3.61 \(\int \frac{x^{3/2}}{a+b \sec (c+d \sqrt{x})} \, dx\)

Optimal. Leaf size=653 \[ \frac{8 b x^{3/2} \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^2 \sqrt{b^2-a^2}}-\frac{8 b x^{3/2} \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{24 i b x \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{24 i b x \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{48 b \sqrt{x} \text{PolyLog}\left (4,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^4 \sqrt{b^2-a^2}}+\frac{48 b \sqrt{x} \text{PolyLog}\left (4,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^4 \sqrt{b^2-a^2}}-\frac{48 i b \text{PolyLog}\left (5,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^5 \sqrt{b^2-a^2}}+\frac{48 i b \text{PolyLog}\left (5,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^5 \sqrt{b^2-a^2}}+\frac{2 i b x^2 \log \left (1+\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d \sqrt{b^2-a^2}}-\frac{2 i b x^2 \log \left (1+\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d \sqrt{b^2-a^2}}+\frac{2 x^{5/2}}{5 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.04858, antiderivative size = 653, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {4204, 4191, 3321, 2264, 2190, 2531, 6609, 2282, 6589} \[ \frac{8 b x^{3/2} \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^2 \sqrt{b^2-a^2}}-\frac{8 b x^{3/2} \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{24 i b x \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{24 i b x \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{48 b \sqrt{x} \text{PolyLog}\left (4,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^4 \sqrt{b^2-a^2}}+\frac{48 b \sqrt{x} \text{PolyLog}\left (4,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^4 \sqrt{b^2-a^2}}-\frac{48 i b \text{PolyLog}\left (5,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^5 \sqrt{b^2-a^2}}+\frac{48 i b \text{PolyLog}\left (5,-\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^5 \sqrt{b^2-a^2}}+\frac{2 i b x^2 \log \left (1+\frac{a e^{i \left (c+d \sqrt{x}\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d \sqrt{b^2-a^2}}-\frac{2 i b x^2 \log \left (1+\frac{a e^{i \left (c+d \sqrt{x}\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d \sqrt{b^2-a^2}}+\frac{2 x^{5/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(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 3321

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

Rubi steps

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

Mathematica [A]  time = 2.07245, size = 725, normalized size = 1.11 \[ \frac{2 \sec \left (c+d \sqrt{x}\right ) \left (a \cos \left (c+d \sqrt{x}\right )+b\right ) \left (x^{5/2}+\frac{5 i b e^{i c} \left (-4 i d^3 x^{3/2} \text{PolyLog}\left (2,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )+4 i d^3 x^{3/2} \text{PolyLog}\left (2,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )+12 d^2 x \text{PolyLog}\left (3,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )-12 d^2 x \text{PolyLog}\left (3,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )+24 i d \sqrt{x} \text{PolyLog}\left (4,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )-24 i d \sqrt{x} \text{PolyLog}\left (4,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )-24 \text{PolyLog}\left (5,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )+24 \text{PolyLog}\left (5,-\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )+d^4 x^2 \log \left (1+\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )-d^4 x^2 \log \left (1+\frac{a e^{i \left (2 c+d \sqrt{x}\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )\right )}{d^5 \sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )}{5 a \left (a+b \sec \left (c+d \sqrt{x}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + a*Cos[c + d*Sqrt[x]])*(x^(5/2) + ((5*I)*b*E^(I*c)*(d^4*x^2*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I
*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - d^4*x^2*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 +
b^2)*E^((2*I)*c)])] - (4*I)*d^3*x^(3/2)*PolyLog[2, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^
2)*E^((2*I)*c)]))] + (4*I)*d^3*x^(3/2)*PolyLog[2, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)
*E^((2*I)*c)]))] + 12*d^2*x*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c
)]))] - 12*d^2*x*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + (24
*I)*d*Sqrt[x]*PolyLog[4, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - (24*I)
*d*Sqrt[x]*PolyLog[4, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 24*PolyLo
g[5, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 24*PolyLog[5, -((a*E^(I*(2
*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))]))/(d^5*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))*Sec[c
+ d*Sqrt[x]])/(5*a*(a + b*Sec[c + d*Sqrt[x]]))

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Maple [F]  time = 0.083, size = 0, normalized size = 0. \begin{align*} \int{{x}^{{\frac{3}{2}}} \left ( a+b\sec \left ( c+d\sqrt{x} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{\frac{3}{2}}}{b \sec \left (d \sqrt{x} + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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