {\left (a^{3} - a b^{2}\right )} d^{3}} \]

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^2 - b^2)*d^3*x^6 + 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) +
(a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) - 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2
)*dilog(-(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^
2) + a)/a + 1) - 3*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cosh(d*x^2 + c) + 2*a*sinh(d*x^2 + c) + 2*a*sqrt(-(a
^2 - b^2)/a^2) + 2*b) + 3*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cosh(d*x^2 + c) + 2*a*sinh(d*x^2 + c) - 2*a*s
qrt(-(a^2 - b^2)/a^2) + 2*b) - 6*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c)
+ (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) + 6*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3
, -(b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a)
+ 3*(a*b*d^2*x^4 - a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log((b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) + (a*cosh(d*x^2
+ c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a) - 3*(a*b*d^2*x^4 - a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*l
og((b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) - (a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a
)/a))/((a^3 - a*b^2)*d^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{b \operatorname {sech}\left (d x^{2} + c\right ) + a}\,{d x} \]

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(x^5/(b*sech(d*x^2 + c) + a), x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{a +b \,\mathrm {sech}\left (d \,x^{2}+c \right )}\, dx \]

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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-b>0)', see `assume?` for mor
e details)Is a-b positive or negative?

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{a + b \operatorname {sech}{\left (c + d x^{2} \right )}}\, dx \]

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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