\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx\) [76]

Optimal result

Integrand size = 48, antiderivative size = 280 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\frac {5 (c d f-a e g)^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 g^3 \sqrt {d+e x}}-\frac {5 (c d f-a e g) \sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2}}+\frac {\sqrt {f+g x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2}}-\frac {5 (c d f-a e g)^3 \text {arctanh}\left (\frac {\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {f+g x}}\right )}{8 \sqrt {c} \sqrt {d} g^{7/2}} \] Output:

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)-5/12*(-a*e*g+c*d*f)*(g*x+f)^(1/2)*(a*d*e+(a*e^2+c*d^2)* 
x+c*d*e*x^2)^(3/2)/g^2/(e*x+d)^(3/2)+1/3*(g*x+f)^(1/2)*(a*d*e+(a*e^2+c*d^2 
)*x+c*d*e*x^2)^(5/2)/g/(e*x+d)^(5/2)-5/8*(-a*e*g+c*d*f)^3*arctanh(g^(1/2)* 
(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^(1/2)/d^(1/2)/(e*x+d)^(1/2)/(g*x 
+f)^(1/2))/c^(1/2)/d^(1/2)/g^(7/2)
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\frac {((a e+c d x) (d+e x))^{5/2} \left (\frac {\sqrt {g} \sqrt {f+g x} \left (33 a^2 e^2 g^2+2 a c d e g (-20 f+13 g x)+c^2 d^2 \left (15 f^2-10 f g x+8 g^2 x^2\right )\right )}{(a e+c d x)^2}-\frac {15 (c d f-a e g)^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {f+g x}}{\sqrt {g} \sqrt {a e+c d x}}\right )}{\sqrt {c} \sqrt {d} (a e+c d x)^{5/2}}\right )}{24 g^{7/2} (d+e x)^{5/2}} \] 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))
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.11, 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)
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(497\) vs. \(2(236)=472\).

Time = 2.68 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.78

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

Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(1/2),x, 
method=_RETURNVERBOSE)
 

Output:

1/48*((e*x+d)*(c*d*x+a*e))^(1/2)*(g*x+f)^(1/2)*(15*ln(1/2*(2*c*d*g*x+a*e*g 
+d*f*c+2*((c*d*x+a*e)*(g*x+f))^(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+d*f*c+2*((c*d*x+a*e)*(g*x+f))^(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+d*f*c 
+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*c*d*g*x+a*e*g+d*f*c+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+16*c^2*d^2*g^2*x^2*((c*d*x+a*e)*(g*x+f 
))^(1/2)*(c*d*g)^(1/2)+52*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2)*a*c*d* 
e*g^2*x-20*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2)*c^2*d^2*f*g*x+66*((c* 
d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2)*a^2*e^2*g^2-80*a*c*d*e*f*g*((c*d*x+a 
*e)*(g*x+f))^(1/2)*(c*d*g)^(1/2)+30*((c*d*x+a*e)*(g*x+f))^(1/2)*(c*d*g)^(1 
/2)*c^2*d^2*f^2)/(e*x+d)^(1/2)/g^3/((c*d*x+a*e)*(g*x+f))^(1/2)/(c*d*g)^(1/ 
2)
 

Fricas [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 898, normalized size of antiderivative = 3.21 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx =\text {Too large to display} \] 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="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(1/2*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)/(c^2 
*d^2*e*g^2*x^3 + a*c*d^2*e*f*g + (c^2*d^2*e*f*g + (c^2*d^3 + a*c*d*e^2)*g^ 
2)*x^2 + (a*c*d^2*e*g^2 + (c^2*d^3 + a*c*d*e^2)*f*g)*x)))/(c*d*e*g^4*x + c 
*d^2*g^4)]
 

