3.57 \(\int x^{3/2} (a+b \text {sech}(c+d \sqrt {x}))^2 \, dx\)

Optimal. Leaf size=407 \[ \frac {2}{5} a^2 x^{5/2}+\frac {96 i a b \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 i a b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {6 b^2 \text {Li}_4\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 b^2 x \text {Li}_2\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {8 b^2 x^{3/2} \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {2 b^2 x^2}{d} \]

[Out]

2*b^2*x^2/d+2/5*a^2*x^(5/2)+8*a*b*x^2*arctan(exp(c+d*x^(1/2)))/d-8*b^2*x^(3/2)*ln(1+exp(2*c+2*d*x^(1/2)))/d^2+
16*I*a*b*x^(3/2)*polylog(2,I*exp(c+d*x^(1/2)))/d^2-16*I*a*b*x^(3/2)*polylog(2,-I*exp(c+d*x^(1/2)))/d^2-12*b^2*
x*polylog(2,-exp(2*c+2*d*x^(1/2)))/d^3-96*I*a*b*polylog(4,-I*exp(c+d*x^(1/2)))*x^(1/2)/d^4-96*I*a*b*polylog(5,
I*exp(c+d*x^(1/2)))/d^5-6*b^2*polylog(4,-exp(2*c+2*d*x^(1/2)))/d^5-48*I*a*b*x*polylog(3,I*exp(c+d*x^(1/2)))/d^
3+96*I*a*b*polylog(5,-I*exp(c+d*x^(1/2)))/d^5+12*b^2*polylog(3,-exp(2*c+2*d*x^(1/2)))*x^(1/2)/d^4+96*I*a*b*pol
ylog(4,I*exp(c+d*x^(1/2)))*x^(1/2)/d^4+48*I*a*b*x*polylog(3,-I*exp(c+d*x^(1/2)))/d^3+2*b^2*x^2*tanh(c+d*x^(1/2
))/d

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 407, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5436, 4190, 4180, 2531, 6609, 2282, 6589, 4184, 3718, 2190} \[ -\frac {16 i a b x^{3/2} \text {PolyLog}\left (2,-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {PolyLog}\left (2,i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {48 i a b x \text {PolyLog}\left (3,-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {PolyLog}\left (3,i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {96 i a b \sqrt {x} \text {PolyLog}\left (4,-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {PolyLog}\left (4,i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \text {PolyLog}\left (5,-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \text {PolyLog}\left (5,i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {12 b^2 x \text {PolyLog}\left (2,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {PolyLog}\left (3,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 b^2 \text {PolyLog}\left (4,-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )}{d^2}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {2 b^2 x^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*x^2)/d + (2*a^2*x^(5/2))/5 + (8*a*b*x^2*ArcTan[E^(c + d*Sqrt[x])])/d - (8*b^2*x^(3/2)*Log[1 + E^(2*(c +
d*Sqrt[x]))])/d^2 - ((16*I)*a*b*x^(3/2)*PolyLog[2, (-I)*E^(c + d*Sqrt[x])])/d^2 + ((16*I)*a*b*x^(3/2)*PolyLog
[2, I*E^(c + d*Sqrt[x])])/d^2 - (12*b^2*x*PolyLog[2, -E^(2*(c + d*Sqrt[x]))])/d^3 + ((48*I)*a*b*x*PolyLog[3, (
-I)*E^(c + d*Sqrt[x])])/d^3 - ((48*I)*a*b*x*PolyLog[3, I*E^(c + d*Sqrt[x])])/d^3 + (12*b^2*Sqrt[x]*PolyLog[3,
-E^(2*(c + d*Sqrt[x]))])/d^4 - ((96*I)*a*b*Sqrt[x]*PolyLog[4, (-I)*E^(c + d*Sqrt[x])])/d^4 + ((96*I)*a*b*Sqrt[
x]*PolyLog[4, I*E^(c + d*Sqrt[x])])/d^4 - (6*b^2*PolyLog[4, -E^(2*(c + d*Sqrt[x]))])/d^5 + ((96*I)*a*b*PolyLog
[5, (-I)*E^(c + d*Sqrt[x])])/d^5 - ((96*I)*a*b*PolyLog[5, I*E^(c + d*Sqrt[x])])/d^5 + (2*b^2*x^2*Tanh[c + d*Sq
rt[x]])/d

