\(\int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [129]

Optimal result

Integrand size = 46, antiderivative size = 270 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \left (c d^2-a e^2\right ) (c d f-a e g)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^4 d^4 \sqrt {d+e x}}-\frac {2 (c d f-a e g) \left (3 a e^2 g-c d (e f+2 d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 c^4 d^4 (d+e x)^{3/2}}-\frac {2 g \left (3 a e^2 g-c d (2 e f+d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c^4 d^4 (d+e x)^{5/2}}+\frac {2 e g^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{7 c^4 d^4 (d+e x)^{7/2}} \] Output:

2*(-a*e^2+c*d^2)*(-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)-2/3*(-a*e*g+c*d*f)*(3*a*e^2*g-c*d*(2*d*g+e*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*e^2*g- 
c*d*(d*g+2*e*f))*(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)+2/7*e*g^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(7/2)/c^4/d^4/(e*x+d)^(7/ 
2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.63 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-48 a^3 e^4 g^2+8 a^2 c d e^2 g (14 e f+7 d g+3 e g x)-2 a c^2 d^2 e \left (14 d g (5 f+g x)+e \left (35 f^2+28 f g x+9 g^2 x^2\right )\right )+c^3 d^3 \left (7 d \left (15 f^2+10 f g x+3 g^2 x^2\right )+e x \left (35 f^2+42 f g x+15 g^2 x^2\right )\right )\right )}{105 c^4 d^4 \sqrt {d+e x}} \] Input:

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

Output:

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-48*a^3*e^4*g^2 + 8*a^2*c*d*e^2*g*(14*e* 
f + 7*d*g + 3*e*g*x) - 2*a*c^2*d^2*e*(14*d*g*(5*f + g*x) + e*(35*f^2 + 28* 
f*g*x + 9*g^2*x^2)) + c^3*d^3*(7*d*(15*f^2 + 10*f*g*x + 3*g^2*x^2) + e*x*( 
35*f^2 + 42*f*g*x + 15*g^2*x^2))))/(105*c^4*d^4*Sqrt[d + e*x])
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1258, 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.

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

\(\Big \downarrow \) 1258

\(\displaystyle \frac {1}{7} \left (-\frac {6 a e^2}{c d}+7 d-\frac {e f}{g}\right ) \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx+\frac {2 e (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d g \sqrt {d+e x}}\)

\(\Big \downarrow \) 1253

\(\displaystyle \frac {1}{7} \left (-\frac {6 a e^2}{c d}+7 d-\frac {e f}{g}\right ) \left (\frac {4 (c d f-a e g) \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{5 c d}+\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}}\right )+\frac {2 e (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d g \sqrt {d+e x}}\)

\(\Big \downarrow \) 1221

\(\displaystyle \frac {1}{7} \left (-\frac {6 a e^2}{c d}+7 d-\frac {e f}{g}\right ) \left (\frac {4 (c d f-a e g) \left (\frac {1}{3} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right ) \int \frac {\sqrt {d+e x}}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{5 c d}+\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}}\right )+\frac {2 e (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d g \sqrt {d+e x}}\)

\(\Big \downarrow \) 1122

\(\displaystyle \frac {2 e (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d g \sqrt {d+e x}}+\frac {1}{7} \left (-\frac {6 a e^2}{c d}+7 d-\frac {e f}{g}\right ) \left (\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}}+\frac {4 (c d f-a e g) \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right )}{3 c d \sqrt {d+e x}}+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{5 c d}\right )\)

Input:

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

Output:

(2*e*(f + g*x)^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(7*c*d*g*Sqr 
t[d + e*x]) + ((7*d - (6*a*e^2)/(c*d) - (e*f)/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)))/7
 

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 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])
 

