Integrand size = 31, antiderivative size = 61 \[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}-\frac {8 b^2 (e x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},b^2 x^4\right )}{7 e^5} \] Output:
-2*(-b^2*x^4+1)^(1/2)/e/(e*x)^(1/2)-8/7*b^2*(e*x)^(7/2)*hypergeom([1/2, 7/ 8],[15/8],b^2*x^4)/e^5
Time = 10.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=-\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{8},\frac {7}{8},b^2 x^4\right )}{(e x)^{3/2}} \] Input:
Integrate[(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2])/(e*x)^(3/2),x]
Output:
(-2*x*Hypergeometric2F1[-1/2, -1/8, 7/8, b^2*x^4])/(e*x)^(3/2)
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {335, 809, 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.
\(\displaystyle \int \frac {\sqrt {1-b x^2} \sqrt {b x^2+1}}{(e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \frac {\sqrt {1-b^2 x^4}}{(e x)^{3/2}}dx\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -\frac {4 b^2 \int \frac {(e x)^{5/2}}{\sqrt {1-b^2 x^4}}dx}{e^4}-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle -\frac {8 b^2 \int \frac {e^3 x^3}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{e^5}-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle -\frac {8 b^2 (e x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},b^2 x^4\right )}{7 e^5}-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}\) |
Input:
Int[(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2])/(e*x)^(3/2),x]
Output:
(-2*Sqrt[1 - b^2*x^4])/(e*Sqrt[e*x]) - (8*b^2*(e*x)^(7/2)*Hypergeometric2F 1[1/2, 7/8, 15/8, b^2*x^4])/(7*e^5)
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/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[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])
\[\int \frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}{\left (e x \right )^{\frac {3}{2}}}d x\]
Input:
int((-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(e*x)^(3/2),x)
Output:
int((-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(e*x)^(3/2),x)
\[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=\int { \frac {\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}}{\left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(e*x)^(3/2),x, algorithm="frica s")
Output:
integral(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)/(e^2*x^2), x)
\[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=\int \frac {\sqrt {- b x^{2} + 1} \sqrt {b x^{2} + 1}}{\left (e x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((-b*x**2+1)**(1/2)*(b*x**2+1)**(1/2)/(e*x)**(3/2),x)
Output:
Integral(sqrt(-b*x**2 + 1)*sqrt(b*x**2 + 1)/(e*x)**(3/2), x)
\[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=\int { \frac {\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}}{\left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(e*x)^(3/2),x, algorithm="maxim a")
Output:
integrate(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)/(e*x)^(3/2), x)
\[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=\int { \frac {\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}}{\left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(e*x)^(3/2),x, algorithm="giac" )
Output:
integrate(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)/(e*x)^(3/2), x)
Timed out. \[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=\int \frac {\sqrt {1-b\,x^2}\,\sqrt {b\,x^2+1}}{{\left (e\,x\right )}^{3/2}} \,d x \] Input:
int(((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2))/(e*x)^(3/2),x)
Output:
int(((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2))/(e*x)^(3/2), x)
\[ \int \frac {\sqrt {1-b x^2} \sqrt {1+b x^2}}{(e x)^{3/2}} \, dx=\frac {2 \sqrt {e}\, \left (\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}-2 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} x^{6}-x^{2}}d x \right )\right )}{3 \sqrt {x}\, e^{2}} \] Input:
int((-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(e*x)^(3/2),x)
Output:
(2*sqrt(e)*(sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1) - 2*sqrt(x)*int((sqrt(x)* sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**2*x**6 - x**2),x)))/(3*sqrt(x)*e **2)