3.4.8 \(\int \frac {x^3}{(\frac {e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\) [308]

3.4.8.1 Optimal result

Integrand size = 26, antiderivative size = 202 \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {a (b c-a d)}{b^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^3 e^2}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b^2 e^2}+\frac {3 (b c-5 a d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{7/2} \sqrt {d} e^{3/2}} \]

output

3/8*(-5*a*d+b*c)*(-a*d+b*c)*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/ 
b^(1/2)/e^(1/2))/b^(7/2)/e^(3/2)/d^(1/2)+a*(-a*d+b*c)/b^3/e/(e*(b*x^2+a)/( 
d*x^2+c))^(1/2)+1/8*(-7*a*d+3*b*c)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2) 
/b^3/e^2+1/4*(d*x^2+c)^2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^2/e^2
 

3.4.8.2 Mathematica [A] (verified)

Time = 2.61 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {\sqrt {b} \sqrt {d} \sqrt {c+d x^2} \left (-15 a^2 d+a b \left (13 c-5 d x^2\right )+b^2 x^2 \left (5 c+2 d x^2\right )\right )+3 \left (b^2 c^2-6 a b c d+5 a^2 d^2\right ) \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{7/2} \sqrt {d} e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}} \]

input

Integrate[x^3/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]
 

output

(Sqrt[b]*Sqrt[d]*Sqrt[c + d*x^2]*(-15*a^2*d + a*b*(13*c - 5*d*x^2) + b^2*x 
^2*(5*c + 2*d*x^2)) + 3*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*Sqrt[a + b*x^2]* 
ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(8*b^(7/2)*S 
qrt[d]*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[c + d*x^2])
 

3.4.8.3 Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {2053, 2052, 25, 361, 25, 27, 361, 25, 27, 359, 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.

\(\displaystyle \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2053

\(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (\frac {e \left (b x^2+a\right )}{d x^2+c}\right )^{3/2}}dx^2\)

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 361

\(\displaystyle e (b c-a d) \left (\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e \left (b e-d x^4\right )^2}-\frac {-\frac {1}{2} \int -\frac {8 a e-\left (3 c-\frac {7 a d}{b}\right ) x^4}{e x^4 \left (b e-d x^4\right )}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}-\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}}{4 b^2 e}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle e (b c-a d) \left (\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e \left (b e-d x^4\right )^2}-\frac {\frac {1}{2} \int \frac {8 a e-\left (3 c-\frac {7 a d}{b}\right ) x^4}{e x^4 \left (b e-d x^4\right )}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}-\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}}{4 b^2 e}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle e (b c-a d) \left (\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e \left (b e-d x^4\right )^2}-\frac {\frac {\int \frac {8 a e-\left (3 c-\frac {7 a d}{b}\right ) x^4}{x^4 \left (b e-d x^4\right )}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 e}-\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}}{4 b^2 e}\right )\)

\(\Big \downarrow \) 359

\(\displaystyle e (b c-a d) \left (\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e \left (b e-d x^4\right )^2}-\frac {\frac {-\frac {3 (b c-5 a d) \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{b}-\frac {8 a}{b x^2}}{2 e}-\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}}{4 b^2 e}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle e (b c-a d) \left (\frac {(b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e \left (b e-d x^4\right )^2}-\frac {\frac {-\frac {3 (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{b^{3/2} \sqrt {d} \sqrt {e}}-\frac {8 a}{b x^2}}{2 e}-\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e \left (b e-d x^4\right )}}{4 b^2 e}\right )\)

input

Int[x^3/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]
 

output

(b*c - a*d)*e*(((b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*b^2*e*(b 
*e - d*x^4)^2) - (-1/2*((3*b*c - 7*a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) 
/(b*e*(b*e - d*x^4)) + ((-8*a)/(b*x^2) - (3*(b*c - 5*a*d)*ArcTanh[(Sqrt[d] 
*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(b^(3/2)*Sqrt[d]*S 
qrt[e]))/(2*e))/(4*b^2*e))
 

3.4.8.3.1 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 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 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 361

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[x^m*(a + b*x^2)^(p + 1)*E 
xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c 
- a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], 
 x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 
2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

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]]
 

