\(\int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [130]

Optimal result

Integrand size = 44, antiderivative size = 186 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \left (c d^2-a e^2\right ) (c d f-a e g) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}-\frac {2 \left (2 a e^2 g-c d (e f+d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 c^3 d^3 (d+e x)^{3/2}}+\frac {2 e g \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c^3 d^3 (d+e x)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.52 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (8 a^2 e^3 g-2 a c d e (5 e f+5 d g+2 e g x)+c^2 d^2 (5 d (3 f+g x)+e x (5 f+3 g x))\right )}{15 c^3 d^3 \sqrt {d+e x}} \] Input:

Integrate[((d + e*x)^(3/2)*(f + g*x))/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)]*(8*a^2*e^3*g - 2*a*c*d*e*(5*e*f + 5*d*g + 
 2*e*g*x) + c^2*d^2*(5*d*(3*f + g*x) + e*x*(5*f + 3*g*x))))/(15*c^3*d^3*Sq 
rt[d + e*x])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1221, 1128, 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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1128

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

\(\Big \downarrow \) 1122

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

Input:

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

Output:

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

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 1128

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*(m + 2*p + 
 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1)))   Int[(d 
 + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[Simplify[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]
 

Maple [A] (verified)

Time = 3.56 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {2 \sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (3 e g \,x^{2} d^{2} c^{2}-4 a c d \,e^{2} g x +5 c^{2} d^{3} g x +5 c^{2} d^{2} e f x +8 a^{2} e^{3} g -10 a c \,d^{2} e g -10 a c d \,e^{2} f +15 d^{3} f \,c^{2}\right )}{15 \sqrt {e x +d}\, d^{3} c^{3}}\)	\(113\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (3 e g \,x^{2} d^{2} c^{2}-4 a c d \,e^{2} g x +5 c^{2} d^{3} g x +5 c^{2} d^{2} e f x +8 a^{2} e^{3} g -10 a c \,d^{2} e g -10 a c d \,e^{2} f +15 d^{3} f \,c^{2}\right ) \sqrt {e x +d}}{15 d^{3} c^{3} \sqrt {c d \,x^{2} e +a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(131\)
orering	\(\frac {2 \left (3 e g \,x^{2} d^{2} c^{2}-4 a c d \,e^{2} g x +5 c^{2} d^{3} g x +5 c^{2} d^{2} e f x +8 a^{2} e^{3} g -10 a c \,d^{2} e g -10 a c d \,e^{2} f +15 d^{3} f \,c^{2}\right ) \left (c d x +a e \right ) \sqrt {e x +d}}{15 d^{3} c^{3} \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e}}\)	\(132\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.76 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (3 \, c^{2} d^{2} e g x^{2} + 5 \, {\left (3 \, c^{2} d^{3} - 2 \, a c d e^{2}\right )} f - 2 \, {\left (5 \, a c d^{2} e - 4 \, a^{2} e^{3}\right )} g + {\left (5 \, c^{2} d^{2} e f + {\left (5 \, c^{2} d^{3} - 4 \, a c d e^{2}\right )} g\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{15 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (c^{2} d^{2} e x^{2} + 3 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (3 \, c^{2} d^{3} - a c d e^{2}\right )} x\right )} f}{3 \, \sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (3 \, c^{3} d^{3} e x^{3} - 10 \, a^{2} c d^{2} e^{2} + 8 \, a^{3} e^{4} + {\left (5 \, c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} x^{2} - {\left (5 \, a c^{2} d^{3} e - 4 \, a^{2} c d e^{3}\right )} x\right )} g}{15 \, \sqrt {c d x + a e} c^{3} d^{3}} \] Input:

integrate((e*x+d)^(3/2)*(g*x+f)/(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/(sqrt(c*d*x + a*e)*c^2*d^2) + 2/15*(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)*g/(sqrt(c*d*x + a*e)*c^3*d^3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.13 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, e {\left (\frac {15 \, {\left (c^{2} d^{3} f - a c d e^{2} f - a c d^{2} e g + a^{2} e^{3} g\right )} \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{c^{3} d^{3} e} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d e^{2} f + 5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d^{2} e g - 10 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} g + 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} g}{c^{3} d^{3} e^{4}}\right )}}{15 \, {\left | e \right |}} \] Input:

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

Output:

2/15*e*(15*(c^2*d^3*f - a*c*d*e^2*f - a*c*d^2*e*g + a^2*e^3*g)*sqrt((e*x + 
 d)*c*d*e - c*d^2*e + a*e^3)/(c^3*d^3*e) + (5*((e*x + d)*c*d*e - c*d^2*e + 
 a*e^3)^(3/2)*c*d*e^2*f + 5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c*d^ 
2*e*g - 10*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*e^3*g + 3*((e*x + d 
)*c*d*e - c*d^2*e + a*e^3)^(5/2)*g)/(c^3*d^3*e^4))/abs(e)
 

Mupad [B] (verification not implemented)

Time = 6.65 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (16\,g\,a^2\,e^3-20\,g\,a\,c\,d^2\,e-20\,f\,a\,c\,d\,e^2+30\,f\,c^2\,d^3\right )}{15\,c^3\,d^3\,e}+\frac {2\,g\,x^2\,\sqrt {d+e\,x}}{5\,c\,d}+\frac {2\,x\,\sqrt {d+e\,x}\,\left (5\,c\,g\,d^2+5\,c\,f\,d\,e-4\,a\,g\,e^2\right )}{15\,c^2\,d^2\,e}\right )}{x+\frac {d}{e}} \] Input:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.53 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {c d x +a e}\, \left (3 c^{2} d^{2} e g \,x^{2}-4 a c d \,e^{2} g x +5 c^{2} d^{3} g x +5 c^{2} d^{2} e f x +8 a^{2} e^{3} g -10 a c \,d^{2} e g -10 a c d \,e^{2} f +15 c^{2} d^{3} f \right )}{15 c^{3} d^{3}} \] Input:

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