Integrand size = 21, antiderivative size = 310 \[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {(b+a c)^2 \left (c+d x^2\right )^3}{a b^2 d^3 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}+\frac {\left (35 b^2+60 a b c+24 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{16 a^4 d^3}-\frac {\left (35 b^2+60 a b c+24 a^2 c^2\right ) \left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{24 a^3 b d^3}+\frac {\left (7 b^2+12 a b c+6 a^2 c^2\right ) \left (c+d x^2\right )^3 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{6 a^2 b^2 d^3}-\frac {b \left (35 b^2+60 a b c+24 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{16 a^{9/2} d^3} \]
-1/16*b*(24*a^2*c^2+60*a*b*c+35*b^2)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^( 1/2)/a^(1/2))/a^(9/2)/d^3-(a*c+b)^2*(d*x^2+c)^3/a/b^2/d^3/((a*d*x^2+a*c+b) /(d*x^2+c))^(1/2)+1/16*(24*a^2*c^2+60*a*b*c+35*b^2)*(d*x^2+c)*((a*d*x^2+a* c+b)/(d*x^2+c))^(1/2)/a^4/d^3-1/24*(24*a^2*c^2+60*a*b*c+35*b^2)*(d*x^2+c)^ 2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^3/b/d^3+1/6*(6*a^2*c^2+12*a*b*c+7*b^ 2)*(d*x^2+c)^3*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^2/b^2/d^3
Time = 0.33 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.59 \[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {\frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (105 b^3+5 a b^2 \left (43 c+7 d x^2\right )+2 a^2 b \left (59 c^2+16 c d x^2-7 d^2 x^4\right )+8 a^3 \left (c^3+d^3 x^6\right )\right )}{b+a \left (c+d x^2\right )}-3 b \left (35 b^2+60 a b c+24 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{48 a^{9/2} d^3} \]
((Sqrt[a]*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(105*b^3 + 5*a *b^2*(43*c + 7*d*x^2) + 2*a^2*b*(59*c^2 + 16*c*d*x^2 - 7*d^2*x^4) + 8*a^3* (c^3 + d^3*x^6)))/(b + a*(c + d*x^2)) - 3*b*(35*b^2 + 60*a*b*c + 24*a^2*c^ 2)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(48*a^(9/2)*d^3 )
Time = 0.43 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.82, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2057, 2053, 2052, 27, 365, 298, 215, 215, 219}
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.
| \(\displaystyle \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int \frac {x^5}{\left (\frac {a c+a d x^2+b}{c+d x^2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int \frac {x^4}{\left (\frac {a d x^2+b+a c}{d x^2+c}\right )^{3/2}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int \frac {\left (-c x^4+b+a c\right )^2}{d^4 x^4 \left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b \int \frac {\left (-c x^4+b+a c\right )^2}{x^4 \left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{d^3}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle -\frac {b \left (\frac {\int \frac {a c^2 x^4+(b+a c) (7 b+5 a c)}{\left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{a}-\frac {(a c+b)^2}{a x^2 \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle -\frac {b \left (\frac {\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \int \frac {1}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{6 a}+\frac {\left (6 a^2 c^2+12 a b c+7 b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}}{a}-\frac {(a c+b)^2}{a x^2 \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle -\frac {b \left (\frac {\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {3 \int \frac {1}{\left (a-x^4\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{4 a}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}\right )}{6 a}+\frac {\left (6 a^2 c^2+12 a b c+7 b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}}{a}-\frac {(a c+b)^2}{a x^2 \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle -\frac {b \left (\frac {\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {3 \left (\frac {\int \frac {1}{a-x^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{2 a}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )}{4 a}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}\right )}{6 a}+\frac {\left (6 a^2 c^2+12 a b c+7 b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}}{a}-\frac {(a c+b)^2}{a x^2 \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {b \left (\frac {\frac {\left (6 a^2 c^2+12 a b c+7 b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}+\frac {\left (24 a^2 c^2+60 a b c+35 b^2\right ) \left (\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )}{4 a}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}\right )}{6 a}}{a}-\frac {(a c+b)^2}{a x^2 \left (a-x^4\right )^3}\right )}{d^3}\) |
-((b*(-((b + a*c)^2/(a*x^2*(a - x^4)^3)) + (((7*b^2 + 12*a*b*c + 6*a^2*c^2 )*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(6*a*(a - x^4)^3) + ((35*b^2 + 60 *a*b*c + 24*a^2*c^2)*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/(4*a*(a - x^4) ^2) + (3*(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/(2*a*(a - x^4)) + ArcTanh[ Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]]/(2*a^(3/2))))/(4*a)))/(6*a) )/a))/d^3)
3.4.55.3.1 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[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
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[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* ((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
Time = 0.23 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(\frac {\left (8 a^{2} d^{2} x^{4}-8 a^{2} c d \,x^{2}-22 a b d \,x^{2}+8 a^{2} c^{2}+62 a b c +57 b^{2}\right ) \left (a d \,x^{2}+a c +b \right )}{48 d^{3} a^{4} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}-\frac {b \left (\frac {\left (24 a^{2} c^{2}+60 a b c +35 b^{2}\right ) \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right )}{2 \sqrt {a \,d^{2}}}-\frac {16 \left (a^{2} c^{2}+2 a b c +b^{2}\right ) \left (d \,x^{2}+c \right )}{d \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}\right ) \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{16 a^{4} d^{2} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\) | \(303\) |
| default | \(\text {Expression too large to display}\) | \(1240\) |
1/48/d^3*(8*a^2*d^2*x^4-8*a^2*c*d*x^2-22*a*b*d*x^2+8*a^2*c^2+62*a*b*c+57*b ^2)*(a*d*x^2+a*c+b)/a^4/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)-1/16*b/a^4/d^2*( 1/2*(24*a^2*c^2+60*a*b*c+35*b^2)*ln((a*c*d+1/2*b*d+a*d^2*x^2)/(a*d^2)^(1/2 )+(a*c^2+b*c+(2*a*c*d+b*d)*x^2+a*d^2*x^4)^(1/2))/(a*d^2)^(1/2)-16*(a^2*c^2 +2*a*b*c+b^2)*(d*x^2+c)/d/(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)) /((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*((a*d*x^2+a*c+b)*(d*x^2+c))^(1/2)/(d*x^ 2+c)
Time = 0.41 (sec) , antiderivative size = 675, normalized size of antiderivative = 2.18 \[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (24 \, a^{3} b c^{3} + 84 \, a^{2} b^{2} c^{2} + 95 \, a b^{3} c + 35 \, b^{4} + {\left (24 \, a^{3} b c^{2} + 60 \, a^{2} b^{2} c + 35 \, a b^{3}\right )} d x^{2}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (8 \, a^{4} d^{4} x^{8} + 2 \, {\left (4 \, a^{4} c - 7 \, a^{3} b\right )} d^{3} x^{6} + 8 \, a^{4} c^{4} + 118 \, a^{3} b c^{3} + {\left (18 \, a^{3} b c + 35 \, a^{2} b^{2}\right )} d^{2} x^{4} + 215 \, a^{2} b^{2} c^{2} + 105 \, a b^{3} c + {\left (8 \, a^{4} c^{3} + 150 \, a^{3} b c^{2} + 250 \, a^{2} b^{2} c + 105 \, a b^{3}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{192 \, {\left (a^{6} d^{4} x^{2} + {\left (a^{6} c + a^{5} b\right )} d^{3}\right )}}, \frac {3 \, {\left (24 \, a^{3} b c^{3} + 84 \, a^{2} b^{2} c^{2} + 