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 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 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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int x^{3/2} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^4 (a+b \text {sech}(c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^4+2 a b x^4 \text {sech}(c+d x)+b^2 x^4 \text {sech}^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}+(4 a b) \operatorname {Subst}\left (\int x^4 \text {sech}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^4 \text {sech}^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}-\frac {(16 i a b) \operatorname {Subst}\left (\int x^3 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(16 i a b) \operatorname {Subst}\left (\int x^3 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (8 b^2\right ) \operatorname {Subst}\left (\int x^3 \tanh (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {(48 i a b) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(48 i a b) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (16 b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 (c+d x)} x^3}{1+e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {48 i a b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}-\frac {(96 i a b) \operatorname {Subst}\left (\int x \text {Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(96 i a b) \operatorname {Subst}\left (\int x \text {Li}_3\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {\left (24 b^2\right ) \operatorname {Subst}\left (\int x^2 \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {12 b^2 x \text {Li}_2\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 i a b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {(96 i a b) \operatorname {Subst}\left (\int \text {Li}_4\left (-i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(96 i a b) \operatorname {Subst}\left (\int \text {Li}_4\left (i e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (24 b^2\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {12 b^2 x \text {Li}_2\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 i a b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}+\frac {(96 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_4(-i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(96 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_4(i x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {12 b^2 x \text {Li}_2\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 i a b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ &=\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}+\frac {8 a b x^2 \tan ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {8 b^2 x^{3/2} \log \left (1+e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )}{d^2}-\frac {12 b^2 x \text {Li}_2\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 i a b x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 i a b x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )}{d^4}-\frac {6 b^2 \text {Li}_4\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 i a b \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 i a b \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )}{d^5}+\frac {2 b^2 x^2 \tanh \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 6.34, size = 487, normalized size = 1.20 \[ \frac {2 \cosh \left (c+d \sqrt {x}\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2 \left (a^2 x^{5/2} \cosh \left (c+d \sqrt {x}\right )+\frac {5 b \cosh \left (c+d \sqrt {x}\right ) \left (\frac {2 b e^{2 c} d^4 x^2}{e^{2 c}+1}+i \left (2 a d^4 x^2 \log \left (1-i e^{c+d \sqrt {x}}\right )-2 a d^4 x^2 \log \left (1+i e^{c+d \sqrt {x}}\right )-8 a d^3 x^{3/2} \text {Li}_2\left (-i e^{c+d \sqrt {x}}\right )+8 a d^3 x^{3/2} \text {Li}_2\left (i e^{c+d \sqrt {x}}\right )+24 a d^2 x \text {Li}_3\left (-i e^{c+d \sqrt {x}}\right )-24 a d^2 x \text {Li}_3\left (i e^{c+d \sqrt {x}}\right )-48 a d \sqrt {x} \text {Li}_4\left (-i e^{c+d \sqrt {x}}\right )+48 a d \sqrt {x} \text {Li}_4\left (i e^{c+d \sqrt {x}}\right )+48 a \text {Li}_5\left (-i e^{c+d \sqrt {x}}\right )-48 a \text {Li}_5\left (i e^{c+d \sqrt {x}}\right )+4 i b d^3 x^{3/2} \log \left (e^{2 \left (c+d \sqrt {x}\right )}+1\right )+6 i b d^2 x \text {Li}_2\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )-6 i b d \sqrt {x} \text {Li}_3\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )+3 i b \text {Li}_4\left (-e^{2 \left (c+d \sqrt {x}\right )}\right )\right )\right )}{d^5}+\frac {5 b^2 x^2 \text {sech}(c) \sinh \left (d \sqrt {x}\right )}{d}\right )}{5 \left (a \cosh \left (c+d \sqrt {x}\right )+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cosh[c + d*Sqrt[x]]*(a + b*Sech[c + d*Sqrt[x]])^2*(a^2*x^(5/2)*Cosh[c + d*Sqrt[x]] + (5*b*Cosh[c + d*Sqrt[x
]]*((2*b*d^4*E^(2*c)*x^2)/(1 + E^(2*c)) + I*(2*a*d^4*x^2*Log[1 - I*E^(c + d*Sqrt[x])] - 2*a*d^4*x^2*Log[1 + I*
E^(c + d*Sqrt[x])] + (4*I)*b*d^3*x^(3/2)*Log[1 + E^(2*(c + d*Sqrt[x]))] - 8*a*d^3*x^(3/2)*PolyLog[2, (-I)*E^(c
+ d*Sqrt[x])] + 8*a*d^3*x^(3/2)*PolyLog[2, I*E^(c + d*Sqrt[x])] + (6*I)*b*d^2*x*PolyLog[2, -E^(2*(c + d*Sqrt[
x]))] + 24*a*d^2*x*PolyLog[3, (-I)*E^(c + d*Sqrt[x])] - 24*a*d^2*x*PolyLog[3, I*E^(c + d*Sqrt[x])] - (6*I)*b*d
*Sqrt[x]*PolyLog[3, -E^(2*(c + d*Sqrt[x]))] - 48*a*d*Sqrt[x]*PolyLog[4, (-I)*E^(c + d*Sqrt[x])] + 48*a*d*Sqrt[
x]*PolyLog[4, I*E^(c + d*Sqrt[x])] + (3*I)*b*PolyLog[4, -E^(2*(c + d*Sqrt[x]))] + 48*a*PolyLog[5, (-I)*E^(c +
d*Sqrt[x])] - 48*a*PolyLog[5, I*E^(c + d*Sqrt[x])])))/d^5 + (5*b^2*x^2*Sech[c]*Sinh[d*Sqrt[x]])/d))/(5*(b + a*
Cosh[c + d*Sqrt[x]])^2)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{\frac {3}{2}} \operatorname {sech}\left (d \sqrt {x} + c\right )^{2} + 2 \, a b x^{\frac {3}{2}} \operatorname {sech}\left (d \sqrt {x} + c\right ) + a^{2} x^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^(3/2)*sech(d*sqrt(x) + c)^2 + 2*a*b*x^(3/2)*sech(d*sqrt(x) + c) + a^2*x^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {sech}\left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="giac")

