Integrand size = 28, antiderivative size = 377 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}+\frac {7 b (11 b B-9 A c) x^{3/2} \left (b+c x^2\right )}{15 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}-\frac {7 b^{5/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{15/4} \sqrt {b x^2+c x^4}}+\frac {7 b^{5/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{30 c^{15/4} \sqrt {b x^2+c x^4}} \]
-(-A*c+B*b)*x^(13/2)/b/c/(c*x^4+b*x^2)^(1/2)+7/15*b*(-9*A*c+11*B*b)*x^(3/2 )*(c*x^2+b)/c^(7/2)/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)+1/9*(-9*A*c+11 *B*b)*x^(5/2)*(c*x^4+b*x^2)^(1/2)/b/c^2-7/45*(-9*A*c+11*B*b)*x^(1/2)*(c*x^ 4+b*x^2)^(1/2)/c^3-7/15*b^(5/4)*(-9*A*c+11*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)))*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^(15/4)/(c*x^4+b*x^2)^(1/2)+7/30*b^ (5/4)*(-9*A*c+11*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/c os(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(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^(15/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.29 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {2 x^{5/2} \left (77 b^2 B+c^2 x^2 \left (9 A+5 B x^2\right )-b c \left (63 A+11 B x^2\right )+7 b (-11 b B+9 A c) \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{45 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \]
(2*x^(5/2)*(77*b^2*B + c^2*x^2*(9*A + 5*B*x^2) - b*c*(63*A + 11*B*x^2) + 7 *b*(-11*b*B + 9*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[3/4, 3/2, 7/4, -((c*x^2)/b)]))/(45*c^3*Sqrt[x^2*(b + c*x^2)])
Time = 0.51 (sec) , antiderivative size = 370, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1943, 1429, 1429, 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.
| \(\displaystyle \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1943 |
| \(\displaystyle \frac {(11 b B-9 A c) \int \frac {x^{11/2}}{\sqrt {c x^4+b x^2}}dx}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \int \frac {x^{7/2}}{\sqrt {c x^4+b x^2}}dx}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {(11 b B-9 A c) \left (\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\right )}{2 b c}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}}\) |
-(((b*B - A*c)*x^(13/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + ((11*b*B - 9*A*c)*(( 2*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(9*c) - (7*b*((2*Sqrt[x]*Sqrt[b*x^2 + c*x^4 ])/(5*c) - (6*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]*EllipticE[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*c*Sqrt[b*x^2 + c*x^4])))/(9*c) ))/(2*b*c)
3.3.58.3.1 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^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[m + 2*p - 1, 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[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1))) Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 ] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
Time = 2.49 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.11
| method | result | size |
| default | \(-\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (-20 B \,c^{3} x^{6}+378 A \,b^{2} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-189 A \,b^{2} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-462 B \,b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, E\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+231 B \,b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-36 A \,c^{3} x^{4}+44 B b \,c^{2} x^{4}-126 A b \,c^{2} x^{2}+154 B \,b^{2} c \,x^{2}\right )}{90 \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} c^{4}}\) | \(420\) |
| risch | \(\frac {2 x^{\frac {5}{2}} \left (5 B c \,x^{2}+9 A c -16 B b \right ) \left (c \,x^{2}+b \right )}{45 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \left (\frac {\left (24 A c -31 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{c \sqrt {c \,x^{3}+b x}}-15 b \left (A c -B b \right ) \left (\frac {x^{2}}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}-\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(440\) |
-1/90/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(-20*B*c^3*x^6+378*A*b^2*c*((c *x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^( 1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c) ^(1/2))^(1/2),1/2*2^(1/2))-189*A*b^2*c*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^( 1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^ (1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-462*B *b^3*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/ (-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2) )/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+231*B*b^3*((c*x+(-b*c)^(1/2))/(-b*c)^(1 /2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^( 1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2)) -36*A*c^3*x^4+44*B*b*c^2*x^4-126*A*b*c^2*x^2+154*B*b^2*c*x^2)/c^4
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.36 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {21 \, {\left (11 \, B b^{3} - 9 \, A b^{2} c + {\left (11 \, B b^{2} c - 9 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - {\left (10 \, B c^{3} x^{4} - 77 \, B b^{2} c + 63 \, A b c^{2} - 2 \, {\left (11 \, B b c^{2} - 9 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{45 \, {\left (c^{5} x^{2} + b c^{4}\right )}} \]
-1/45*(21*(11*B*b^3 - 9*A*b^2*c + (11*B*b^2*c - 9*A*b*c^2)*x^2)*sqrt(c)*we ierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) - (10*B*c^3*x^ 4 - 77*B*b^2*c + 63*A*b*c^2 - 2*(11*B*b*c^2 - 9*A*c^3)*x^2)*sqrt(c*x^4 + b *x^2)*sqrt(x))/(c^5*x^2 + b*c^4)
Timed out. \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {15}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} x^{\frac {15}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{15/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]