Integrand size = 21, antiderivative size = 225 \[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {\left (5 b^2+12 a b c+8 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{16 a^3 d^3}-\frac {(5 b+8 a c) \left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{24 a^2 d^3}+\frac {x^2 \left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{6 a d^2}-\frac {b \left (5 b^2+12 a b c+8 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{16 a^{7/2} d^3} \]
-1/16*b*(8*a^2*c^2+12*a*b*c+5*b^2)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/ 2)/a^(1/2))/a^(7/2)/d^3+1/16*(8*a^2*c^2+12*a*b*c+5*b^2)*(d*x^2+c)*((a*d*x^ 2+a*c+b)/(d*x^2+c))^(1/2)/a^3/d^3-1/24*(8*a*c+5*b)*(d*x^2+c)^2*((a*d*x^2+a *c+b)/(d*x^2+c))^(1/2)/a^2/d^3+1/6*x^2*(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2 +c))^(1/2)/a/d^2
Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.66 \[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (15 b^2+2 a b \left (13 c-5 d x^2\right )+8 a^2 \left (c^2-c d x^2+d^2 x^4\right )\right )-3 b \left (5 b^2+12 a b c+8 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{48 a^{7/2} d^3} \]
(Sqrt[a]*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(15*b^2 + 2*a*b *(13*c - 5*d*x^2) + 8*a^2*(c^2 - c*d*x^2 + d^2*x^4)) - 3*b*(5*b^2 + 12*a*b *c + 8*a^2*c^2)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(4 8*a^(7/2)*d^3)
Time = 0.42 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.01, 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, 315, 25, 298, 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.
| \(\displaystyle \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int \frac {x^5}{\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int \frac {x^4}{\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int \frac {\left (-c x^4+b+a c\right )^2}{d^4 \left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b \int \frac {\left (-c x^4+b+a c\right )^2}{\left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{d^3}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle -\frac {b \left (\frac {b \left (a c+b-c x^4\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}-\frac {\int -\frac {(b+a c) (5 b+6 a c)-3 c (b+2 a c) x^4}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{6 a}\right )}{d^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {b \left (\frac {\int \frac {(b+a c) (5 b+6 a c)-3 c (b+2 a c) x^4}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{6 a}+\frac {b \left (a c+b-c x^4\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle -\frac {b \left (\frac {\frac {3 \left (8 a^2 c^2+12 a b c+5 b^2\right ) \int \frac {1}{\left (a-x^4\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{4 a}+\frac {b (8 a c+5 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}}{6 a}+\frac {b \left (a c+b-c x^4\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle -\frac {b \left (\frac {\frac {3 \left (8 a^2 c^2+12 a b c+5 b^2\right ) \left (\frac {\int \frac {1}{a-x^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{2 a}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )}{4 a}+\frac {b (8 a c+5 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}}{6 a}+\frac {b \left (a c+b-c x^4\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}\right )}{d^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {b \left (\frac {\frac {3 \left (8 a^2 c^2+12 a b c+5 b^2\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )}{4 a}+\frac {b (8 a c+5 b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 a \left (a-x^4\right )^2}}{6 a}+\frac {b \left (a c+b-c x^4\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{6 a \left (a-x^4\right )^3}\right )}{d^3}\) |
-((b*((b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(b + a*c - c*x^4))/(6*a*(a - x^4)^3) + ((b*(5*b + 8*a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(4*a* (a - x^4)^2) + (3*(5*b^2 + 12*a*b*c + 8*a^2*c^2)*(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)))/d^3)
3.4.44.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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 = 0.14 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.06
| method | result | size |
| risch | \(\frac {\left (8 a^{2} d^{2} x^{4}-8 a^{2} c d \,x^{2}-10 a b d \,x^{2}+8 a^{2} c^{2}+26 a b c +15 b^{2}\right ) \left (a d \,x^{2}+a c +b \right )}{48 d^{3} a^{3} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}-\frac {b \left (8 a^{2} c^{2}+12 a b c +5 b^{2}\right ) \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right ) \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{32 d^{2} a^{3} \sqrt {a \,d^{2}}\, \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\) | \(239\) |
