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

Optimal. Leaf size=222 \[ -\frac {3 \sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{g^{5/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {3 c d \sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^2 \sqrt {d+e x}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{g (d+e x)^{3/2} \sqrt {f+g x}} \]

[Out]

-2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/g/(e*x+d)^(3/2)/(g*x+f)^(1/2)-3*(-a*e*g+c*d*f)*arctanh(g^(1/2)*(c*d
*x+a*e)^(1/2)/c^(1/2)/d^(1/2)/(g*x+f)^(1/2))*c^(1/2)*d^(1/2)*(c*d*x+a*e)^(1/2)*(e*x+d)^(1/2)/g^(5/2)/(a*d*e+(a
*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+3*c*d*(g*x+f)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/g^2/(e*x+d)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 222, normalized size of antiderivative = 1.00, 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 = {862, 864, 891, 63, 217, 206} \[ \frac {3 c d \sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^2 \sqrt {d+e x}}-\frac {3 \sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{g^{5/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{g (d+e x)^{3/2} \sqrt {f+g x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])

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 862

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

Rule 864

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

Rubi steps

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

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Mathematica [C]  time = 0.16, size = 102, normalized size = 0.46 \[ \frac {2 ((d+e x) (a e+c d x))^{5/2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{3/2} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {g (a e+c d x)}{a e g-c d f}\right )}{5 c d (d+e x)^{5/2} (f+g x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*((c*d*(f + g*x))/(c*d*f - a*e*g))^(3/2)*Hypergeometric2F1[3/2, 5/2, 7/2, (g
*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/(5*c*d*(d + e*x)^(5/2)*(f + g*x)^(3/2))

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fricas [A]  time = 2.07, size = 663, normalized size = 2.99 \[ \left [\frac {4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d g x + 3 \, c d f - 2 \, a e g\right )} \sqrt {e x + d} \sqrt {g x + f} - 3 \, {\left (c d^{2} f^{2} - a d e f g + {\left (c d e f g - a e^{2} g^{2}\right )} x^{2} + {\left (c d e f^{2} - a d e g^{2} + {\left (c d^{2} - a e^{2}\right )} f g\right )} x\right )} \sqrt {\frac {c d}{g}} \log \left (-\frac {8 \, c^{2} d^{2} e g^{2} x^{3} + c^{2} d^{3} f^{2} + 6 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} + 4 \, {\left (2 \, c d g^{2} x + c d f g + a e g^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f} \sqrt {\frac {c d}{g}} + 8 \, {\left (c^{2} d^{2} e f g + {\left (c^{2} d^{3} + a c d e^{2}\right )} g^{2}\right )} x^{2} + {\left (c^{2} d^{2} e f^{2} + 2 \, {\left (4 \, c^{2} d^{3} + 3 \, a c d e^{2}\right )} f g + {\left (8 \, a c d^{2} e + a^{2} e^{3}\right )} g^{2}\right )} x}{e x + d}\right )}{4 \, {\left (e g^{3} x^{2} + d f g^{2} + {\left (e f g^{2} + d g^{3}\right )} x\right )}}, \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d g x + 3 \, c d f - 2 \, a e g\right )} \sqrt {e x + d} \sqrt {g x + f} + 3 \, {\left (c d^{2} f^{2} - a d e f g + {\left (c d e f g - a e^{2} g^{2}\right )} x^{2} + {\left (c d e f^{2} - a d e g^{2} + {\left (c d^{2} - a e^{2}\right )} f g\right )} x\right )} \sqrt {-\frac {c d}{g}} \arctan \left (\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f} \sqrt {-\frac {c d}{g}} g}{2 \, c d e g x^{2} + c d^{2} f + a d e g + {\left (c d e f + {\left (2 \, c d^{2} + a e^{2}\right )} g\right )} x}\right )}{2 \, {\left (e g^{3} x^{2} + d f g^{2} + {\left (e f g^{2} + d g^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^(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 + 3*c*d*f - 2*a*e*g)*sqrt(e*x + d)*sqrt(g*x + f)
- 3*(c*d^2*f^2 - a*d*e*f*g + (c*d*e*f*g - a*e^2*g^2)*x^2 + (c*d*e*f^2 - a*d*e*g^2 + (c*d^2 - a*e^2)*f*g)*x)*sq
rt(c*d/g)*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 + 4*(2*c*d*g^2*x + c*d*f*g
+ a*e*g^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)*sqrt(c*d/g) + 8*(c^2*d^2*e
*f*g + (c^2*d^3 + a*c*d*e^2)*g^2)*x^2 + (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)))/(e*g^3*x^2 + d*f*g^2 + (e*f*g^2 + d*g^3)*x), 1/2*(2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 +
a*e^2)*x)*(c*d*g*x + 3*c*d*f - 2*a*e*g)*sqrt(e*x + d)*sqrt(g*x + f) + 3*(c*d^2*f^2 - a*d*e*f*g + (c*d*e*f*g -
a*e^2*g^2)*x^2 + (c*d*e*f^2 - a*d*e*g^2 + (c*d^2 - a*e^2)*f*g)*x)*sqrt(-c*d/g)*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)*sqrt(-c*d/g)*g/(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)))/(e*g^3*x^2 + d*f*g^2 + (e*f*g^2 + d*g^3)*x)]

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

[Out]

Timed out

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maple [B]  time = 0.03, size = 383, normalized size = 1.73 \[ \frac {\left (3 a c d e \,g^{2} x \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )-3 c^{2} d^{2} f g x \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )+3 a c d e f g \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )-3 c^{2} d^{2} f^{2} \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )+2 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c d g x -4 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, a e g +6 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c d f \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, \sqrt {g x +f}\, \sqrt {e x +d}\, g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {3}{2}} {\left (g x + f\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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