Integrand size = 25, antiderivative size = 155 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=-\frac {2 (b e-a f) (d e-c f) (f g-e h)}{f^4 \sqrt {e+f x}}-\frac {2 (b d e (2 f g-3 e h)-b c f (f g-2 e h)-a f (d f g-2 d e h+c f h)) \sqrt {e+f x}}{f^4}+\frac {2 (a d f h+b (d f g-3 d e h+c f h)) (e+f x)^{3/2}}{3 f^4}+\frac {2 b d h (e+f x)^{5/2}}{5 f^4} \] Output:
-2*(-a*f+b*e)*(-c*f+d*e)*(-e*h+f*g)/f^4/(f*x+e)^(1/2)-2*(b*d*e*(-3*e*h+2*f *g)-b*c*f*(-2*e*h+f*g)-a*f*(c*f*h-2*d*e*h+d*f*g))*(f*x+e)^(1/2)/f^4+2/3*(a *d*f*h+b*(c*f*h-3*d*e*h+d*f*g))*(f*x+e)^(3/2)/f^4+2/5*b*d*h*(f*x+e)^(5/2)/ f^4
Time = 0.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {2 \left (5 a f \left (3 c f (-f g+2 e h+f h x)+d \left (-8 e^2 h+e f (6 g-4 h x)+f^2 x (3 g+h x)\right )\right )+b \left (5 c f \left (-8 e^2 h+e f (6 g-4 h x)+f^2 x (3 g+h x)\right )+d \left (48 e^3 h-8 e^2 f (5 g-3 h x)+f^3 x^2 (5 g+3 h x)-2 e f^2 x (10 g+3 h x)\right )\right )\right )}{15 f^4 \sqrt {e+f x}} \] Input:
Integrate[((a + b*x)*(c + d*x)*(g + h*x))/(e + f*x)^(3/2),x]
Output:
(2*(5*a*f*(3*c*f*(-(f*g) + 2*e*h + f*h*x) + d*(-8*e^2*h + e*f*(6*g - 4*h*x ) + f^2*x*(3*g + h*x))) + b*(5*c*f*(-8*e^2*h + e*f*(6*g - 4*h*x) + f^2*x*( 3*g + h*x)) + d*(48*e^3*h - 8*e^2*f*(5*g - 3*h*x) + f^3*x^2*(5*g + 3*h*x) - 2*e*f^2*x*(10*g + 3*h*x)))))/(15*f^4*Sqrt[e + f*x])
Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {159, 2009}
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.
| \(\displaystyle \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 159 |
| \(\displaystyle \int \left (\frac {\sqrt {e+f x} (a d f h+b (c f h-3 d e h+d f g))}{f^3}+\frac {a f (c f h-2 d e h+d f g)+b c f (f g-2 e h)-b d e (2 f g-3 e h)}{f^3 \sqrt {e+f x}}+\frac {(a f-b e) (c f-d e) (f g-e h)}{f^3 (e+f x)^{3/2}}+\frac {b d h (e+f x)^{3/2}}{f^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 (e+f x)^{3/2} (a d f h+b (c f h-3 d e h+d f g))}{3 f^4}-\frac {2 \sqrt {e+f x} (-a f (c f h-2 d e h+d f g)-b c f (f g-2 e h)+b d e (2 f g-3 e h))}{f^4}-\frac {2 (b e-a f) (d e-c f) (f g-e h)}{f^4 \sqrt {e+f x}}+\frac {2 b d h (e+f x)^{5/2}}{5 f^4}\) |
Input:
Int[((a + b*x)*(c + d*x)*(g + h*x))/(e + f*x)^(3/2),x]
Output:
(-2*(b*e - a*f)*(d*e - c*f)*(f*g - e*h))/(f^4*Sqrt[e + f*x]) - (2*(b*d*e*( 2*f*g - 3*e*h) - b*c*f*(f*g - 2*e*h) - a*f*(d*f*g - 2*d*e*h + c*f*h))*Sqrt [e + f*x])/f^4 + (2*(a*d*f*h + b*(d*f*g - 3*d*e*h + c*f*h))*(e + f*x)^(3/2 ))/(3*f^4) + (2*b*d*h*(e + f*x)^(5/2))/(5*f^4)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n *(e + f*x)*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && (IGtQ [m, 0] || IntegersQ[m, n])
Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97
| method | result | size |
| pseudoelliptic | \(\frac {\left (\left (\left (6 h \,x^{3}+10 g \,x^{2}\right ) d +30 x \left (\frac {h x}{3}+g \right ) c \right ) b -30 a \left (\left (-\frac {1}{3} h \,x^{2}-g x \right ) d +c \left (-h x +g \right )\right )\right ) f^{3}+60 \left (\left (\left (-\frac {1}{5} h \,x^{2}-\frac {2}{3} g x \right ) d +c \left (-\frac {2 h x}{3}+g \right )\right ) b +a \left (\left (-\frac {2 h x}{3}+g \right ) d +c h \right )\right ) e \,f^{2}-80 \left (\left (\left (-\frac {3 h x}{5}+g \right ) d +c h \right ) b +a d h \right ) e^{2} f +96 b d \,e^{3} h}{15 \sqrt {f x +e}\, f^{4}}\) | \(150\) |
| risch | \(\frac {2 \left (3 f^{2} b d h \,x^{2}+5 f^{2} a d h x +5 f^{2} b c h x -9 e f b d h x +5 f^{2} b d g x +15 a c h \,f^{2}-25 a d e f h +15 a d g \,f^{2}-25 b c e f h +15 b c \,f^{2} g +33 b d \,e^{2} h -25 b d e f g \right ) \sqrt {f x +e}}{15 f^{4}}+\frac {2 a c e \,f^{2} h -2 a c g \,f^{3}-2 a d \,e^{2} f h +2 a d e \,f^{2} g -2 b c \,e^{2} f h +2 b c e \,f^{2} g +2 b d \,e^{3} h -2 b d \,e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(193\) |
| gosper | \(\frac {\frac {2}{5} b d h \,x^{3} f^{3}+\frac {2}{3} a d \,f^{3} h \,x^{2}+\frac {2}{3} b c \,f^{3} h \,x^{2}-\frac {4}{5} b d e \,f^{2} h \,x^{2}+\frac {2}{3} b d \,f^{3} g \,x^{2}+2 a c \,f^{3} h x -\frac {8}{3} a d e \,f^{2} h x +2 a d \,f^{3} g x -\frac {8}{3} b c e \,f^{2} h x +2 b c \,f^{3} g x +\frac {16}{5} b d \,e^{2} f h x -\frac {8}{3} b d e \,f^{2} g x +4 a c e \,f^{2} h -2 a c g \,f^{3}-\frac {16}{3} a d \,e^{2} f h +4 a d e \,f^{2} g -\frac {16}{3} b c \,e^{2} f h +4 b c e \,f^{2} g +\frac {32}{5} b d \,e^{3} h -\frac {16}{3} b d \,e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(207\) |
| trager | \(\frac {\frac {2}{5} b d h \,x^{3} f^{3}+\frac {2}{3} a d \,f^{3} h \,x^{2}+\frac {2}{3} b c \,f^{3} h \,x^{2}-\frac {4}{5} b d e \,f^{2} h \,x^{2}+\frac {2}{3} b d \,f^{3} g \,x^{2}+2 a c \,f^{3} h x -\frac {8}{3} a d e \,f^{2} h x +2 a d \,f^{3} g x -\frac {8}{3} b c e \,f^{2} h x +2 b c \,f^{3} g x +\frac {16}{5} b d \,e^{2} f h x -\frac {8}{3} b d e \,f^{2} g x +4 a c e \,f^{2} h -2 a c g \,f^{3}-\frac {16}{3} a d \,e^{2} f h +4 a d e \,f^{2} g -\frac {16}{3} b c \,e^{2} f h +4 b c e \,f^{2} g +\frac {32}{5} b d \,e^{3} h -\frac {16}{3} b d \,e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(207\) |
