Integrand size = 24, antiderivative size = 70 \[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}}-\frac {4 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{5 e^4 \sqrt [4]{1-x^2}} \] Output:
-2/5*(-x^2+1)^(3/4)/e/(e*x)^(5/2)-4/5*(1-1/x^2)^(1/4)*(e*x)^(1/2)*Elliptic E(sin(1/2*arccsc(x)),2^(1/2))/e^4/(-x^2+1)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=-\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},-\frac {1}{4},x^2\right )}{5 (e x)^{7/2}} \] Input:
Integrate[1/((1 - x)^(1/4)*(e*x)^(7/2)*(1 + x)^(1/4)),x]
Output:
(-2*x*Hypergeometric2F1[-5/4, 1/4, -1/4, x^2])/(5*(e*x)^(7/2))
Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {135, 264, 258, 858, 226}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt [4]{1-x} \sqrt [4]{x+1} (e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 135 |
\(\displaystyle \int \frac {1}{\sqrt [4]{1-x^2} (e x)^{7/2}}dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2 \int \frac {1}{(e x)^{3/2} \sqrt [4]{1-x^2}}dx}{5 e^2}-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}}\) |
\(\Big \downarrow \) 258 |
\(\displaystyle \frac {2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} \int \frac {1}{\sqrt [4]{1-\frac {1}{x^2}} x^2}dx}{5 e^4 \sqrt [4]{1-x^2}}-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} \int \frac {1}{\sqrt [4]{1-\frac {1}{x^2}}}d\frac {1}{x}}{5 e^4 \sqrt [4]{1-x^2}}-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}}\) |
\(\Big \downarrow \) 226 |
\(\displaystyle -\frac {4 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \arcsin \left (\frac {1}{x}\right )\right |2\right )}{5 e^4 \sqrt [4]{1-x^2}}-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}}\) |
Input:
Int[1/((1 - x)^(1/4)*(e*x)^(7/2)*(1 + x)^(1/4)),x]
Output:
(-2*(1 - x^2)^(3/4))/(5*e*(e*x)^(5/2)) - (4*(1 - x^(-2))^(1/4)*Sqrt[e*x]*E llipticE[ArcSin[x^(-1)]/2, 2])/(5*e^4*(1 - x^2)^(1/4))
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> 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]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] ))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Simp[S qrt[c*x]*((1 + a/(b*x^2))^(1/4)/(c^2*(a + b*x^2)^(1/4))) Int[1/(x^2*(1 + a/(b*x^2))^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && NegQ[b/a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
\[\int \frac {1}{\left (1-x \right )^{\frac {1}{4}} \left (e x \right )^{\frac {7}{2}} \left (1+x \right )^{\frac {1}{4}}}d x\]
Input:
int(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x)
Output:
int(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x)
\[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=\int { \frac {1}{\left (e x\right )^{\frac {7}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x, algorithm="fricas")
Output:
integral(-sqrt(e*x)*(x + 1)^(3/4)*(-x + 1)^(3/4)/(e^4*x^6 - e^4*x^4), x)
Timed out. \[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=\text {Timed out} \] Input:
integrate(1/(1-x)**(1/4)/(e*x)**(7/2)/(1+x)**(1/4),x)
Output:
Timed out
\[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=\int { \frac {1}{\left (e x\right )^{\frac {7}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((e*x)^(7/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)
\[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=\int { \frac {1}{\left (e x\right )^{\frac {7}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x, algorithm="giac")
Output:
integrate(1/((e*x)^(7/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)
Timed out. \[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=\int \frac {1}{{\left (e\,x\right )}^{7/2}\,{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \,d x \] Input:
int(1/((e*x)^(7/2)*(1 - x)^(1/4)*(x + 1)^(1/4)),x)
Output:
int(1/((e*x)^(7/2)*(1 - x)^(1/4)*(x + 1)^(1/4)), x)
Time = 0.44 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx=\frac {2 \sqrt {e}\, \left (x +1\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {1}{4}} \left (x^{2}-1\right )}{5 \sqrt {x}\, \sqrt {x +1}\, \sqrt {1-x}\, e^{4} x^{2}} \] Input:
int(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x)
Output:
(2*sqrt(e)*(x + 1)**(1/4)*( - x + 1)**(1/4)*(x**2 - 1))/(5*sqrt(x)*sqrt(x + 1)*sqrt( - x + 1)*e**4*x**2)