rule 1258

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
+ (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e^2*(d + e*x)^(m - 2)*(f + g*x)^(n 
+ 1)*((a + b*x + c*x^2)^(p + 1)/(c*g*(n + p + 2))), x] - Simp[(b*e*g*(n + 1 
) + c*e*f*(p + 1) - c*d*g*(2*n + p + 3))/(c*g*(n + p + 2))   Int[(d + e*x)^ 
(m - 1)*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, 
 g, m, n, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p - 1, 0] && 
!LtQ[n, -1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 3.57 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {2 \sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (-15 e \,g^{2} x^{3} d^{3} c^{3}+18 a \,c^{2} d^{2} e^{2} g^{2} x^{2}-21 c^{3} d^{4} g^{2} x^{2}-42 c^{3} d^{3} e f g \,x^{2}-24 a^{2} c d \,e^{3} g^{2} x +28 a \,c^{2} d^{3} e \,g^{2} x +56 a \,c^{2} d^{2} e^{2} f g x -70 c^{3} d^{4} f g x -35 c^{3} d^{3} e \,f^{2} x +48 a^{3} e^{4} g^{2}-56 a^{2} c \,d^{2} e^{2} g^{2}-112 a^{2} c d \,e^{3} f g +140 a \,c^{2} d^{3} e f g +70 a \,c^{2} d^{2} e^{2} f^{2}-105 d^{4} f^{2} c^{3}\right )}{105 \sqrt {e x +d}\, d^{4} c^{4}}\)	\(237\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-15 e \,g^{2} x^{3} d^{3} c^{3}+18 a \,c^{2} d^{2} e^{2} g^{2} x^{2}-21 c^{3} d^{4} g^{2} x^{2}-42 c^{3} d^{3} e f g \,x^{2}-24 a^{2} c d \,e^{3} g^{2} x +28 a \,c^{2} d^{3} e \,g^{2} x +56 a \,c^{2} d^{2} e^{2} f g x -70 c^{3} d^{4} f g x -35 c^{3} d^{3} e \,f^{2} x +48 a^{3} e^{4} g^{2}-56 a^{2} c \,d^{2} e^{2} g^{2}-112 a^{2} c d \,e^{3} f g +140 a \,c^{2} d^{3} e f g +70 a \,c^{2} d^{2} e^{2} f^{2}-105 d^{4} f^{2} c^{3}\right ) \sqrt {e x +d}}{105 d^{4} c^{4} \sqrt {c d \,x^{2} e +a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(255\)
orering	\(-\frac {2 \left (-15 e \,g^{2} x^{3} d^{3} c^{3}+18 a \,c^{2} d^{2} e^{2} g^{2} x^{2}-21 c^{3} d^{4} g^{2} x^{2}-42 c^{3} d^{3} e f g \,x^{2}-24 a^{2} c d \,e^{3} g^{2} x +28 a \,c^{2} d^{3} e \,g^{2} x +56 a \,c^{2} d^{2} e^{2} f g x -70 c^{3} d^{4} f g x -35 c^{3} d^{3} e \,f^{2} x +48 a^{3} e^{4} g^{2}-56 a^{2} c \,d^{2} e^{2} g^{2}-112 a^{2} c d \,e^{3} f g +140 a \,c^{2} d^{3} e f g +70 a \,c^{2} d^{2} e^{2} f^{2}-105 d^{4} f^{2} c^{3}\right ) \left (c d x +a e \right ) \sqrt {e x +d}}{105 d^{4} c^{4} \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e}}\)	\(256\)

Input:

int((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2),x,meth 
od=_RETURNVERBOSE)
 

Output:

-2/105/(e*x+d)^(1/2)*((e*x+d)*(c*d*x+a*e))^(1/2)*(-15*c^3*d^3*e*g^2*x^3+18 
*a*c^2*d^2*e^2*g^2*x^2-21*c^3*d^4*g^2*x^2-42*c^3*d^3*e*f*g*x^2-24*a^2*c*d* 
e^3*g^2*x+28*a*c^2*d^3*e*g^2*x+56*a*c^2*d^2*e^2*f*g*x-70*c^3*d^4*f*g*x-35* 
c^3*d^3*e*f^2*x+48*a^3*e^4*g^2-56*a^2*c*d^2*e^2*g^2-112*a^2*c*d*e^3*f*g+14 
0*a*c^2*d^3*e*f*g+70*a*c^2*d^2*e^2*f^2-105*c^3*d^4*f^2)/d^4/c^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (15 \, c^{3} d^{3} e g^{2} x^{3} + 35 \, {\left (3 \, c^{3} d^{4} - 2 \, a c^{2} d^{2} e^{2}\right )} f^{2} - 28 \, {\left (5 \, a c^{2} d^{3} e - 4 \, a^{2} c d e^{3}\right )} f g + 8 \, {\left (7 \, a^{2} c d^{2} e^{2} - 6 \, a^{3} e^{4}\right )} g^{2} + 3 \, {\left (14 \, c^{3} d^{3} e f g + {\left (7 \, c^{3} d^{4} - 6 \, a c^{2} d^{2} e^{2}\right )} g^{2}\right )} x^{2} + {\left (35 \, c^{3} d^{3} e f^{2} + 14 \, {\left (5 \, c^{3} d^{4} - 4 \, a c^{2} d^{2} e^{2}\right )} f g - 4 \, {\left (7 \, a c^{2} d^{3} e - 6 \, a^{2} c d e^{3}\right )} g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{105 \, {\left (c^{4} d^{4} e x + c^{4} d^{5}\right )}} \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2), 
x, algorithm="fricas")
 

Output:

2/105*(15*c^3*d^3*e*g^2*x^3 + 35*(3*c^3*d^4 - 2*a*c^2*d^2*e^2)*f^2 - 28*(5 
*a*c^2*d^3*e - 4*a^2*c*d*e^3)*f*g + 8*(7*a^2*c*d^2*e^2 - 6*a^3*e^4)*g^2 + 
3*(14*c^3*d^3*e*f*g + (7*c^3*d^4 - 6*a*c^2*d^2*e^2)*g^2)*x^2 + (35*c^3*d^3 
*e*f^2 + 14*(5*c^3*d^4 - 4*a*c^2*d^2*e^2)*f*g - 4*(7*a*c^2*d^3*e - 6*a^2*c 
*d*e^3)*g^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/ 
(c^4*d^4*e*x + c^4*d^5)
 

Sympy [F]

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

integrate((e*x+d)**(3/2)*(g*x+f)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)** 
(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.14 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (c^{2} d^{2} e x^{2} + 3 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (3 \, c^{2} d^{3} - a c d e^{2}\right )} x\right )} f^{2}}{3 \, \sqrt {c d x + a e} c^{2} d^{2}} + \frac {4 \, {\left (3 \, c^{3} d^{3} e x^{3} - 10 \, a^{2} c d^{2} e^{2} + 8 \, a^{3} e^{4} + {\left (5 \, c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} x^{2} - {\left (5 \, a c^{2} d^{3} e - 4 \, a^{2} c d e^{3}\right )} x\right )} f g}{15 \, \sqrt {c d x + a e} c^{3} d^{3}} + \frac {2 \, {\left (15 \, c^{4} d^{4} e x^{4} + 56 \, a^{3} c d^{2} e^{3} - 48 \, a^{4} e^{5} + 3 \, {\left (7 \, c^{4} d^{5} - a c^{3} d^{3} e^{2}\right )} x^{3} - {\left (7 \, a c^{3} d^{4} e - 6 \, a^{2} c^{2} d^{2} e^{3}\right )} x^{2} + 4 \, {\left (7 \, a^{2} c^{2} d^{3} e^{2} - 6 \, a^{3} c d e^{4}\right )} x\right )} g^{2}}{105 \, \sqrt {c d x + a e} c^{4} d^{4}} \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2), 
x, algorithm="maxima")
 

Output:

2/3*(c^2*d^2*e*x^2 + 3*a*c*d^2*e - 2*a^2*e^3 + (3*c^2*d^3 - a*c*d*e^2)*x)* 
f^2/(sqrt(c*d*x + a*e)*c^2*d^2) + 4/15*(3*c^3*d^3*e*x^3 - 10*a^2*c*d^2*e^2 
 + 8*a^3*e^4 + (5*c^3*d^4 - a*c^2*d^2*e^2)*x^2 - (5*a*c^2*d^3*e - 4*a^2*c* 
d*e^3)*x)*f*g/(sqrt(c*d*x + a*e)*c^3*d^3) + 2/105*(15*c^4*d^4*e*x^4 + 56*a 
^3*c*d^2*e^3 - 48*a^4*e^5 + 3*(7*c^4*d^5 - a*c^3*d^3*e^2)*x^3 - (7*a*c^3*d 
^4*e - 6*a^2*c^2*d^2*e^3)*x^2 + 4*(7*a^2*c^2*d^3*e^2 - 6*a^3*c*d*e^4)*x)*g 
^2/(sqrt(c*d*x + a*e)*c^4*d^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, e {\left (\frac {105 \, {\left (c^{3} d^{4} f^{2} - a c^{2} d^{2} e^{2} f^{2} - 2 \, a c^{2} d^{3} e f g + 2 \, a^{2} c d e^{3} f g + a^{2} c d^{2} e^{2} g^{2} - a^{3} e^{4} g^{2}\right )} \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{c^{4} d^{4} e} + \frac {35 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{2} d^{2} e^{4} f^{2} + 70 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{2} d^{3} e^{3} f g - 140 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a c d e^{5} f g - 70 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a c d^{2} e^{4} g^{2} + 105 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a^{2} e^{6} g^{2} + 42 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} c d e^{2} f g + 21 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} c d^{2} e g^{2} - 63 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} a e^{3} g^{2} + 15 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {7}{2}} g^{2}}{c^{4} d^{4} e^{6}}\right )}}{105 \, {\left | e \right |}} \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2), 
x, algorithm="giac")
 

Output:

2/105*e*(105*(c^3*d^4*f^2 - a*c^2*d^2*e^2*f^2 - 2*a*c^2*d^3*e*f*g + 2*a^2* 
c*d*e^3*f*g + a^2*c*d^2*e^2*g^2 - a^3*e^4*g^2)*sqrt((e*x + d)*c*d*e - c*d^ 
2*e + a*e^3)/(c^4*d^4*e) + (35*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c 
^2*d^2*e^4*f^2 + 70*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c^2*d^3*e^3* 
f*g - 140*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*c*d*e^5*f*g - 70*((e 
*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*c*d^2*e^4*g^2 + 105*((e*x + d)*c* 
d*e - c*d^2*e + a*e^3)^(3/2)*a^2*e^6*g^2 + 42*((e*x + d)*c*d*e - c*d^2*e + 
 a*e^3)^(5/2)*c*d*e^2*f*g + 21*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*c 
*d^2*e*g^2 - 63*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*a*e^3*g^2 + 15*( 
(e*x + d)*c*d*e - c*d^2*e + a*e^3)^(7/2)*g^2)/(c^4*d^4*e^6))/abs(e)
 

