3.721 \(\int \frac{(d+e x)^{3/2} (f+g x)^{3/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=227 \[ \frac{3 g \sqrt{f+g x} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{3 \sqrt{g} \sqrt{d+e x} \sqrt{a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{c^{5/2} d^{5/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^(3/2))/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3*g*Sqrt[f + g*x]*Sqrt
[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c^2*d^2*Sqrt[d + e*x]) + (3*Sqrt[g]*(c*d*f - a*e*g)*Sqrt[a*e + c*d*x
]*Sqrt[d + e*x]*ArcTanh[(Sqrt[g]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[f + g*x])])/(c^(5/2)*d^(5/2)*Sqrt[a*
d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

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Rubi [A]  time = 0.314217, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {866, 870, 891, 63, 217, 206} \[ \frac{3 g \sqrt{f+g x} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{3 \sqrt{g} \sqrt{d+e x} \sqrt{a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{c^{5/2} d^{5/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^(3/2))/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3*g*Sqrt[f + g*x]*Sqrt
[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c^2*d^2*Sqrt[d + e*x]) + (3*Sqrt[g]*(c*d*f - a*e*g)*Sqrt[a*e + c*d*x
]*Sqrt[d + e*x]*ArcTanh[(Sqrt[g]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[f + g*x])])/(c^(5/2)*d^(5/2)*Sqrt[a*
d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[(e*(d + e*x)^(m - 1)*(f + g*x)^n*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e*g*n)/(c*(p + 1)), I
nt[(d + e*x)^(m - 1)*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] &
& LtQ[p, -1] && GtQ[n, 0]

Rule 870

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
-Simp[(e*(d + e*x)^(m - 1)*(f + g*x)^n*(a + b*x + c*x^2)^(p + 1))/(c*(m - n - 1)), x] - Dist[(n*(c*e*f + c*d*g
- b*e*g))/(c*e*(m - n - 1)), Int[(d + e*x)^m*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Integ
erQ[p] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (IntegerQ[2*p] || IntegerQ[n])

Rule 891

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^{3/2} (f+g x)^{3/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{(3 g) \int \frac{\sqrt{d+e x} \sqrt{f+g x}}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d}\\ &=-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 g \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{(3 g (c d f-a e g)) \int \frac{\sqrt{d+e x}}{\sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 c^2 d^2}\\ &=-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 g \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{\left (3 g (c d f-a e g) \sqrt{a e+c d x} \sqrt{d+e x}\right ) \int \frac{1}{\sqrt{a e+c d x} \sqrt{f+g x}} \, dx}{2 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 g \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{\left (3 g (c d f-a e g) \sqrt{a e+c d x} \sqrt{d+e x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{f-\frac{a e g}{c d}+\frac{g x^2}{c d}}} \, dx,x,\sqrt{a e+c d x}\right )}{c^3 d^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 g \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{\left (3 g (c d f-a e g) \sqrt{a e+c d x} \sqrt{d+e x}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{g x^2}{c d}} \, dx,x,\frac{\sqrt{a e+c d x}}{\sqrt{f+g x}}\right )}{c^3 d^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac{2 \sqrt{d+e x} (f+g x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 g \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^2 d^2 \sqrt{d+e x}}+\frac{3 \sqrt{g} (c d f-a e g) \sqrt{a e+c d x} \sqrt{d+e x} \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{c^{5/2} d^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end{align*}

Mathematica [C]  time = 0.087196, size = 100, normalized size = 0.44 \[ -\frac{2 \sqrt{d+e x} (f+g x)^{3/2} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{g (a e+c d x)}{a e g-c d f}\right )}{c d \sqrt{(d+e x) (a e+c d x)} \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^(3/2)*Hypergeometric2F1[-3/2, -1/2, 1/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/(c
*d*Sqrt[(a*e + c*d*x)*(d + e*x)]*((c*d*(f + g*x))/(c*d*f - a*e*g))^(3/2))

