Integrand size = 28, antiderivative size = 387 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c-3 a d) \sqrt {c+d x^2}}{5 c^2 e^3 \sqrt {e x}}+\frac {2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \sqrt {e x} \sqrt {c+d x^2}}{5 c^2 \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}+\frac {\left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}} \]
-2/5*a^2*(d*x^2+c)^(1/2)/c/e/(e*x)^(5/2)-2/5*a*(-3*a*d+10*b*c)*(d*x^2+c)^( 1/2)/c^2/e^3/(e*x)^(1/2)+2/5*(-3*a^2*d^2+10*a*b*c*d+5*b^2*c^2)*(e*x)^(1/2) *(d*x^2+c)^(1/2)/c^2/e^4/d^(1/2)/(c^(1/2)+x*d^(1/2))-2/5*(-3*a^2*d^2+10*a* b*c*d+5*b^2*c^2)*(cos(2*arctan(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)))*EllipticE(sin(2*arc tan(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^(7/4)/d^(3/4)/e^(7/2)/(d*x^2+c) ^(1/2)+1/5*(-3*a^2*d^2+10*a*b*c*d+5*b^2*c^2)*(cos(2*arctan(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^(7/ 4)/d^(3/4)/e^(7/2)/(d*x^2+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 11.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.30 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=\frac {x \left (-2 a \left (c+d x^2\right ) \left (10 b c x^2+a \left (c-3 d x^2\right )\right )+2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \sqrt {1+\frac {c}{d x^2}} x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {c}{d x^2}\right )\right )}{5 c^2 (e x)^{7/2} \sqrt {c+d x^2}} \]
(x*(-2*a*(c + d*x^2)*(10*b*c*x^2 + a*(c - 3*d*x^2)) + 2*(5*b^2*c^2 + 10*a* b*c*d - 3*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^4*Hypergeometric2F1[-1/4, 1/2, 3/ 4, -(c/(d*x^2))]))/(5*c^2*(e*x)^(7/2)*Sqrt[c + d*x^2])
Time = 0.50 (sec) , antiderivative size = 383, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {365, 27, 359, 266, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx\) |
\(\Big \downarrow \) 365 |
\(\displaystyle \frac {2 \int \frac {5 b^2 c x^2+a (10 b c-3 a d)}{2 (e x)^{3/2} \sqrt {d x^2+c}}dx}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {5 b^2 c x^2+a (10 b c-3 a d)}{(e x)^{3/2} \sqrt {d x^2+c}}dx}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {\left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \int \frac {\sqrt {e x}}{\sqrt {d x^2+c}}dx}{c e^2}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{c e \sqrt {e x}}}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\frac {2 \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \int \frac {e x}{\sqrt {d x^2+c}}d\sqrt {e x}}{c e^3}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{c e \sqrt {e x}}}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {\frac {2 \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt {c} e \int \frac {1}{\sqrt {d x^2+c}}d\sqrt {e x}}{\sqrt {d}}-\frac {\sqrt {c} e \int \frac {\sqrt {c} e-\sqrt {d} e x}{\sqrt {c} e \sqrt {d x^2+c}}d\sqrt {e x}}{\sqrt {d}}\right )}{c e^3}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{c e \sqrt {e x}}}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt {c} e \int \frac {1}{\sqrt {d x^2+c}}d\sqrt {e x}}{\sqrt {d}}-\frac {\int \frac {\sqrt {c} e-\sqrt {d} e x}{\sqrt {d x^2+c}}d\sqrt {e x}}{\sqrt {d}}\right )}{c e^3}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{c e \sqrt {e x}}}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\frac {2 \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt [4]{c} \sqrt {e} \left (\sqrt {c} e+\sqrt {d} e x\right ) \sqrt {\frac {c e^2+d e^2 x^2}{\left (\sqrt {c} e+\sqrt {d} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 d^{3/4} \sqrt {c+d x^2}}-\frac {\int \frac {\sqrt {c} e-\sqrt {d} e x}{\sqrt {d x^2+c}}d\sqrt {e x}}{\sqrt {d}}\right )}{c e^3}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{c e \sqrt {e x}}}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {\frac {2 \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \left (\frac {\sqrt [4]{c} \sqrt {e} \left (\sqrt {c} e+\sqrt {d} e x\right ) \sqrt {\frac {c e^2+d e^2 x^2}{\left (\sqrt {c} e+\sqrt {d} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 d^{3/4} \sqrt {c+d x^2}}-\frac {\frac {\sqrt [4]{c} \sqrt {e} \left (\sqrt {c} e+\sqrt {d} e x\right ) \sqrt {\frac {c e^2+d e^2 x^2}{\left (\sqrt {c} e+\sqrt {d} e x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{\sqrt [4]{d} \sqrt {c+d x^2}}-\frac {e^2 \sqrt {e x} \sqrt {c+d x^2}}{\sqrt {c} e+\sqrt {d} e x}}{\sqrt {d}}\right )}{c e^3}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{c e \sqrt {e x}}}{5 c e^2}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}\) |
