Integrand size = 26, antiderivative size = 180 \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=-\frac {7 a (10 b c-11 a d) e^3 (e x)^{3/2}}{60 b^3 \sqrt [4]{a+b x^2}}+\frac {(10 b c-11 a d) e (e x)^{7/2}}{30 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}-\frac {7 a^{3/2} (10 b c-11 a d) e^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 b^{7/2} \sqrt [4]{a+b x^2}} \]
-7/60*a*(-11*a*d+10*b*c)*e^3*(e*x)^(3/2)/b^3/(b*x^2+a)^(1/4)+1/30*(-11*a*d +10*b*c)*e*(e*x)^(7/2)/b^2/(b*x^2+a)^(1/4)+1/5*d*(e*x)^(11/2)/b/e/(b*x^2+a )^(1/4)-7/20*a^(3/2)*(-11*a*d+10*b*c)*e^4*(1+a/b/x^2)^(1/4)*(cos(1/2*arcco t(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*Elliptic E(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*(e*x)^(1/2)/b^(7/2)/(b*x^2+a )^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.62 \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {e^3 (e x)^{3/2} \left (77 a^2 d+4 b^2 x^2 \left (5 c+3 d x^2\right )-2 a b \left (35 c+11 d x^2\right )+7 a (10 b c-11 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{60 b^3 \sqrt [4]{a+b x^2}} \]
(e^3*(e*x)^(3/2)*(77*a^2*d + 4*b^2*x^2*(5*c + 3*d*x^2) - 2*a*b*(35*c + 11* d*x^2) + 7*a*(10*b*c - 11*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4 , 5/4, 7/4, -((b*x^2)/a)]))/(60*b^3*(a + b*x^2)^(1/4))
Time = 0.29 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {363, 250, 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.
| \(\displaystyle \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(10 b c-11 a d) \int \frac {(e x)^{9/2}}{\left (b x^2+a\right )^{5/4}}dx}{10 b}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 250 |
| \(\displaystyle \frac {(10 b c-11 a d) \left (\frac {e (e x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {7 a e^2 \int \frac {(e x)^{5/2}}{\left (b x^2+a\right )^{5/4}}dx}{6 b}\right )}{10 b}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 250 |
| \(\displaystyle \frac {(10 b c-11 a d) \left (\frac {e (e x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {7 a e^2 \left (\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a e^2 \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx}{2 b}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 249 |
| \(\displaystyle \frac {(10 b c-11 a d) \left (\frac {e (e x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {7 a e^2 \left (\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}-\frac {3 a e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b^2 \sqrt [4]{a+b x^2}}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle \frac {(10 b c-11 a d) \left (\frac {e (e x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {7 a e^2 \left (\frac {3 a e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {(10 b c-11 a d) \left (\frac {e (e x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {7 a e^2 \left (\frac {3 \sqrt {a} e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2}}{b \sqrt [4]{a+b x^2}}\right )}{6 b}\right )}{10 b}+\frac {d (e x)^{11/2}}{5 b e \sqrt [4]{a+b x^2}}\) |
(d*(e*x)^(11/2))/(5*b*e*(a + b*x^2)^(1/4)) + ((10*b*c - 11*a*d)*((e*(e*x)^ (7/2))/(3*b*(a + b*x^2)^(1/4)) - (7*a*e^2*((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[S qrt[a]/(Sqrt[b]*x)]/2, 2])/(b^(3/2)*(a + b*x^2)^(1/4))))/(6*b)))/(10*b)
3.12.10.3.1 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[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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]
\[\int \frac {\left (e x \right )^{\frac {9}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
\[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
integral((d*e^4*x^6 + c*e^4*x^4)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(b^2*x^4 + 2* a*b*x^2 + a^2), x)
Timed out. \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\text {Timed out} \]
\[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{9/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (e\,x\right )}^{9/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]