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

Optimal. Leaf size=217 \[ -\frac{2 i a b x^2 \text{PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b x^2 \text{PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b \text{PolyLog}\left (3,-i e^{c+d x^2}\right )}{d^3}-\frac{2 i a b \text{PolyLog}\left (3,i e^{c+d x^2}\right )}{d^3}-\frac{b^2 \text{PolyLog}\left (2,-e^{2 \left (c+d x^2\right )}\right )}{2 d^3}+\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{b^2 x^2 \log \left (e^{2 \left (c+d x^2\right )}+1\right )}{d^2}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac{b^2 x^4}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.357863, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {5436, 4190, 4180, 2531, 2282, 6589, 4184, 3718, 2190, 2279, 2391} \[ -\frac{2 i a b x^2 \text{PolyLog}\left (2,-i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b x^2 \text{PolyLog}\left (2,i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b \text{PolyLog}\left (3,-i e^{c+d x^2}\right )}{d^3}-\frac{2 i a b \text{PolyLog}\left (3,i e^{c+d x^2}\right )}{d^3}-\frac{b^2 \text{PolyLog}\left (2,-e^{2 \left (c+d x^2\right )}\right )}{2 d^3}+\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{b^2 x^2 \log \left (e^{2 \left (c+d x^2\right )}+1\right )}{d^2}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac{b^2 x^4}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sech[c + d*x^2])^2,x]

[Out]

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

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 4190

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

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]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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 x^5 \left (a+b \text{sech}\left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b \text{sech}(c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \text{sech}(c+d x)+b^2 x^2 \text{sech}^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 x^6}{6}+(a b) \operatorname{Subst}\left (\int x^2 \text{sech}(c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int x^2 \text{sech}^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}-\frac{(2 i a b) \operatorname{Subst}\left (\int x \log \left (1-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}+\frac{(2 i a b) \operatorname{Subst}\left (\int x \log \left (1+i e^{c+d x}\right ) \, dx,x,x^2\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int x \tanh (c+d x) \, dx,x,x^2\right )}{d}\\ &=\frac{b^2 x^4}{2 d}+\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{2 i a b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac{(2 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{c+d x}\right ) \, dx,x,x^2\right )}{d^2}-\frac{(2 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{c+d x}\right ) \, dx,x,x^2\right )}{d^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 (c+d x)} x}{1+e^{2 (c+d x)}} \, dx,x,x^2\right )}{d}\\ &=\frac{b^2 x^4}{2 d}+\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{b^2 x^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac{(2 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}-\frac{(2 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{c+d x^2}\right )}{d^3}+\frac{b^2 \operatorname{Subst}\left (\int \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}\\ &=\frac{b^2 x^4}{2 d}+\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{b^2 x^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b \text{Li}_3\left (-i e^{c+d x^2}\right )}{d^3}-\frac{2 i a b \text{Li}_3\left (i e^{c+d x^2}\right )}{d^3}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \left (c+d x^2\right )}\right )}{2 d^3}\\ &=\frac{b^2 x^4}{2 d}+\frac{a^2 x^6}{6}+\frac{2 a b x^4 \tan ^{-1}\left (e^{c+d x^2}\right )}{d}-\frac{b^2 x^2 \log \left (1+e^{2 \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (-i e^{c+d x^2}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (i e^{c+d x^2}\right )}{d^2}-\frac{b^2 \text{Li}_2\left (-e^{2 \left (c+d x^2\right )}\right )}{2 d^3}+\frac{2 i a b \text{Li}_3\left (-i e^{c+d x^2}\right )}{d^3}-\frac{2 i a b \text{Li}_3\left (i e^{c+d x^2}\right )}{d^3}+\frac{b^2 x^4 \tanh \left (c+d x^2\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 4.49302, size = 294, normalized size = 1.35 \[ \frac{\cosh \left (c+d x^2\right ) \left (a+b \text{sech}\left (c+d x^2\right )\right )^2 \left (\frac{3 b \cosh \left (c+d x^2\right ) \left (-4 i a d x^2 \text{PolyLog}\left (2,-i e^{c+d x^2}\right )+4 i a d x^2 \text{PolyLog}\left (2,i e^{c+d x^2}\right )+4 i a \text{PolyLog}\left (3,-i e^{c+d x^2}\right )-4 i a \text{PolyLog}\left (3,i e^{c+d x^2}\right )-b \text{PolyLog}\left (2,-e^{2 \left (c+d x^2\right )}\right )+2 i a d^2 x^4 \log \left (1-i e^{c+d x^2}\right )-2 i a d^2 x^4 \log \left (1+i e^{c+d x^2}\right )+\frac{2 b e^{2 c} d^2 x^4}{e^{2 c}+1}-2 b d x^2 \log \left (e^{2 \left (c+d x^2\right )}+1\right )\right )}{d^3}+a^2 x^6 \cosh \left (c+d x^2\right )+\frac{3 b^2 x^4 \text{sech}(c) \sinh \left (d x^2\right )}{d}\right )}{6 \left (a \cosh \left (c+d x^2\right )+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \left ( a+b{\rm sech} \left (d{x}^{2}+c\right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, a^{2} x^{6} - \frac{b^{2} x^{4}}{d e^{\left (2 \, d x^{2} + 2 \, c\right )} + d} + \int \frac{4 \,{\left (a b d x^{5} e^{\left (d x^{2} + c\right )} + b^{2} x^{3}\right )}}{d e^{\left (2 \, d x^{2} + 2 \, c\right )} + d}\,{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))^2,x, algorithm="maxima")

