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

Optimal. Leaf size=202 \[ \frac {a (b c-a d)}{b^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {(3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^3 e^2}+\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b^2 e^2}+\frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{8 b^{7/2} \sqrt {d} e^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1981, 1980, 467, 464, 214} \begin {gather*} \frac {3 (b c-5 a d) (b c-a d) \tanh ^{-1}\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}}+\frac {\left (c+d x^2\right ) (3 b c-7 a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{8 b^3 e^2}+\frac {a (b c-a d)}{b^3 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {\left (c+d x^2\right )^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 467

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] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[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 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[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 1981

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Dist[1/n, Sub
st[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]]

Rubi steps

\begin {align*} \int \frac {x^3}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \text {Subst}\left (\int \frac {-a e+c x^2}{x^2 \left (b e-d x^2\right )^3} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\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 {1}{4} ((b c-a d) e) \text {Subst}\left (\int \frac {\frac {4 a}{b}-\frac {3 (b c-a d) x^2}{b^2 e}}{x^2 \left (b e-d x^2\right )^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\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 {1}{8} ((b c-a d) e) \text {Subst}\left (\int \frac {-\frac {8 a}{b^2 e}+\frac {(3 b c-7 a d) x^2}{b^3 e^2}}{x^2 \left (b e-d x^2\right )} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\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 {Subst}\left (\int \frac {1}{b e-d x^2} \, dx,x,\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 b^3 e}\\ &=\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) \tanh ^{-1}\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}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 2.21, size = 179, normalized size = 0.89 \begin {gather*} \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} \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(678\) vs. \(2(176)=352\).
time = 0.11, size = 679, normalized size = 3.36

method	result	size
risch	\(-\frac {\left (-2 b d \,x^{2}+7 a d -5 b c \right ) \left (b \,x^{2}+a \right )}{8 b^{3} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}+\frac {\left (\frac {15 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a^{2} d^{2}}{16 b^{3} \sqrt {d e b}}-\frac {9 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) a c d}{8 b^{2} \sqrt {d e b}}+\frac {3 \ln \left (\frac {\frac {1}{2} a d e +\frac {1}{2} b c e +d e b \,x^{2}}{\sqrt {d e b}}+\sqrt {d e b \,x^{4}+\left (a d e +b c e \right ) x^{2}+a c e}\right ) c^{2}}{16 b \sqrt {d e b}}-\frac {a^{3} x^{2} d^{3}}{b^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 a^{2} x^{2} c \,d^{2}}{b^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {a \,x^{2} c^{2} d}{b \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {a^{3} c \,d^{2}}{b^{3} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}+\frac {2 a^{2} c^{2} d}{b^{2} \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}-\frac {a \,c^{3}}{b \left (a d -b c \right ) \sqrt {d e b \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\)	\(615\)
default	\(-\frac {\left (-4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} d \,x^{4}-15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{2} x^{2}+18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c d \,x^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{2} x^{2}+10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b d \,x^{2}-10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, b^{2} c \,x^{2}-15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{2}+18 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c d -3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2}+14 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a^{2} d -10 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, a b c +16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a^{2} d -16 \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a b c \right ) \left (b \,x^{2}+a \right )}{16 b^{3} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}\)	\(679\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^2*d*x^4-15*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*d^2*x^2+18*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x
^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c*d*x^2-3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a
*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^2*x^2+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b
*d*x^2-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^2*c*x^2-15*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*d^2+18*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c*d-3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c^2+14*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a^2*d-10*(b*d*x
^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*c+16*(b*d)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a^2*d-16*(b*d)^(1/2
)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*b*c)/b^3*(b*x^2+a)/(b*d)^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/(d*x^2+c)/(e*(b*x^2
+a)/(d*x^2+c))^(3/2)

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 281, normalized size = 1.39 \begin {gather*} \frac {1}{16} \, {\left (\frac {2 \, {\left (8 \, a b^{3} c - 8 \, a^{2} b^{2} d - \frac {3 \, {\left (b^{2} c^{2} d - 6 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} {\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{2}} + \frac {5 \, {\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} {\left (b x^{2} + a\right )}}{d x^{2} + c}\right )}}{b^{3} d^{2} \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {5}{2}} - 2 \, b^{4} d \left (\frac {b x^{2} + a}{d x^{2} + c}\right )^{\frac {3}{2}} + b^{5} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}} - \frac {3 \, {\left (b^{2} c^{2} - 6 \, a b c d + 5 \, a^{2} d^{2}\right )} \log \left (\frac {d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {b d}}{d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {b d}}\right )}{\sqrt {b d} b^{3}}\right )} e^{\left (-\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*(8*a*b^3*c - 8*a^2*b^2*d - 3*(b^2*c^2*d - 6*a*b*c*d^2 + 5*a^2*d^3)*(b*x^2 + a)^2/(d*x^2 + c)^2 + 5*(b^
3*c^2 - 6*a*b^2*c*d + 5*a^2*b*d^2)*(b*x^2 + a)/(d*x^2 + c))/(b^3*d^2*((b*x^2 + a)/(d*x^2 + c))^(5/2) - 2*b^4*d
*((b*x^2 + a)/(d*x^2 + c))^(3/2) + b^5*sqrt((b*x^2 + a)/(d*x^2 + c))) - 3*(b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*lo
g((d*sqrt((b*x^2 + a)/(d*x^2 + c)) - sqrt(b*d))/(d*sqrt((b*x^2 + a)/(d*x^2 + c)) + sqrt(b*d)))/(sqrt(b*d)*b^3)
)*e^(-3/2)

________________________________________________________________________________________

Fricas [A]
time = 0.59, size = 554, normalized size = 2.74 \begin {gather*} \left [\frac {{\left (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} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 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} \sqrt {\frac {b x^{2} + a}{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 x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{32 \, {\left (b^{5} d x^{2} + a b^{4} d\right )}}, -\frac {{\left (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} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{2 \, {\left (b^{2} d x^{2} + a b d\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 x^{2} + a}{d x^{2} + c}}\right )} e^{\left (-\frac {3}{2}\right )}}{16 \, {\left (b^{5} d x^{2} + a b^{4} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3}{{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________