3.8.53 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx\) [753]

3.8.53.1 Optimal result
3.8.53.2 Mathematica [A] (verified)
3.8.53.3 Rubi [A] (verified)
3.8.53.4 Maple [A] (verified)
3.8.53.5 Fricas [A] (verification not implemented)
3.8.53.6 Sympy [F(-1)]
3.8.53.7 Maxima [F]
3.8.53.8 Giac [B] (verification not implemented)
3.8.53.9 Mupad [F(-1)]

3.8.53.1 Optimal result

Integrand size = 48, antiderivative size = 304 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\frac {5 (c d f-a e g)^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 g^3 \sqrt {d+e x}}-\frac {5 (c d f-a e g) \sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2}}+\frac {\sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g)^3 \sqrt {a e+c d x} \sqrt {d+e x} \text {arctanh}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{8 \sqrt {c} \sqrt {d} g^{7/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

output
-5/12*(-a*e*g+c*d*f)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*(g*x+f)^(1/2) 
/g^2/(e*x+d)^(3/2)+1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*(g*x+f)^(1/ 
2)/g/(e*x+d)^(5/2)-5/8*(-a*e*g+c*d*f)^3*arctanh(g^(1/2)*(c*d*x+a*e)^(1/2)/ 
c^(1/2)/d^(1/2)/(g*x+f)^(1/2))*(c*d*x+a*e)^(1/2)*(e*x+d)^(1/2)/g^(7/2)/c^( 
1/2)/d^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5/8*(-a*e*g+c*d*f)^2* 
(g*x+f)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/g^3/(e*x+d)^(1/2)
 
3.8.53.2 Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.62 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\frac {((a e+c d x) (d+e x))^{5/2} \left (\frac {\sqrt {g} \sqrt {f+g x} \left (33 a^2 e^2 g^2+2 a c d e g (-20 f+13 g x)+c^2 d^2 \left (15 f^2-10 f g x+8 g^2 x^2\right )\right )}{(a e+c d x)^2}-\frac {15 (c d f-a e g)^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {f+g x}}{\sqrt {g} \sqrt {a e+c d x}}\right )}{\sqrt {c} \sqrt {d} (a e+c d x)^{5/2}}\right )}{24 g^{7/2} (d+e x)^{5/2}} \]

input
Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*S 
qrt[f + g*x]),x]
 
output
(((a*e + c*d*x)*(d + e*x))^(5/2)*((Sqrt[g]*Sqrt[f + g*x]*(33*a^2*e^2*g^2 + 
 2*a*c*d*e*g*(-20*f + 13*g*x) + c^2*d^2*(15*f^2 - 10*f*g*x + 8*g^2*x^2)))/ 
(a*e + c*d*x)^2 - (15*(c*d*f - a*e*g)^3*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[f + 
g*x])/(Sqrt[g]*Sqrt[a*e + c*d*x])])/(Sqrt[c]*Sqrt[d]*(a*e + c*d*x)^(5/2))) 
)/(24*g^(7/2)*(d + e*x)^(5/2))
 
3.8.53.3 Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1250, 1250, 1250, 1268, 66, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1250

\(\displaystyle \frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g) \int \frac {\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/2}}{(d+e x)^{3/2} \sqrt {f+g x}}dx}{6 g}\)

\(\Big \downarrow \) 1250

\(\displaystyle \frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g) \left (\frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 g (d+e x)^{3/2}}-\frac {3 (c d f-a e g) \int \frac {\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {d+e x} \sqrt {f+g x}}dx}{4 g}\right )}{6 g}\)

\(\Big \downarrow \) 1250

\(\displaystyle \frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g) \left (\frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 g (d+e x)^{3/2}}-\frac {3 (c d f-a e g) \left (\frac {\sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x}}-\frac {(c d f-a e g) \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{2 g}\right )}{4 g}\right )}{6 g}\)

