3.3.96 \(\int \frac {x^5}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\) [296]

3.3.96.1 Optimal result
3.3.96.2 Mathematica [A] (verified)
3.3.96.3 Rubi [A] (warning: unable to verify)
3.3.96.4 Maple [A] (verified)
3.3.96.5 Fricas [A] (verification not implemented)
3.3.96.6 Sympy [F(-1)]
3.3.96.7 Maxima [F(-2)]
3.3.96.8 Giac [A] (verification not implemented)
3.3.96.9 Mupad [F(-1)]

3.3.96.1 Optimal result

Integrand size = 26, antiderivative size = 281 \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\frac {\left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{16 b^3 d^2 e}-\frac {(3 b c+5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 b^2 d^2 e}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^3 \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}{6 b d (b c-a d) e}+\frac {(b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{7/2} d^{5/2} \sqrt {e}} \]

output
1/16*(-a*d+b*c)*(5*a^2*d^2+2*a*b*c*d+b^2*c^2)*arctanh(d^(1/2)*(e*(b*x^2+a) 
/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/b^(7/2)/d^(5/2)/e^(1/2)+1/16*(5*a^2*d^2 
+2*a*b*c*d+b^2*c^2)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^3/d^2/e-1/24 
*(5*a*d+3*b*c)*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^2/d^2/e-1/6*(d* 
x^2+c)^3*(a-c*(b*x^2+a)/(d*x^2+c))*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b/d/(-a*d 
+b*c)/e
 
3.3.96.2 Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\frac {\sqrt {a+b x^2} \left (\sqrt {d} \sqrt {a+b x^2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \left (15 a^2 d^2-2 a b d \left (2 c+5 d x^2\right )+b^2 \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )\right )+3 \sqrt {b c-a d} \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )\right )}{48 b^3 d^{5/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}}} \]

input
Integrate[x^5/Sqrt[(e*(a + b*x^2))/(c + d*x^2)],x]
 
output
(Sqrt[a + b*x^2]*(Sqrt[d]*Sqrt[a + b*x^2]*Sqrt[(b*(c + d*x^2))/(b*c - a*d) 
]*(15*a^2*d^2 - 2*a*b*d*(2*c + 5*d*x^2) + b^2*(-3*c^2 + 2*c*d*x^2 + 8*d^2* 
x^4)) + 3*Sqrt[b*c - a*d]*(b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*ArcSinh[(Sqrt[ 
d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]]))/(48*b^3*d^(5/2)*Sqrt[(e*(a + b*x^2) 
)/(c + d*x^2)]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)])
 
3.3.96.3 Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2053, 2052, 315, 25, 27, 298, 215, 221}

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.

\(\displaystyle \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx\)

\(\Big \downarrow \) 2053

\(\displaystyle \frac {1}{2} \int \frac {x^4}{\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}dx^2\)

\(\Big \downarrow \) 2052

\(\displaystyle e (b c-a d) \int \frac {\left (a e-c x^4\right )^2}{\left (b e-d x^4\right )^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\)

\(\Big \downarrow \) 315

\(\displaystyle e (b c-a d) \left (-\frac {\int -\frac {e \left (a (b c+5 a d) e-3 c (b c+a d) x^4\right )}{\left (b e-d x^4\right )^3}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{6 b d e}-\frac {(b c-a d) \left (a e-c x^4\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d \left (b e-d x^4\right )^3}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle e (b c-a d) \left (\frac {\int \frac {e \left (a (b c+5 a d) e-3 c (b c+a d) x^4\right )}{\left (b e-d x^4\right )^3}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{6 b d e}-\frac {(b c-a d) \left (a e-c x^4\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d \left (b e-d x^4\right )^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle e (b c-a d) \left (\frac {\int \frac {a (b c+5 a d) e-3 c (b c+a d) x^4}{\left (b e-d x^4\right )^3}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{6 b d}-\frac {(b c-a d) \left (a e-c x^4\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d \left (b e-d x^4\right )^3}\right )\)

