Integrand size = 31, antiderivative size = 88 \[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {8 (e x)^{5/2} \sqrt {1-b^2 x^4}}{51 e}+\frac {2 (e x)^{5/2} \left (1-b^2 x^4\right )^{3/2}}{17 e}+\frac {32 (e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{255 e} \] Output:
8/51*(e*x)^(5/2)*(-b^2*x^4+1)^(1/2)/e+2/17*(e*x)^(5/2)*(-b^2*x^4+1)^(3/2)/ e+32/255*(e*x)^(5/2)*hypergeom([1/2, 5/8],[13/8],b^2*x^4)/e
Time = 10.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.33 \[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {2}{5} x (e x)^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right ) \] Input:
Integrate[(e*x)^(3/2)*(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2),x]
Output:
(2*x*(e*x)^(3/2)*Hypergeometric2F1[-3/2, 5/8, 13/8, b^2*x^4])/5
Time = 0.24 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {335, 811, 811, 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.
\(\displaystyle \int \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2} (e x)^{3/2} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \left (1-b^2 x^4\right )^{3/2} (e x)^{3/2}dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {12}{17} \int (e x)^{3/2} \sqrt {1-b^2 x^4}dx+\frac {2 \left (1-b^2 x^4\right )^{3/2} (e x)^{5/2}}{17 e}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {12}{17} \left (\frac {4}{9} \int \frac {(e x)^{3/2}}{\sqrt {1-b^2 x^4}}dx+\frac {2 \sqrt {1-b^2 x^4} (e x)^{5/2}}{9 e}\right )+\frac {2 \left (1-b^2 x^4\right )^{3/2} (e x)^{5/2}}{17 e}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {12}{17} \left (\frac {8 \int \frac {e^2 x^2}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{9 e}+\frac {2 \sqrt {1-b^2 x^4} (e x)^{5/2}}{9 e}\right )+\frac {2 \left (1-b^2 x^4\right )^{3/2} (e x)^{5/2}}{17 e}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {12}{17} \left (\frac {8 (e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{45 e}+\frac {2 \sqrt {1-b^2 x^4} (e x)^{5/2}}{9 e}\right )+\frac {2 \left (1-b^2 x^4\right )^{3/2} (e x)^{5/2}}{17 e}\) |
Input:
Int[(e*x)^(3/2)*(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2),x]
Output:
(2*(e*x)^(5/2)*(1 - b^2*x^4)^(3/2))/(17*e) + (12*((2*(e*x)^(5/2)*Sqrt[1 - b^2*x^4])/(9*e) + (8*(e*x)^(5/2)*Hypergeometric2F1[1/2, 5/8, 13/8, b^2*x^4 ])/(45*e)))/17
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 + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && 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])
\[\int \left (e x \right )^{\frac {3}{2}} \left (-b \,x^{2}+1\right )^{\frac {3}{2}} \left (b \,x^{2}+1\right )^{\frac {3}{2}}d x\]
Input:
int((e*x)^(3/2)*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
Output:
int((e*x)^(3/2)*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
\[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int { {\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((e*x)^(3/2)*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="frica s")
Output:
integral(-(b^2*e*x^5 - e*x)*sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x), x)
Timed out. \[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate((e*x)**(3/2)*(-b*x**2+1)**(3/2)*(b*x**2+1)**(3/2),x)
Output:
Timed out
\[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int { {\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((e*x)^(3/2)*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="maxim a")
Output:
integrate((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2)*(e*x)^(3/2), x)
Exception generated. \[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((e*x)^(3/2)*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="giac" )
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. \[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int {\left (e\,x\right )}^{3/2}\,{\left (1-b\,x^2\right )}^{3/2}\,{\left (b\,x^2+1\right )}^{3/2} \,d x \] Input:
int((e*x)^(3/2)*(1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2),x)
Output:
int((e*x)^(3/2)*(1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2), x)
\[ \int (e x)^{3/2} \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {2 \sqrt {e}\, e \left (-3 \sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, b^{2} x^{6}+7 \sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x^{2}-8 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x}{b^{2} x^{4}-1}d x \right )\right )}{51} \] Input:
int((e*x)^(3/2)*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
Output:
(2*sqrt(e)*e*( - 3*sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*b**2*x**6 + 7*sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x**2 - 8*int((sqrt(x)*sqr t(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x)/(b**2*x**4 - 1),x)))/51