\(\int (a+b x) (c+d x) (e+f x) (g+h x) \, dx\) [1]

Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=a c e g x+\frac {1}{2} (b c e g+a (d e g+c f g+c e h)) x^2+\frac {1}{3} (b (d e g+c f g+c e h)+a (d f g+d e h+c f h)) x^3+\frac {1}{4} (a d f h+b (d f g+d e h+c f h)) x^4+\frac {1}{5} b d f h x^5 \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {147} \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=\frac {1}{4} x^4 (a d f h+b (c f h+d e h+d f g))+\frac {1}{3} x^3 (a (c f h+d e h+d f g)+b (c e h+c f g+d e g))+\frac {1}{2} x^2 (a (c e h+c f g+d e g)+b c e g)+a c e g x+\frac {1}{5} b d f h x^5 \]

[In]

Int[(a + b*x)*(c + d*x)*(e + f*x)*(g + h*x),x]

[Out]

a*c*e*g*x + ((b*c*e*g + a*(d*e*g + c*f*g + c*e*h))*x^2)/2 + ((b*(d*e*g + c*f*g + c*e*h) + a*(d*f*g + d*e*h + c
*f*h))*x^3)/3 + ((a*d*f*h + b*(d*f*g + d*e*h + c*f*h))*x^4)/4 + (b*d*f*h*x^5)/5

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> 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])

Rubi steps \begin{align*} \text {integral}& = \int \left (a c e g+(b c e g+a (d e g+c f g+c e h)) x+(b (d e g+c f g+c e h)+a (d f g+d e h+c f h)) x^2+(a d f h+b (d f g+d e h+c f h)) x^3+b d f h x^4\right ) \, dx \\ & = a c e g x+\frac {1}{2} (b c e g+a (d e g+c f g+c e h)) x^2+\frac {1}{3} (b (d e g+c f g+c e h)+a (d f g+d e h+c f h)) x^3+\frac {1}{4} (a d f h+b (d f g+d e h+c f h)) x^4+\frac {1}{5} b d f h x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=a c e g x+\frac {1}{2} (b c e g+a d e g+a c f g+a c e h) x^2+\frac {1}{3} (b d e g+b c f g+a d f g+b c e h+a d e h+a c f h) x^3+\frac {1}{4} (b d f g+b d e h+b c f h+a d f h) x^4+\frac {1}{5} b d f h x^5 \]

[In]

Integrate[(a + b*x)*(c + d*x)*(e + f*x)*(g + h*x),x]

[Out]

a*c*e*g*x + ((b*c*e*g + a*d*e*g + a*c*f*g + a*c*e*h)*x^2)/2 + ((b*d*e*g + b*c*f*g + a*d*f*g + b*c*e*h + a*d*e*
h + a*c*f*h)*x^3)/3 + ((b*d*f*g + b*d*e*h + b*c*f*h + a*d*f*h)*x^4)/4 + (b*d*f*h*x^5)/5