| orering | \(\frac {\frac {2}{5} b d h \,x^{3} f^{3}+\frac {2}{3} a d \,f^{3} h \,x^{2}+\frac {2}{3} b c \,f^{3} h \,x^{2}-\frac {4}{5} b d e \,f^{2} h \,x^{2}+\frac {2}{3} b d \,f^{3} g \,x^{2}+2 a c \,f^{3} h x -\frac {8}{3} a d e \,f^{2} h x +2 a d \,f^{3} g x -\frac {8}{3} b c e \,f^{2} h x +2 b c \,f^{3} g x +\frac {16}{5} b d \,e^{2} f h x -\frac {8}{3} b d e \,f^{2} g x +4 a c e \,f^{2} h -2 a c g \,f^{3}-\frac {16}{3} a d \,e^{2} f h +4 a d e \,f^{2} g -\frac {16}{3} b c \,e^{2} f h +4 b c e \,f^{2} g +\frac {32}{5} b d \,e^{3} h -\frac {16}{3} b d \,e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(207\) |
| derivativedivides | \(\frac {\frac {2 d b h \left (f x +e \right )^{\frac {5}{2}}}{5}+\frac {2 a d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {2 b c f h \left (f x +e \right )^{\frac {3}{2}}}{3}-2 b d e h \left (f x +e \right )^{\frac {3}{2}}+\frac {2 b d f g \left (f x +e \right )^{\frac {3}{2}}}{3}+2 a c \,f^{2} h \sqrt {f x +e}-4 a d e f h \sqrt {f x +e}+2 a d \,f^{2} g \sqrt {f x +e}-4 b c e f h \sqrt {f x +e}+2 b c \,f^{2} g \sqrt {f x +e}+6 b d \,e^{2} h \sqrt {f x +e}-4 b d e f g \sqrt {f x +e}-\frac {2 \left (-a c e \,f^{2} h +a c g \,f^{3}+a d \,e^{2} f h -a d e \,f^{2} g +b c \,e^{2} f h -b c e \,f^{2} g -b d \,e^{3} h +b d \,e^{2} f g \right )}{\sqrt {f x +e}}}{f^{4}}\) | \(246\) |
| default | \(\frac {\frac {2 d b h \left (f x +e \right )^{\frac {5}{2}}}{5}+\frac {2 a d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {2 b c f h \left (f x +e \right )^{\frac {3}{2}}}{3}-2 b d e h \left (f x +e \right )^{\frac {3}{2}}+\frac {2 b d f g \left (f x +e \right )^{\frac {3}{2}}}{3}+2 a c \,f^{2} h \sqrt {f x +e}-4 a d e f h \sqrt {f x +e}+2 a d \,f^{2} g \sqrt {f x +e}-4 b c e f h \sqrt {f x +e}+2 b c \,f^{2} g \sqrt {f x +e}+6 b d \,e^{2} h \sqrt {f x +e}-4 b d e f g \sqrt {f x +e}-\frac {2 \left (-a c e \,f^{2} h +a c g \,f^{3}+a d \,e^{2} f h -a d e \,f^{2} g +b c \,e^{2} f h -b c e \,f^{2} g -b d \,e^{3} h +b d \,e^{2} f g \right )}{\sqrt {f x +e}}}{f^{4}}\) | \(246\) |
Input:
int((b*x+a)*(d*x+c)*(h*x+g)/(f*x+e)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/15*((((6*h*x^3+10*g*x^2)*d+30*x*(1/3*h*x+g)*c)*b-30*a*((-1/3*h*x^2-g*x)* d+c*(-h*x+g)))*f^3+60*(((-1/5*h*x^2-2/3*g*x)*d+c*(-2/3*h*x+g))*b+a*((-2/3* h*x+g)*d+c*h))*e*f^2-80*(((-3/5*h*x+g)*d+c*h)*b+a*d*h)*e^2*f+96*b*d*e^3*h) /(f*x+e)^(1/2)/f^4
