\(\int \frac {(d+e x)^{3/2} (f+g x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [42]

Optimal result

Integrand size = 46, antiderivative size = 228 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 (c d f-a e g)^3 \sqrt {d+e x}}{c^4 d^4 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {6 g (c d f-a e g)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^4 d^4 \sqrt {d+e x}}+\frac {2 g^2 (c d f-a e g) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{c^4 d^4 (d+e x)^{3/2}}+\frac {2 g^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c^4 d^4 (d+e x)^{5/2}} \] Output:

-2*(-a*e*g+c*d*f)^3*(e*x+d)^(1/2)/c^4/d^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2 
)^(1/2)+6*g*(-a*e*g+c*d*f)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^4/d 
^4/(e*x+d)^(1/2)+2*g^2*(-a*e*g+c*d*f)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3 
/2)/c^4/d^4/(e*x+d)^(3/2)+2/5*g^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/ 
c^4/d^4/(e*x+d)^(5/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.59 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d+e x} \left (16 a^3 e^3 g^3+8 a^2 c d e^2 g^2 (-5 f+g x)-2 a c^2 d^2 e g \left (-15 f^2+10 f g x+g^2 x^2\right )+c^3 d^3 \left (-5 f^3+15 f^2 g x+5 f g^2 x^2+g^3 x^3\right )\right )}{5 c^4 d^4 \sqrt {(a e+c d x) (d+e x)}} \] Input:

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

Output:

(2*Sqrt[d + e*x]*(16*a^3*e^3*g^3 + 8*a^2*c*d*e^2*g^2*(-5*f + g*x) - 2*a*c^ 
2*d^2*e*g*(-15*f^2 + 10*f*g*x + g^2*x^2) + c^3*d^3*(-5*f^3 + 15*f^2*g*x + 
5*f*g^2*x^2 + g^3*x^3)))/(5*c^4*d^4*Sqrt[(a*e + c*d*x)*(d + e*x)])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1251, 1253, 1221, 1122}

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.

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

\(\Big \downarrow \) 1251

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

\(\Big \downarrow \) 1253

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1122

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1122

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), 
 x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && 
EqQ[m + p, 0]
 

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

rule 1251

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] - Simp[e*g*(n/(c*(p + 1)))   Int[( 
d + e*x)^(m - 1)*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 
 0] && LtQ[p, -1] && GtQ[n, 0]
 

rule 1253

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] - Simp[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] && EqQ[c*d^2 - b*d* 
e + a*e^2, 0] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (Intege 
rQ[2*p] || IntegerQ[n])
 

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {2 \sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (x^{3} g^{3} d^{3} c^{3}-2 a \,c^{2} d^{2} e \,g^{3} x^{2}+5 c^{3} d^{3} f \,g^{2} x^{2}+8 a^{2} c d \,e^{2} g^{3} x -20 a \,c^{2} d^{2} e f \,g^{2} x +15 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-40 a^{2} c d \,e^{2} f \,g^{2}+30 a \,c^{2} d^{2} e \,f^{2} g -5 f^{3} d^{3} c^{3}\right )}{5 \sqrt {e x +d}\, \left (c d x +a e \right ) c^{4} d^{4}}\)	\(179\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (x^{3} g^{3} d^{3} c^{3}-2 a \,c^{2} d^{2} e \,g^{3} x^{2}+5 c^{3} d^{3} f \,g^{2} x^{2}+8 a^{2} c d \,e^{2} g^{3} x -20 a \,c^{2} d^{2} e f \,g^{2} x +15 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-40 a^{2} c d \,e^{2} f \,g^{2}+30 a \,c^{2} d^{2} e \,f^{2} g -5 f^{3} d^{3} c^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{5 d^{4} c^{4} \left (c d \,x^{2} e +a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}\)	\(187\)
orering	\(\frac {2 \left (x^{3} g^{3} d^{3} c^{3}-2 a \,c^{2} d^{2} e \,g^{3} x^{2}+5 c^{3} d^{3} f \,g^{2} x^{2}+8 a^{2} c d \,e^{2} g^{3} x -20 a \,c^{2} d^{2} e f \,g^{2} x +15 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-40 a^{2} c d \,e^{2} f \,g^{2}+30 a \,c^{2} d^{2} e \,f^{2} g -5 f^{3} d^{3} c^{3}\right ) \left (c d x +a e \right ) \left (e x +d \right )^{\frac {3}{2}}}{5 d^{4} c^{4} {\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{\frac {3}{2}}}\)	\(188\)