[Out]

1/6*a^2*x^6 - b^2*x^4/(d*e^(2*d*x^2 + 2*c) + d) + integrate(4*(a*b*d*x^5*e^(d*x^2 + c) + b^2*x^3)/(d*e^(2*d*x^
2 + 2*c) + d), x)

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Fricas [C]  time = 2.60872, size = 3021, normalized size = 13.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(a^2*d^3*x^6 - 6*b^2*c^2 + (a^2*d^3*x^6 + 6*b^2*d^2*x^4 - 6*b^2*c^2)*cosh(d*x^2 + c)^2 + 2*(a^2*d^3*x^6 +
6*b^2*d^2*x^4 - 6*b^2*c^2)*cosh(d*x^2 + c)*sinh(d*x^2 + c) + (a^2*d^3*x^6 + 6*b^2*d^2*x^4 - 6*b^2*c^2)*sinh(d*
x^2 + c)^2 + (12*I*a*b*d*x^2 - 6*(-2*I*a*b*d*x^2 + b^2)*cosh(d*x^2 + c)^2 - 12*(-2*I*a*b*d*x^2 + b^2)*cosh(d*x
^2 + c)*sinh(d*x^2 + c) - 6*(-2*I*a*b*d*x^2 + b^2)*sinh(d*x^2 + c)^2 - 6*b^2)*dilog(I*cosh(d*x^2 + c) + I*sinh
(d*x^2 + c)) - 6*(2*I*a*b*d*x^2 + (2*I*a*b*d*x^2 + b^2)*cosh(d*x^2 + c)^2 + 2*(2*I*a*b*d*x^2 + b^2)*cosh(d*x^2
+ c)*sinh(d*x^2 + c) + (2*I*a*b*d*x^2 + b^2)*sinh(d*x^2 + c)^2 + b^2)*dilog(-I*cosh(d*x^2 + c) - I*sinh(d*x^2
+ c)) + (6*I*a*b*c^2 + 6*b^2*c + (6*I*a*b*c^2 + 6*b^2*c)*cosh(d*x^2 + c)^2 + (12*I*a*b*c^2 + 12*b^2*c)*cosh(d
*x^2 + c)*sinh(d*x^2 + c) + (6*I*a*b*c^2 + 6*b^2*c)*sinh(d*x^2 + c)^2)*log(cosh(d*x^2 + c) + sinh(d*x^2 + c) +
I) + (-6*I*a*b*c^2 + 6*b^2*c + (-6*I*a*b*c^2 + 6*b^2*c)*cosh(d*x^2 + c)^2 + (-12*I*a*b*c^2 + 12*b^2*c)*cosh(d
*x^2 + c)*sinh(d*x^2 + c) + (-6*I*a*b*c^2 + 6*b^2*c)*sinh(d*x^2 + c)^2)*log(cosh(d*x^2 + c) + sinh(d*x^2 + c)
- I) + (-6*I*a*b*d^2*x^4 - 6*b^2*d*x^2 + 6*I*a*b*c^2 - 6*b^2*c + (-6*I*a*b*d^2*x^4 - 6*b^2*d*x^2 + 6*I*a*b*c^2
- 6*b^2*c)*cosh(d*x^2 + c)^2 + (-12*I*a*b*d^2*x^4 - 12*b^2*d*x^2 + 12*I*a*b*c^2 - 12*b^2*c)*cosh(d*x^2 + c)*s
inh(d*x^2 + c) + (-6*I*a*b*d^2*x^4 - 6*b^2*d*x^2 + 6*I*a*b*c^2 - 6*b^2*c)*sinh(d*x^2 + c)^2)*log(I*cosh(d*x^2
+ c) + I*sinh(d*x^2 + c) + 1) + (6*I*a*b*d^2*x^4 - 6*b^2*d*x^2 - 6*I*a*b*c^2 - 6*b^2*c + (6*I*a*b*d^2*x^4 - 6*
b^2*d*x^2 - 6*I*a*b*c^2 - 6*b^2*c)*cosh(d*x^2 + c)^2 + (12*I*a*b*d^2*x^4 - 12*b^2*d*x^2 - 12*I*a*b*c^2 - 12*b^
2*c)*cosh(d*x^2 + c)*sinh(d*x^2 + c) + (6*I*a*b*d^2*x^4 - 6*b^2*d*x^2 - 6*I*a*b*c^2 - 6*b^2*c)*sinh(d*x^2 + c)
^2)*log(-I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c) + 1) + (-12*I*a*b*cosh(d*x^2 + c)^2 - 24*I*a*b*cosh(d*x^2 + c)*
sinh(d*x^2 + c) - 12*I*a*b*sinh(d*x^2 + c)^2 - 12*I*a*b)*polylog(3, I*cosh(d*x^2 + c) + I*sinh(d*x^2 + c)) + (
12*I*a*b*cosh(d*x^2 + c)^2 + 24*I*a*b*cosh(d*x^2 + c)*sinh(d*x^2 + c) + 12*I*a*b*sinh(d*x^2 + c)^2 + 12*I*a*b)
*polylog(3, -I*cosh(d*x^2 + c) - I*sinh(d*x^2 + c)))/(d^3*cosh(d*x^2 + c)^2 + 2*d^3*cosh(d*x^2 + c)*sinh(d*x^2
+ c) + d^3*sinh(d*x^2 + c)^2 + 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 )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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