(-2*a^2*Sqrt[c + d*x^2])/(5*c*e*(e*x)^(5/2)) + ((-2*a*(10*b*c - 3*a*d)*Sqr t[c + d*x^2])/(c*e*Sqrt[e*x]) + (2*(5*b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*(- ((-((e^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(Sqrt[c]*e + Sqrt[d]*e*x)) + (c^(1/4)* Sqrt[e]*(Sqrt[c]*e + Sqrt[d]*e*x)*Sqrt[(c*e^2 + d*e^2*x^2)/(Sqrt[c]*e + Sq rt[d]*e*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1 /2])/(d^(1/4)*Sqrt[c + d*x^2]))/Sqrt[d]) + (c^(1/4)*Sqrt[e]*(Sqrt[c]*e + S qrt[d]*e*x)*Sqrt[(c*e^2 + d*e^2*x^2)/(Sqrt[c]*e + Sqrt[d]*e*x)^2]*Elliptic F[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(2*d^(3/4)*Sqrt[c + d*x^2])))/(c*e^3))/(5*c*e^2)
3.9.45.3.1 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] :> 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[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[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -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]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 3.12 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.67
method | result | size |
risch | \(-\frac {2 \sqrt {d \,x^{2}+c}\, a \left (-3 a d \,x^{2}+10 c b \,x^{2}+a c \right )}{5 c^{2} x^{2} e^{3} \sqrt {e x}}-\frac {\left (3 a^{2} d^{2}-10 a b c d -5 b^{2} c^{2}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right ) \sqrt {e x \left (d \,x^{2}+c \right )}}{5 c^{2} d \sqrt {d e \,x^{3}+c e x}\, e^{3} \sqrt {e x}\, \sqrt {d \,x^{2}+c}}\) | \(261\) |
elliptic | \(\frac {\sqrt {e x \left (d \,x^{2}+c \right )}\, \left (-\frac {2 a^{2} \sqrt {d e \,x^{3}+c e x}}{5 e^{4} c \,x^{3}}+\frac {2 \left (d e \,x^{2}+c e \right ) a \left (3 a d -10 b c \right )}{5 e^{4} c^{2} \sqrt {x \left (d e \,x^{2}+c e \right )}}+\frac {\left (\frac {b^{2}}{e^{3}}-\frac {d a \left (3 a d -10 b c \right )}{5 c^{2} e^{3}}\right ) \sqrt {-c d}\, \sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, \left (-\frac {2 \sqrt {-c d}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}+\frac {\sqrt {-c d}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-c d}}{d}\right ) d}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{d}\right )}{d \sqrt {d e \,x^{3}+c e x}}\right )}{\sqrt {e x}\, \sqrt {d \,x^{2}+c}}\) | \(284\) |
default | \(-\frac {6 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{2}-20 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{2}-10 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, E\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{2}-3 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a^{2} c \,d^{2} x^{2}+10 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) a b \,c^{2} d \,x^{2}+5 \sqrt {2}\, \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {x d}{\sqrt {-c d}}}\, F\left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c^{3} x^{2}-6 a^{2} d^{3} x^{4}+20 a b c \,d^{2} x^{4}-4 a^{2} c \,d^{2} x^{2}+20 a b \,c^{2} d \,x^{2}+2 a^{2} c^{2} d}{5 x^{2} \sqrt {d \,x^{2}+c}\, d \,e^{3} \sqrt {e x}\, c^{2}}\) | \(626\) |
-2/5*(d*x^2+c)^(1/2)*a*(-3*a*d*x^2+10*b*c*x^2+a*c)/c^2/x^2/e^3/(e*x)^(1/2) -1/5*(3*a^2*d^2-10*a*b*c*d-5*b^2*c^2)/c^2*(-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)*(-2*(-c*d)^(1/2)/d*EllipticE(( (x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2),1/2*2^(1/2))+(-c*d)^(1/2)/d*Ellip ticF(((x+(-c*d)^(1/2)/d)/(-c*d)^(1/2)*d)^(1/2),1/2*2^(1/2)))/e^3*(e*x*(d*x ^2+c))^(1/2)/(e*x)^(1/2)/(d*x^2+c)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.27 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=-\frac {2 \, {\left ({\left (5 \, b^{2} c^{2} + 10 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d e} x^{3} {\rm weierstrassZeta}\left (-\frac {4 \, c}{d}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, c}{d}, 0, x\right )\right ) + {\left (a^{2} c d + {\left (10 \, a b c d - 3 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}\right )}}{5 \, c^{2} d e^{4} x^{3}} \]
-2/5*((5*b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*sqrt(d*e)*x^3*weierstrassZeta(- 4*c/d, 0, weierstrassPInverse(-4*c/d, 0, x)) + (a^2*c*d + (10*a*b*c*d - 3* a^2*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(e*x))/(c^2*d*e^4*x^3)
Result contains complex when optimal does not.
Time = 22.64 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=\frac {a^{2} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {a b \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {b^{2} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} \]
a**2*gamma(-5/4)*hyper((-5/4, 1/2), (-1/4,), d*x**2*exp_polar(I*pi)/c)/(2* sqrt(c)*e**(7/2)*x**(5/2)*gamma(-1/4)) + a*b*gamma(-1/4)*hyper((-1/4, 1/2) , (3/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*e**(7/2)*sqrt(x)*gamma(3/4)) + b**2*x**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), d*x**2*exp_polar(I*pi )/c)/(2*sqrt(c)*e**(7/2)*gamma(7/4))
\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{2}}{\sqrt {d x^{2} + c} \left (e x\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2}{{\left (e\,x\right )}^{7/2}\,\sqrt {d\,x^2+c}} \,d x \]