3.755 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} (f+g x)^{5/2}} \, dx\)

Optimal. Leaf size=284 \[ \frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{5 c^{3/2} d^{3/2} \sqrt{d+e x} \sqrt{a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{g^{7/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{10 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}} \]

[Out]

(5*c^2*d^2*Sqrt[f + g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g^3*Sqrt[d + e*x]) - (10*c*d*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(3*g^2*(d + e*x)^(3/2)*Sqrt[f + g*x]) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c
*d*e*x^2)^(5/2))/(3*g*(d + e*x)^(5/2)*(f + g*x)^(3/2)) - (5*c^(3/2)*d^(3/2)*(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^(7/2)*Sqrt[a*d*e + (c*d
^2 + a*e^2)*x + c*d*e*x^2])

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Rubi [A]  time = 0.403685, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{5 c^{3/2} d^{3/2} \sqrt{d+e x} \sqrt{a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{g^{7/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{10 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g*x)^(5/2)),x]

[Out]

(5*c^2*d^2*Sqrt[f + g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g^3*Sqrt[d + e*x]) - (10*c*d*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(3*g^2*(d + e*x)^(3/2)*Sqrt[f + g*x]) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c
*d*e*x^2)^(5/2))/(3*g*(d + e*x)^(5/2)*(f + g*x)^(3/2)) - (5*c^(3/2)*d^(3/2)*(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^(7/2)*Sqrt[a*d*e + (c*d
^2 + a*e^2)*x + c*d*e*x^2])

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]

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{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^{5/2}} \, dx &=-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}+\frac{(5 c d) \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}{3 g}\\ &=-\frac{10 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}+\frac{\left (5 c^2 d^2\right ) \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^2}\\ &=\frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{10 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}-\frac{\left (5 c^2 d^2 (c d f-a e g)\right ) \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^3}\\ &=\frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{10 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}-\frac{\left (5 c^2 d^2 (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^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{10 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}-\frac{\left (5 c d (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^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{10 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}-\frac{\left (5 c d (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^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{5 c^2 d^2 \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt{d+e x}}-\frac{10 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} \sqrt{f+g x}}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^{3/2}}-\frac{5 c^{3/2} d^{3/2} (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^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end{align*}

Mathematica [C]  time = 0.126918, size = 112, normalized size = 0.39 \[ \frac{2 (a e+c d x)^3 \sqrt{(d+e x) (a e+c d x)} \left (\frac{c d (f+g x)}{c d f-a e g}\right )^{5/2} \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};\frac{g (a e+c d x)}{a e g-c d f}\right )}{7 c d \sqrt{d+e x} (f+g x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g*x)^(5/2)),x]

[Out]

(2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*((c*d*(f + g*x))/(c*d*f - a*e*g))^(5/2)*Hypergeometric2F1[5/2
, 7/2, 9/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/(7*c*d*Sqrt[d + e*x]*(f + g*x)^(5/2))

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Maple [B]  time = 0.385, size = 638, normalized size = 2.3 \begin{align*}{\frac{1}{6\,{g}^{3}} \left ( 15\,\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}^{2}a{c}^{2}{d}^{2}e{g}^{3}-15\,\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}^{2}{c}^{3}{d}^{3}f{g}^{2}+30\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}}{\sqrt{cdg}}} \right ) xa{c}^{2}{d}^{2}ef{g}^{2}-30\,\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}^{3}{d}^{3}{f}^{2}g+15\,\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{c}^{2}{d}^{2}e{f}^{2}g-15\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}}{\sqrt{cdg}}} \right ){c}^{3}{d}^{3}{f}^{3}+6\,{x}^{2}{c}^{2}{d}^{2}{g}^{2}\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}-28\,\sqrt{cdg}\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }xacde{g}^{2}+40\,\sqrt{cdg}\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }x{c}^{2}{d}^{2}fg-4\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}{a}^{2}{e}^{2}{g}^{2}-20\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}acdefg+30\,\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }\sqrt{cdg}{c}^{2}{d}^{2}{f}^{2} \right ) \sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade}{\frac{1}{\sqrt{ \left ( cdx+ae \right ) \left ( gx+f \right ) }}}{\frac{1}{\sqrt{cdg}}} \left ( gx+f \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(5/2),x)