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
 

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)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\frac {{\left (\sqrt {c^{2} d^{2} e^{2} f - a c d e^{3} g + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} c d g} \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} {\left (2 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} {\left (\frac {4 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} {\left | e \right |}}{c d e^{3} g} - \frac {5 \, {\left (c d e^{2} f g^{3} {\left | e \right |} - a e^{3} g^{4} {\left | e \right |}\right )}}{c d e^{3} g^{5}}\right )} + \frac {15 \, {\left (c^{2} d^{2} e^{4} f^{2} g^{2} {\left | e \right |} - 2 \, a c d e^{5} f g^{3} {\left | e \right |} + a^{2} e^{6} g^{4} {\left | e \right |}\right )}}{c d e^{3} g^{5}}\right )} + \frac {15 \, {\left (c^{3} d^{3} e^{3} f^{3} {\left | e \right |} - 3 \, a c^{2} d^{2} e^{4} f^{2} g {\left | e \right |} + 3 \, a^{2} c d e^{5} f g^{2} {\left | e \right |} - a^{3} e^{6} g^{3} {\left | e \right |}\right )} \log \left ({\left | -\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} \sqrt {c d g} + \sqrt {c^{2} d^{2} e^{2} f - a c d e^{3} g + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} c d g} \right |}\right )}{\sqrt {c d g} g^{3}}\right )} {\left | c \right |} {\left | d \right |}}{24 \, c d e^{3} {\left | e \right |}} \] 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="giac")
 

Output:

1/24*(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^3*abs(e))
 

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)
 

Reduce [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} \sqrt {f+g x}} \, dx=\frac {33 \sqrt {g x +f}\, \sqrt {c d x +a e}\, a^{2} c d \,e^{2} g^{3}-40 \sqrt {g x +f}\, \sqrt {c d x +a e}\, a \,c^{2} d^{2} e f \,g^{2}+26 \sqrt {g x +f}\, \sqrt {c d x +a e}\, a \,c^{2} d^{2} e \,g^{3} x +15 \sqrt {g x +f}\, \sqrt {c d x +a e}\, c^{3} d^{3} f^{2} g -10 \sqrt {g x +f}\, \sqrt {c d x +a e}\, c^{3} d^{3} f \,g^{2} x +8 \sqrt {g x +f}\, \sqrt {c d x +a e}\, c^{3} d^{3} g^{3} x^{2}+15 \sqrt {g}\, \sqrt {d}\, \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {g}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {g x +f}}{\sqrt {a e g -c d f}}\right ) a^{3} e^{3} g^{3}-45 \sqrt {g}\, \sqrt {d}\, \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {g}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {g x +f}}{\sqrt {a e g -c d f}}\right ) a^{2} c d \,e^{2} f \,g^{2}+45 \sqrt {g}\, \sqrt {d}\, \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {g}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {g x +f}}{\sqrt {a e g -c d f}}\right ) a \,c^{2} d^{2} e \,f^{2} g -15 \sqrt {g}\, \sqrt {d}\, \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {g}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {g x +f}}{\sqrt {a e g -c d f}}\right ) c^{3} d^{3} f^{3}}{24 c d \,g^{4}} \] 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)
 

Output:

(33*sqrt(f + g*x)*sqrt(a*e + c*d*x)*a**2*c*d*e**2*g**3 - 40*sqrt(f + g*x)* 
sqrt(a*e + c*d*x)*a*c**2*d**2*e*f*g**2 + 26*sqrt(f + g*x)*sqrt(a*e + c*d*x 
)*a*c**2*d**2*e*g**3*x + 15*sqrt(f + g*x)*sqrt(a*e + c*d*x)*c**3*d**3*f**2 
*g - 10*sqrt(f + g*x)*sqrt(a*e + c*d*x)*c**3*d**3*f*g**2*x + 8*sqrt(f + g* 
x)*sqrt(a*e + c*d*x)*c**3*d**3*g**3*x**2 + 15*sqrt(g)*sqrt(d)*sqrt(c)*log( 
(sqrt(g)*sqrt(a*e + c*d*x) + sqrt(d)*sqrt(c)*sqrt(f + g*x))/sqrt(a*e*g - c 
*d*f))*a**3*e**3*g**3 - 45*sqrt(g)*sqrt(d)*sqrt(c)*log((sqrt(g)*sqrt(a*e + 
 c*d*x) + sqrt(d)*sqrt(c)*sqrt(f + g*x))/sqrt(a*e*g - c*d*f))*a**2*c*d*e** 
2*f*g**2 + 45*sqrt(g)*sqrt(d)*sqrt(c)*log((sqrt(g)*sqrt(a*e + c*d*x) + sqr 
t(d)*sqrt(c)*sqrt(f + g*x))/sqrt(a*e*g - c*d*f))*a*c**2*d**2*e*f**2*g - 15 
*sqrt(g)*sqrt(d)*sqrt(c)*log((sqrt(g)*sqrt(a*e + c*d*x) + sqrt(d)*sqrt(c)* 
sqrt(f + g*x))/sqrt(a*e*g - c*d*f))*c**3*d**3*f**3)/(24*c*d*g**4)