Mupad [B] (verification not implemented)

Time = 6.75 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.03 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,g^2\,x^3\,\sqrt {d+e\,x}}{7\,c\,d}-\frac {\sqrt {d+e\,x}\,\left (96\,a^3\,e^4\,g^2-112\,a^2\,c\,d^2\,e^2\,g^2-224\,a^2\,c\,d\,e^3\,f\,g+280\,a\,c^2\,d^3\,e\,f\,g+140\,a\,c^2\,d^2\,e^2\,f^2-210\,c^3\,d^4\,f^2\right )}{105\,c^4\,d^4\,e}+\frac {x\,\sqrt {d+e\,x}\,\left (48\,a^2\,c\,d\,e^3\,g^2-56\,a\,c^2\,d^3\,e\,g^2-112\,a\,c^2\,d^2\,e^2\,f\,g+140\,c^3\,d^4\,f\,g+70\,c^3\,d^3\,e\,f^2\right )}{105\,c^4\,d^4\,e}+\frac {2\,g\,x^2\,\sqrt {d+e\,x}\,\left (7\,c\,g\,d^2+14\,c\,f\,d\,e-6\,a\,g\,e^2\right )}{35\,c^2\,d^2\,e}\right )}{x+\frac {d}{e}} \] Input:

int(((f + g*x)^2*(d + e*x)^(3/2))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^ 
(1/2),x)
 

Output:

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*g^2*x^3*(d + e*x)^(1/2) 
)/(7*c*d) - ((d + e*x)^(1/2)*(96*a^3*e^4*g^2 - 210*c^3*d^4*f^2 + 140*a*c^2 
*d^2*e^2*f^2 - 112*a^2*c*d^2*e^2*g^2 + 280*a*c^2*d^3*e*f*g - 224*a^2*c*d*e 
^3*f*g))/(105*c^4*d^4*e) + (x*(d + e*x)^(1/2)*(70*c^3*d^3*e*f^2 + 140*c^3* 
d^4*f*g - 56*a*c^2*d^3*e*g^2 + 48*a^2*c*d*e^3*g^2 - 112*a*c^2*d^2*e^2*f*g) 
)/(105*c^4*d^4*e) + (2*g*x^2*(d + e*x)^(1/2)*(7*c*d^2*g - 6*a*e^2*g + 14*c 
*d*e*f))/(35*c^2*d^2*e)))/(x + d/e)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {c d x +a e}\, \left (15 c^{3} d^{3} e \,g^{2} x^{3}-18 a \,c^{2} d^{2} e^{2} g^{2} x^{2}+21 c^{3} d^{4} g^{2} x^{2}+42 c^{3} d^{3} e f g \,x^{2}+24 a^{2} c d \,e^{3} g^{2} x -28 a \,c^{2} d^{3} e \,g^{2} x -56 a \,c^{2} d^{2} e^{2} f g x +70 c^{3} d^{4} f g x +35 c^{3} d^{3} e \,f^{2} x -48 a^{3} e^{4} g^{2}+56 a^{2} c \,d^{2} e^{2} g^{2}+112 a^{2} c d \,e^{3} f g -140 a \,c^{2} d^{3} e f g -70 a \,c^{2} d^{2} e^{2} f^{2}+105 c^{3} d^{4} f^{2}\right )}{105 c^{4} d^{4}} \] Input:

int((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)
 

Output:

(2*sqrt(a*e + c*d*x)*( - 48*a**3*e**4*g**2 + 56*a**2*c*d**2*e**2*g**2 + 11 
2*a**2*c*d*e**3*f*g + 24*a**2*c*d*e**3*g**2*x - 140*a*c**2*d**3*e*f*g - 28 
*a*c**2*d**3*e*g**2*x - 70*a*c**2*d**2*e**2*f**2 - 56*a*c**2*d**2*e**2*f*g 
*x - 18*a*c**2*d**2*e**2*g**2*x**2 + 105*c**3*d**4*f**2 + 70*c**3*d**4*f*g 
*x + 21*c**3*d**4*g**2*x**2 + 35*c**3*d**3*e*f**2*x + 42*c**3*d**3*e*f*g*x 
**2 + 15*c**3*d**3*e*g**2*x**3))/(105*c**4*d**4)