[Out]

1/6*(15*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^2*a*c^2*d^
2*e*g^3-15*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^2*c^3*d
^3*f*g^2+30*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^2*
d^2*e*f*g^2-30*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^3
*d^3*f^2*g+15*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^2*
d^2*e*f^2*g-15*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))*c^3*d
^3*f^3+6*x^2*c^2*d^2*g^2*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2)-28*(c*d*g)^(1/2)*((c*d*x+a*e)*(g*x+f))^(1/2
)*x*a*c*d*e*g^2+40*(c*d*g)^(1/2)*((c*d*x+a*e)*(g*x+f))^(1/2)*x*c^2*d^2*f*g-4*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*
g)^(1/2)*a^2*e^2*g^2-20*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2)*a*c*d*e*f*g+30*((c*d*x+a*e)*(g*x+f))^(1/2)*(
c*d*g)^(1/2)*c^2*d^2*f^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)/((c*d*x+a*e)*(g*x+f))^(1/2)/(c*d*g)^(1/2)/g^
3/(g*x+f)^(3/2)/(e*x+d)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{\frac{5}{2}}{\left (g x + f\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((e*x + d)^(5/2)*(g*x + f)^(5/2)), x)

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Fricas [A]  time = 5.88059, size = 2059, normalized size = 7.25 \begin{align*} \text{result too large to display} \end{align*}

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)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(4*(3*c^2*d^2*g^2*x^2 + 15*c^2*d^2*f^2 - 10*a*c*d*e*f*g - 2*a^2*e^2*g^2 + 2*(10*c^2*d^2*f*g - 7*a*c*d*e*
g^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*sqrt(g*x + f) - 15*(c^2*d^3*f^3 - a*c*d^2*e*
f^2*g + (c^2*d^2*e*f*g^2 - a*c*d*e^2*g^3)*x^3 + (2*c^2*d^2*e*f^2*g - a*c*d^2*e*g^3 + (c^2*d^3 - 2*a*c*d*e^2)*f
*g^2)*x^2 + (c^2*d^2*e*f^3 - 2*a*c*d^2*e*f*g^2 + (2*c^2*d^3 - a*c*d*e^2)*f^2*g)*x)*sqrt(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^5*x^3 + d*f^2*g^3 + (2*e*f*g^4 + d*g^5)*x^2 + (e*f^2*g^3 + 2*d*f*g^4)*x), 1/6*(2*(3*c^2*d^2*g^2*x^2 + 15*
c^2*d^2*f^2 - 10*a*c*d*e*f*g - 2*a^2*e^2*g^2 + 2*(10*c^2*d^2*f*g - 7*a*c*d*e*g^2)*x)*sqrt(c*d*e*x^2 + a*d*e +
(c*d^2 + a*e^2)*x)*sqrt(e*x + d)*sqrt(g*x + f) + 15*(c^2*d^3*f^3 - a*c*d^2*e*f^2*g + (c^2*d^2*e*f*g^2 - a*c*d*
e^2*g^3)*x^3 + (2*c^2*d^2*e*f^2*g - a*c*d^2*e*g^3 + (c^2*d^3 - 2*a*c*d*e^2)*f*g^2)*x^2 + (c^2*d^2*e*f^3 - 2*a*
c*d^2*e*f*g^2 + (2*c^2*d^3 - a*c*d*e^2)*f^2*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^5*x^3 + d*f^2*g^3 + (2*e*f*g^4 + d*g^5)*x^2 + (e*f^2*g^3 + 2*d*f*g^4)*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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(5/2)/(g*x+f)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{\frac{5}{2}}{\left (g x + f\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(5/2),x, algorithm="giac")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((e*x + d)^(5/2)*(g*x + f)^(5/2)), x)