\(\int x^5 \sqrt {\frac {e (a+b x^2)}{c+d x^2}} \, dx\) [42]

Optimal result

Integrand size = 26, antiderivative size = 287 \[ \int x^5 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {\left (11 b^2 c^2-2 a b c d-a^2 d^2\right ) \left (c+d x^2\right ) \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{16 b^2 d^3}-\frac {(13 b c-a d) \left (c+d x^2\right )^2 \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{24 b d^3}+\frac {\left (c+d x^2\right )^3 \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{6 d^3}-\frac {(b c-a d) \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{\sqrt {b} \sqrt {e}}\right )}{16 b^{5/2} d^{7/2}} \] Output:

1/16*(-a^2*d^2-2*a*b*c*d+11*b^2*c^2)*(d*x^2+c)*(b*e/d-(-a*d+b*c)*e/d/(d*x^ 
2+c))^(1/2)/b^2/d^3-1/24*(-a*d+13*b*c)*(d*x^2+c)^2*(b*e/d-(-a*d+b*c)*e/d/( 
d*x^2+c))^(1/2)/b/d^3+1/6*(d*x^2+c)^3*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/ 
2)/d^3-1/16*(-a*d+b*c)*(a^2*d^2+2*a*b*c*d+5*b^2*c^2)*e^(1/2)*arctanh(d^(1/ 
2)*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/b^(5/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.69 \[ \int x^5 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (-b \sqrt {d} \left (c+d x^2\right ) \left (3 a^2 d^2-2 a b d \left (-2 c+d x^2\right )+b^2 \left (-15 c^2+10 c d x^2-8 d^2 x^4\right )\right )-\frac {3 (b c-a d)^{3/2} \left (5 b^2 c^2+2 a b c d+a^2 d^2\right ) \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {a+b x^2}}\right )}{48 b^3 d^{7/2}} \] Input:

Integrate[x^5*Sqrt[(e*(a + b*x^2))/(c + d*x^2)],x]
 

Output:

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(-(b*Sqrt[d]*(c + d*x^2)*(3*a^2*d^2 - 2 
*a*b*d*(-2*c + d*x^2) + b^2*(-15*c^2 + 10*c*d*x^2 - 8*d^2*x^4))) - (3*(b*c 
 - a*d)^(3/2)*(5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*Sqrt[(b*(c + d*x^2))/(b*c 
- a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/Sqrt[a + b*x^2 
]))/(48*b^3*d^(7/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.81 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2053, 2052, 366, 27, 360, 27, 298, 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.

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

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 366

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 360

\(\displaystyle e (b c-a d) \left (\frac {e x^6 (b c-a d)^2}{6 b d^2 \left (b e-d x^4\right )^3}-\frac {\frac {e (b c-a d) (a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d \left (b e-d x^4\right )^2}-\frac {\int \frac {d \left (8 b c^2 d x^4+(b c-a d) (3 b c+a d) e\right )}{\left (b e-d x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{4 d^2}}{2 b d^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle e (b c-a d) \left (\frac {e x^6 (b c-a d)^2}{6 b d^2 \left (b e-d x^4\right )^3}-\frac {\frac {e (b c-a d) (a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d \left (b e-d x^4\right )^2}-\frac {\int \frac {8 b c^2 d x^4+(b c-a d) (3 b c+a d) e}{\left (b e-d x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{4 d}}{2 b d^2}\right )\)

\(\Big \downarrow \) 298

\(\displaystyle e (b c-a d) \left (\frac {e x^6 (b c-a d)^2}{6 b d^2 \left (b e-d x^4\right )^3}-\frac {\frac {e (b c-a d) (a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d \left (b e-d x^4\right )^2}-\frac {\frac {\left (-a^2 d^2-2 a b c d+11 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b \left (b e-d x^4\right )}-\frac {\left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 b}}{4 d}}{2 b d^2}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle e (b c-a d) \left (\frac {e x^6 (b c-a d)^2}{6 b d^2 \left (b e-d x^4\right )^3}-\frac {\frac {e (b c-a d) (a d+3 b c) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d \left (b e-d x^4\right )^2}-\frac {\frac {\left (-a^2 d^2-2 a b c d+11 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b \left (b e-d x^4\right )}-\frac {\left (a^2 d^2+2 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{3/2} \sqrt {d} \sqrt {e}}}{4 d}}{2 b d^2}\right )\)

Input:

Int[x^5*Sqrt[(e*(a + b*x^2))/(c + d*x^2)],x]
 

Output:

(b*c - a*d)*e*(((b*c - a*d)^2*e*x^6)/(6*b*d^2*(b*e - d*x^4)^3) - (((b*c - 
a*d)*(3*b*c + a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*d*(b*e - d*x^4) 
^2) - (((11*b^2*c^2 - 2*a*b*c*d - a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2 
)])/(2*b*(b*e - d*x^4)) - ((5*b^2*c^2 + 2*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt 
[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(2*b^(3/2)*Sqrt 
[d]*Sqrt[e]))/(4*d))/(2*b*d^2))
 

Defintions of rubi rules used

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 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 360

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])
 

rule 366

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]
 

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]]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {\left (-8 b^{2} d^{2} x^{4}-2 a b \,d^{2} x^{2}+10 b^{2} c \,x^{2} d +3 a^{2} d^{2}+4 a b c d -15 b^{2} c^{2}\right ) \left (d \,x^{2}+c \right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{48 b^{2} d^{3}}+\frac {\left (a^{3} d^{3}+a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +b d \,x^{2} e}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right ) e}}{32 b^{2} d^{3} \sqrt {b d e}\, \left (b \,x^{2}+a \right )}\)	\(241\)
default	\(\frac {\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (-12 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b \,d^{2} x^{2}-36 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}+9 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d -15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+16 \left (d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} b d \sqrt {b d}-6 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} d^{2}-24 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b c d +30 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c^{2}\right )}{96 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, d^{3} b^{2} \sqrt {b d}}\)	\(527\)