\(\Big \downarrow \) 1268

\(\displaystyle \frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g) \left (\frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 g (d+e x)^{3/2}}-\frac {3 (c d f-a e g) \left (\frac {\sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x}}-\frac {\sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g) \int \frac {1}{\sqrt {a e+c d x} \sqrt {f+g x}}dx}{2 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{4 g}\right )}{6 g}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g) \left (\frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 g (d+e x)^{3/2}}-\frac {3 (c d f-a e g) \left (\frac {\sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x}}-\frac {\sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g) \int \frac {1}{c d-\frac {g (a e+c d x)}{f+g x}}d\frac {\sqrt {a e+c d x}}{\sqrt {f+g x}}}{g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{4 g}\right )}{6 g}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g) \left (\frac {\sqrt {f+g x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 g (d+e x)^{3/2}}-\frac {3 (c d f-a e g) \left (\frac {\sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g \sqrt {d+e x}}-\frac {\sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g) \text {arctanh}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{\sqrt {c} \sqrt {d} g^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{4 g}\right )}{6 g}\)

input
Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*Sqrt[f 
+ g*x]),x]
 
output
(Sqrt[f + g*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(3*g*(d + e* 
x)^(5/2)) - (5*(c*d*f - a*e*g)*((Sqrt[f + g*x]*(a*d*e + (c*d^2 + a*e^2)*x 
+ c*d*e*x^2)^(3/2))/(2*g*(d + e*x)^(3/2)) - (3*(c*d*f - a*e*g)*((Sqrt[f + 
g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g*Sqrt[d + e*x]) - ((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])])/(Sqrt[c]*Sqrt[d]*g^(3/2)*Sqrt[a*d 
*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])))/(4*g)))/(6*g)
 

3.8.53.3.1 Defintions of rubi rules used

rule 66
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 221
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 1250
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] - Simp[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] && EqQ[c*d^2 - b*d*e + 
 a*e^2, 0] && 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 1268
Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(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/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
 
3.8.53.4 Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.64

method result size
default \(\frac {\sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \sqrt {g x +f}\, \left (15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right ) a^{3} e^{3} g^{3}-45 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right ) a^{2} c d \,e^{2} f \,g^{2}+45 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right ) a \,c^{2} d^{2} e \,f^{2} g -15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right ) c^{3} d^{3} f^{3}+16 c^{2} d^{2} g^{2} x^{2} \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}+52 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, a c d e \,g^{2} x -20 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c^{2} d^{2} f g x +66 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, a^{2} e^{2} g^{2}-80 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, a c d e f g +30 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, c^{2} d^{2} f^{2}\right )}{48 \sqrt {e x +d}\, g^{3} \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}\) \(498\)

input
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)^(1/2),x, 
method=_RETURNVERBOSE)
 
output
1/48*((c*d*x+a*e)*(e*x+d))^(1/2)*(g*x+f)^(1/2)*(15*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))*a^3*e^3 
*g^3-45*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))*a^2*c*d*e^2*f*g^2+45*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))*a*c^2*d^2*e*f 
^2*g-15*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))*c^3*d^3*f^3+16*c^2*d^2*g^2*x^2*(c*d*g)^(1/2)*((g*x 
+f)*(c*d*x+a*e))^(1/2)+52*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*a*c*d* 
e*g^2*x-20*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*c^2*d^2*f*g*x+66*((g* 
x+f)*(c*d*x+a*e))^(1/2)*(c*d*g)^(1/2)*a^2*e^2*g^2-80*((g*x+f)*(c*d*x+a*e)) 
^(1/2)*(c*d*g)^(1/2)*a*c*d*e*f*g+30*((g*x+f)*(c*d*x+a*e))^(1/2)*(c*d*g)^(1 
/2)*c^2*d^2*f^2)/(e*x+d)^(1/2)/g^3/((g*x+f)*(c*d*x+a*e))^(1/2)/(c*d*g)^(1/ 
2)
 
3.8.53.5 Fricas [A] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 837, normalized size of antiderivative = 2.75 \[ \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} \sqrt {f+g x}} \, dx=\left [\frac {4 \, {\left (8 \, c^{3} d^{3} g^{3} x^{2} + 15 \, c^{3} d^{3} f^{2} g - 40 \, a c^{2} d^{2} e f g^{2} + 33 \, a^{2} c d e^{2} g^{3} - 2 \, {\left (5 \, c^{3} d^{3} f g^{2} - 13 \, a c^{2} d^{2} e g^{3}\right )} x\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} - 15 \, {\left (c^{3} d^{4} f^{3} - 3 \, a c^{2} d^{3} e f^{2} g + 3 \, a^{2} c d^{2} e^{2} f g^{2} - a^{3} d e^{3} g^{3} + {\left (c^{3} d^{3} e f^{3} - 3 \, a c^{2} d^{2} e^{2} f^{2} g + 3 \, a^{2} c d e^{3} f g^{2} - a^{3} e^{4} g^{3}\right )} x\right )} \sqrt {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 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d g x + c d f + a e g\right )} \sqrt {c d g} \sqrt {e x + d} \sqrt {g x + f} + 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 )}{96 \, {\left (c d e g^{4} x + c d^{2} g^{4}\right )}}, \frac {2 \, {\left (8 \, c^{3} d^{3} g^{3} x^{2} + 15 \, c^{3} d^{3} f^{2} g - 40 \, a c^{2} d^{2} e f g^{2} + 33 \, a^{2} c d e^{2} g^{3} - 2 \, {\left (5 \, c^{3} d^{3} f g^{2} - 13 \, a c^{2} d^{2} e g^{3}\right )} x\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} + 15 \, {\left (c^{3} d^{4} f^{3} - 3 \, a c^{2} d^{3} e f^{2} g + 3 \, a^{2} c d^{2} e^{2} f g^{2} - a^{3} d e^{3} g^{3} + {\left (c^{3} d^{3} e f^{3} - 3 \, a c^{2} d^{2} e^{2} f^{2} g + 3 \, a^{2} c d e^{3} f g^{2} - a^{3} e^{4} g^{3}\right )} x\right )} \sqrt {-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 {-c d g} \sqrt {e x + d} \sqrt {g x + f}}{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 )}{48 \, {\left (c d e g^{4} x + c d^{2} g^{4}\right )}}\right ] \]

input
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)^(1 
/2),x, algorithm="fricas")
 
output
[1/96*(4*(8*c^3*d^3*g^3*x^2 + 15*c^3*d^3*f^2*g - 40*a*c^2*d^2*e*f*g^2 + 33 
*a^2*c*d*e^2*g^3 - 2*(5*c^3*d^3*f*g^2 - 13*a*c^2*d^2*e*g^3)*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^3*d^4 
*f^3 - 3*a*c^2*d^3*e*f^2*g + 3*a^2*c*d^2*e^2*f*g^2 - a^3*d*e^3*g^3 + (c^3* 
d^3*e*f^3 - 3*a*c^2*d^2*e^2*f^2*g + 3*a^2*c*d*e^3*f*g^2 - a^3*e^4*g^3)*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*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*g*x + c 
*d*f + a*e*g)*sqrt(c*d*g)*sqrt(e*x + d)*sqrt(g*x + f) + 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)))/(c*d*e*g^4*x + c*d^ 
2*g^4), 1/48*(2*(8*c^3*d^3*g^3*x^2 + 15*c^3*d^3*f^2*g - 40*a*c^2*d^2*e*f*g 
^2 + 33*a^2*c*d*e^2*g^3 - 2*(5*c^3*d^3*f*g^2 - 13*a*c^2*d^2*e*g^3)*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^3*d^4*f^3 - 3*a*c^2*d^3*e*f^2*g + 3*a^2*c*d^2*e^2*f*g^2 - a^3*d*e^3*g^3 
+ (c^3*d^3*e*f^3 - 3*a*c^2*d^2*e^2*f^2*g + 3*a^2*c*d*e^3*f*g^2 - a^3*e^4*g 
^3)*x)*sqrt(-c*d*g)*arctan(2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*s 
qrt(-c*d*g)*sqrt(e*x + d)*sqrt(g*x + f)/(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*d*e*g^4*x + c*d^2*g^4)]
 
3.8.53.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\text {Timed out} \]

input
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)**(1/2),x)
 
output
Timed out
 
3.8.53.7 Maxima [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}} \sqrt {g x + f}} \,d x } \]

input
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)^(1 
/2),x, algorithm="maxima")
 
output
integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((e*x + d)^(5/2)*s 
qrt(g*x + f)), x)
 
3.8.53.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1098 vs. \(2 (256) = 512\).

