\(\int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx\) [908]

Optimal result

Integrand size = 24, antiderivative size = 244 \[ \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=-\frac {1}{2} e \sqrt {e x} \left (1-x^2\right )^{3/4}-\frac {e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}+\frac {e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}-\frac {e^{3/2} \log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}} \]

[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)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {126, 327, 335, 246, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=-\frac {e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{4 \sqrt {2}}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}+1\right )}{4 \sqrt {2}}-\frac {e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{8 \sqrt {2}}+\frac {e^{3/2} \log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{8 \sqrt {2}}-\frac {1}{2} e \left (1-x^2\right )^{3/4} \sqrt {e x} \]

[In]

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

[Out]

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

Rule 126

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[n, m] && GtQ[a, 0] && GtQ[c,
0]

Rule 210

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

Rule 217

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 246

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 327

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 335

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 631

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 642

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 1176

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 1179

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*} \text {integral}& = \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 \text {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 \text {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} \text {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} \text {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} \text {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} \text {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 \text {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 \text {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} \text {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} \text {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*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.51 \[ \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\frac {(e x)^{3/2} \left (-4 \sqrt {x} \left (1-x^2\right )^{3/4}+\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {x} \sqrt [4]{1-x^2}}{-x+\sqrt {1-x^2}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {x} \sqrt [4]{1-x^2}}{x+\sqrt {1-x^2}}\right )\right )}{8 x^{3/2}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {3}{2}}}{\left (1-x \right )^{\frac {1}{4}} \left (1+x \right )^{\frac {1}{4}}}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.94 \[ \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=-\frac {1}{2} \, \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \frac {1}{8} \, \left (-e^{6}\right )^{\frac {1}{4}} \log \left (\frac {\sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + \left (-e^{6}\right )^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{x^{2} - 1}\right ) + \frac {1}{8} \, \left (-e^{6}\right )^{\frac {1}{4}} \log \left (\frac {\sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - \left (-e^{6}\right )^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{x^{2} - 1}\right ) - \frac {1}{8} i \, \left (-e^{6}\right )^{\frac {1}{4}} \log \left (\frac {\sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + \left (-e^{6}\right )^{\frac {1}{4}} {\left (i \, x^{2} - i\right )}}{x^{2} - 1}\right ) + \frac {1}{8} i \, \left (-e^{6}\right )^{\frac {1}{4}} \log \left (\frac {\sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + \left (-e^{6}\right )^{\frac {1}{4}} {\left (-i \, x^{2} + i\right )}}{x^{2} - 1}\right ) \]

[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) - 1/8*(-e^6)^(1/4)*log((sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4)
+ (-e^6)^(1/4)*(x^2 - 1))/(x^2 - 1)) + 1/8*(-e^6)^(1/4)*log((sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) - (-e^6
)^(1/4)*(x^2 - 1))/(x^2 - 1)) - 1/8*I*(-e^6)^(1/4)*log((sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) + (-e^6)^(1/4
)*(I*x^2 - I))/(x^2 - 1)) + 1/8*I*(-e^6)^(1/4)*log((sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4) + (-e^6)^(1/4)*(-
I*x^2 + I))/(x^2 - 1))

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{3/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}}{{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \,d x \]

[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)