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Maple [B]  time = 0.379, size = 396, normalized size = 1.7 \begin{align*} -{\frac{1}{ \left ( 2\,cdx+2\,ae \right ){c}^{2}{d}^{2}} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}}{\sqrt{cdg}}} \right ) xacde{g}^{2}-3\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}}{\sqrt{cdg}}} \right ) x{c}^{2}{d}^{2}fg+3\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}}{\sqrt{cdg}}} \right ){a}^{2}{e}^{2}{g}^{2}-3\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}}{\sqrt{cdg}}} \right ) acdefg-2\,\sqrt{cdg}\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }xcdg-6\,\sqrt{cdg}\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }aeg+4\,\sqrt{cdg}\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }cdf \right ) \sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade}\sqrt{gx+f}{\frac{1}{\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }}}{\frac{1}{\sqrt{cdg}}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*ln(1/2*(2*x*c*d*g+a*e*g+c*d*f+2*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2))/(c*d*g)^(1/2))*x*a*c*d*e*g^
2-3*ln(1/2*(2*x*c*d*g+a*e*g+c*d*f+2*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2))/(c*d*g)^(1/2))*x*c^2*d^2*f*g+3*
ln(1/2*(2*x*c*d*g+a*e*g+c*d*f+2*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2))/(c*d*g)^(1/2))*a^2*e^2*g^2-3*ln(1/2
*(2*x*c*d*g+a*e*g+c*d*f+2*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2))/(c*d*g)^(1/2))*a*c*d*e*f*g-2*(c*d*g)^(1/2
)*((c*d*x+a*e)*(g*x+f))^(1/2)*x*c*d*g-6*(c*d*g)^(1/2)*((c*d*x+a*e)*(g*x+f))^(1/2)*a*e*g+4*(c*d*g)^(1/2)*((c*d*
x+a*e)*(g*x+f))^(1/2)*c*d*f)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(g*x+f)^(1/2)/((c*d*x+a*e)*(g*x+f))^(1/2)
/(c*d*x+a*e)/(c*d*g)^(1/2)/c^2/d^2/(e*x+d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{3}{2}}{\left (g x + f\right )}^{\frac{3}{2}}}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 7.00522, size = 1544, normalized size = 6.8 \begin{align*} \left [\frac{4 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d g x - 2 \, c d f + 3 \, a e g\right )} \sqrt{e x + d} \sqrt{g x + f} - 3 \,{\left (a c d^{2} e f - a^{2} d e^{2} g +{\left (c^{2} d^{2} e f - a c d e^{2} g\right )} x^{2} +{\left ({\left (c^{2} d^{3} + a c d e^{2}\right )} f -{\left (a c d^{2} e + a^{2} e^{3}\right )} g\right )} x\right )} \sqrt{\frac{g}{c d}} \log \left (-\frac{8 \, c^{2} d^{2} e g^{2} x^{3} + c^{2} d^{3} f^{2} + 6 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} + 8 \,{\left (c^{2} d^{2} e f g +{\left (c^{2} d^{3} + a c d e^{2}\right )} g^{2}\right )} x^{2} - 4 \,{\left (2 \, c^{2} d^{2} g x + c^{2} d^{2} f + a c d e g\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} \sqrt{g x + f} \sqrt{\frac{g}{c d}} +{\left (c^{2} d^{2} e f^{2} + 2 \,{\left (4 \, c^{2} d^{3} + 3 \, a c d e^{2}\right )} f g +{\left (8 \, a c d^{2} e + a^{2} e^{3}\right )} g^{2}\right )} x}{e x + d}\right )}{4 \,{\left (c^{3} d^{3} e x^{2} + a c^{2} d^{3} e +{\left (c^{3} d^{4} + a c^{2} d^{2} e^{2}\right )} x\right )}}, \frac{2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d g x - 2 \, c d f + 3 \, a e g\right )} \sqrt{e x + d} \sqrt{g x + f} - 3 \,{\left (a c d^{2} e f - a^{2} d e^{2} g +{\left (c^{2} d^{2} e f - a c d e^{2} g\right )} x^{2} +{\left ({\left (c^{2} d^{3} + a c d e^{2}\right )} f -{\left (a c d^{2} e + a^{2} e^{3}\right )} g\right )} x\right )} \sqrt{-\frac{g}{c d}} \arctan \left (\frac{2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} \sqrt{g x + f} c d \sqrt{-\frac{g}{c d}}}{2 \, c d e g x^{2} + c d^{2} f + a d e g +{\left (c d e f +{\left (2 \, c d^{2} + a e^{2}\right )} g\right )} x}\right )}{2 \,{\left (c^{3} d^{3} e x^{2} + a c^{2} d^{3} e +{\left (c^{3} d^{4} + a c^{2} d^{2} e^{2}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x