\(\Big \downarrow \) 298

\(\displaystyle e (b c-a d) \left (\frac {\frac {3}{4} \left (\frac {5 a^2 d}{b}+2 a c+\frac {b c^2}{d}\right ) \int \frac {1}{\left (b e-d x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}-\frac {(b c-a d) (5 a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b d \left (b e-d x^4\right )^2}}{6 b d}-\frac {(b c-a d) \left (a e-c x^4\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d \left (b e-d x^4\right )^3}\right )\)

\(\Big \downarrow \) 215

\(\displaystyle e (b c-a d) \left (\frac {\frac {3}{4} \left (\frac {5 a^2 d}{b}+2 a c+\frac {b c^2}{d}\right ) \left (\frac {\int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 b e}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}\right )-\frac {(b c-a d) (5 a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b d \left (b e-d x^4\right )^2}}{6 b d}-\frac {(b c-a d) \left (a e-c x^4\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d \left (b e-d x^4\right )^3}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle e (b c-a d) \left (\frac {\frac {3}{4} \left (\frac {5 a^2 d}{b}+2 a c+\frac {b c^2}{d}\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{3/2} \sqrt {d} e^{3/2}}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}\right )-\frac {(b c-a d) (5 a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b d \left (b e-d x^4\right )^2}}{6 b d}-\frac {(b c-a d) \left (a e-c x^4\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{6 b d \left (b e-d x^4\right )^3}\right )\)

input
Int[x^5/Sqrt[(e*(a + b*x^2))/(c + d*x^2)],x]
 
output
(b*c - a*d)*e*(-1/6*((b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(a*e - 
c*x^4))/(b*d*(b*e - d*x^4)^3) + (-1/4*((b*c - a*d)*(3*b*c + 5*a*d)*Sqrt[(e 
*(a + b*x^2))/(c + d*x^2)])/(b*d*(b*e - d*x^4)^2) + (3*(2*a*c + (b*c^2)/d 
+ (5*a^2*d)/b)*(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/(2*b*e*(b*e - d*x^4)) + 
ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])]/(2* 
b^(3/2)*Sqrt[d]*e^(3/2))))/4)/(6*b*d))
 

3.3.96.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 215
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])
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 298
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])
 

rule 315
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]
 

rule 2052
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]
 

rule 2053
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]]
 
3.3.96.4 Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.86

method result size
risch \(\frac {\left (8 b^{2} d^{2} x^{4}-10 a b \,d^{2} x^{2}+2 b^{2} c d \,x^{2}+15 a^{2} d^{2}-4 a b c d -3 b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{48 b^{3} d^{2} \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}-\frac {\left (5 a^{3} d^{3}-3 a^{2} b c \,d^{2}-d \,c^{2} b^{2} a -b^{3} c^{3}\right ) \ln \left (\frac {\frac {1}{2} e d a +\frac {1}{2} e b c +b d e \,x^{2}}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (e d a +e b c \right ) x^{2}+a c e}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{32 b^{3} d^{2} \sqrt {b d e}\, \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\) \(243\)
default \(\frac {\left (b \,x^{2}+a \right ) \left (-36 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a b \,d^{2} x^{2}-12 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b^{2} c d \,x^{2}-15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}+9 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+16 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} b d \sqrt {b d}+30 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} d^{2}-24 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a b c d -6 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b^{2} c^{2}\right )}{96 \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b^{3} d^{2} \sqrt {b d}}\) \(527\)

input
int(x^5/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
1/48*(8*b^2*d^2*x^4-10*a*b*d^2*x^2+2*b^2*c*d*x^2+15*a^2*d^2-4*a*b*c*d-3*b^ 
2*c^2)*(b*x^2+a)/b^3/d^2/(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/32*(5*a^3*d^3-3*a 
^2*b*c*d^2-a*b^2*c^2*d-b^3*c^3)/b^3/d^2*ln((1/2*e*d*a+1/2*e*b*c+b*d*e*x^2) 
/(b*d*e)^(1/2)+(b*d*e*x^4+(a*d*e+b*c*e)*x^2+a*c*e)^(1/2))/(b*d*e)^(1/2)/(e 
*(b*x^2+a)/(d*x^2+c))^(1/2)*((d*x^2+c)*e*(b*x^2+a))^(1/2)/(d*x^2+c)
 