Time = 0.69 (sec) , antiderivative size = 1098, normalized size of antiderivative = 3.61 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\text {Too large to display} \]

input
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)^(1 
/2),x, algorithm="giac")
 
output
1/24*e*((sqrt(c^2*d^2*e^2*f - a*c*d*e^3*g + ((e*x + d)*c*d*e - c*d^2*e + a 
*e^3)*c*d*g)*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*(2*((e*x + d)*c*d*e - 
 c*d^2*e + a*e^3)*(4*((e*x + d)*c*d*e - c*d^2*e + a*e^3)*abs(e)/(c*d*e^3*g 
) - 5*(c*d*e^2*f*g^3*abs(e) - a*e^3*g^4*abs(e))/(c*d*e^3*g^5)) + 15*(c^2*d 
^2*e^4*f^2*g^2*abs(e) - 2*a*c*d*e^5*f*g^3*abs(e) + a^2*e^6*g^4*abs(e))/(c* 
d*e^3*g^5)) + 15*(c^3*d^3*e^3*f^3*abs(e) - 3*a*c^2*d^2*e^4*f^2*g*abs(e) + 
3*a^2*c*d*e^5*f*g^2*abs(e) - a^3*e^6*g^3*abs(e))*log(abs(-sqrt((e*x + d)*c 
*d*e - c*d^2*e + a*e^3)*sqrt(c*d*g) + sqrt(c^2*d^2*e^2*f - a*c*d*e^3*g + ( 
(e*x + d)*c*d*e - c*d^2*e + a*e^3)*c*d*g)))/(sqrt(c*d*g)*g^3))*abs(c)*abs( 
d)/(c*d*e^4) - (15*c^4*d^4*e^4*f^3*abs(c)*abs(d)*abs(e)*log(abs(-sqrt(-c*d 
^2*e + a*e^3)*sqrt(c*d*g) + sqrt(c^2*d^2*e^2*f - c^2*d^3*e*g))) - 45*a*c^3 
*d^3*e^5*f^2*g*abs(c)*abs(d)*abs(e)*log(abs(-sqrt(-c*d^2*e + a*e^3)*sqrt(c 
*d*g) + sqrt(c^2*d^2*e^2*f - c^2*d^3*e*g))) + 45*a^2*c^2*d^2*e^6*f*g^2*abs 
(c)*abs(d)*abs(e)*log(abs(-sqrt(-c*d^2*e + a*e^3)*sqrt(c*d*g) + sqrt(c^2*d 
^2*e^2*f - c^2*d^3*e*g))) - 15*a^3*c*d*e^7*g^3*abs(c)*abs(d)*abs(e)*log(ab 
s(-sqrt(-c*d^2*e + a*e^3)*sqrt(c*d*g) + sqrt(c^2*d^2*e^2*f - c^2*d^3*e*g)) 
) + 15*sqrt(c^2*d^2*e^2*f - c^2*d^3*e*g)*sqrt(-c*d^2*e + a*e^3)*sqrt(c*d*g 
)*c^2*d^2*e^2*f^2*abs(c)*abs(d)*abs(e) + 10*sqrt(c^2*d^2*e^2*f - c^2*d^3*e 
*g)*sqrt(-c*d^2*e + a*e^3)*sqrt(c*d*g)*c^2*d^3*e*f*g*abs(c)*abs(d)*abs(e) 
- 40*sqrt(c^2*d^2*e^2*f - c^2*d^3*e*g)*sqrt(-c*d^2*e + a*e^3)*sqrt(c*d*...
 
3.8.53.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^{5/2}} \,d x \]

input
int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/((f + g*x)^(1/2)*(d + e* 
x)^(5/2)),x)
 
output
int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/((f + g*x)^(1/2)*(d + e* 
x)^(5/2)), x)