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

Optimal. Leaf size=319 \[ -\frac{12 i a b x \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{24 i a b \sqrt{x} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \sqrt{x} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{24 i a b \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{6 b^2 \sqrt{x} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{3 b^2 \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 b^2 x \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x^{3/2}}{d} \]

[Out]

(2*b^2*x^(3/2))/d + (a^2*x^2)/2 + (8*a*b*x^(3/2)*ArcTan[E^(c + d*Sqrt[x])])/d - (6*b^2*x*Log[1 + E^(2*(c + d*S
qrt[x]))])/d^2 - ((12*I)*a*b*x*PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((12*I)*a*b*x*PolyLog[2, I*E^(c + d*S
qrt[x])])/d^2 - (6*b^2*Sqrt[x]*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^3 + ((24*I)*a*b*Sqrt[x]*PolyLog[3, (-I)*E
^(c + d*Sqrt[x])])/d^3 - ((24*I)*a*b*Sqrt[x]*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (3*b^2*PolyLog[3, -E^(2*(c
+ d*Sqrt[x]))])/d^4 - ((24*I)*a*b*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((24*I)*a*b*PolyLog[4, I*E^(c + d
*Sqrt[x])])/d^4 + (2*b^2*x^(3/2)*Tanh[c + d*Sqrt[x]])/d

________________________________________________________________________________________

Rubi [A]  time = 0.433306, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {5436, 4190, 4180, 2531, 6609, 2282, 6589, 4184, 3718, 2190} \[ -\frac{12 i a b x \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{24 i a b \sqrt{x} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \sqrt{x} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{24 i a b \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}-\frac{6 b^2 \sqrt{x} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{3 b^2 \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 b^2 x \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sech[c + d*Sqrt[x]])^2,x]

[Out]

(2*b^2*x^(3/2))/d + (a^2*x^2)/2 + (8*a*b*x^(3/2)*ArcTan[E^(c + d*Sqrt[x])])/d - (6*b^2*x*Log[1 + E^(2*(c + d*S
qrt[x]))])/d^2 - ((12*I)*a*b*x*PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((12*I)*a*b*x*PolyLog[2, I*E^(c + d*S
qrt[x])])/d^2 - (6*b^2*Sqrt[x]*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^3 + ((24*I)*a*b*Sqrt[x]*PolyLog[3, (-I)*E
^(c + d*Sqrt[x])])/d^3 - ((24*I)*a*b*Sqrt[x]*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (3*b^2*PolyLog[3, -E^(2*(c
+ d*Sqrt[x]))])/d^4 - ((24*I)*a*b*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((24*I)*a*b*PolyLog[4, I*E^(c + d
*Sqrt[x])])/d^4 + (2*b^2*x^(3/2)*Tanh[c + d*Sqrt[x]])/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 6609

