Integrand size = 21, antiderivative size = 216 \[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=-\frac {\left (b^2+4 a b c-8 a^2 c^2\right ) \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{16 a^2 d^3}-\frac {(b+4 a c) \left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{8 a d^3}+\frac {\left (c+d x^2\right )^3 \left (\frac {b+a c+a d x^2}{c+d x^2}\right )^{3/2}}{6 a d^3}+\frac {b \left (b^2+4 a b c+8 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{16 a^{5/2} d^3} \]
1/6*(d*x^2+c)^3*((a*d*x^2+a*c+b)/(d*x^2+c))^(3/2)/a/d^3+1/16*b*(8*a^2*c^2+ 4*a*b*c+b^2)*arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^(1/2))/a^(5/2)/d^ 3-1/16*(-8*a^2*c^2+4*a*b*c+b^2)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2 )/a^2/d^3-1/8*(4*a*c+b)*(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a/d^ 3
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.67 \[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (-3 b^2+2 a b \left (-5 c+d x^2\right )+8 a^2 \left (c^2-c d x^2+d^2 x^4\right )\right )+3 b \left (b^2+4 a b c+8 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{48 a^{5/2} d^3} \]
(Sqrt[a]*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(-3*b^2 + 2*a*b *(-5*c + d*x^2) + 8*a^2*(c^2 - c*d*x^2 + d^2*x^4)) + 3*b*(b^2 + 4*a*b*c + 8*a^2*c^2)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(48*a^( 5/2)*d^3)
Time = 0.42 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.94, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2057, 2053, 2052, 27, 366, 27, 360, 25, 298, 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.
| \(\displaystyle \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int x^5 \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int x^4 \sqrt {\frac {a d x^2+b+a c}{d x^2+c}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int \frac {x^4 \left (-c x^4+b+a c\right )^2}{d^4 \left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b \int \frac {x^4 \left (-c x^4+b+a c\right )^2}{\left (a-x^4\right )^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{d^3}\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle -\frac {b \left (\frac {b^2 x^6}{6 a \left (a-x^4\right )^3}-\frac {\int \frac {3 x^4 \left (2 a c^2 x^4+b^2-2 (b+a c)^2\right )}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{6 a}\right )}{d^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b \left (\frac {b^2 x^6}{6 a \left (a-x^4\right )^3}-\frac {\int \frac {x^4 \left (2 a c^2 x^4+b^2-2 (b+a c)^2\right )}{\left (a-x^4\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{2 a}\right )}{d^3}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle -\frac {b \left (\frac {b^2 x^6}{6 a \left (a-x^4\right )^3}-\frac {-\frac {1}{4} \int -\frac {b (b+4 a c)-8 a c^2 x^4}{\left (a-x^4\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}-\frac {b (4 a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}}{2 a}\right )}{d^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {b \left (\frac {b^2 x^6}{6 a \left (a-x^4\right )^3}-\frac {\frac {1}{4} \int \frac {b (b+4 a c)-8 a c^2 x^4}{\left (a-x^4\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}-\frac {b (4 a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}}{2 a}\right )}{d^3}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle -\frac {b \left (\frac {b^2 x^6}{6 a \left (a-x^4\right )^3}-\frac {\frac {1}{4} \left (\frac {\left (8 a^2 c^2+4 a b c+b^2\right ) \int \frac {1}{a-x^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{2 a}+\frac {\left (-8 a^2 c^2+4 a b c+b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}\right )-\frac {b (4 a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}}{2 a}\right )}{d^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {b \left (\frac {b^2 x^6}{6 a \left (a-x^4\right )^3}-\frac {\frac {1}{4} \left (\frac {\left (-8 a^2 c^2+4 a b c+b^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 a \left (a-x^4\right )}+\frac {\left (8 a^2 c^2+4 a b c+b^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}\right )-\frac {b (4 a c+b) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 \left (a-x^4\right )^2}}{2 a}\right )}{d^3}\) |
-((b*((b^2*x^6)/(6*a*(a - x^4)^3) - (-1/4*(b*(b + 4*a*c)*Sqrt[(b + a*c + a *d*x^2)/(c + d*x^2)])/(a - x^4)^2 + (((b^2 + 4*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)) + ((b^2 + 4*a*b*c + 8*a^2*c ^2)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]])/(2*a^(3/2)))/4 )/(2*a)))/d^3)
3.4.18.3.1 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)^(-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[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -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 = 237, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(\frac {\left (8 a^{2} d^{2} x^{4}-8 a^{2} c d \,x^{2}+2 a b d \,x^{2}+8 a^{2} c^{2}-10 a b c -3 b^{2}\right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{48 d^{3} a^{2}}+\frac {b \left (8 a^{2} c^{2}+4 a b c +b^{2}\right ) \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{32 d^{2} a^{2} \sqrt {a \,d^{2}}\, \left (a d \,x^{2}+a c +b \right )}\) | \(237\) |