| default | \(\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-48 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a^{2} c d \,x^{2}-36 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a b d \,x^{2}-24 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a^{2} b \,c^{2} d -36 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a \,b^{2} c d +16 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c \right )^{\frac {3}{2}} a \sqrt {a \,d^{2}}+36 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a b c -15 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{3} d +30 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b^{2}\right )}{96 a^{3} d^{3} \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a \,d^{2}}}\) | \(533\) |
1/48/d^3*(8*a^2*d^2*x^4-8*a^2*c*d*x^2-10*a*b*d*x^2+8*a^2*c^2+26*a*b*c+15*b ^2)*(a*d*x^2+a*c+b)/a^3/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)-1/32*b/d^2*(8*a^ 2*c^2+12*a*b*c+5*b^2)/a^3*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)/((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 = 0.30 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.89 \[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\left [\frac {3 \, {\left (8 \, a^{2} b c^{2} + 12 \, a b^{2} c + 5 \, b^{3}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (8 \, a^{3} d^{3} x^{6} - 10 \, a^{2} b d^{2} x^{4} + 8 \, a^{3} c^{3} + 26 \, a^{2} b c^{2} + 15 \, a b^{2} c + {\left (16 \, a^{2} b c + 15 \, a b^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{192 \, a^{4} d^{3}}, \frac {3 \, {\left (8 \, a^{2} b c^{2} + 12 \, a b^{2} c + 5 \, b^{3}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \, {\left (8 \, a^{3} d^{3} x^{6} - 10 \, a^{2} b d^{2} x^{4} + 8 \, a^{3} c^{3} + 26 \, a^{2} b c^{2} + 15 \, a b^{2} c + {\left (16 \, a^{2} b c + 15 \, a b^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{96 \, a^{4} d^{3}}\right ] \]
[1/192*(3*(8*a^2*b*c^2 + 12*a*b^2*c + 5*b^3)*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^3*d^3*x^6 - 10*a^2*b*d^2*x^4 + 8*a^3*c^3 + 26*a^2*b*c^2 + 15* a*b^2*c + (16*a^2*b*c + 15*a*b^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*d^3), 1/96*(3*(8*a^2*b*c^2 + 12*a*b^2*c + 5*b^3)*sqrt(-a)*arcta n(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-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^3*d^3*x^6 - 10*a^2*b*d^2*x^4 + 8*a^ 3*c^3 + 26*a^2*b*c^2 + 15*a*b^2*c + (16*a^2*b*c + 15*a*b^2)*d*x^2)*sqrt((a *d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*d^3)]
\[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\int \frac {x^{5}}{\sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.51 \[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=-\frac {3 \, {\left (8 \, a^{2} b c^{2} + 12 \, a b^{2} c + 5 \, b^{3}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {5}{2}} - 8 \, {\left (6 \, a^{3} b c^{2} + 12 \, a^{2} b^{2} c + 5 \, a b^{3}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} + 3 \, {\left (8 \, a^{4} b c^{2} + 20 \, a^{3} b^{2} c + 11 \, a^{2} b^{3}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{48 \, {\left (a^{6} d^{3} - \frac {3 \, {\left (a d x^{2} + a c + b\right )} a^{5} d^{3}}{d x^{2} + c} + \frac {3 \, {\left (a d x^{2} + a c + b\right )}^{2} a^{4} d^{3}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (a d x^{2} + a c + b\right )}^{3} a^{3} d^{3}}{{\left (d x^{2} + c\right )}^{3}}\right )}} + \frac {{\left (8 \, a^{2} c^{2} + 12 \, a b c + 5 \, b^{2}\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{32 \, a^{\frac {7}{2}} d^{3}} \]
-1/48*(3*(8*a^2*b*c^2 + 12*a*b^2*c + 5*b^3)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(5/2) - 8*(6*a^3*b*c^2 + 12*a^2*b^2*c + 5*a*b^3)*((a*d*x^2 + a*c + b)/ (d*x^2 + c))^(3/2) + 3*(8*a^4*b*c^2 + 20*a^3*b^2*c + 11*a^2*b^3)*sqrt((a*d *x^2 + a*c + b)/(d*x^2 + c)))/(a^6*d^3 - 3*(a*d*x^2 + a*c + b)*a^5*d^3/(d* x^2 + c) + 3*(a*d*x^2 + a*c + b)^2*a^4*d^3/(d*x^2 + c)^2 - (a*d*x^2 + a*c + b)^3*a^3*d^3/(d*x^2 + c)^3) + 1/32*(8*a^2*c^2 + 12*a*b*c + 5*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^(7/2)*d^3)
Time = 0.39 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\frac {2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{a d} - \frac {4 \, a^{2} c d^{3} + 5 \, a b d^{3}}{a^{3} d^{5}}\right )} + \frac {8 \, a^{2} c^{2} d^{2} + 26 \, a b c d^{2} + 15 \, b^{2} d^{2}}{a^{3} d^{5}}\right )} + \frac {3 \, {\left (8 \, a^{2} b c^{2} + 12 \, a b^{2} c + 5 \, b^{3}\right )} \log \left ({\left | 2 \, a c d + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} {\left | d \right |} + b d \right |}\right )}{a^{\frac {7}{2}} d^{2} {\left | d \right |}}}{96 \, \mathrm {sgn}\left (d x^{2} + c\right )} \]
1/96*(2*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*d) - (4*a^2*c*d^3 + 5*a*b*d^3)/(a^3*d^5)) + (8*a^2*c^2*d^2 + 26*a*b*c *d^2 + 15*b^2*d^2)/(a^3*d^5)) + 3*(8*a^2*b*c^2 + 12*a*b^2*c + 5*b^3)*log(a bs(2*a*c*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))*sqrt(a)*abs(d) + b*d))/(a^(7/2)*d^2*abs(d)))/sgn(d*x^2 + c)
Timed out. \[ \int \frac {x^5}{\sqrt {a+\frac {b}{c+d x^2}}} \, dx=\int \frac {x^5}{\sqrt {a+\frac {b}{d\,x^2+c}}} \,d x \]