[Out]

integrate((b*sech(d*sqrt(x) + c) + a)^2*x^(3/2), x)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \left (a +b \,\mathrm {sech}\left (c +d \sqrt {x}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x)

[Out]

int(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (a^{2} d x^{\frac {5}{2}} e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + a^{2} d x^{\frac {5}{2}} - 10 \, b^{2} x^{2}\right )}}{5 \, {\left (d e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + d\right )}} + \int \frac {4 \, {\left (a b d x^{\frac {5}{2}} e^{\left (d \sqrt {x} + c\right )} + 2 \, b^{2} x^{2}\right )}}{d x e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + d x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sech(c+d*x^(1/2)))^2,x, algorithm="maxima")

[Out]

2/5*(a^2*d*x^(5/2)*e^(2*d*sqrt(x) + 2*c) + a^2*d*x^(5/2) - 10*b^2*x^2)/(d*e^(2*d*sqrt(x) + 2*c) + d) + integra
te(4*(a*b*d*x^(5/2)*e^(d*sqrt(x) + c) + 2*b^2*x^2)/(d*x*e^(2*d*sqrt(x) + 2*c) + d*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^{3/2}\,{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b/cosh(c + d*x^(1/2)))^2,x)

[Out]

int(x^(3/2)*(a + b/cosh(c + d*x^(1/2)))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(a+b*sech(c+d*x**(1/2)))**2,x)

[Out]

Integral(x**(3/2)*(a + b*sech(c + d*sqrt(x)))**2, x)

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