3.908 \(\int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx\)

Optimal. Leaf size=244 \[ -\frac {e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{8 \sqrt {2}}-\frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}+1\right )}{4 \sqrt {2}}-\frac {1}{2} e \left (1-x^2\right )^{3/4} \sqrt {e x} \]

[Out]

-1/8*e^(3/2)*arctan(1-2^(1/2)*(e*x)^(1/2)/(-x^2+1)^(1/4)/e^(1/2))*2^(1/2)+1/8*e^(3/2)*arctan(1+2^(1/2)*(e*x)^(
1/2)/(-x^2+1)^(1/4)/e^(1/2))*2^(1/2)-1/16*e^(3/2)*ln(e^(1/2)-2^(1/2)*(e*x)^(1/2)/(-x^2+1)^(1/4)+x*e^(1/2)/(-x^
2+1)^(1/2))*2^(1/2)+1/16*e^(3/2)*ln(e^(1/2)+2^(1/2)*(e*x)^(1/2)/(-x^2+1)^(1/4)+x*e^(1/2)/(-x^2+1)^(1/2))*2^(1/
2)-1/2*e*(-x^2+1)^(3/4)*(e*x)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {125, 321, 329, 240, 211, 1165, 628, 1162, 617, 204} \[ -\frac {e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{8 \sqrt {2}}-\frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}+1\right )}{4 \sqrt {2}}-\frac {1}{2} e \left (1-x^2\right )^{3/4} \sqrt {e x} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(3/2)/((1 - x)^(1/4)*(1 + x)^(1/4)),x]

[Out]

-(e*Sqrt[e*x]*(1 - x^2)^(3/4))/2 - (e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*x])/(Sqrt[e]*(1 - x^2)^(1/4))])/(4*Sqrt
[2]) + (e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*x])/(Sqrt[e]*(1 - x^2)^(1/4))])/(4*Sqrt[2]) - (e^(3/2)*Log[Sqrt[e]
+ (Sqrt[e]*x)/Sqrt[1 - x^2] - (Sqrt[2]*Sqrt[e*x])/(1 - x^2)^(1/4)])/(8*Sqrt[2]) + (e^(3/2)*Log[Sqrt[e] + (Sqrt
[e]*x)/Sqrt[1 - x^2] + (Sqrt[2]*Sqrt[e*x])/(1 - x^2)^(1/4)])/(8*Sqrt[2])

Rule 125

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[(a*c + b*d*x^2)
^m*(f*x)^p, x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 0] && GtQ[a, 0] && GtQ
[c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx &=\int \frac {(e x)^{3/2}}{\sqrt [4]{1-x^2}} \, dx\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}+\frac {1}{4} e^2 \int \frac {1}{\sqrt {e x} \sqrt [4]{1-x^2}} \, dx\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}+\frac {1}{2} e \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}+\frac {1}{2} e \operatorname {Subst}\left (\int \frac {1}{1+\frac {x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {e-x^2}{1+\frac {x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {e+x^2}{1+\frac {x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}-\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}-\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}+\frac {1}{8} e^2 \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )+\frac {1}{8} e^2 \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}-\frac {e^{3/2} \log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}-\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}\\ &=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}-\frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}+\frac {e^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}-\frac {e^{3/2} \log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 190, normalized size = 0.78 \[ -\frac {(e x)^{3/2} \left (8 \sqrt {x} \left (1-x^2\right )^{3/4}+\sqrt {2} \log \left (\frac {x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}+1\right )-\sqrt {2} \log \left (\frac {x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}+1\right )+2 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}\right )-2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}+1\right )\right )}{16 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^(3/2)/((1 - x)^(1/4)*(1 + x)^(1/4)),x]

[Out]

-1/16*((e*x)^(3/2)*(8*Sqrt[x]*(1 - x^2)^(3/4) + 2*Sqrt[2]*ArcTan[1 - (Sqrt[2]*Sqrt[x])/(1 - x^2)^(1/4)] - 2*Sq
rt[2]*ArcTan[1 + (Sqrt[2]*Sqrt[x])/(1 - x^2)^(1/4)] + Sqrt[2]*Log[1 + x/Sqrt[1 - x^2] - (Sqrt[2]*Sqrt[x])/(1 -
x^2)^(1/4)] - Sqrt[2]*Log[1 + x/Sqrt[1 - x^2] + (Sqrt[2]*Sqrt[x])/(1 - x^2)^(1/4)]))/x^(3/2)