Time = 0.11 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, b d f^{3} h x^{3} + {\left (5 \, b d f^{3} g - {\left (6 \, b d e f^{2} - 5 \, {\left (b c + a d\right )} f^{3}\right )} h\right )} x^{2} - 5 \, {\left (8 \, b d e^{2} f + 3 \, a c f^{3} - 6 \, {\left (b c + a d\right )} e f^{2}\right )} g + 2 \, {\left (24 \, b d e^{3} + 15 \, a c e f^{2} - 20 \, {\left (b c + a d\right )} e^{2} f\right )} h - {\left (5 \, {\left (4 \, b d e f^{2} - 3 \, {\left (b c + a d\right )} f^{3}\right )} g - {\left (24 \, b d e^{2} f + 15 \, a c f^{3} - 20 \, {\left (b c + a d\right )} e f^{2}\right )} h\right )} x\right )} \sqrt {f x + e}}{15 \, {\left (f^{5} x + e f^{4}\right )}} \] Input:
integrate((b*x+a)*(d*x+c)*(h*x+g)/(f*x+e)^(3/2),x, algorithm="fricas")
Output:
2/15*(3*b*d*f^3*h*x^3 + (5*b*d*f^3*g - (6*b*d*e*f^2 - 5*(b*c + a*d)*f^3)*h )*x^2 - 5*(8*b*d*e^2*f + 3*a*c*f^3 - 6*(b*c + a*d)*e*f^2)*g + 2*(24*b*d*e^ 3 + 15*a*c*e*f^2 - 20*(b*c + a*d)*e^2*f)*h - (5*(4*b*d*e*f^2 - 3*(b*c + a* d)*f^3)*g - (24*b*d*e^2*f + 15*a*c*f^3 - 20*(b*c + a*d)*e*f^2)*h)*x)*sqrt( f*x + e)/(f^5*x + e*f^4)
Time = 6.32 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b d h \left (e + f x\right )^{\frac {5}{2}}}{5 f^{3}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (a d f h + b c f h - 3 b d e h + b d f g\right )}{3 f^{3}} + \frac {\sqrt {e + f x} \left (a c f^{2} h - 2 a d e f h + a d f^{2} g - 2 b c e f h + b c f^{2} g + 3 b d e^{2} h - 2 b d e f g\right )}{f^{3}} + \frac {\left (a f - b e\right ) \left (c f - d e\right ) \left (e h - f g\right )}{f^{3} \sqrt {e + f x}}\right )}{f} & \text {for}\: f \neq 0 \\\frac {a c g x + \frac {b d h x^{4}}{4} + \frac {x^{3} \left (a d h + b c h + b d g\right )}{3} + \frac {x^{2} \left (a c h + a d g + b c g\right )}{2}}{e^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*(d*x+c)*(h*x+g)/(f*x+e)**(3/2),x)
Output:
Piecewise((2*(b*d*h*(e + f*x)**(5/2)/(5*f**3) + (e + f*x)**(3/2)*(a*d*f*h + b*c*f*h - 3*b*d*e*h + b*d*f*g)/(3*f**3) + sqrt(e + f*x)*(a*c*f**2*h - 2* a*d*e*f*h + a*d*f**2*g - 2*b*c*e*f*h + b*c*f**2*g + 3*b*d*e**2*h - 2*b*d*e *f*g)/f**3 + (a*f - b*e)*(c*f - d*e)*(e*h - f*g)/(f**3*sqrt(e + f*x)))/f, Ne(f, 0)), ((a*c*g*x + b*d*h*x**4/4 + x**3*(a*d*h + b*c*h + b*d*g)/3 + x** 2*(a*c*h + a*d*g + b*c*g)/2)/e**(3/2), True))