95 \, a b^{3} c + 35 \, b^{4} + {\left (24 \, a^{3} b c^{2} + 60 \, a^{2} b^{2} c + 35 \, a b^{3}\right )} d x^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (8 \, a^{4} d^{4} x^{8} + 2 \, {\left (4 \, a^{4} c - 7 \, a^{3} b\right )} d^{3} x^{6} + 8 \, a^{4} c^{4} + 118 \, a^{3} b c^{3} + {\left (18 \, a^{3} b c + 35 \, a^{2} b^{2}\right )} d^{2} x^{4} + 215 \, a^{2} b^{2} c^{2} + 105 \, a b^{3} c + {\left (8 \, a^{4} c^{3} + 150 \, a^{3} b c^{2} + 250 \, a^{2} b^{2} c + 105 \, a b^{3}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{96 \, {\left (a^{6} d^{4} x^{2} + {\left (a^{6} c + a^{5} b\right )} d^{3}\right )}}\right ] \]
[1/192*(3*(24*a^3*b*c^3 + 84*a^2*b^2*c^2 + 95*a*b^3*c + 35*b^4 + (24*a^3*b *c^2 + 60*a^2*b^2*c + 35*a*b^3)*d*x^2)*sqrt(a)*log(8*a^2*d^2*x^4 + 8*a^2*c ^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 - 4*(2*a*d^2*x^4 + (4*a*c + b )*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 4*(8*a^4*d^4*x^8 + 2*(4*a^4*c - 7*a^3*b)*d^3*x^6 + 8*a^4*c^4 + 118*a^3*b*c ^3 + (18*a^3*b*c + 35*a^2*b^2)*d^2*x^4 + 215*a^2*b^2*c^2 + 105*a*b^3*c + ( 8*a^4*c^3 + 150*a^3*b*c^2 + 250*a^2*b^2*c + 105*a*b^3)*d*x^2)*sqrt((a*d*x^ 2 + a*c + b)/(d*x^2 + c)))/(a^6*d^4*x^2 + (a^6*c + a^5*b)*d^3), 1/96*(3*(2 4*a^3*b*c^3 + 84*a^2*b^2*c^2 + 95*a*b^3*c + 35*b^4 + (24*a^3*b*c^2 + 60*a^ 2*b^2*c + 35*a*b^3)*d*x^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqr t(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) + 2 *(8*a^4*d^4*x^8 + 2*(4*a^4*c - 7*a^3*b)*d^3*x^6 + 8*a^4*c^4 + 118*a^3*b*c^ 3 + (18*a^3*b*c + 35*a^2*b^2)*d^2*x^4 + 215*a^2*b^2*c^2 + 105*a*b^3*c + (8 *a^4*c^3 + 150*a^3*b*c^2 + 250*a^2*b^2*c + 105*a*b^3)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^6*d^4*x^2 + (a^6*c + a^5*b)*d^3)]
\[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.25 \[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {48 \, a^{5} b c^{2} + 96 \, a^{4} b^{2} c + 48 \, a^{3} b^{3} - \frac {3 \, {\left (24 \, a^{2} b c^{2} + 60 \, a b^{2} c + 35 \, b^{3}\right )} {\left (a d x^{2} + a c + b\right )}^{3}}{{\left (d x^{2} + c\right )}^{3}} + \frac {8 \, {\left (24 \, a^{3} b c^{2} + 60 \, a^{2} b^{2} c + 35 \, a b^{3}\right )} {\left (a d x^{2} + a c + b\right )}^{2}}{{\left (d x^{2} + c\right )}^{2}} - \frac {3 \, {\left (56 \, a^{4} b c^{2} + 132 \, a^{3} b^{2} c + 77 \, a^{2} b^{3}\right )} {\left (a d x^{2} + a c + b\right )}}{d x^{2} + c}}{48 \, {\left (a^{7} d^{3} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - 3 \, a^{6} d^{3} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} + 3 \, a^{5} d^{3} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {5}{2}} - a^{4} d^{3} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {7}{2}}\right )}} + \frac {{\left (24 \, a^{2} c^{2} + 60 \, a b c + 35 \, b^{2}\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{32 \, a^{\frac {9}{2}} d^{3}} \]
1/48*(48*a^5*b*c^2 + 96*a^4*b^2*c + 48*a^3*b^3 - 3*(24*a^2*b*c^2 + 60*a*b^ 2*c + 35*b^3)*(a*d*x^2 + a*c + b)^3/(d*x^2 + c)^3 + 8*(24*a^3*b*c^2 + 60*a ^2*b^2*c + 35*a*b^3)*(a*d*x^2 + a*c + b)^2/(d*x^2 + c)^2 - 3*(56*a^4*b*c^2 + 132*a^3*b^2*c + 77*a^2*b^3)*(a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^7*d^3*s qrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) - 3*a^6*d^3*((a*d*x^2 + a*c + b)/(d*x ^2 + c))^(3/2) + 3*a^5*d^3*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(5/2) - a^4*d ^3*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(7/2)) + 1/32*(24*a^2*c^2 + 60*a*b*c + 35*b^2)*b*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(sqrt(a ) + sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))))/(a^(9/2)*d^3)
Leaf count of result is larger than twice the leaf count of optimal. 663 vs. \(2 (288) = 576\).