Input:

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

Output:

2/5/(e*x+d)^(1/2)*((e*x+d)*(c*d*x+a*e))^(1/2)*(c^3*d^3*g^3*x^3-2*a*c^2*d^2 
*e*g^3*x^2+5*c^3*d^3*f*g^2*x^2+8*a^2*c*d*e^2*g^3*x-20*a*c^2*d^2*e*f*g^2*x+ 
15*c^3*d^3*f^2*g*x+16*a^3*e^3*g^3-40*a^2*c*d*e^2*f*g^2+30*a*c^2*d^2*e*f^2* 
g-5*c^3*d^3*f^3)/(c*d*x+a*e)/c^4/d^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (c^{3} d^{3} g^{3} x^{3} - 5 \, c^{3} d^{3} f^{3} + 30 \, a c^{2} d^{2} e f^{2} g - 40 \, a^{2} c d e^{2} f g^{2} + 16 \, a^{3} e^{3} g^{3} + {\left (5 \, c^{3} d^{3} f g^{2} - 2 \, a c^{2} d^{2} e g^{3}\right )} x^{2} + {\left (15 \, c^{3} d^{3} f^{2} g - 20 \, a c^{2} d^{2} e f g^{2} + 8 \, a^{2} c d e^{2} g^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{5 \, {\left (c^{5} d^{5} e x^{2} + a c^{4} d^{5} e + {\left (c^{5} d^{6} + a c^{4} d^{4} e^{2}\right )} x\right )}} \] Input:

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

Output:

2/5*(c^3*d^3*g^3*x^3 - 5*c^3*d^3*f^3 + 30*a*c^2*d^2*e*f^2*g - 40*a^2*c*d*e 
^2*f*g^2 + 16*a^3*e^3*g^3 + (5*c^3*d^3*f*g^2 - 2*a*c^2*d^2*e*g^3)*x^2 + (1 
5*c^3*d^3*f^2*g - 20*a*c^2*d^2*e*f*g^2 + 8*a^2*c*d*e^2*g^3)*x)*sqrt(c*d*e* 
x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^5*d^5*e*x^2 + a*c^4*d^5* 
e + (c^5*d^6 + a*c^4*d^4*e^2)*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \, f^{3}}{\sqrt {c d x + a e} c d} + \frac {6 \, {\left (c d x + 2 \, a e\right )} f^{2} g}{\sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (c^{2} d^{2} x^{2} - 4 \, a c d e x - 8 \, a^{2} e^{2}\right )} f g^{2}}{\sqrt {c d x + a e} c^{3} d^{3}} + \frac {2 \, {\left (c^{3} d^{3} x^{3} - 2 \, a c^{2} d^{2} e x^{2} + 8 \, a^{2} c d e^{2} x + 16 \, a^{3} e^{3}\right )} g^{3}}{5 \, \sqrt {c d x + a e} c^{4} d^{4}} \] Input:

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

Output:

-2*f^3/(sqrt(c*d*x + a*e)*c*d) + 6*(c*d*x + 2*a*e)*f^2*g/(sqrt(c*d*x + a*e 
)*c^2*d^2) + 2*(c^2*d^2*x^2 - 4*a*c*d*e*x - 8*a^2*e^2)*f*g^2/(sqrt(c*d*x + 
 a*e)*c^3*d^3) + 2/5*(c^3*d^3*x^3 - 2*a*c^2*d^2*e*x^2 + 8*a^2*c*d*e^2*x + 
16*a^3*e^3)*g^3/(sqrt(c*d*x + a*e)*c^4*d^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (c^{3} d^{3} f^{3} - 3 \, a c^{2} d^{2} e f^{2} g + 3 \, a^{2} c d e^{2} f g^{2} - a^{3} e^{3} g^{3}\right )}}{\sqrt {c d x + a e} c^{4} d^{4}} + \frac {2 \, {\left (15 \, \sqrt {c d x + a e} c^{18} d^{18} f^{2} g - 30 \, \sqrt {c d x + a e} a c^{17} d^{17} e f g^{2} + 15 \, \sqrt {c d x + a e} a^{2} c^{16} d^{16} e^{2} g^{3} + 5 \, {\left (c d x + a e\right )}^{\frac {3}{2}} c^{17} d^{17} f g^{2} - 5 \, {\left (c d x + a e\right )}^{\frac {3}{2}} a c^{16} d^{16} e g^{3} + {\left (c d x + a e\right )}^{\frac {5}{2}} c^{16} d^{16} g^{3}\right )}}{5 \, c^{20} d^{20}} \] Input:

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

Output:

-2*(c^3*d^3*f^3 - 3*a*c^2*d^2*e*f^2*g + 3*a^2*c*d*e^2*f*g^2 - a^3*e^3*g^3) 
/(sqrt(c*d*x + a*e)*c^4*d^4) + 2/5*(15*sqrt(c*d*x + a*e)*c^18*d^18*f^2*g - 
 30*sqrt(c*d*x + a*e)*a*c^17*d^17*e*f*g^2 + 15*sqrt(c*d*x + a*e)*a^2*c^16* 
d^16*e^2*g^3 + 5*(c*d*x + a*e)^(3/2)*c^17*d^17*f*g^2 - 5*(c*d*x + a*e)^(3/ 
2)*a*c^16*d^16*e*g^3 + (c*d*x + a*e)^(5/2)*c^16*d^16*g^3)/(c^20*d^20)
 

Mupad [B] (verification not implemented)

Time = 6.50 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.11 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (32\,a^3\,e^3\,g^3-80\,a^2\,c\,d\,e^2\,f\,g^2+60\,a\,c^2\,d^2\,e\,f^2\,g-10\,c^3\,d^3\,f^3\right )}{5\,c^5\,d^5\,e}+\frac {2\,g^3\,x^3\,\sqrt {d+e\,x}}{5\,c^2\,d^2\,e}-\frac {2\,g^2\,x^2\,\left (2\,a\,e\,g-5\,c\,d\,f\right )\,\sqrt {d+e\,x}}{5\,c^3\,d^3\,e}+\frac {2\,g\,x\,\sqrt {d+e\,x}\,\left (8\,a^2\,e^2\,g^2-20\,a\,c\,d\,e\,f\,g+15\,c^2\,d^2\,f^2\right )}{5\,c^4\,d^4\,e}\right )}{\frac {a}{c}+x^2+\frac {x\,\left (5\,c^5\,d^6+5\,a\,c^4\,d^4\,e^2\right )}{5\,c^5\,d^5\,e}} \] Input:

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

Output:

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(((d + e*x)^(1/2)*(32*a^3*e 
^3*g^3 - 10*c^3*d^3*f^3 + 60*a*c^2*d^2*e*f^2*g - 80*a^2*c*d*e^2*f*g^2))/(5 
*c^5*d^5*e) + (2*g^3*x^3*(d + e*x)^(1/2))/(5*c^2*d^2*e) - (2*g^2*x^2*(2*a* 
e*g - 5*c*d*f)*(d + e*x)^(1/2))/(5*c^3*d^3*e) + (2*g*x*(d + e*x)^(1/2)*(8* 
a^2*e^2*g^2 + 15*c^2*d^2*f^2 - 20*a*c*d*e*f*g))/(5*c^4*d^4*e)))/(a/c + x^2 
 + (x*(5*c^5*d^6 + 5*a*c^4*d^4*e^2))/(5*c^5*d^5*e))
 

Reduce [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {\frac {2}{5} c^{3} d^{3} g^{3} x^{3}-\frac {4}{5} a \,c^{2} d^{2} e \,g^{3} x^{2}+2 c^{3} d^{3} f \,g^{2} x^{2}+\frac {16}{5} a^{2} c d \,e^{2} g^{3} x -8 a \,c^{2} d^{2} e f \,g^{2} x +6 c^{3} d^{3} f^{2} g x +\frac {32}{5} a^{3} e^{3} g^{3}-16 a^{2} c d \,e^{2} f \,g^{2}+12 a \,c^{2} d^{2} e \,f^{2} g -2 c^{3} d^{3} f^{3}}{\sqrt {c d x +a e}\, c^{4} d^{4}} \] Input:

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

Output:

(2*(16*a**3*e**3*g**3 - 40*a**2*c*d*e**2*f*g**2 + 8*a**2*c*d*e**2*g**3*x + 
 30*a*c**2*d**2*e*f**2*g - 20*a*c**2*d**2*e*f*g**2*x - 2*a*c**2*d**2*e*g** 
3*x**2 - 5*c**3*d**3*f**3 + 15*c**3*d**3*f**2*g*x + 5*c**3*d**3*f*g**2*x** 
2 + c**3*d**3*g**3*x**3))/(5*sqrt(a*e + c*d*x)*c**4*d**4)