Integrand size = 31, antiderivative size = 60 \[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\frac {1}{2 e \sqrt {e x} \sqrt {1-b^2 x^4}}-\frac {5 \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {1}{2},\frac {7}{8},b^2 x^4\right )}{2 e \sqrt {e x}} \] Output:
1/2/e/(e*x)^(1/2)/(-b^2*x^4+1)^(1/2)-5/2*hypergeom([-1/8, 1/2],[7/8],b^2*x ^4)/e/(e*x)^(1/2)
Time = 10.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.45 \[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=-\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {3}{2},\frac {7}{8},b^2 x^4\right )}{(e x)^{3/2}} \] Input:
Integrate[1/((e*x)^(3/2)*(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2)),x]
Output:
(-2*x*Hypergeometric2F1[-1/8, 3/2, 7/8, b^2*x^4])/(e*x)^(3/2)
Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.57, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {335, 819, 847, 851, 888}
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.
\(\displaystyle \int \frac {1}{\left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2} (e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \frac {1}{\left (1-b^2 x^4\right )^{3/2} (e x)^{3/2}}dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {5}{4} \int \frac {1}{(e x)^{3/2} \sqrt {1-b^2 x^4}}dx+\frac {1}{2 e \sqrt {1-b^2 x^4} \sqrt {e x}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {5}{4} \left (-\frac {3 b^2 \int \frac {(e x)^{5/2}}{\sqrt {1-b^2 x^4}}dx}{e^4}-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}\right )+\frac {1}{2 e \sqrt {1-b^2 x^4} \sqrt {e x}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {5}{4} \left (-\frac {6 b^2 \int \frac {e^3 x^3}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{e^5}-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}\right )+\frac {1}{2 e \sqrt {1-b^2 x^4} \sqrt {e x}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {5}{4} \left (-\frac {6 b^2 (e x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},b^2 x^4\right )}{7 e^5}-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}\right )+\frac {1}{2 e \sqrt {1-b^2 x^4} \sqrt {e x}}\) |
Input:
Int[1/((e*x)^(3/2)*(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2)),x]
Output:
1/(2*e*Sqrt[e*x]*Sqrt[1 - b^2*x^4]) + (5*((-2*Sqrt[1 - b^2*x^4])/(e*Sqrt[e *x]) - (6*b^2*(e*x)^(7/2)*Hypergeometric2F1[1/2, 7/8, 15/8, b^2*x^4])/(7*e ^5)))/4
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p _.), x_Symbol] :> Int[(e*x)^m*(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, e , m, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] ))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
\[\int \frac {1}{\left (e x \right )^{\frac {3}{2}} \left (-b \,x^{2}+1\right )^{\frac {3}{2}} \left (b \,x^{2}+1\right )^{\frac {3}{2}}}d x\]
Input:
int(1/(e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
Output:
int(1/(e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
\[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="fri cas")
Output:
integral(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)/(b^4*e^2*x^10 - 2*b^2* e^2*x^6 + e^2*x^2), x)
Result contains complex when optimal does not.
Time = 94.64 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.73 \[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=- \frac {i \sqrt [4]{b} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {11}{8}, \frac {15}{8}, 1 & \frac {9}{8}, \frac {17}{8}, \frac {21}{8} \\\frac {11}{8}, \frac {13}{8}, \frac {15}{8}, \frac {17}{8}, \frac {21}{8} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{4 \pi ^{\frac {3}{2}} e^{\frac {3}{2}}} + \frac {\sqrt [4]{b} {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{8}, \frac {5}{8}, \frac {7}{8}, \frac {9}{8}, \frac {11}{8}, 1 & \\\frac {7}{8}, \frac {11}{8} & \frac {1}{8}, \frac {5}{8}, \frac {13}{8}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {3 i \pi }{4}}}{4 \pi ^{\frac {3}{2}} e^{\frac {3}{2}}} \] Input:
integrate(1/(e*x)**(3/2)/(-b*x**2+1)**(3/2)/(b*x**2+1)**(3/2),x)
Output:
-I*b**(1/4)*meijerg(((11/8, 15/8, 1), (9/8, 17/8, 21/8)), ((11/8, 13/8, 15 /8, 17/8, 21/8), (0,)), 1/(b**2*x**4))/(4*pi**(3/2)*e**(3/2)) + b**(1/4)*m eijerg(((1/8, 5/8, 7/8, 9/8, 11/8, 1), ()), ((7/8, 11/8), (1/8, 5/8, 13/8, 0)), exp_polar(-2*I*pi)/(b**2*x**4))*exp(-3*I*pi/4)/(4*pi**(3/2)*e**(3/2) )
\[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="max ima")
Output:
integrate(1/((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2)*(e*x)^(3/2)), x)
Exception generated. \[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="gia c")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (e\,x\right )}^{3/2}\,{\left (1-b\,x^2\right )}^{3/2}\,{\left (b\,x^2+1\right )}^{3/2}} \,d x \] Input:
int(1/((e*x)^(3/2)*(1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2)),x)
Output:
int(1/((e*x)^(3/2)*(1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2)), x)
\[ \int \frac {1}{(e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{4} x^{10}-2 b^{2} x^{6}+x^{2}}d x \right )}{e^{2}} \] Input:
int(1/(e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
Output:
(sqrt(e)*int((sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**4*x**10 - 2*b**2*x**6 + x**2),x))/e**2