\(\int (a+b x) (A+B x) (d+e x)^3 \, dx\) [1012]

Optimal result

Integrand size = 18, antiderivative size = 77 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=\frac {(b d-a e) (B d-A e) (d+e x)^4}{4 e^3}-\frac {(2 b B d-A b e-a B e) (d+e x)^5}{5 e^3}+\frac {b B (d+e x)^6}{6 e^3} \]

[Out]

1/4*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^4/e^3-1/5*(-A*b*e-B*a*e+2*B*b*d)*(e*x+d)^5/e^3+1/6*b*B*(e*x+d)^6/e^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=-\frac {(d+e x)^5 (-a B e-A b e+2 b B d)}{5 e^3}+\frac {(d+e x)^4 (b d-a e) (B d-A e)}{4 e^3}+\frac {b B (d+e x)^6}{6 e^3} \]

[In]

Int[(a + b*x)*(A + B*x)*(d + e*x)^3,x]

[Out]

((b*d - a*e)*(B*d - A*e)*(d + e*x)^4)/(4*e^3) - ((2*b*B*d - A*b*e - a*B*e)*(d + e*x)^5)/(5*e^3) + (b*B*(d + e*
x)^6)/(6*e^3)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b d+a e) (-B d+A e) (d+e x)^3}{e^2}+\frac {(-2 b B d+A b e+a B e) (d+e x)^4}{e^2}+\frac {b B (d+e x)^5}{e^2}\right ) \, dx \\ & = \frac {(b d-a e) (B d-A e) (d+e x)^4}{4 e^3}-\frac {(2 b B d-A b e-a B e) (d+e x)^5}{5 e^3}+\frac {b B (d+e x)^6}{6 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.69 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=a A d^3 x+\frac {1}{2} d^2 (A b d+a B d+3 a A e) x^2+\frac {1}{3} d (3 a e (B d+A e)+b d (B d+3 A e)) x^3+\frac {1}{4} e (3 b d (B d+A e)+a e (3 B d+A e)) x^4+\frac {1}{5} e^2 (3 b B d+A b e+a B e) x^5+\frac {1}{6} b B e^3 x^6 \]

[In]

Integrate[(a + b*x)*(A + B*x)*(d + e*x)^3,x]

[Out]

a*A*d^3*x + (d^2*(A*b*d + a*B*d + 3*a*A*e)*x^2)/2 + (d*(3*a*e*(B*d + A*e) + b*d*(B*d + 3*A*e))*x^3)/3 + (e*(3*
b*d*(B*d + A*e) + a*e*(3*B*d + A*e))*x^4)/4 + (e^2*(3*b*B*d + A*b*e + a*B*e)*x^5)/5 + (b*B*e^3*x^6)/6

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75

method	result	size
default	\(\frac {b B \,e^{3} x^{6}}{6}+\frac {\left (\left (A b +B a \right ) e^{3}+3 b B d \,e^{2}\right ) x^{5}}{5}+\frac {\left (A a \,e^{3}+3 \left (A b +B a \right ) d \,e^{2}+3 b B \,d^{2} e \right ) x^{4}}{4}+\frac {\left (3 A a d \,e^{2}+3 \left (A b +B a \right ) d^{2} e +b B \,d^{3}\right ) x^{3}}{3}+\frac {\left (3 A a \,d^{2} e +\left (A b +B a \right ) d^{3}\right ) x^{2}}{2}+A a \,d^{3} x\)	\(135\)
norman	\(\frac {b B \,e^{3} x^{6}}{6}+\left (\frac {1}{5} A b \,e^{3}+\frac {1}{5} B a \,e^{3}+\frac {3}{5} b B d \,e^{2}\right ) x^{5}+\left (\frac {1}{4} A a \,e^{3}+\frac {3}{4} A b d \,e^{2}+\frac {3}{4} B a d \,e^{2}+\frac {3}{4} b B \,d^{2} e \right ) x^{4}+\left (A a d \,e^{2}+A b \,d^{2} e +B a \,d^{2} e +\frac {1}{3} b B \,d^{3}\right ) x^{3}+\left (\frac {3}{2} A a \,d^{2} e +\frac {1}{2} A b \,d^{3}+\frac {1}{2} B a \,d^{3}\right ) x^{2}+A a \,d^{3} x\)	\(142\)
gosper	\(\frac {1}{6} b B \,e^{3} x^{6}+\frac {1}{5} x^{5} A b \,e^{3}+\frac {1}{5} x^{5} B a \,e^{3}+\frac {3}{5} x^{5} b B d \,e^{2}+\frac {1}{4} x^{4} A a \,e^{3}+\frac {3}{4} x^{4} A b d \,e^{2}+\frac {3}{4} x^{4} B a d \,e^{2}+\frac {3}{4} x^{4} b B \,d^{2} e +x^{3} A a d \,e^{2}+x^{3} A b \,d^{2} e +x^{3} B a \,d^{2} e +\frac {1}{3} x^{3} b B \,d^{3}+\frac {3}{2} x^{2} A a \,d^{2} e +\frac {1}{2} x^{2} A b \,d^{3}+\frac {1}{2} x^{2} B a \,d^{3}+A a \,d^{3} x\)	\(164\)
risch	\(\frac {1}{6} b B \,e^{3} x^{6}+\frac {1}{5} x^{5} A b \,e^{3}+\frac {1}{5} x^{5} B a \,e^{3}+\frac {3}{5} x^{5} b B d \,e^{2}+\frac {1}{4} x^{4} A a \,e^{3}+\frac {3}{4} x^{4} A b d \,e^{2}+\frac {3}{4} x^{4} B a d \,e^{2}+\frac {3}{4} x^{4} b B \,d^{2} e +x^{3} A a d \,e^{2}+x^{3} A b \,d^{2} e +x^{3} B a \,d^{2} e +\frac {1}{3} x^{3} b B \,d^{3}+\frac {3}{2} x^{2} A a \,d^{2} e +\frac {1}{2} x^{2} A b \,d^{3}+\frac {1}{2} x^{2} B a \,d^{3}+A a \,d^{3} x\)	\(164\)
parallelrisch	\(\frac {1}{6} b B \,e^{3} x^{6}+\frac {1}{5} x^{5} A b \,e^{3}+\frac {1}{5} x^{5} B a \,e^{3}+\frac {3}{5} x^{5} b B d \,e^{2}+\frac {1}{4} x^{4} A a \,e^{3}+\frac {3}{4} x^{4} A b d \,e^{2}+\frac {3}{4} x^{4} B a d \,e^{2}+\frac {3}{4} x^{4} b B \,d^{2} e +x^{3} A a d \,e^{2}+x^{3} A b \,d^{2} e +x^{3} B a \,d^{2} e +\frac {1}{3} x^{3} b B \,d^{3}+\frac {3}{2} x^{2} A a \,d^{2} e +\frac {1}{2} x^{2} A b \,d^{3}+\frac {1}{2} x^{2} B a \,d^{3}+A a \,d^{3} x\)	\(164\)