3.3.96.5 Fricas [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 545, normalized size of antiderivative = 1.94 \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\left [-\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b d e} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} - {\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{192 \, b^{4} d^{3} e}, -\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-b d e} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{4} x^{6} - 3 \, b^{3} c^{3} d - 4 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} + 10 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )} x^{4} - {\left (b^{3} c^{2} d^{2} + 14 \, a b^{2} c d^{3} - 15 \, a^{2} b d^{4}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{96 \, b^{4} d^{3} e}\right ] \]

input
integrate(x^5/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="fricas")
 
output
[-1/192*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(b*d*e) 
*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d 
+ a^2*d^2)*e - 4*(2*b*d^2*x^4 + b*c^2 + a*c*d + (3*b*c*d + a*d^2)*x^2)*sqr 
t(b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))) - 4*(8*b^3*d^4*x^6 - 3*b^3*c^3 
*d - 4*a*b^2*c^2*d^2 + 15*a^2*b*c*d^3 + 10*(b^3*c*d^3 - a*b^2*d^4)*x^4 - ( 
b^3*c^2*d^2 + 14*a*b^2*c*d^3 - 15*a^2*b*d^4)*x^2)*sqrt((b*e*x^2 + a*e)/(d* 
x^2 + c)))/(b^4*d^3*e), -1/96*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 
5*a^3*d^3)*sqrt(-b*d*e)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*d*e)*sq 
rt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*d*e*x^2 + a*b*d*e)) - 2*(8*b^3*d^4*x^ 
6 - 3*b^3*c^3*d - 4*a*b^2*c^2*d^2 + 15*a^2*b*c*d^3 + 10*(b^3*c*d^3 - a*b^2 
*d^4)*x^4 - (b^3*c^2*d^2 + 14*a*b^2*c*d^3 - 15*a^2*b*d^4)*x^2)*sqrt((b*e*x 
^2 + a*e)/(d*x^2 + c)))/(b^4*d^3*e)]
 
3.3.96.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\text {Timed out} \]

input
integrate(x**5/(e*(b*x**2+a)/(d*x**2+c))**(1/2),x)
 
output
Timed out
 
3.3.96.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\text {Exception raised: ValueError} \]

input
integrate(x^5/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 
3.3.96.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.84 \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\frac {2 \, \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{b e} + \frac {b^{2} c d e - 5 \, a b d^{2} e}{b^{3} d^{2} e^{2}}\right )} - \frac {3 \, b^{2} c^{2} e + 4 \, a b c d e - 15 \, a^{2} d^{2} e}{b^{3} d^{2} e^{2}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -b c e - a d e - 2 \, \sqrt {b d e} {\left (\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}\right )} \right |}\right )}{\sqrt {b d e} b^{3} d^{2}}}{96 \, \mathrm {sgn}\left (d x^{2} + c\right )} \]

input
integrate(x^5/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="giac")
 
output
1/96*(2*sqrt(b*d*e*x^4 + b*c*e*x^2 + a*d*e*x^2 + a*c*e)*(2*x^2*(4*x^2/(b*e 
) + (b^2*c*d*e - 5*a*b*d^2*e)/(b^3*d^2*e^2)) - (3*b^2*c^2*e + 4*a*b*c*d*e 
- 15*a^2*d^2*e)/(b^3*d^2*e^2)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 
- 5*a^3*d^3)*log(abs(-b*c*e - a*d*e - 2*sqrt(b*d*e)*(sqrt(b*d*e)*x^2 - sqr 
t(b*d*e*x^4 + b*c*e*x^2 + a*d*e*x^2 + a*c*e))))/(sqrt(b*d*e)*b^3*d^2))/sgn 
(d*x^2 + c)
 
3.3.96.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\int \frac {x^5}{\sqrt {\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}}} \,d x \]

input
int(x^5/((e*(a + b*x^2))/(c + d*x^2))^(1/2),x)
 
output
int(x^5/((e*(a + b*x^2))/(c + d*x^2))^(1/2), x)