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

Rubi steps

\begin{align*} \int x \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^3 (a+b \text{sech}(c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^3+2 a b x^3 \text{sech}(c+d x)+b^2 x^3 \text{sech}^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^2}{2}+(4 a b) \operatorname{Subst}\left (\int x^3 \text{sech}(c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^3 \text{sech}^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(12 i a b) \operatorname{Subst}\left (\int x^2 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(12 i a b) \operatorname{Subst}\left (\int x^2 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int x^2 \tanh (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2 b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{12 i a b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(24 i a b) \operatorname{Subst}\left (\int x \text{Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(24 i a b) \operatorname{Subst}\left (\int x \text{Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 (c+d x)} x^2}{1+e^{2 (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2 b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 b^2 x \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i a b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{24 i a b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(24 i a b) \operatorname{Subst}\left (\int \text{Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{(24 i a b) \operatorname{Subst}\left (\int \text{Li}_3\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int x \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{2 b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 b^2 x \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i a b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{6 b^2 \sqrt{x} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{24 i a b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(24 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{(24 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{2 b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 b^2 x \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i a b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{6 b^2 \sqrt{x} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{24 i a b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{24 i a b \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}\\ &=\frac{2 b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+\frac{8 a b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 b^2 x \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i a b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i a b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{6 b^2 \sqrt{x} \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{24 i a b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{24 i a b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{3 b^2 \text{Li}_3\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{24 i a b \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{24 i a b \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{2 b^2 x^{3/2} \tanh \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 8.31505, size = 459, normalized size = 1.44 \[ \frac{\cosh \left (c+d \sqrt{x}\right ) \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \left (\frac{2 b \cosh \left (c+d \sqrt{x}\right ) \left (\frac{4 b e^{2 c} d^3 x^{3/2}}{e^{2 c}+1}+i \left (-12 \left (a d^2 x-i b d \sqrt{x}\right ) \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )+12 \left (a d^2 x+i b d \sqrt{x}\right ) \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )+24 a d \sqrt{x} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )-24 a d \sqrt{x} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )-24 a \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )+24 a \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )-3 i b \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )+4 a d^3 x^{3/2} \log \left (1-i e^{c+d \sqrt{x}}\right )-4 a d^3 x^{3/2} \log \left (1+i e^{c+d \sqrt{x}}\right )+12 i b d^2 x \log \left (1-i e^{c+d \sqrt{x}}\right )+12 i b d^2 x \log \left (1+i e^{c+d \sqrt{x}}\right )-6 i b d^2 x \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )\right )\right )}{d^4}+a^2 x^2 \cosh \left (c+d \sqrt{x}\right )+\frac{4 b^2 x^{3/2} \text{sech}(c) \sinh \left (d \sqrt{x}\right )}{d}\right )}{2 \left (a \cosh \left (c+d \sqrt{x}\right )+b\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*(a^2*x^2*Cosh[c + d*Sqrt[x]] + (2*b*Cosh[c + d*Sqrt[x]]*((4
*b*d^3*E^(2*c)*x^(3/2))/(1 + E^(2*c)) + I*((12*I)*b*d^2*x*Log[1 - I*E^(c + d*Sqrt[x])] + 4*a*d^3*x^(3/2)*Log[1
- I*E^(c + d*Sqrt[x])] + (12*I)*b*d^2*x*Log[1 + I*E^(c + d*Sqrt[x])] - 4*a*d^3*x^(3/2)*Log[1 + I*E^(c + d*Sqr
t[x])] - (6*I)*b*d^2*x*Log[1 + E^(2*(c + d*Sqrt[x]))] - 12*((-I)*b*d*Sqrt[x] + a*d^2*x)*PolyLog[2, (-I)*E^(c +
d*Sqrt[x])] + 12*(I*b*d*Sqrt[x] + a*d^2*x)*PolyLog[2, I*E^(c + d*Sqrt[x])] + 24*a*d*Sqrt[x]*PolyLog[3, (-I)*E
^(c + d*Sqrt[x])] - 24*a*d*Sqrt[x]*PolyLog[3, I*E^(c + d*Sqrt[x])] - (3*I)*b*PolyLog[3, -E^(2*(c + d*Sqrt[x]))
] - 24*a*PolyLog[4, (-I)*E^(c + d*Sqrt[x])] + 24*a*PolyLog[4, I*E^(c + d*Sqrt[x])])))/d^4 + (4*b^2*x^(3/2)*Sec
h[c]*Sinh[d*Sqrt[x]])/d))/(2*(b + a*Cosh[c + d*Sqrt[x]])^2)

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Maple [F]  time = 0.064, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b{\rm sech} \left (c+d\sqrt{x}\right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*sech(c+d*x^(1/2)))^2,x)

[Out]

int(x*(a+b*sech(c+d*x^(1/2)))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} d x^{2} e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + a^{2} d x^{2} - 8 \, b^{2} x^{\frac{3}{2}}}{2 \,{\left (d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d\right )}} + \int \frac{2 \,{\left (2 \, a b d x e^{\left (d \sqrt{x} + c\right )} + 3 \, b^{2} \sqrt{x}\right )}}{d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="maxima")

[Out]

1/2*(a^2*d*x^2*e^(2*d*sqrt(x) + 2*c) + a^2*d*x^2 - 8*b^2*x^(3/2))/(d*e^(2*d*sqrt(x) + 2*c) + d) + integrate(2*
(2*a*b*d*x*e^(d*sqrt(x) + c) + 3*b^2*sqrt(x))/(d*e^(2*d*sqrt(x) + 2*c) + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} x \operatorname{sech}\left (d \sqrt{x} + c\right )^{2} + 2 \, a b x \operatorname{sech}\left (d \sqrt{x} + c\right ) + a^{2} x, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="fricas")

[Out]

integral(b^2*x*sech(d*sqrt(x) + c)^2 + 2*a*b*x*sech(d*sqrt(x) + c) + a^2*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*sech(c+d*x**(1/2)))**2,x)

[Out]

Integral(x*(a + b*sech(c + d*sqrt(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 \sqrt{x} + c\right ) + a\right )}^{2} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="giac")

[Out]

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