3.1.61 \(\int \frac {x^{3/2}}{a+b \sec (c+d \sqrt {x})} \, dx\) [61]

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

[Out]

2/5*x^(5/2)/a+2*I*b*x^2*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/d/(-a^2+b^2)^(1/2)-2*I*b*x^2*ln(1+
a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/d/(-a^2+b^2)^(1/2)+8*b*x^(3/2)*polylog(2,-a*exp(I*(c+d*x^(1/2))
)/(b-(-a^2+b^2)^(1/2)))/a/d^2/(-a^2+b^2)^(1/2)-8*b*x^(3/2)*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/
2)))/a/d^2/(-a^2+b^2)^(1/2)+24*I*b*x*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^
(1/2)-24*I*b*x*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)-48*I*b*polylog(5
,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a/d^5/(-a^2+b^2)^(1/2)+48*I*b*polylog(5,-a*exp(I*(c+d*x^(1/2)))
/(b+(-a^2+b^2)^(1/2)))/a/d^5/(-a^2+b^2)^(1/2)-48*b*polylog(4,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))*x^(
1/2)/a/d^4/(-a^2+b^2)^(1/2)+48*b*polylog(4,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))*x^(1/2)/a/d^4/(-a^2+b
^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.75, antiderivative size = 653, normalized size of antiderivative = 1.00, 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 = {4289, 4276, 3402, 2296, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {48 i b \text {Li}_5\left (-\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 {Li}_5\left (-\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 b \sqrt {x} \text {Li}_4\left (-\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 {Li}_4\left (-\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 {24 i b x \text {Li}_3\left (-\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 {Li}_3\left (-\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 {8 b x^{3/2} \text {Li}_2\left (-\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 {Li}_2\left (-\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 {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} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 3402

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 4276

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 4289

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 6724

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 6744

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 \frac {x^{3/2}}{a+b \sec \left (c+d \sqrt {x}\right )} \, dx &=2 \text {Subst}\left (\int \frac {x^4}{a+b \sec (c+d x)} \, dx,x,\sqrt {x}\right )\\ &=2 \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 2.25, size = 725, normalized size = 1.11 \begin {gather*} \frac {2 \left (b+a \cos \left (c+d \sqrt {x}\right )\right ) \left (x^{5/2}+\frac {5 i b e^{i c} \left (d^4 x^2 \log \left (1+\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{b e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 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 {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-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 {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+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 {\left (-a^2+b^2\right ) e^{2 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 {\left (-a^2+b^2\right ) e^{2 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 {\left (-a^2+b^2\right ) e^{2 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 {\left (-a^2+b^2\right ) e^{2 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 {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-24 \text {PolyLog}\left (5,-\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{b e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+24 \text {PolyLog}\left (5,-\frac {a e^{i \left (2 c+d \sqrt {x}\right )}}{b e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )\right )}{d^5 \sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right ) \sec \left (c+d \sqrt {x}\right )}{5 a \left (a+b \sec \left (c+d \sqrt {x}\right )\right )} \end {gather*}

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {3}{2}}}{a + b \sec {\left (c + d \sqrt {x} \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{3/2}}{a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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