[In]

int((b*x+a)*(B*x+A)*(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

1/6*b*B*e^3*x^6+1/5*((A*b+B*a)*e^3+3*b*B*d*e^2)*x^5+1/4*(A*a*e^3+3*(A*b+B*a)*d*e^2+3*b*B*d^2*e)*x^4+1/3*(3*A*a
*d*e^2+3*(A*b+B*a)*d^2*e+b*B*d^3)*x^3+1/2*(3*A*a*d^2*e+(A*b+B*a)*d^3)*x^2+A*a*d^3*x

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.74 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=\frac {1}{6} \, B b e^{3} x^{6} + A a d^{3} x + \frac {1}{5} \, {\left (3 \, B b d e^{2} + {\left (B a + A b\right )} e^{3}\right )} x^{5} + \frac {1}{4} \, {\left (3 \, B b d^{2} e + A a e^{3} + 3 \, {\left (B a + A b\right )} d e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b d^{3} + 3 \, A a d e^{2} + 3 \, {\left (B a + A b\right )} d^{2} e\right )} x^{3} + \frac {1}{2} \, {\left (3 \, A a d^{2} e + {\left (B a + A b\right )} d^{3}\right )} x^{2} \]

[In]

integrate((b*x+a)*(B*x+A)*(e*x+d)^3,x, algorithm="fricas")

[Out]

1/6*B*b*e^3*x^6 + A*a*d^3*x + 1/5*(3*B*b*d*e^2 + (B*a + A*b)*e^3)*x^5 + 1/4*(3*B*b*d^2*e + A*a*e^3 + 3*(B*a +
A*b)*d*e^2)*x^4 + 1/3*(B*b*d^3 + 3*A*a*d*e^2 + 3*(B*a + A*b)*d^2*e)*x^3 + 1/2*(3*A*a*d^2*e + (B*a + A*b)*d^3)*
x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (73) = 146\).

Time = 0.03 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.18 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=A a d^{3} x + \frac {B b e^{3} x^{6}}{6} + x^{5} \left (\frac {A b e^{3}}{5} + \frac {B a e^{3}}{5} + \frac {3 B b d e^{2}}{5}\right ) + x^{4} \left (\frac {A a e^{3}}{4} + \frac {3 A b d e^{2}}{4} + \frac {3 B a d e^{2}}{4} + \frac {3 B b d^{2} e}{4}\right ) + x^{3} \left (A a d e^{2} + A b d^{2} e + B a d^{2} e + \frac {B b d^{3}}{3}\right ) + x^{2} \cdot \left (\frac {3 A a d^{2} e}{2} + \frac {A b d^{3}}{2} + \frac {B a d^{3}}{2}\right ) \]

[In]

integrate((b*x+a)*(B*x+A)*(e*x+d)**3,x)

[Out]

A*a*d**3*x + B*b*e**3*x**6/6 + x**5*(A*b*e**3/5 + B*a*e**3/5 + 3*B*b*d*e**2/5) + x**4*(A*a*e**3/4 + 3*A*b*d*e*
*2/4 + 3*B*a*d*e**2/4 + 3*B*b*d**2*e/4) + x**3*(A*a*d*e**2 + A*b*d**2*e + B*a*d**2*e + B*b*d**3/3) + x**2*(3*A
*a*d**2*e/2 + A*b*d**3/2 + B*a*d**3/2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.74 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=\frac {1}{6} \, B b e^{3} x^{6} + A a d^{3} x + \frac {1}{5} \, {\left (3 \, B b d e^{2} + {\left (B a + A b\right )} e^{3}\right )} x^{5} + \frac {1}{4} \, {\left (3 \, B b d^{2} e + A a e^{3} + 3 \, {\left (B a + A b\right )} d e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b d^{3} + 3 \, A a d e^{2} + 3 \, {\left (B a + A b\right )} d^{2} e\right )} x^{3} + \frac {1}{2} \, {\left (3 \, A a d^{2} e + {\left (B a + A b\right )} d^{3}\right )} x^{2} \]

[In]

integrate((b*x+a)*(B*x+A)*(e*x+d)^3,x, algorithm="maxima")

[Out]

1/6*B*b*e^3*x^6 + A*a*d^3*x + 1/5*(3*B*b*d*e^2 + (B*a + A*b)*e^3)*x^5 + 1/4*(3*B*b*d^2*e + A*a*e^3 + 3*(B*a +
A*b)*d*e^2)*x^4 + 1/3*(B*b*d^3 + 3*A*a*d*e^2 + 3*(B*a + A*b)*d^2*e)*x^3 + 1/2*(3*A*a*d^2*e + (B*a + A*b)*d^3)*
x^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (71) = 142\).

Time = 0.28 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.12 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=\frac {1}{6} \, B b e^{3} x^{6} + \frac {3}{5} \, B b d e^{2} x^{5} + \frac {1}{5} \, B a e^{3} x^{5} + \frac {1}{5} \, A b e^{3} x^{5} + \frac {3}{4} \, B b d^{2} e x^{4} + \frac {3}{4} \, B a d e^{2} x^{4} + \frac {3}{4} \, A b d e^{2} x^{4} + \frac {1}{4} \, A a e^{3} x^{4} + \frac {1}{3} \, B b d^{3} x^{3} + B a d^{2} e x^{3} + A b d^{2} e x^{3} + A a d e^{2} x^{3} + \frac {1}{2} \, B a d^{3} x^{2} + \frac {1}{2} \, A b d^{3} x^{2} + \frac {3}{2} \, A a d^{2} e x^{2} + A a d^{3} x \]

[In]

integrate((b*x+a)*(B*x+A)*(e*x+d)^3,x, algorithm="giac")

[Out]

1/6*B*b*e^3*x^6 + 3/5*B*b*d*e^2*x^5 + 1/5*B*a*e^3*x^5 + 1/5*A*b*e^3*x^5 + 3/4*B*b*d^2*e*x^4 + 3/4*B*a*d*e^2*x^
4 + 3/4*A*b*d*e^2*x^4 + 1/4*A*a*e^3*x^4 + 1/3*B*b*d^3*x^3 + B*a*d^2*e*x^3 + A*b*d^2*e*x^3 + A*a*d*e^2*x^3 + 1/
2*B*a*d^3*x^2 + 1/2*A*b*d^3*x^2 + 3/2*A*a*d^2*e*x^2 + A*a*d^3*x

Mupad [B] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.83 \[ \int (a+b x) (A+B x) (d+e x)^3 \, dx=x^2\,\left (\frac {A\,b\,d^3}{2}+\frac {B\,a\,d^3}{2}+\frac {3\,A\,a\,d^2\,e}{2}\right )+x^5\,\left (\frac {A\,b\,e^3}{5}+\frac {B\,a\,e^3}{5}+\frac {3\,B\,b\,d\,e^2}{5}\right )+x^3\,\left (\frac {B\,b\,d^3}{3}+A\,a\,d\,e^2+A\,b\,d^2\,e+B\,a\,d^2\,e\right )+x^4\,\left (\frac {A\,a\,e^3}{4}+\frac {3\,A\,b\,d\,e^2}{4}+\frac {3\,B\,a\,d\,e^2}{4}+\frac {3\,B\,b\,d^2\,e}{4}\right )+A\,a\,d^3\,x+\frac {B\,b\,e^3\,x^6}{6} \]

[In]

int((A + B*x)*(a + b*x)*(d + e*x)^3,x)

[Out]

x^2*((A*b*d^3)/2 + (B*a*d^3)/2 + (3*A*a*d^2*e)/2) + x^5*((A*b*e^3)/5 + (B*a*e^3)/5 + (3*B*b*d*e^2)/5) + x^3*((
B*b*d^3)/3 + A*a*d*e^2 + A*b*d^2*e + B*a*d^2*e) + x^4*((A*a*e^3)/4 + (3*A*b*d*e^2)/4 + (3*B*a*d*e^2)/4 + (3*B*
b*d^2*e)/4) + A*a*d^3*x + (B*b*e^3*x^6)/6