\(\int x^3 (\frac {e (a+b x^2)}{c+d x^2})^{3/2} \, dx\) [56]

Optimal result

Integrand size = 26, antiderivative size = 247 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=-\frac {c (b c-a d) e \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{d^3}-\frac {(9 b c-5 a d) e \left (c+d x^2\right ) \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{8 d^3}+\frac {b e \left (c+d x^2\right )^2 \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{4 d^3}+\frac {3 (b c-a d) (5 b c-a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}{\sqrt {b} \sqrt {e}}\right )}{8 \sqrt {b} d^{7/2}} \] Output:

-c*(-a*d+b*c)*e*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/d^3-1/8*(-5*a*d+9*b 
*c)*e*(d*x^2+c)*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/d^3+1/4*b*e*(d*x^2+ 
c)^2*(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)/d^3+3/8*(-a*d+b*c)*(-a*d+5*b*c 
)*e^(3/2)*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^(1/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 2.96 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.72 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {b} \sqrt {d} \sqrt {a+b x^2} \left (a d \left (13 c+5 d x^2\right )+b \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )\right )+3 \left (5 b^2 c^2-6 a b c d+a^2 d^2\right ) \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )\right )}{8 \sqrt {b} d^{7/2} \sqrt {a+b x^2}} \] Input:

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

Output:

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b]*Sqrt[d]*Sqrt[a + b*x^2]*(a*d 
*(13*c + 5*d*x^2) + b*(-15*c^2 - 5*c*d*x^2 + 2*d^2*x^4)) + 3*(5*b^2*c^2 - 
6*a*b*c*d + a^2*d^2)*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sq 
rt[d]*Sqrt[a + b*x^2])]))/(8*Sqrt[b]*d^(7/2)*Sqrt[a + b*x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2053, 2052, 25, 360, 1471, 27, 299, 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.

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

\(\Big \downarrow \) 2053

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

\(\Big \downarrow \) 2052

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 1471

\(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}-\frac {\int \frac {b e \left (8 c d x^4+(7 b c-3 a d) e\right )}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 b e}}{4 d^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}-\frac {1}{2} \int \frac {8 c d x^4+(7 b c-3 a d) e}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{4 d^3}\right )\)

\(\Big \downarrow \) 299

\(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {1}{2} \left (8 c \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}-3 e (5 b c-a d) \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\right )+\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}}{4 d^3}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle e (b c-a d) \left (\frac {b e^2 (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3 \left (b e-d x^4\right )^2}-\frac {\frac {1}{2} \left (8 c \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}-\frac {3 \sqrt {e} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{\sqrt {b} \sqrt {d}}\right )+\frac {e (9 b c-5 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 \left (b e-d x^4\right )}}{4 d^3}\right )\)

Input:

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

Output:

(b*c - a*d)*e*((b*(b*c - a*d)*e^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*d^ 
3*(b*e - d*x^4)^2) - (((9*b*c - 5*a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)] 
)/(2*(b*e - d*x^4)) + (8*c*Sqrt[(e*(a + b*x^2))/(c + d*x^2)] - (3*(5*b*c - 
 a*d)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b] 
*Sqrt[e])])/(Sqrt[b]*Sqrt[d]))/2)/(4*d^3))
 

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 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 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -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 = 0.20 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {\left (2 b d \,x^{2}+5 a d -7 b c \right ) \left (d \,x^{2}+c \right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{8 d^{3}}+\frac {\left (\frac {3 \left (a d -5 b c \right ) \left (a d -b c \right ) \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +b d \,x^{2} e}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right )}{2 \sqrt {b d e}}+\frac {8 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right ) e}}{8 d^{3} \left (b \,x^{2}+a \right )}\)	\(252\)
default	\(\frac {\left (4 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,d^{2} x^{4}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{3} x^{2}-18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c \,d^{2} x^{2}+15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2} d \,x^{2}+10 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a \,d^{2} x^{2}-10 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b c d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} c \,d^{2}-18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,c^{2} d +15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{3}+16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a c d -16 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b \,c^{2}+10 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a c d -14 \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b \,c^{2}\right ) \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}{16 d^{3} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right )}\)	\(679\)

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

Fricas [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.69 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt {\frac {e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} + {\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {e}{b d}}\right ) + 4 \, {\left (2 \, b d^{2} e x^{4} - 5 \, {\left (b c d - a d^{2}\right )} e x^{2} - {\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{32 \, d^{3}}, -\frac {3 \, {\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt {-\frac {e}{b d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {e}{b d}}}{2 \, {\left (b e x^{2} + a e\right )}}\right ) - 2 \, {\left (2 \, b d^{2} e x^{4} - 5 \, {\left (b c d - a d^{2}\right )} e x^{2} - {\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{16 \, d^{3}}\right ] \] Input:

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

Output:

[1/32*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*e*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)*sqr 
t((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) + 4*(2*b*d^2*e*x^4 - 5*(b*c* 
d - a*d^2)*e*x^2 - (15*b*c^2 - 13*a*c*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + 
c)))/d^3, -1/16*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*e*sqrt(-e/(b*d))*arct 
an(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*(2*b*d^2*e*x^4 - 5*(b*c*d - a*d^2)*e*x^2 - (15* 
b*c^2 - 13*a*c*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^3]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

Giac [F(-2)]

Exception generated. \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] 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,[0,4,0]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-4,[0,3,0]%%%} 
,0]:[1,0,
 

Mupad [F(-1)]

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

Reduce [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.63 \[ \int x^3 \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\frac {\sqrt {e}\, e \left (13 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a b c \,d^{2}+5 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a b \,d^{3} x^{2}-15 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{2} c^{2} d -5 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{2} c \,d^{2} x^{2}+2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, b^{2} d^{3} x^{4}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{2} c \,d^{2}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a^{2} d^{3} x^{2}-18 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a b \,c^{2} d -18 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) a b c \,d^{2} x^{2}+15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{2} c^{3}+15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (-\sqrt {b}\, \sqrt {b \,x^{2}+a}\, d -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b \right ) b^{2} c^{2} d \,x^{2}\right )}{8 b \,d^{4} \left (d \,x^{2}+c \right )} \] Input:

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

Output:

(sqrt(e)*e*(13*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b*c*d**2 + 5*sqrt(c + d 
*x**2)*sqrt(a + b*x**2)*a*b*d**3*x**2 - 15*sqrt(c + d*x**2)*sqrt(a + b*x** 
2)*b**2*c**2*d - 5*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**2*c*d**2*x**2 + 2* 
sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**2*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**2*c*d**2 + 3 
*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d - sqrt(d)*sqrt(c + d*x* 
*2)*b)*a**2*d**3*x**2 - 18*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2) 
*d - sqrt(d)*sqrt(c + d*x**2)*b)*a*b*c**2*d - 18*sqrt(d)*sqrt(b)*log( - sq 
rt(b)*sqrt(a + b*x**2)*d - sqrt(d)*sqrt(c + d*x**2)*b)*a*b*c*d**2*x**2 + 1 
5*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d - sqrt(d)*sqrt(c + d*x 
**2)*b)*b**2*c**3 + 15*sqrt(d)*sqrt(b)*log( - sqrt(b)*sqrt(a + b*x**2)*d - 
 sqrt(d)*sqrt(c + d*x**2)*b)*b**2*c**2*d*x**2))/(8*b*d**4*(c + d*x**2))