Integrand size = 26, antiderivative size = 155 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 (b c-a d) (e x)^{7/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-7 a d) e (e x)^{3/2}}{5 a b^2 \sqrt [4]{a+b x^2}}-\frac {3 (2 b c-7 a d) e^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {a} b^{5/2} \sqrt [4]{a+b x^2}} \]
2/5*(-a*d+b*c)*(e*x)^(7/2)/a/b/e/(b*x^2+a)^(5/4)-1/5*(-7*a*d+2*b*c)*e*(e*x )^(3/2)/a/b^2/(b*x^2+a)^(1/4)-3/5*(-7*a*d+2*b*c)*e^2*(1+a/b/x^2)^(1/4)*(co s(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)/b^(5/ 2)/(b*x^2+a)^(1/4)/a^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.63 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {e (e x)^{3/2} \left (a \left (-2 b c+7 a d+2 b d x^2\right )+(2 b c-7 a d) \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {9}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{2 a b^2 \left (a+b x^2\right )^{5/4}} \]
(e*(e*x)^(3/2)*(a*(-2*b*c + 7*a*d + 2*b*d*x^2) + (2*b*c - 7*a*d)*(a + b*x^ 2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 9/4, 7/4, -((b*x^2)/a)]))/ (2*a*b^2*(a + b*x^2)^(5/4))
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {362, 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.
\(\displaystyle \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 362 |
\(\displaystyle \frac {2 (e x)^{7/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-7 a d) \int \frac {(e x)^{5/2}}{\left (b x^2+a\right )^{5/4}}dx}{5 a b}\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {2 (e x)^{7/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-7 a d) \left (\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a e^2 \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx}{2 b}\right )}{5 a b}\) |
\(\Big \downarrow \) 249 |
\(\displaystyle \frac {2 (e x)^{7/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-7 a d) \left (\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b^2 \sqrt [4]{a+b x^2}}\right )}{5 a b}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {2 (e x)^{7/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-7 a d) \left (\frac {3 a e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}\right )}{5 a b}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {2 (e x)^{7/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-7 a d) \left (\frac {3 \sqrt {a} e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}\right )}{5 a b}\) |
(2*(b*c - a*d)*(e*x)^(7/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - ((2*b*c - 7*a*d) *((e*(e*x)^(3/2))/(b*(a + b*x^2)^(1/4)) + (3*Sqrt[a]*e^2*(1 + a/(b*x^2))^( 1/4)*Sqrt[e*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(b^(3/2)*(a + b*x^2)^(1/4))))/(5*a*b)
3.12.34.3.1 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)^(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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e *(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1)) I nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(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]
\[\int \frac {\left (e x \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]
integral((d*e^2*x^4 + c*e^2*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(b^3*x^6 + 3* a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)
Result contains complex when optimal does not.
Time = 154.71 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.61 \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {9}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{4}} \Gamma \left (\frac {15}{4}\right )} \]
c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((7/4, 9/4), (11/4,), b*x**2*exp_polar (I*pi)/a)/(2*a**(9/4)*gamma(11/4)) + d*e**(5/2)*x**(11/2)*gamma(11/4)*hype r((9/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(9/4)*gamma(15/4))
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]
\[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {{\left (e\,x\right )}^{5/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \]