3.3.79 \(\int \frac {(\frac {e (a+b x^2)}{c+d x^2})^{3/2}}{x} \, dx\) [279]

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

[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 = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1981, 1980, 490, 536, 214} \begin {gather*} -\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} \end {gather*}

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 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 490

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 536

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 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 {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x} \, dx &=((b c-a d) e) \text {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) \text {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 ) \text {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 ) \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 )}{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 = 1.89, size = 187, normalized size = 1.24 \begin {gather*} \frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (-a^{3/2} d^{3/2} \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )+\sqrt {c} \left (\sqrt {d} (-b c+a d) \sqrt {a+b x^2}+b^{3/2} c \sqrt {c+d x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )\right )}{c^{3/2} d^{3/2} \sqrt {a+b x^2}} \end {gather*}

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(121)=242\).
time = 0.08, size = 401, normalized size = 2.66

method	result	size
default	\(\frac {\left (\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{2} c d \,x^{2}-\sqrt {b d}\, \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} d^{2} x^{2}+\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{2} c^{2}-\sqrt {b d}\, \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} c d +2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, a d -2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, b c \right ) \left (d \,x^{2}+c \right ) \left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )^{\frac {3}{2}}}{2 c d \sqrt {a c}\, \sqrt {b d}\, \left (b \,x^{2}+a \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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*c)^(1/2)*b^2
*c*d*x^2-(b*d)^(1/2)*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*a^2*d^2
*x^2+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*c)^(1/2)*b^2
*c^2-(b*d)^(1/2)*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*a^2*c*d+2*(
b*d)^(1/2)*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*d-2*(b*d)^(1/2)*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*b
*c)/c/d*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(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 = 0.50, size = 185, normalized size = 1.23 \begin {gather*} \frac {1}{2} \, {\left (\frac {a^{2} \log \left (\frac {c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {a c}}{c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {a c}}\right )}{\sqrt {a c} c} - \frac {b^{2} \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} d} - \frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}}}{c d}\right )} e^{\frac {3}{2}} \end {gather*}

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*log((c*sqrt((b*x^2 + a)/(d*x^2 + c)) - sqrt(a*c))/(c*sqrt((b*x^2 + a)/(d*x^2 + c)) + sqrt(a*c)))/(sqr
t(a*c)*c) - b^2*log((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)*d) - 2*(b*c - a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))/(c*d))*e^(3/2)

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Fricas [A]
time = 0.73, size = 983, normalized size = 6.51 \begin {gather*} \left [\frac {b c \sqrt {\frac {b}{d}} e^{\frac {3}{2}} \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^{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 x^{2} + a}{d x^{2} + c}} \sqrt {\frac {b}{d}}\right ) + a d \sqrt {\frac {a}{c}} e^{\frac {3}{2}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} 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 {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {a}{c}}}{x^{4}}\right ) - 4 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{4 \, c d}, -\frac {2 \, b c \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} x^{2} + a b\right )}}\right ) e^{\frac {3}{2}} - a d \sqrt {\frac {a}{c}} e^{\frac {3}{2}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} 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 {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {a}{c}}}{x^{4}}\right ) + 4 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{4 \, c d}, \frac {2 \, a d \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b x^{2} + a^{2}\right )}}\right ) e^{\frac {3}{2}} + b c \sqrt {\frac {b}{d}} e^{\frac {3}{2}} \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^{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 x^{2} + a}{d x^{2} + c}} \sqrt {\frac {b}{d}}\right ) - 4 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{4 \, c d}, \frac {a d \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b x^{2} + a^{2}\right )}}\right ) e^{\frac {3}{2}} - b c \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} x^{2} + a b\right )}}\right ) e^{\frac {3}{2}} - 2 \, {\left (b c - a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} e^{\frac {3}{2}}}{2 \, c d}\right ] \end {gather*}

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/d)*e^(3/2)*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^3*x^4 + b*c^2*d + a*c*d^2 + (3*b*c*d^2 + a*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(b/d)) + a*d*sqr
t(a/c)*e^(3/2)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*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((b*x^2 + a)/(d*x^2 + c))*sqrt(a/c))/x^4) - 4*(b*c - a
*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/(c*d), -1/4*(2*b*c*sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sq
rt((b*x^2 + a)/(d*x^2 + c))*sqrt(-b/d)/(b^2*x^2 + a*b))*e^(3/2) - a*d*sqrt(a/c)*e^(3/2)*log(((b^2*c^2 + 6*a*b*
c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*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((b*x^2 + a)/(d*x^2 + c))*sqrt(a/c))/x^4) + 4*(b*c - a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(
3/2))/(c*d), 1/4*(2*a*d*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-a/
c)/(a*b*x^2 + a^2))*e^(3/2) + b*c*sqrt(b/d)*e^(3/2)*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^3*x^4 + b*c^2*d + a*c*d^2 + (3*b*c*d^2 + a*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 +
c))*sqrt(b/d)) - 4*(b*c - a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/(c*d), 1/2*(a*d*sqrt(-a/c)*arctan(1/2*((
b*c + a*d)*x^2 + 2*a*c)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-a/c)/(a*b*x^2 + a^2))*e^(3/2) - b*c*sqrt(-b/d)*arc
tan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-b/d)/(b^2*x^2 + a*b))*e^(3/2) - 2*(b*c - a
*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*e^(3/2))/(c*d)]

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

[Out]

Timed out

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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((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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

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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