Integrand size = 31, antiderivative size = 31 \[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {2 (e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{5 e} \] Output:
2/5*(e*x)^(5/2)*hypergeom([1/2, 5/8],[13/8],b^2*x^4)/e
Time = 10.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {2}{5} x (e x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right ) \] Input:
Integrate[(e*x)^(3/2)/(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
Output:
(2*x*(e*x)^(3/2)*Hypergeometric2F1[1/2, 5/8, 13/8, b^2*x^4])/5
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {335, 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.
\(\displaystyle \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {b x^2+1}} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \frac {(e x)^{3/2}}{\sqrt {1-b^2 x^4}}dx\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {2 \int \frac {e^2 x^2}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{e}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 (e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{5 e}\) |
Input:
Int[(e*x)^(3/2)/(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
Output:
(2*(e*x)^(5/2)*Hypergeometric2F1[1/2, 5/8, 13/8, b^2*x^4])/(5*e)
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] :> 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])
\[\int \frac {\left (e x \right )^{\frac {3}{2}}}{\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}d x\]
Input:
int((e*x)^(3/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
Output:
int((e*x)^(3/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
\[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{\frac {3}{2}}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:
integrate((e*x)^(3/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="frica s")
Output:
integral(-sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)*e*x/(b^2*x^4 - 1), x)
Result contains complex when optimal does not.
Time = 6.41 (sec) , antiderivative size = 116, normalized size of antiderivative = 3.74 \[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {i e^{\frac {3}{2}} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{8}, \frac {5}{8}, 1 & \frac {3}{8}, \frac {3}{8}, \frac {7}{8} \\- \frac {1}{8}, \frac {1}{8}, \frac {3}{8}, \frac {5}{8}, \frac {7}{8} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}} b^{\frac {5}{4}}} + \frac {e^{\frac {3}{2}} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {5}{8}, - \frac {3}{8}, - \frac {1}{8}, \frac {1}{8}, \frac {3}{8}, 1 & \\- \frac {3}{8}, \frac {1}{8} & - \frac {5}{8}, - \frac {1}{8}, - \frac {1}{8}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {i \pi }{4}}}{8 \pi ^{\frac {3}{2}} b^{\frac {5}{4}}} \] Input:
integrate((e*x)**(3/2)/(-b*x**2+1)**(1/2)/(b*x**2+1)**(1/2),x)
Output:
I*e**(3/2)*meijerg(((1/8, 5/8, 1), (3/8, 3/8, 7/8)), ((-1/8, 1/8, 3/8, 5/8 , 7/8), (0,)), 1/(b**2*x**4))/(8*pi**(3/2)*b**(5/4)) + e**(3/2)*meijerg((( -5/8, -3/8, -1/8, 1/8, 3/8, 1), ()), ((-3/8, 1/8), (-5/8, -1/8, -1/8, 0)), exp_polar(-2*I*pi)/(b**2*x**4))*exp(-I*pi/4)/(8*pi**(3/2)*b**(5/4))
\[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{\frac {3}{2}}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:
integrate((e*x)^(3/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="maxim a")
Output:
integrate((e*x)^(3/2)/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)), x)
\[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{\frac {3}{2}}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:
integrate((e*x)^(3/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="giac" )
Output:
integrate((e*x)^(3/2)/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)), x)
Timed out. \[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}}{\sqrt {1-b\,x^2}\,\sqrt {b\,x^2+1}} \,d x \] Input:
int((e*x)^(3/2)/((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2)),x)
Output:
int((e*x)^(3/2)/((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2)), x)
\[ \int \frac {(e x)^{3/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=-\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x}{b^{2} x^{4}-1}d x \right ) e \] Input:
int((e*x)^(3/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
Output:
- sqrt(e)*int((sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x)/(b**2*x**4 - 1),x)*e