Time = 0.68 (sec) , antiderivative size = 663, normalized size of antiderivative = 2.14 \[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {1}{48} \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{a^{2} d \mathrm {sgn}\left (d x^{2} + c\right )} - \frac {4 \, a^{11} c d^{6} \mathrm {sgn}\left (d x^{2} + c\right ) + 11 \, a^{10} b d^{6} \mathrm {sgn}\left (d x^{2} + c\right )}{a^{13} d^{8}}\right )} + \frac {8 \, a^{11} c^{2} d^{5} \mathrm {sgn}\left (d x^{2} + c\right ) + 62 \, a^{10} b c d^{5} \mathrm {sgn}\left (d x^{2} + c\right ) + 57 \, a^{9} b^{2} d^{5} \mathrm {sgn}\left (d x^{2} + c\right )}{a^{13} d^{8}}\right )} + \frac {{\left (24 \, a^{2} b c^{2} + 60 \, a b^{2} c + 35 \, b^{3}\right )} \log \left ({\left | 2 \, a^{3} c^{3} d + 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{\frac {5}{2}} c^{2} {\left | d \right |} + 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a^{2} c d + 5 \, a^{2} b c^{2} d + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{3} a^{\frac {3}{2}} {\left | d \right |} + 10 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{\frac {3}{2}} b c {\left | d \right |} + 5 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a b d + 4 \, a b^{2} c d + 4 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} b^{2} {\left | d \right |} + b^{3} d \right |}\right )}{96 \, a^{\frac {9}{2}} d^{2} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right )} + \frac {{\left (24 \, a^{\frac {13}{2}} b c^{2} d^{3} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right ) + 60 \, a^{\frac {11}{2}} b^{2} c d^{3} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right ) + 35 \, a^{\frac {9}{2}} b^{3} d^{3} {\left | d \right |} \mathrm {sgn}\left (d x^{2} + c\right )\right )} \log \left ({\left | a \right |}\right )}{96 \, a^{9} d^{7}} \]
1/48*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)*(2*x^2*(4*x^2/( a^2*d*sgn(d*x^2 + c)) - (4*a^11*c*d^6*sgn(d*x^2 + c) + 11*a^10*b*d^6*sgn(d *x^2 + c))/(a^13*d^8)) + (8*a^11*c^2*d^5*sgn(d*x^2 + c) + 62*a^10*b*c*d^5* sgn(d*x^2 + c) + 57*a^9*b^2*d^5*sgn(d*x^2 + c))/(a^13*d^8)) + 1/96*(24*a^2 *b*c^2 + 60*a*b^2*c + 35*b^3)*log(abs(2*a^3*c^3*d + 6*(sqrt(a*d^2)*x^2 - s qrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a^(5/2)*c^2*abs(d) + 6*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c ))^2*a^2*c*d + 5*a^2*b*c^2*d + 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c *d*x^2 + b*d*x^2 + a*c^2 + b*c))^3*a^(3/2)*abs(d) + 10*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a^(3/2)*b*c*abs(d) + 5*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b* c))^2*a*b*d + 4*a*b^2*c*d + 4*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d* x^2 + b*d*x^2 + a*c^2 + b*c))*sqrt(a)*b^2*abs(d) + b^3*d))/(a^(9/2)*d^2*ab s(d)*sgn(d*x^2 + c)) + 1/96*(24*a^(13/2)*b*c^2*d^3*abs(d)*sgn(d*x^2 + c) + 60*a^(11/2)*b^2*c*d^3*abs(d)*sgn(d*x^2 + c) + 35*a^(9/2)*b^3*d^3*abs(d)*s gn(d*x^2 + c))*log(abs(a))/(a^9*d^7)
Timed out. \[ \int \frac {x^5}{\left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \,d x \]