Time = 0.04 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (f x + e\right )}^{\frac {5}{2}} b d h + 5 \, {\left (b d f g - {\left (3 \, b d e - {\left (b c + a d\right )} f\right )} h\right )} {\left (f x + e\right )}^{\frac {3}{2}} - 15 \, {\left ({\left (2 \, b d e f - {\left (b c + a d\right )} f^{2}\right )} g - {\left (3 \, b d e^{2} + a c f^{2} - 2 \, {\left (b c + a d\right )} e f\right )} h\right )} \sqrt {f x + e}}{f^{3}} - \frac {15 \, {\left ({\left (b d e^{2} f + a c f^{3} - {\left (b c + a d\right )} e f^{2}\right )} g - {\left (b d e^{3} + a c e f^{2} - {\left (b c + a d\right )} e^{2} f\right )} h\right )}}{\sqrt {f x + e} f^{3}}\right )}}{15 \, f} \] Input:
integrate((b*x+a)*(d*x+c)*(h*x+g)/(f*x+e)^(3/2),x, algorithm="maxima")
Output:
2/15*((3*(f*x + e)^(5/2)*b*d*h + 5*(b*d*f*g - (3*b*d*e - (b*c + a*d)*f)*h) *(f*x + e)^(3/2) - 15*((2*b*d*e*f - (b*c + a*d)*f^2)*g - (3*b*d*e^2 + a*c* f^2 - 2*(b*c + a*d)*e*f)*h)*sqrt(f*x + e))/f^3 - 15*((b*d*e^2*f + a*c*f^3 - (b*c + a*d)*e*f^2)*g - (b*d*e^3 + a*c*e*f^2 - (b*c + a*d)*e^2*f)*h)/(sqr t(f*x + e)*f^3))/f
Time = 0.13 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.76 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=-\frac {2 \, {\left (b d e^{2} f g - b c e f^{2} g - a d e f^{2} g + a c f^{3} g - b d e^{3} h + b c e^{2} f h + a d e^{2} f h - a c e f^{2} h\right )}}{\sqrt {f x + e} f^{4}} + \frac {2 \, {\left (5 \, {\left (f x + e\right )}^{\frac {3}{2}} b d f^{17} g - 30 \, \sqrt {f x + e} b d e f^{17} g + 15 \, \sqrt {f x + e} b c f^{18} g + 15 \, \sqrt {f x + e} a d f^{18} g + 3 \, {\left (f x + e\right )}^{\frac {5}{2}} b d f^{16} h - 15 \, {\left (f x + e\right )}^{\frac {3}{2}} b d e f^{16} h + 45 \, \sqrt {f x + e} b d e^{2} f^{16} h + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b c f^{17} h + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} a d f^{17} h - 30 \, \sqrt {f x + e} b c e f^{17} h - 30 \, \sqrt {f x + e} a d e f^{17} h + 15 \, \sqrt {f x + e} a c f^{18} h\right )}}{15 \, f^{20}} \] Input:
integrate((b*x+a)*(d*x+c)*(h*x+g)/(f*x+e)^(3/2),x, algorithm="giac")
Output:
-2*(b*d*e^2*f*g - b*c*e*f^2*g - a*d*e*f^2*g + a*c*f^3*g - b*d*e^3*h + b*c* e^2*f*h + a*d*e^2*f*h - a*c*e*f^2*h)/(sqrt(f*x + e)*f^4) + 2/15*(5*(f*x + e)^(3/2)*b*d*f^17*g - 30*sqrt(f*x + e)*b*d*e*f^17*g + 15*sqrt(f*x + e)*b*c *f^18*g + 15*sqrt(f*x + e)*a*d*f^18*g + 3*(f*x + e)^(5/2)*b*d*f^16*h - 15* (f*x + e)^(3/2)*b*d*e*f^16*h + 45*sqrt(f*x + e)*b*d*e^2*f^16*h + 5*(f*x + e)^(3/2)*b*c*f^17*h + 5*(f*x + e)^(3/2)*a*d*f^17*h - 30*sqrt(f*x + e)*b*c* e*f^17*h - 30*sqrt(f*x + e)*a*d*e*f^17*h + 15*sqrt(f*x + e)*a*c*f^18*h)/f^ 20