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^4-2*a*b*d^2*x^2+10*b^2*c*d*x^2+3*a^2*d^2+4*a*b*c*d-15* 
b^2*c^2)*(d*x^2+c)/b^2/d^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2)+1/32*(a^3*d^3+a^2 
*b*c*d^2+3*a*b^2*c^2*d-5*b^3*c^3)/b^2/d^3*ln((1/2*a*d*e+1/2*b*c*e+b*d*x^2* 
e)/(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)*(b*x^2+a)*e)^(1/2)/(b*x^2+a)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 541, normalized size of antiderivative = 1.89 \[ \int x^5 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) - 4 \, {\left (8 \, b^{2} d^{3} x^{6} + 15 \, b^{2} c^{3} - 4 \, a b c^{2} d - 3 \, a^{2} c d^{2} - 2 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{4} + {\left (5 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{192 \, b^{2} d^{3}}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) + 2 \, {\left (8 \, b^{2} d^{3} x^{6} + 15 \, b^{2} c^{3} - 4 \, a b c^{2} d - 3 \, a^{2} c d^{2} - 2 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{4} + {\left (5 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{96 \, b^{2} d^{3}}\right ] \] Input:

integrate(x^5*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="fricas")
 

Output:

[-1/192*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 - a^3*d^3)*sqrt(e/(b*d 
))*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^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*c*d^2 + 
 a*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) - 4*(8*b^2 
*d^3*x^6 + 15*b^2*c^3 - 4*a*b*c^2*d - 3*a^2*c*d^2 - 2*(b^2*c*d^2 - a*b*d^3 
)*x^4 + (5*b^2*c^2*d - 2*a*b*c*d^2 - 3*a^2*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/ 
(d*x^2 + c)))/(b^2*d^3), 1/96*(3*(5*b^3*c^3 - 3*a*b^2*c^2*d - a^2*b*c*d^2 
- a^3*d^3)*sqrt(-e/(b*d))*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 
 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^2 + a*e)) + 2*(8*b^2*d^3*x^6 + 
15*b^2*c^3 - 4*a*b*c^2*d - 3*a^2*c*d^2 - 2*(b^2*c*d^2 - a*b*d^3)*x^4 + (5* 
b^2*c^2*d - 2*a*b*c*d^2 - 3*a^2*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c) 
))/(b^2*d^3)]
 

Sympy [F(-1)]

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

integrate(x**5*(e*(b*x**2+a)/(d*x**2+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.79 \[ \int x^5 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {1}{96} \, {\left (2 \, \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{d} - \frac {5 \, b^{2} c d - a b d^{2}}{b^{2} d^{3}}\right )} + \frac {15 \, b^{2} c^{2} - 4 \, a b c d - 3 \, a^{2} d^{2}}{b^{2} d^{3}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} e - 3 \, a b^{2} c^{2} d e - a^{2} b c d^{2} e - a^{3} d^{3} e\right )} \log \left ({\left | -b c e - a d e - 2 \, \sqrt {b d e} {\left (\sqrt {b d e} x^{2} - \sqrt {b d e x^{4} + b c e x^{2} + a d e x^{2} + a c e}\right )} \right |}\right )}{\sqrt {b d e} b^{2} d^{3}}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \] 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/d - 
(5*b^2*c*d - a*b*d^2)/(b^2*d^3)) + (15*b^2*c^2 - 4*a*b*c*d - 3*a^2*d^2)/(b 
^2*d^3)) + 3*(5*b^3*c^3*e - 3*a*b^2*c^2*d*e - a^2*b*c*d^2*e - a^3*d^3*e)*l 
og(abs(-b*c*e - a*d*e - 2*sqrt(b*d*e)*(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))))/(sqrt(b*d*e)*b^2*d^3))*sgn(d*x^2 + c)
 

Mupad [F(-1)]

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

Reduce [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.16 \[ \int x^5 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx=\frac {\sqrt {e}\, \left (-3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3}-4 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2}+2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{3} x^{2}+15 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} c^{2} d -10 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} c \,d^{2} x^{2}+8 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{3} d^{3} x^{4}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{3} d^{3}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{2} b c \,d^{2}+9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a \,b^{2} c^{2} d -15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{3} c^{3}\right )}{48 b^{3} d^{4}} \] Input:

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

Output:

(sqrt(e)*( - 3*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a**2*b*d**3 - 4*sqrt(c + 
d*x**2)*sqrt(a + b*x**2)*a*b**2*c*d**2 + 2*sqrt(c + d*x**2)*sqrt(a + b*x** 
2)*a*b**2*d**3*x**2 + 15*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**3*c**2*d - 1 
0*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**3*c*d**2*x**2 + 8*sqrt(c + d*x**2)* 
sqrt(a + b*x**2)*b**3*d**3*x**4 + 3*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a 
+ b*x**2)*d - sqrt(d)*sqrt(c + d*x**2)*b)*a**3*d**3 + 3*sqrt(d)*sqrt(b)*lo 
g( - sqrt(b)*sqrt(a + b*x**2)*d - sqrt(d)*sqrt(c + d*x**2)*b)*a**2*b*c*d** 
2 + 9*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d - sqrt(d)*sqrt(c + 
 d*x**2)*b)*a*b**2*c**2*d - 15*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x 
**2)*d - sqrt(d)*sqrt(c + d*x**2)*b)*b**3*c**3))/(48*b**3*d**4)