3.4.8.4 Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.27

method	result	size
risch	\(-\frac {\left (-2 b d \,x^{2}+7 a d -5 b c \right ) \left (b \,x^{2}+a \right )}{8 b^{3} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (\frac {3 \left (5 a d -b c \right ) \left (a d -b c \right ) \ln \left (\frac {\frac {1}{2} e d a +\frac {1}{2} e b c +b d e \,x^{2}}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (e d a +e b c \right ) x^{2}+a c e}\right )}{2 \sqrt {b d e}}-\frac {8 a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (d \,x^{2}+c \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{8 b^{3} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(257\)
default	\(-\frac {\left (-4 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b^{2} d \,x^{4}-15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{2} x^{2}+18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2}+10 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a b d \,x^{2}-10 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b^{2} c \,x^{2}-15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{2}+18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c d -3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2}+16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a^{2} d -16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a b c +14 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a^{2} d -10 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a b c \right ) \left (b \,x^{2}+a \right )}{16 b^{3} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}\)	\(679\)

input

int(x^3/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x,method=_RETURNVERBOSE)
 

output

-1/8*(-2*b*d*x^2+7*a*d-5*b*c)*(b*x^2+a)/b^3/e/(e*(b*x^2+a)/(d*x^2+c))^(1/2 
)+1/8/b^3*(3/2*(5*a*d-b*c)*(a*d-b*c)*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)-8*a*( 
a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x^2+c)/(a*d-b*c)/(b*d*e*x^4+a*d*e*x^2+b*c*e* 
x^2+a*c*e)^(1/2))/e/(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)
 

3.4.8.5 Fricas [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 585, normalized size of antiderivative = 2.90 \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {b d e} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) + 4 \, {\left (2 \, b^{3} d^{3} x^{6} + 13 \, a b^{2} c^{2} d - 15 \, a^{2} b c d^{2} + {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{4} + {\left (5 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, {\left (b^{5} d e^{2} x^{2} + a b^{4} d e^{2}\right )}}, -\frac {3 \, {\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-b d e} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \, {\left (2 \, b^{3} d^{3} x^{6} + 13 \, a b^{2} c^{2} d - 15 \, a^{2} b c d^{2} + {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{4} + {\left (5 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, {\left (b^{5} d e^{2} x^{2} + a b^{4} d e^{2}\right )}}\right ] \]

input

integrate(x^3/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")
 

output

[1/32*(3*(a*b^2*c^2 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5 
*a^2*b*d^2)*x^2)*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)*sqrt(b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))) + 
4*(2*b^3*d^3*x^6 + 13*a*b^2*c^2*d - 15*a^2*b*c*d^2 + (7*b^3*c*d^2 - 5*a*b^ 
2*d^3)*x^4 + (5*b^3*c^2*d + 8*a*b^2*c*d^2 - 15*a^2*b*d^3)*x^2)*sqrt((b*e*x 
^2 + a*e)/(d*x^2 + c)))/(b^5*d*e^2*x^2 + a*b^4*d*e^2), -1/16*(3*(a*b^2*c^2 
 - 6*a^2*b*c*d + 5*a^3*d^2 + (b^3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*x^2)*sq 
rt(-b*d*e)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*d*e)*sqrt((b*e*x^2 + 
 a*e)/(d*x^2 + c))/(b^2*d*e*x^2 + a*b*d*e)) - 2*(2*b^3*d^3*x^6 + 13*a*b^2* 
c^2*d - 15*a^2*b*c*d^2 + (7*b^3*c*d^2 - 5*a*b^2*d^3)*x^4 + (5*b^3*c^2*d + 
8*a*b^2*c*d^2 - 15*a^2*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^5 
*d*e^2*x^2 + a*b^4*d*e^2)]
 

3.4.8.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Timed out} \]

input

integrate(x**3/(e*(b*x**2+a)/(d*x**2+c))**(3/2),x)
 

output

Timed out
 

3.4.8.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

input

integrate(x^3/(e*(b*x^2+a)/(d*x^2+c))^(3/2),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
 

3.4.8.8 Giac [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(x^3/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{2,[1,0,0]%%%},[2,1,0]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1 
,1,1]%%%}
 

3.4.8.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {x^3}{{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}} \,d x \]

input

int(x^3/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x)
 

output

int(x^3/((e*(a + b*x^2))/(c + d*x^2))^(3/2), x)