Time = 0.07 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {\sqrt {e+f\,x}\,\left (2\,a\,c\,f^2\,h+2\,a\,d\,f^2\,g+2\,b\,c\,f^2\,g+6\,b\,d\,e^2\,h-4\,a\,d\,e\,f\,h-4\,b\,c\,e\,f\,h-4\,b\,d\,e\,f\,g\right )}{f^4}+\frac {{\left (e+f\,x\right )}^{3/2}\,\left (2\,a\,d\,f\,h+2\,b\,c\,f\,h-6\,b\,d\,e\,h+2\,b\,d\,f\,g\right )}{3\,f^4}-\frac {2\,a\,c\,f^3\,g-2\,b\,d\,e^3\,h-2\,a\,c\,e\,f^2\,h-2\,a\,d\,e\,f^2\,g-2\,b\,c\,e\,f^2\,g+2\,a\,d\,e^2\,f\,h+2\,b\,c\,e^2\,f\,h+2\,b\,d\,e^2\,f\,g}{f^4\,\sqrt {e+f\,x}}+\frac {2\,b\,d\,h\,{\left (e+f\,x\right )}^{5/2}}{5\,f^4} \] Input:
int(((g + h*x)*(a + b*x)*(c + d*x))/(e + f*x)^(3/2),x)
Output:
((e + f*x)^(1/2)*(2*a*c*f^2*h + 2*a*d*f^2*g + 2*b*c*f^2*g + 6*b*d*e^2*h - 4*a*d*e*f*h - 4*b*c*e*f*h - 4*b*d*e*f*g))/f^4 + ((e + f*x)^(3/2)*(2*a*d*f* h + 2*b*c*f*h - 6*b*d*e*h + 2*b*d*f*g))/(3*f^4) - (2*a*c*f^3*g - 2*b*d*e^3 *h - 2*a*c*e*f^2*h - 2*a*d*e*f^2*g - 2*b*c*e*f^2*g + 2*a*d*e^2*f*h + 2*b*c *e^2*f*h + 2*b*d*e^2*f*g)/(f^4*(e + f*x)^(1/2)) + (2*b*d*h*(e + f*x)^(5/2) )/(5*f^4)
Time = 0.26 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x) (c+d x) (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {\frac {2}{5} b d \,f^{3} h \,x^{3}+\frac {2}{3} a d \,f^{3} h \,x^{2}+\frac {2}{3} b c \,f^{3} h \,x^{2}-\frac {4}{5} b d e \,f^{2} h \,x^{2}+\frac {2}{3} b d \,f^{3} g \,x^{2}+2 a c \,f^{3} h x -\frac {8}{3} a d e \,f^{2} h x +2 a d \,f^{3} g x -\frac {8}{3} b c e \,f^{2} h x +2 b c \,f^{3} g x +\frac {16}{5} b d \,e^{2} f h x -\frac {8}{3} b d e \,f^{2} g x +4 a c e \,f^{2} h -2 a c \,f^{3} g -\frac {16}{3} a d \,e^{2} f h +4 a d e \,f^{2} g -\frac {16}{3} b c \,e^{2} f h +4 b c e \,f^{2} g +\frac {32}{5} b d \,e^{3} h -\frac {16}{3} b d \,e^{2} f g}{\sqrt {f x +e}\, f^{4}} \] Input:
int((b*x+a)*(d*x+c)*(h*x+g)/(f*x+e)^(3/2),x)
Output:
(2*(30*a*c*e*f**2*h - 15*a*c*f**3*g + 15*a*c*f**3*h*x - 40*a*d*e**2*f*h + 30*a*d*e*f**2*g - 20*a*d*e*f**2*h*x + 15*a*d*f**3*g*x + 5*a*d*f**3*h*x**2 - 40*b*c*e**2*f*h + 30*b*c*e*f**2*g - 20*b*c*e*f**2*h*x + 15*b*c*f**3*g*x + 5*b*c*f**3*h*x**2 + 48*b*d*e**3*h - 40*b*d*e**2*f*g + 24*b*d*e**2*f*h*x - 20*b*d*e*f**2*g*x - 6*b*d*e*f**2*h*x**2 + 5*b*d*f**3*g*x**2 + 3*b*d*f**3 *h*x**3))/(15*sqrt(e + f*x)*f**4)