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fricas [B]  time = 1.01, size = 478, normalized size = 1.96 \[ -\frac {1}{2} \, \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + \frac {1}{4} \, \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \arctan \left (-\frac {e^{6} x^{2} - e^{6} + \sqrt {2} {\left (e^{6}\right )}^{\frac {3}{4}} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \sqrt {2} {\left (e^{6}\right )}^{\frac {3}{4}} {\left (x^{2} - 1\right )} \sqrt {-\frac {e^{3} \sqrt {x + 1} x \sqrt {-x + 1} - \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \sqrt {e^{6}} {\left (x^{2} - 1\right )}}{x^{2} - 1}}}{e^{6} x^{2} - e^{6}}\right ) + \frac {1}{4} \, \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \arctan \left (\frac {e^{6} x^{2} - e^{6} - \sqrt {2} {\left (e^{6}\right )}^{\frac {3}{4}} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + \sqrt {2} {\left (e^{6}\right )}^{\frac {3}{4}} {\left (x^{2} - 1\right )} \sqrt {-\frac {e^{3} \sqrt {x + 1} x \sqrt {-x + 1} + \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \sqrt {e^{6}} {\left (x^{2} - 1\right )}}{x^{2} - 1}}}{e^{6} x^{2} - e^{6}}\right ) + \frac {1}{16} \, \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \log \left (-\frac {e^{3} \sqrt {x + 1} x \sqrt {-x + 1} + \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \sqrt {e^{6}} {\left (x^{2} - 1\right )}}{x^{2} - 1}\right ) - \frac {1}{16} \, \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \log \left (-\frac {e^{3} \sqrt {x + 1} x \sqrt {-x + 1} - \sqrt {2} {\left (e^{6}\right )}^{\frac {1}{4}} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \sqrt {e^{6}} {\left (x^{2} - 1\right )}}{x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)/(1-x)^(1/4)/(1+x)^(1/4),x, algorithm="fricas")

[Out]

-1/2*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) + 1/4*sqrt(2)*(e^6)^(1/4)*arctan(-(e^6*x^2 - e^6 + sqrt(2)*(e^6)
^(3/4)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) - sqrt(2)*(e^6)^(3/4)*(x^2 - 1)*sqrt(-(e^3*sqrt(x + 1)*x*sqrt(
-x + 1) - sqrt(2)*(e^6)^(1/4)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) - sqrt(e^6)*(x^2 - 1))/(x^2 - 1)))/(e^6
*x^2 - e^6)) + 1/4*sqrt(2)*(e^6)^(1/4)*arctan((e^6*x^2 - e^6 - sqrt(2)*(e^6)^(3/4)*sqrt(e*x)*e*(x + 1)^(3/4)*(
-x + 1)^(3/4) + sqrt(2)*(e^6)^(3/4)*(x^2 - 1)*sqrt(-(e^3*sqrt(x + 1)*x*sqrt(-x + 1) + sqrt(2)*(e^6)^(1/4)*sqrt
(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) - sqrt(e^6)*(x^2 - 1))/(x^2 - 1)))/(e^6*x^2 - e^6)) + 1/16*sqrt(2)*(e^6)^
(1/4)*log(-(e^3*sqrt(x + 1)*x*sqrt(-x + 1) + sqrt(2)*(e^6)^(1/4)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) - sq
rt(e^6)*(x^2 - 1))/(x^2 - 1)) - 1/16*sqrt(2)*(e^6)^(1/4)*log(-(e^3*sqrt(x + 1)*x*sqrt(-x + 1) - sqrt(2)*(e^6)^
(1/4)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) - sqrt(e^6)*(x^2 - 1))/(x^2 - 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x\right )^{\frac {3}{2}}}{{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)/(1-x)^(1/4)/(1+x)^(1/4),x, algorithm="giac")

[Out]

integrate((e*x)^(3/2)/((x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{\frac {3}{2}}}{\left (-x +1\right )^{\frac {1}{4}} \left (x +1\right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)/(-x+1)^(1/4)/(x+1)^(1/4),x)

[Out]

int((e*x)^(3/2)/(-x+1)^(1/4)/(x+1)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x\right )^{\frac {3}{2}}}{{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)/(1-x)^(1/4)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

integrate((e*x)^(3/2)/((x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^{3/2}}{{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)/((1 - x)^(1/4)*(x + 1)^(1/4)),x)

[Out]

int((e*x)^(3/2)/((1 - x)^(1/4)*(x + 1)^(1/4)), x)

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sympy [C]  time = 14.18, size = 114, normalized size = 0.47 \[ - \frac {i e^{\frac {3}{2}} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{8}, - \frac {1}{8} & - \frac {1}{2}, - \frac {1}{4}, 0, 1 \\-1, - \frac {5}{8}, - \frac {1}{2}, - \frac {1}{8}, 0, 0 & \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{x^{2}}} \right )} e^{\frac {i \pi }{4}}}{4 \pi \Gamma \left (\frac {1}{4}\right )} - \frac {e^{\frac {3}{2}} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {5}{4}, - \frac {9}{8}, - \frac {3}{4}, - \frac {5}{8}, - \frac {1}{4}, 1 & \\- \frac {9}{8}, - \frac {5}{8} & - \frac {5}{4}, -1, - \frac {3}{4}, 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{4 \pi \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)/(1-x)**(1/4)/(1+x)**(1/4),x)

[Out]

-I*e**(3/2)*meijerg(((-5/8, -1/8), (-1/2, -1/4, 0, 1)), ((-1, -5/8, -1/2, -1/8, 0, 0), ()), exp_polar(-2*I*pi)
/x**2)*exp(I*pi/4)/(4*pi*gamma(1/4)) - e**(3/2)*meijerg(((-5/4, -9/8, -3/4, -5/8, -1/4, 1), ()), ((-9/8, -5/8)
, (-5/4, -1, -3/4, 0)), x**(-2))/(4*pi*gamma(1/4))

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