Integrand size = 28, antiderivative size = 248 \[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {(b c-a d)^2 (e x)^{5/2}}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac {\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e \sqrt {e x}}{6 c d^3 \sqrt {c+d x^2}}+\frac {2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt {c+d x^2}}-\frac {\left (15 b^2 c^2-10 a b c d-a^2 d^2\right ) e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{12 c^{5/4} d^{13/4} \sqrt {c+d x^2}} \]
1/3*(-a*d+b*c)^2*(e*x)^(5/2)/c/d^2/e/(d*x^2+c)^(3/2)+2/3*b^2*(e*x)^(5/2)/d ^2/e/(d*x^2+c)^(1/2)+1/6*(-a^2*d^2-10*a*b*c*d+15*b^2*c^2)*e*(e*x)^(1/2)/c/ d^3/(d*x^2+c)^(1/2)-1/12*(-a^2*d^2-10*a*b*c*d+15*b^2*c^2)*e^(3/2)*(cos(2*a rctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)* (e*x)^(1/2)/c^(1/4)/e^(1/2)))*EllipticF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c ^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*((d*x^2+c)/(c^(1/2)+x*d^ (1/2))^2)^(1/2)/c^(5/4)/d^(13/4)/(d*x^2+c)^(1/2)
Result contains complex when optimal does not.
Time = 11.21 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.82 \[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {(e x)^{3/2} \left (\frac {\sqrt {x} \left (a^2 d^2 \left (-c+d x^2\right )-2 a b c d \left (5 c+7 d x^2\right )+b^2 c \left (15 c^2+21 c d x^2+4 d^2 x^4\right )\right )}{c d^3 \left (c+d x^2\right )}+\frac {i \left (-15 b^2 c^2+10 a b c d+a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right ),-1\right )}{c \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}} d^3}\right )}{6 x^{3/2} \sqrt {c+d x^2}} \]
((e*x)^(3/2)*((Sqrt[x]*(a^2*d^2*(-c + d*x^2) - 2*a*b*c*d*(5*c + 7*d*x^2) + b^2*c*(15*c^2 + 21*c*d*x^2 + 4*d^2*x^4)))/(c*d^3*(c + d*x^2)) + (I*(-15*b ^2*c^2 + 10*a*b*c*d + a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[S qrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(c*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d^3)) )/(6*x^(3/2)*Sqrt[c + d*x^2])
Time = 0.33 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {366, 27, 363, 252, 266, 761}
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.
\(\displaystyle \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 366 |
\(\displaystyle \frac {(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}-\frac {\int -\frac {(e x)^{3/2} \left (6 a^2 d^2+6 b^2 c x^2 d-5 (b c-a d)^2\right )}{2 \left (d x^2+c\right )^{3/2}}dx}{3 c d^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(e x)^{3/2} \left (6 a^2 d^2+6 b^2 c x^2 d-5 (b c-a d)^2\right )}{\left (d x^2+c\right )^{3/2}}dx}{6 c d^2}+\frac {(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {\frac {4 b^2 c (e x)^{5/2}}{e \sqrt {c+d x^2}}-\left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {(e x)^{3/2}}{\left (d x^2+c\right )^{3/2}}dx}{6 c d^2}+\frac {(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {4 b^2 c (e x)^{5/2}}{e \sqrt {c+d x^2}}-\left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) \left (\frac {e^2 \int \frac {1}{\sqrt {e x} \sqrt {d x^2+c}}dx}{2 d}-\frac {e \sqrt {e x}}{d \sqrt {c+d x^2}}\right )}{6 c d^2}+\frac {(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\frac {4 b^2 c (e x)^{5/2}}{e \sqrt {c+d x^2}}-\left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) \left (\frac {e \int \frac {1}{\sqrt {d x^2+c}}d\sqrt {e x}}{d}-\frac {e \sqrt {e x}}{d \sqrt {c+d x^2}}\right )}{6 c d^2}+\frac {(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\frac {4 b^2 c (e x)^{5/2}}{e \sqrt {c+d x^2}}-\left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) \left (\frac {\sqrt {e} \left (\sqrt {c} e+\sqrt {d} e x\right ) \sqrt {\frac {c e^2+d e^2 x^2}{\left (\sqrt {c} e+\sqrt {d} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{c} d^{5/4} \sqrt {c+d x^2}}-\frac {e \sqrt {e x}}{d \sqrt {c+d x^2}}\right )}{6 c d^2}+\frac {(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}\) |
((b*c - a*d)^2*(e*x)^(5/2))/(3*c*d^2*e*(c + d*x^2)^(3/2)) + ((4*b^2*c*(e*x )^(5/2))/(e*Sqrt[c + d*x^2]) - (15*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*(-((e*S qrt[e*x])/(d*Sqrt[c + d*x^2])) + (Sqrt[e]*(Sqrt[c]*e + Sqrt[d]*e*x)*Sqrt[( c*e^2 + d*e^2*x^2)/(Sqrt[c]*e + Sqrt[d]*e*x)^2]*EllipticF[2*ArcTan[(d^(1/4 )*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(2*c^(1/4)*d^(5/4)*Sqrt[c + d*x^2]) ))/(6*c*d^2)
3.9.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Time = 4.58 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.33