| default | \(\frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-48 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a^{2} c d \,x^{2}-12 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a b d \,x^{2}+24 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a^{2} b \,c^{2} d +12 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a \,b^{2} c d +16 \left (a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c \right )^{\frac {3}{2}} a \sqrt {a \,d^{2}}-36 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, a b c +3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{3} d -6 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\, b^{2}\right )}{96 d^{3} \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}\, a^{2} \sqrt {a \,d^{2}}}\) | \(533\) |
1/48/d^3*(8*a^2*d^2*x^4-8*a^2*c*d*x^2+2*a*b*d*x^2+8*a^2*c^2-10*a*b*c-3*b^2 )*(d*x^2+c)/a^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)+1/32*b/d^2*(8*a^2*c^2+4* a*b*c+b^2)/a^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)*((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)/(a*d*x^2+a*c+b)
Time = 0.33 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.96 \[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\left [\frac {3 \, {\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \sqrt {a} \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (8 \, a^{3} d^{3} x^{6} + 2 \, a^{2} b d^{2} x^{4} + 8 \, a^{3} c^{3} - 10 \, a^{2} b c^{2} - 3 \, a b^{2} c - {\left (8 \, a^{2} b c + 3 \, a b^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{192 \, a^{3} d^{3}}, -\frac {3 \, {\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - 2 \, {\left (8 \, a^{3} d^{3} x^{6} + 2 \, a^{2} b d^{2} x^{4} + 8 \, a^{3} c^{3} - 10 \, a^{2} b c^{2} - 3 \, a b^{2} c - {\left (8 \, a^{2} b c + 3 \, a b^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{96 \, a^{3} d^{3}}\right ] \]
[1/192*(3*(8*a^2*b*c^2 + 4*a*b^2*c + 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 + 2*a^2*b*d^2*x^4 + 8*a^3*c^3 - 10*a^2*b*c^2 - 3*a*b^2 *c - (8*a^2*b*c + 3*a*b^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/( a^3*d^3), -1/96*(3*(8*a^2*b*c^2 + 4*a*b^2*c + b^3)*sqrt(-a)*arctan(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 + 2*a^2*b*d^2*x^4 + 8*a^3*c^3 - 10 *a^2*b*c^2 - 3*a*b^2*c - (8*a^2*b*c + 3*a*b^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^3*d^3)]
\[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\int x^{5} \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.52 \[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=-\frac {3 \, {\left (8 \, a^{2} b c^{2} - 4 \, a b^{2} c - b^{3}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {5}{2}} - 8 \, {\left (6 \, a^{3} b c^{2} - a b^{3}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} + 3 \, {\left (8 \, a^{4} b c^{2} + 4 \, a^{3} b^{2} c + a^{2} b^{3}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{48 \, {\left (a^{5} d^{3} - \frac {3 \, {\left (a d x^{2} + a c + b\right )} a^{4} d^{3}}{d x^{2} + c} + \frac {3 \, {\left (a d x^{2} + a c + b\right )}^{2} a^{3} d^{3}}{{\left (d x^{2} + c\right )}^{2}} - \frac {{\left (a d x^{2} + a c + b\right )}^{3} a^{2} d^{3}}{{\left (d x^{2} + c\right )}^{3}}\right )}} - \frac {{\left (8 \, a^{2} c^{2} + 4 \, a b c + b^{2}\right )} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{32 \, a^{\frac {5}{2}} d^{3}} \]
-1/48*(3*(8*a^2*b*c^2 - 4*a*b^2*c - b^3)*((a*d*x^2 + a*c + b)/(d*x^2 + c)) ^(5/2) - 8*(6*a^3*b*c^2 - a*b^3)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3/2) + 3*(8*a^4*b*c^2 + 4*a^3*b^2*c + a^2*b^3)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^5*d^3 - 3*(a*d*x^2 + a*c + b)*a^4*d^3/(d*x^2 + c) + 3*(a*d*x^2 + a*c + b)^2*a^3*d^3/(d*x^2 + c)^2 - (a*d*x^2 + a*c + b)^3*a^2*d^3/(d*x^2 + c)^3) - 1/32*(8*a^2*c^2 + 4*a*b*c + 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^(5/2)*d^3)
Time = 0.39 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.01 \[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\frac {1}{96} \, {\left (2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{d} - \frac {4 \, a^{2} c d^{3} - a b d^{3}}{a^{2} d^{5}}\right )} + \frac {8 \, a^{2} c^{2} d^{2} - 10 \, a b c d^{2} - 3 \, b^{2} d^{2}}{a^{2} d^{5}}\right )} - \frac {3 \, {\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \log \left ({\left | 2 \, a c d + 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} \sqrt {a} {\left | d \right |} + b d \right |}\right )}{a^{\frac {5}{2}} d^{2} {\left | d \right |}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]
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/d - (4*a^2*c*d^3 - a*b*d^3)/(a^2*d^5)) + (8*a^2*c^2*d^2 - 10*a*b*c*d^2 - 3*b^2*d^2)/(a^2*d^5)) - 3*(8*a^2*b*c^2 + 4*a*b^2*c + b^3)*log(abs(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^(5/2)*d^2*abs(d)))*sgn(d*x^2 + c)
Timed out. \[ \int x^5 \sqrt {a+\frac {b}{c+d x^2}} \, dx=\int x^5\,\sqrt {a+\frac {b}{d\,x^2+c}} \,d x \]