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {b d f h \,x^{5}}{5}+\frac {\left (\left (\left (a d +b c \right ) f +b d e \right ) h +b d f g \right ) x^{4}}{4}+\frac {\left (\left (a c f +\left (a d +b c \right ) e \right ) h +\left (\left (a d +b c \right ) f +b d e \right ) g \right ) x^{3}}{3}+\frac {\left (a c e h +\left (a c f +\left (a d +b c \right ) e \right ) g \right ) x^{2}}{2}+a c e g x\)	\(109\)
norman	\(\frac {b d f h \,x^{5}}{5}+\left (\frac {1}{4} a d f h +\frac {1}{4} b c f h +\frac {1}{4} b d e h +\frac {1}{4} b d f g \right ) x^{4}+\left (\frac {1}{3} a c f h +\frac {1}{3} a d e h +\frac {1}{3} a d f g +\frac {1}{3} b c e h +\frac {1}{3} b c f g +\frac {1}{3} b d e g \right ) x^{3}+\left (\frac {1}{2} a c e h +\frac {1}{2} a c f g +\frac {1}{2} a d e g +\frac {1}{2} b c e g \right ) x^{2}+a c e g x\)	\(116\)
gosper	\(\frac {1}{5} b d f h \,x^{5}+\frac {1}{4} x^{4} a d f h +\frac {1}{4} x^{4} b c f h +\frac {1}{4} x^{4} b d e h +\frac {1}{4} x^{4} b d f g +\frac {1}{3} x^{3} a c f h +\frac {1}{3} x^{3} a d e h +\frac {1}{3} x^{3} a d f g +\frac {1}{3} x^{3} b c e h +\frac {1}{3} x^{3} b c f g +\frac {1}{3} x^{3} b d e g +\frac {1}{2} x^{2} a c e h +\frac {1}{2} x^{2} a c f g +\frac {1}{2} x^{2} a d e g +\frac {1}{2} x^{2} b c e g +a c e g x\)	\(143\)
risch	\(\frac {1}{5} b d f h \,x^{5}+\frac {1}{4} x^{4} a d f h +\frac {1}{4} x^{4} b c f h +\frac {1}{4} x^{4} b d e h +\frac {1}{4} x^{4} b d f g +\frac {1}{3} x^{3} a c f h +\frac {1}{3} x^{3} a d e h +\frac {1}{3} x^{3} a d f g +\frac {1}{3} x^{3} b c e h +\frac {1}{3} x^{3} b c f g +\frac {1}{3} x^{3} b d e g +\frac {1}{2} x^{2} a c e h +\frac {1}{2} x^{2} a c f g +\frac {1}{2} x^{2} a d e g +\frac {1}{2} x^{2} b c e g +a c e g x\)	\(143\)
parallelrisch	\(\frac {1}{5} b d f h \,x^{5}+\frac {1}{4} x^{4} a d f h +\frac {1}{4} x^{4} b c f h +\frac {1}{4} x^{4} b d e h +\frac {1}{4} x^{4} b d f g +\frac {1}{3} x^{3} a c f h +\frac {1}{3} x^{3} a d e h +\frac {1}{3} x^{3} a d f g +\frac {1}{3} x^{3} b c e h +\frac {1}{3} x^{3} b c f g +\frac {1}{3} x^{3} b d e g +\frac {1}{2} x^{2} a c e h +\frac {1}{2} x^{2} a c f g +\frac {1}{2} x^{2} a d e g +\frac {1}{2} x^{2} b c e g +a c e g x\)	\(143\)

[In]

int((b*x+a)*(d*x+c)*(f*x+e)*(h*x+g),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=\frac {1}{5} x^{5} h f d b + \frac {1}{4} x^{4} g f d b + \frac {1}{4} x^{4} h e d b + \frac {1}{4} x^{4} h f c b + \frac {1}{4} x^{4} h f d a + \frac {1}{3} x^{3} g e d b + \frac {1}{3} x^{3} g f c b + \frac {1}{3} x^{3} h e c b + \frac {1}{3} x^{3} g f d a + \frac {1}{3} x^{3} h e d a + \frac {1}{3} x^{3} h f c a + \frac {1}{2} x^{2} g e c b + \frac {1}{2} x^{2} g e d a + \frac {1}{2} x^{2} g f c a + \frac {1}{2} x^{2} h e c a + x g e c a \]

[In]

integrate((b*x+a)*(d*x+c)*(f*x+e)*(h*x+g),x, algorithm="fricas")

[Out]

1/5*x^5*h*f*d*b + 1/4*x^4*g*f*d*b + 1/4*x^4*h*e*d*b + 1/4*x^4*h*f*c*b + 1/4*x^4*h*f*d*a + 1/3*x^3*g*e*d*b + 1/
3*x^3*g*f*c*b + 1/3*x^3*h*e*c*b + 1/3*x^3*g*f*d*a + 1/3*x^3*h*e*d*a + 1/3*x^3*h*f*c*a + 1/2*x^2*g*e*c*b + 1/2*
x^2*g*e*d*a + 1/2*x^2*g*f*c*a + 1/2*x^2*h*e*c*a + x*g*e*c*a

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.32 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=a c e g x + \frac {b d f h x^{5}}{5} + x^{4} \left (\frac {a d f h}{4} + \frac {b c f h}{4} + \frac {b d e h}{4} + \frac {b d f g}{4}\right ) + x^{3} \left (\frac {a c f h}{3} + \frac {a d e h}{3} + \frac {a d f g}{3} + \frac {b c e h}{3} + \frac {b c f g}{3} + \frac {b d e g}{3}\right ) + x^{2} \left (\frac {a c e h}{2} + \frac {a c f g}{2} + \frac {a d e g}{2} + \frac {b c e g}{2}\right ) \]

[In]

integrate((b*x+a)*(d*x+c)*(f*x+e)*(h*x+g),x)

