Integrand size = 26, antiderivative size = 142 \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 c}{a e \sqrt {e x} \left (a+b x^2\right )^{5/4}}-\frac {2 (6 b c-a d) (e x)^{3/2}}{5 a^2 e^3 \left (a+b x^2\right )^{5/4}}+\frac {4 (6 b c-a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} \sqrt {b} e^2 \sqrt [4]{a+b x^2}} \]
-2/5*(-a*d+6*b*c)*(e*x)^(3/2)/a^2/e^3/(b*x^2+a)^(5/4)-2*c/a/e/(b*x^2+a)^(5 /4)/(e*x)^(1/2)+4/5*(-a*d+6*b*c)*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/ 2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2 *arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*(e*x)^(1/2)/a^(5/2)/e^2/(b*x^2+a)^(1/ 4)/b^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.60 \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {2 x \left (-3 a^2 c+(-6 b c+a d) x^2 \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{3 a^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}} \]
(2*x*(-3*a^2*c + (-6*b*c + a*d)*x^2*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hype rgeometric2F1[3/4, 9/4, 7/4, -((b*x^2)/a)]))/(3*a^3*(e*x)^(3/2)*(a + b*x^2 )^(5/4))
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {359, 253, 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.
\(\displaystyle \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(6 b c-a d) \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{9/4}}dx}{a e^2}-\frac {2 c}{a e \sqrt {e x} \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle -\frac {(6 b c-a d) \left (\frac {2 \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx}{5 a}+\frac {2 (e x)^{3/2}}{5 a e \left (a+b x^2\right )^{5/4}}\right )}{a e^2}-\frac {2 c}{a e \sqrt {e x} \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 249 |
\(\displaystyle -\frac {(6 b c-a d) \left (\frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{5 a b \sqrt [4]{a+b x^2}}+\frac {2 (e x)^{3/2}}{5 a e \left (a+b x^2\right )^{5/4}}\right )}{a e^2}-\frac {2 c}{a e \sqrt {e x} \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {(6 b c-a d) \left (\frac {2 (e x)^{3/2}}{5 a e \left (a+b x^2\right )^{5/4}}-\frac {2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{5 a b \sqrt [4]{a+b x^2}}\right )}{a e^2}-\frac {2 c}{a e \sqrt {e x} \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {(6 b c-a d) \left (\frac {2 (e x)^{3/2}}{5 a e \left (a+b x^2\right )^{5/4}}-\frac {4 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}}\right )}{a e^2}-\frac {2 c}{a e \sqrt {e x} \left (a+b x^2\right )^{5/4}}\) |
(-2*c)/(a*e*Sqrt[e*x]*(a + b*x^2)^(5/4)) - ((6*b*c - a*d)*((2*(e*x)^(3/2)) /(5*a*e*(a + b*x^2)^(5/4)) - (4*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(5*a^(3/2)*Sqrt[b]*(a + b*x^2)^(1/4)))) /(a*e^2)
3.12.36.3.1 Defintions of rubi rules used
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)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]
\[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]
\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {3}{2}}} \,d x } \]
integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(e*x)/(b^3*e^2*x^8 + 3*a*b^2*e^ 2*x^6 + 3*a^2*b*e^2*x^4 + a^3*e^2*x^2), x)
Result contains complex when optimal does not.
Time = 119.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {9}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}} e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {d x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}} e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]
c*gamma(-1/4)*hyper((-1/4, 9/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**( 9/4)*e**(3/2)*sqrt(x)*gamma(3/4)) + d*x**(3/2)*gamma(3/4)*hyper((3/4, 9/4) , (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(9/4)*e**(3/2)*gamma(7/4))
\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^{9/4}} \,d x \]