Integrand size = 19, antiderivative size = 90 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c (c x)^{3/2}}{b \sqrt [4]{a+b x^2}}+\frac {3 \sqrt {a} c^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}} \] Output:
c*(c*x)^(3/2)/b/(b*x^2+a)^(1/4)+3*a^(1/2)*c^2*(1+a/b/x^2)^(1/4)*(c*x)^(1/2 )*EllipticE(sin(1/2*arccot(b^(1/2)*x/a^(1/2))),2^(1/2))/b^(3/2)/(b*x^2+a)^ (1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c (c x)^{3/2} \left (1-\sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{b \sqrt [4]{a+b x^2}} \] Input:
Integrate[(c*x)^(5/2)/(a + b*x^2)^(5/4),x]
Output:
(c*(c*x)^(3/2)*(1 - (1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(b*(a + b*x^2)^(1/4))
Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {250, 249, 858, 212}
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.
\(\displaystyle \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {c (c x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a c^2 \int \frac {\sqrt {c x}}{\left (b x^2+a\right )^{5/4}}dx}{2 b}\) |
\(\Big \downarrow \) 249 |
\(\displaystyle \frac {c (c x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a c^2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b^2 \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {3 a c^2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{3/2}}{b \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {3 \sqrt {a} c^2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}+\frac {c (c x)^{3/2}}{b \sqrt [4]{a+b x^2}}\) |
Input:
Int[(c*x)^(5/2)/(a + b*x^2)^(5/4),x]
Output:
(c*(c*x)^(3/2))/(b*(a + b*x^2)^(1/4)) + (3*Sqrt[a]*c^2*(1 + a/(b*x^2))^(1/ 4)*Sqrt[c*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(b^(3/2)*(a + b* x^2)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[Sqrt[c* x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b*x^2)^(1/4))) Int[1/(x^2*(1 + a/(b*x^2 ))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*(( c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( b*(2*m - 3))) Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
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]
\[\int \frac {\left (c x \right )^{\frac {5}{2}}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Input:
int((c*x)^(5/2)/(b*x^2+a)^(5/4),x)
Output:
int((c*x)^(5/2)/(b*x^2+a)^(5/4),x)
\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((c*x)^(5/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*sqrt(c*x)*c^2*x^2/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
Result contains complex when optimal does not.
Time = 3.42 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.49 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate((c*x)**(5/2)/(b*x**2+a)**(5/4),x)
Output:
c**(5/2)*x**(7/2)*gamma(7/4)*hyper((5/4, 7/4), (11/4,), b*x**2*exp_polar(I *pi)/a)/(2*a**(5/4)*gamma(11/4))
\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((c*x)^(5/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate((c*x)^(5/2)/(b*x^2 + a)^(5/4), x)
\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((c*x)^(5/2)/(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate((c*x)^(5/2)/(b*x^2 + a)^(5/4), x)
Timed out. \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (c\,x\right )}^{5/2}}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:
int((c*x)^(5/2)/(a + b*x^2)^(5/4),x)
Output:
int((c*x)^(5/2)/(a + b*x^2)^(5/4), x)
\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx=\sqrt {c}\, \left (\int \frac {\sqrt {x}\, x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c^{2} \] Input:
int((c*x)^(5/2)/(b*x^2+a)^(5/4),x)
Output:
sqrt(c)*int((sqrt(x)*x**2)/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b* x**2),x)*c**2