[Out]

a*c*e*g*x + b*d*f*h*x**5/5 + x**4*(a*d*f*h/4 + b*c*f*h/4 + b*d*e*h/4 + b*d*f*g/4) + x**3*(a*c*f*h/3 + a*d*e*h/
3 + a*d*f*g/3 + b*c*e*h/3 + b*c*f*g/3 + b*d*e*g/3) + x**2*(a*c*e*h/2 + a*c*f*g/2 + a*d*e*g/2 + b*c*e*g/2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=\frac {1}{5} \, b d f h x^{5} + a c e g x + \frac {1}{4} \, {\left (b d f g + {\left (b d e + {\left (b c + a d\right )} f\right )} h\right )} x^{4} + \frac {1}{3} \, {\left ({\left (b d e + {\left (b c + a d\right )} f\right )} g + {\left (a c f + {\left (b c + a d\right )} e\right )} h\right )} x^{3} + \frac {1}{2} \, {\left (a c e h + {\left (a c f + {\left (b c + a d\right )} e\right )} g\right )} x^{2} \]

[In]

integrate((b*x+a)*(d*x+c)*(f*x+e)*(h*x+g),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=\frac {1}{5} \, b d f h x^{5} + \frac {1}{4} \, b d f g x^{4} + \frac {1}{4} \, b d e h x^{4} + \frac {1}{4} \, b c f h x^{4} + \frac {1}{4} \, a d f h x^{4} + \frac {1}{3} \, b d e g x^{3} + \frac {1}{3} \, b c f g x^{3} + \frac {1}{3} \, a d f g x^{3} + \frac {1}{3} \, b c e h x^{3} + \frac {1}{3} \, a d e h x^{3} + \frac {1}{3} \, a c f h x^{3} + \frac {1}{2} \, b c e g x^{2} + \frac {1}{2} \, a d e g x^{2} + \frac {1}{2} \, a c f g x^{2} + \frac {1}{2} \, a c e h x^{2} + a c e g x \]

[In]

integrate((b*x+a)*(d*x+c)*(f*x+e)*(h*x+g),x, algorithm="giac")

[Out]

1/5*b*d*f*h*x^5 + 1/4*b*d*f*g*x^4 + 1/4*b*d*e*h*x^4 + 1/4*b*c*f*h*x^4 + 1/4*a*d*f*h*x^4 + 1/3*b*d*e*g*x^3 + 1/
3*b*c*f*g*x^3 + 1/3*a*d*f*g*x^3 + 1/3*b*c*e*h*x^3 + 1/3*a*d*e*h*x^3 + 1/3*a*c*f*h*x^3 + 1/2*b*c*e*g*x^2 + 1/2*
a*d*e*g*x^2 + 1/2*a*c*f*g*x^2 + 1/2*a*c*e*h*x^2 + a*c*e*g*x

Mupad [B] (verification not implemented)

Time = 2.64 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03 \[ \int (a+b x) (c+d x) (e+f x) (g+h x) \, dx=\frac {b\,d\,f\,h\,x^5}{5}+\left (\frac {a\,d\,f\,h}{4}+\frac {b\,c\,f\,h}{4}+\frac {b\,d\,e\,h}{4}+\frac {b\,d\,f\,g}{4}\right )\,x^4+\left (\frac {a\,c\,f\,h}{3}+\frac {a\,d\,e\,h}{3}+\frac {a\,d\,f\,g}{3}+\frac {b\,c\,e\,h}{3}+\frac {b\,c\,f\,g}{3}+\frac {b\,d\,e\,g}{3}\right )\,x^3+\left (\frac {a\,c\,e\,h}{2}+\frac {a\,c\,f\,g}{2}+\frac {a\,d\,e\,g}{2}+\frac {b\,c\,e\,g}{2}\right )\,x^2+a\,c\,e\,g\,x \]

[In]

int((e + f*x)*(g + h*x)*(a + b*x)*(c + d*x),x)

[Out]

x^3*((a*c*f*h)/3 + (a*d*e*h)/3 + (a*d*f*g)/3 + (b*c*e*h)/3 + (b*c*f*g)/3 + (b*d*e*g)/3) + x^2*((a*c*e*h)/2 + (
a*c*f*g)/2 + (a*d*e*g)/2 + (b*c*e*g)/2) + x^4*((a*d*f*h)/4 + (b*c*f*h)/4 + (b*d*e*h)/4 + (b*d*f*g)/4) + a*c*e*
g*x + (b*d*f*h*x^5)/5