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

Optimal. Leaf size=241 \[ -\frac{b \text{PolyLog}\left (2,-\frac{a e^{c+d x^2}}{b-\sqrt{b^2-a^2}}\right )}{2 a d^2 \sqrt{b^2-a^2}}+\frac{b \text{PolyLog}\left (2,-\frac{a e^{c+d x^2}}{\sqrt{b^2-a^2}+b}\right )}{2 a d^2 \sqrt{b^2-a^2}}-\frac{b x^2 \log \left (\frac{a e^{c+d x^2}}{b-\sqrt{b^2-a^2}}+1\right )}{2 a d \sqrt{b^2-a^2}}+\frac{b x^2 \log \left (\frac{a e^{c+d x^2}}{\sqrt{b^2-a^2}+b}+1\right )}{2 a d \sqrt{b^2-a^2}}+\frac{x^4}{4 a} \]

[Out]

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

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Rubi [A]  time = 0.508309, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {5436, 4191, 3320, 2264, 2190, 2279, 2391} \[ -\frac{b \text{PolyLog}\left (2,-\frac{a e^{c+d x^2}}{b-\sqrt{b^2-a^2}}\right )}{2 a d^2 \sqrt{b^2-a^2}}+\frac{b \text{PolyLog}\left (2,-\frac{a e^{c+d x^2}}{\sqrt{b^2-a^2}+b}\right )}{2 a d^2 \sqrt{b^2-a^2}}-\frac{b x^2 \log \left (\frac{a e^{c+d x^2}}{b-\sqrt{b^2-a^2}}+1\right )}{2 a d \sqrt{b^2-a^2}}+\frac{b x^2 \log \left (\frac{a e^{c+d x^2}}{\sqrt{b^2-a^2}+b}+1\right )}{2 a d \sqrt{b^2-a^2}}+\frac{x^4}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*Sech[c + d*x^2]),x]

[Out]

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

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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 1.67337, size = 843, normalized size = 3.5 \[ \frac{\left (b+a \cosh \left (d x^2+c\right )\right ) \left (x^4+\frac{2 b \left (2 \left (d x^2+c\right ) \tan ^{-1}\left (\frac{(a+b) \coth \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )+2 \left (c-i \cos ^{-1}\left (-\frac{b}{a}\right )\right ) \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )+\left (\cos ^{-1}\left (-\frac{b}{a}\right )+2 \left (\tan ^{-1}\left (\frac{(a+b) \coth \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )+\tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right )\right ) \log \left (\frac{\sqrt{a^2-b^2} e^{-\frac{d x^2}{2}-\frac{c}{2}}}{\sqrt{2} \sqrt{a} \sqrt{b+a \cosh \left (d x^2+c\right )}}\right )+\left (\cos ^{-1}\left (-\frac{b}{a}\right )-2 \left (\tan ^{-1}\left (\frac{(a+b) \coth \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )+\tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right )\right ) \log \left (\frac{\sqrt{a^2-b^2} e^{\frac{1}{2} \left (d x^2+c\right )}}{\sqrt{2} \sqrt{a} \sqrt{b+a \cosh \left (d x^2+c\right )}}\right )-\left (\cos ^{-1}\left (-\frac{b}{a}\right )+2 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right ) \log \left (\frac{(a+b) \left (-a+b+i \sqrt{a^2-b^2}\right ) \left (\tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )-1\right )}{a \left (a+b+i \sqrt{a^2-b^2} \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )-\left (\cos ^{-1}\left (-\frac{b}{a}\right )-2 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right ) \log \left (\frac{(a+b) \left (a-b+i \sqrt{a^2-b^2}\right ) \left (\tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )+1\right )}{a \left (a+b+i \sqrt{a^2-b^2} \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )+i \left (\text{PolyLog}\left (2,\frac{\left (b-i \sqrt{a^2-b^2}\right ) \left (a+b-i \sqrt{a^2-b^2} \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}{a \left (a+b+i \sqrt{a^2-b^2} \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )-\text{PolyLog}\left (2,\frac{\left (b+i \sqrt{a^2-b^2}\right ) \left (a+b-i \sqrt{a^2-b^2} \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}{a \left (a+b+i \sqrt{a^2-b^2} \tanh \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )\right )\right )}{\sqrt{a^2-b^2} d^2}\right ) \text{sech}\left (d x^2+c\right )}{4 a \left (a+b \text{sech}\left (d x^2+c\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cosh[c + d*x^2])*(x^4 + (2*b*(2*(c + d*x^2)*ArcTan[((a + b)*Coth[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + 2*
(c - I*ArcCos[-(b/a)])*ArcTan[((a - b)*Tanh[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + (ArcCos[-(b/a)] + 2*(ArcTan[((a
+ b)*Coth[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + ArcTan[((a - b)*Tanh[(c + d*x^2)/2])/Sqrt[a^2 - b^2]]))*Log[(Sqr
t[a^2 - b^2]*E^(-c/2 - (d*x^2)/2))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cosh[c + d*x^2]])] + (ArcCos[-(b/a)] - 2*(ArcTa
n[((a + b)*Coth[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + ArcTan[((a - b)*Tanh[(c + d*x^2)/2])/Sqrt[a^2 - b^2]]))*Log
[(Sqrt[a^2 - b^2]*E^((c + d*x^2)/2))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cosh[c + d*x^2]])] - (ArcCos[-(b/a)] + 2*ArcT
an[((a - b)*Tanh[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(-a + b + I*Sqrt[a^2 - b^2])*(-1 + Tanh[(c + d
*x^2)/2]))/(a*(a + b + I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^2)/2]))] - (ArcCos[-(b/a)] - 2*ArcTan[((a - b)*Tanh[(c
+ d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(a - b + I*Sqrt[a^2 - b^2])*(1 + Tanh[(c + d*x^2)/2]))/(a*(a + b +
I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^2)/2]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a + b - I*Sqrt[a^2 - b^2]*
Tanh[(c + d*x^2)/2]))/(a*(a + b + I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^2)/2]))] - PolyLog[2, ((b + I*Sqrt[a^2 - b^2
])*(a + b - I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^2)/2]))/(a*(a + b + I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^2)/2]))])))/(S
qrt[a^2 - b^2]*d^2))*Sech[c + d*x^2])/(4*a*(a + b*Sech[c + d*x^2]))

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{a+b{\rm sech} \left (d{x}^{2}+c\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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

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Fricas [B]  time = 2.24301, size = 1230, normalized size = 5.1 \begin{align*} \frac{{\left (a^{2} - b^{2}\right )} d^{2} x^{4} + 2 \, a b c \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 ) - 2 \, a b c \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 ) + 2 \, a b \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 ) - 2 \, a b \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 ) + 2 \,{\left (a b d x^{2} + a b c\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 ) - 2 \,{\left (a b d x^{2} + a b c\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 )}{4 \,{\left (a^{3} - a b^{2}\right )} 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^2 - b^2)*d^2*x^4 + 2*a*b*c*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) - 2*a*b*c*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) + 2*a*b*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) - 2*a*b*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) + 2*(a*b*d*x^2 + a*b*c)*sqrt(-(a^2 - b^2)/a^2)*log((b*cosh(d*x^2 + c) + b*sinh(d*x^2 + c) + (a*co
sh(d*x^2 + c) + a*sinh(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a) - 2*(a*b*d*x^2 + a*b*c)*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))/((a^3 - a*b^2)*d^2)

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