Integrand size = 28, antiderivative size = 356 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\frac {8 b (b B+9 A c) x^{3/2} \left (b+c x^2\right )}{15 \sqrt {c} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {4}{15} (b B+9 A c) \sqrt {x} \sqrt {b x^2+c x^4}+\frac {2 (b B+9 A c) \left (b x^2+c x^4\right )^{3/2}}{9 b x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}-\frac {8 b^{5/4} (b B+9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{3/4} \sqrt {b x^2+c x^4}}+\frac {4 b^{5/4} (b B+9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 c^{3/4} \sqrt {b x^2+c x^4}} \]
2/9*(9*A*c+B*b)*(c*x^4+b*x^2)^(3/2)/b/x^(3/2)-2*A*(c*x^4+b*x^2)^(5/2)/b/x^ (11/2)+8/15*b*(9*A*c+B*b)*x^(3/2)*(c*x^2+b)/c^(1/2)/(b^(1/2)+x*c^(1/2))/(c *x^4+b*x^2)^(1/2)+4/15*(9*A*c+B*b)*x^(1/2)*(c*x^4+b*x^2)^(1/2)-8/15*b^(5/4 )*(9*A*c+B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arc tan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/ 4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1 /2)/c^(3/4)/(c*x^4+b*x^2)^(1/2)+4/15*b^(5/4)*(9*A*c+B*b)*x*(cos(2*arctan(c ^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*E llipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^ (1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.24 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (-\frac {3 A \left (b+c x^2\right )^2}{b}+\frac {(b B+9 A c) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )}{\sqrt {1+\frac {c x^2}{b}}}\right )}{3 x^{3/2}} \]
(2*Sqrt[x^2*(b + c*x^2)]*((-3*A*(b + c*x^2)^2)/b + ((b*B + 9*A*c)*x^2*Hype rgeometric2F1[-3/2, 3/4, 7/4, -((c*x^2)/b)])/Sqrt[1 + (c*x^2)/b]))/(3*x^(3 /2))
Time = 0.48 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1944, 1426, 1426, 1431, 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.
| \(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx\) |
| \(\Big \downarrow \) 1944 |
| \(\displaystyle \frac {(9 A c+b B) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{5/2}}dx}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \int \frac {\sqrt {c x^4+b x^2}}{\sqrt {x}}dx+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {2}{5} b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {2 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {(9 A c+b B) \left (\frac {2}{3} b \left (\frac {4 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 \sqrt {b x^2+c x^4}}+\frac {2}{5} \sqrt {x} \sqrt {b x^2+c x^4}\right )+\frac {2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{b x^{11/2}}\) |
(-2*A*(b*x^2 + c*x^4)^(5/2))/(b*x^(11/2)) + ((b*B + 9*A*c)*((2*(b*x^2 + c* x^4)^(3/2))/(9*x^(3/2)) + (2*b*((2*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/5 + (4*b*x *Sqrt[b + c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b] + Sqrt[c]*x)) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*E llipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^(1/4)*Sqrt[b + c*x^ 2]))/Sqrt[c]) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(2*c^( 3/4)*Sqrt[b + c*x^2])))/(5*Sqrt[b*x^2 + c*x^4])))/3))/b
3.3.39.3.1 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[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_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
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]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 1.85 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.70
| method | result | size |
| risch | \(-\frac {2 \left (-5 B c \,x^{4}-9 A c \,x^{2}-11 b B \,x^{2}+45 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{45 x^{\frac {3}{2}}}+\frac {4 b \left (9 A c +B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 c \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(249\) |
| default | \(\frac {2 \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} \left (5 B \,c^{3} x^{6}+108 A \,b^{2} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-54 A \,b^{2} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+12 B \,b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-6 B \,b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+9 A \,c^{3} x^{4}+16 B b \,c^{2} x^{4}-36 A b \,c^{2} x^{2}+11 B \,b^{2} c \,x^{2}-45 b^{2} A c \right )}{45 x^{\frac {7}{2}} \left (c \,x^{2}+b \right )^{2} c}\) | \(429\) |
-2/45*(-5*B*c*x^4-9*A*c*x^2-11*B*b*x^2+45*A*b)*(x^2*(c*x^2+b))^(1/2)/x^(3/ 2)+4/15*b*(9*A*c+B*b)/c*(-b*c)^(1/2)*((x+1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2)) ^(1/2)*(-2*(x-1/c*(-b*c)^(1/2))*c/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^ (1/2)/(c*x^3+b*x)^(1/2)*(-2/c*(-b*c)^(1/2)*EllipticE(((x+1/c*(-b*c)^(1/2)) *c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+1/c*(-b*c)^(1/2)*EllipticF(((x+1/c*(-b *c)^(1/2))*c/(-b*c)^(1/2))^(1/2),1/2*2^(1/2)))*(x^2*(c*x^2+b))^(1/2)/x^(3/ 2)/(c*x^2+b)*(x*(c*x^2+b))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.26 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=-\frac {2 \, {\left (12 \, {\left (B b^{2} + 9 \, A b c\right )} \sqrt {c} x^{2} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - {\left (5 \, B c^{2} x^{4} - 45 \, A b c + {\left (11 \, B b c + 9 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{45 \, c x^{2}} \]
-2/45*(12*(B*b^2 + 9*A*b*c)*sqrt(c)*x^2*weierstrassZeta(-4*b/c, 0, weierst rassPInverse(-4*b/c, 0, x)) - (5*B*c^2*x^4 - 45*A*b*c + (11*B*b*c + 9*A*c^ 2)*x^2)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c*x^2)
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{\frac {9}{2}}}\, dx \]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{9/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{9/2}} \,d x \]