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

Optimal. Leaf size=151 \[ -\frac {a^{3/2} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{c^{3/2}}+\frac {b^{3/2} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{d^{3/2}}-\frac {e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

[Out]

-a^(3/2)*e^(3/2)*arctanh(c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^(1/2)/e^(1/2))/c^(3/2)+b^(3/2)*e^(3/2)*arctan
h(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))/d^(3/2)-(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/c/
d

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Rubi [A]  time = 0.19, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1960, 479, 522, 208} \[ -\frac {a^{3/2} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{c^{3/2}}+\frac {b^{3/2} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{d^{3/2}}-\frac {e (b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

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

Rule 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 1.28, size = 193, normalized size = 1.28 \[ \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {d} \left (-\frac {a^{3/2} d \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{c^{3/2} \sqrt {a+b x^2}}+\frac {a d}{c}-b\right )+\frac {b \sqrt {b c-a d} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {a+b x^2}}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*((b*Sqrt[b*c - a*d]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sq
rt[a + b*x^2])/Sqrt[b*c - a*d]])/Sqrt[a + b*x^2] + Sqrt[d]*(-b + (a*d)/c - (a^(3/2)*d*Sqrt[c + d*x^2]*ArcTanh[
(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(c^(3/2)*Sqrt[a + b*x^2]))))/d^(3/2)

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fricas [A]  time = 1.30, size = 1049, normalized size = 6.95 \[ \left [\frac {b c \sqrt {\frac {b e}{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^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) + a d \sqrt {\frac {a e}{c}} e \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} - 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left (b c - a d\right )} e \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c d}, -\frac {2 \, b c \sqrt {-\frac {b e}{d}} e \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} e x^{2} + a b e\right )}}\right ) - a d \sqrt {\frac {a e}{c}} e \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} - 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) + 4 \, {\left (b c - a d\right )} e \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c d}, \frac {2 \, a d \sqrt {-\frac {a e}{c}} e \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) + b c \sqrt {\frac {b e}{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^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (b c - a d\right )} e \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c d}, \frac {a d \sqrt {-\frac {a e}{c}} e \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) - b c \sqrt {-\frac {b e}{d}} e \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} e x^{2} + a b e\right )}}\right ) - 2 \, {\left (b c - a d\right )} e \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, c d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x^2*d+c)]Evaluation time: 0.52Error: Bad Argument Type

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maple [B]  time = 0.06, size = 401, normalized size = 2.66 \[ \frac {\left (-\sqrt {b d}\, a^{2} d^{2} x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+\sqrt {a c}\, b^{2} c d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-\sqrt {b d}\, a^{2} c d \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+\sqrt {a c}\, b^{2} c^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, \sqrt {a c}\, a d -2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, \sqrt {a c}\, b c \right ) \left (d \,x^{2}+c \right ) \left (\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{2 \sqrt {a c}\, \sqrt {b d}\, \left (b \,x^{2}+a \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.27, size = 197, normalized size = 1.30 \[ \frac {1}{2} \, {\left (\frac {a^{2} e \log \left (\frac {c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {a c e}}{c \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {a c e}}\right )}{\sqrt {a c e} c} - \frac {b^{2} e \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} d} - \frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{c d}\right )} e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*e*log((c*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - sqrt(a*c*e))/(c*sqrt((b*x^2 + a)*e/(d*x^2 + c)) + sqrt(a*c
*e)))/(sqrt(a*c*e)*c) - b^2*e*log((d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - sqrt(b*d*e))/(d*sqrt((b*x^2 + a)*e/(d*x
^2 + c)) + sqrt(b*d*e)))/(sqrt(b*d*e)*d) - 2*(b*c - a*d)*sqrt((b*x^2 + a)*e/(d*x^2 + c))/(c*d))*e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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