method | result | size |
elliptic | \(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \sqrt {e x}\, \left (-\frac {e \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {d e \,x^{3}+c e x}}{3 d^{5} \left (x^{2}+\frac {c}{d}\right )^{2}}+\frac {e^{2} x \left (a^{2} d^{2}-14 a b c d +13 b^{2} c^{2}\right )}{6 d^{3} c \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {2 b^{2} e \sqrt {d e \,x^{3}+c e x}}{3 d^{3}}+\frac {\left (\frac {2 \left (a d -b c \right ) b \,e^{2}}{d^{3}}+\frac {e^{2} \left (a^{2} d^{2}-14 a b c d +13 b^{2} c^{2}\right )}{12 d^{3} c}-\frac {b^{2} e^{2} c}{3 d^{3}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{e x \sqrt {d \,x^{2}+c}}\) | \(329\) |
risch | \(\frac {2 b^{2} x \sqrt {d \,x^{2}+c}\, e^{2}}{3 d^{3} \sqrt {e x}}+\frac {\left (-\frac {7 b^{2} c \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d \sqrt {d e \,x^{3}+c e x}}+\frac {6 a b \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {d e \,x^{3}+c e x}}+\left (3 a^{2} d^{2}-12 a b c d +9 b^{2} c^{2}\right ) \left (\frac {x}{c \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {\sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{2 c d \sqrt {d e \,x^{3}+c e x}}\right )-3 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (\frac {\sqrt {d e \,x^{3}+c e x}}{3 c e \,d^{2} \left (x^{2}+\frac {c}{d}\right )^{2}}+\frac {5 x}{6 c^{2} \sqrt {\left (x^{2}+\frac {c}{d}\right ) d e x}}+\frac {5 \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{12 c^{2} d \sqrt {d e \,x^{3}+c e x}}\right )\right ) e^{2} \sqrt {e x \left (d \,x^{2}+c \right )}}{3 d^{3} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\) | \(638\) |
default | \(\frac {\left (\sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a^{2} d^{3} x^{2}+10 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, a b c \,d^{2} x^{2}-15 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-c d}\, b^{2} c^{2} d \,x^{2}+\sqrt {-c d}\, \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2}+10 \sqrt {-c d}\, \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d -15 \sqrt {-c d}\, \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3}+8 b^{2} c \,d^{3} x^{5}+2 a^{2} d^{4} x^{3}-28 x^{3} d^{3} b a c +42 b^{2} c^{2} d^{2} x^{3}-2 a^{2} c \,d^{3} x -20 a b \,c^{2} d^{2} x +30 b^{2} d x \,c^{3}\right ) e \sqrt {e x}}{12 x c \,d^{4} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}\) | \(674\) |
(e*x*(d*x^2+c))^(1/2)/e/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)*(-1/3*e/d^5*(a^2*d^2 -2*a*b*c*d+b^2*c^2)*(d*e*x^3+c*e*x)^(1/2)/(x^2+c/d)^2+1/6/d^3*e^2*x/c*(a^2 *d^2-14*a*b*c*d+13*b^2*c^2)/((x^2+c/d)*d*e*x)^(1/2)+2/3*b^2/d^3*e*(d*e*x^3 +c*e*x)^(1/2)+(2*(a*d-b*c)*b*e^2/d^3+1/12/d^3/c*e^2*(a^2*d^2-14*a*b*c*d+13 *b^2*c^2)-1/3*b^2/d^3*e^2*c)*(-c*d)^(1/2)/d*((x+(-c*d)^(1/2)/d)/(-c*d)^(1/ 2)*d)^(1/2)*(-2*(x-(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2)*(-x/(-c*d)^(1/2)* d)^(1/2)/(d*e*x^3+c*e*x)^(1/2)*EllipticF(((x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)* d)^(1/2),1/2*2^(1/2)))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.96 \[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {{\left ({\left (15 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} - a^{2} d^{4}\right )} e x^{4} + 2 \, {\left (15 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} e x^{2} + {\left (15 \, b^{2} c^{4} - 10 \, a b c^{3} d - a^{2} c^{2} d^{2}\right )} e\right )} \sqrt {d e} {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right ) - {\left (4 \, b^{2} c d^{3} e x^{4} + {\left (21 \, b^{2} c^{2} d^{2} - 14 \, a b c d^{3} + a^{2} d^{4}\right )} e x^{2} + {\left (15 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} e\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{6 \, {\left (c d^{6} x^{4} + 2 \, c^{2} d^{5} x^{2} + c^{3} d^{4}\right )}} \]
-1/6*(((15*b^2*c^2*d^2 - 10*a*b*c*d^3 - a^2*d^4)*e*x^4 + 2*(15*b^2*c^3*d - 10*a*b*c^2*d^2 - a^2*c*d^3)*e*x^2 + (15*b^2*c^4 - 10*a*b*c^3*d - a^2*c^2* d^2)*e)*sqrt(d*e)*weierstrassPInverse(-4*c/d, 0, x) - (4*b^2*c*d^3*e*x^4 + (21*b^2*c^2*d^2 - 14*a*b*c*d^3 + a^2*d^4)*e*x^2 + (15*b^2*c^3*d - 10*a*b* c^2*d^2 - a^2*c*d^3)*e)*sqrt(d*x^2 + c)*sqrt(e*x))/(c*d^6*x^4 + 2*c^2*d^5* x^2 + c^3*d^4)
\[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (e x\right )^{\frac {3}{2}} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \]