Integrand size = 21, antiderivative size = 80 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=a A d x+\frac {1}{2} (A b d+a B d+a A e) x^2+\frac {1}{3} (b B d+A c d+A b e+a B e) x^3+\frac {1}{4} (B c d+b B e+A c e) x^4+\frac {1}{5} B c e x^5 \] Output:
a*A*d*x+1/2*(A*a*e+A*b*d+B*a*d)*x^2+1/3*(A*b*e+A*c*d+B*a*e+B*b*d)*x^3+1/4* (A*c*e+B*b*e+B*c*d)*x^4+1/5*B*c*e*x^5
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=a A d x+\frac {1}{2} (A b d+a B d+a A e) x^2+\frac {1}{3} (b B d+A c d+A b e+a B e) x^3+\frac {1}{4} (B c d+b B e+A c e) x^4+\frac {1}{5} B c e x^5 \] Input:
Integrate[(A + B*x)*(d + e*x)*(a + b*x + c*x^2),x]
Output:
a*A*d*x + ((A*b*d + a*B*d + a*A*e)*x^2)/2 + ((b*B*d + A*c*d + A*b*e + a*B* e)*x^3)/3 + ((B*c*d + b*B*e + A*c*e)*x^4)/4 + (B*c*e*x^5)/5
Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1195, 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.
\(\displaystyle \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (x^2 (a B e+A b e+A c d+b B d)+x (a A e+a B d+A b d)+a A d+x^3 (A c e+b B e+B c d)+B c e x^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} x^3 (a B e+A b e+A c d+b B d)+\frac {1}{2} x^2 (a A e+a B d+A b d)+a A d x+\frac {1}{4} x^4 (A c e+b B e+B c d)+\frac {1}{5} B c e x^5\) |
Input:
Int[(A + B*x)*(d + e*x)*(a + b*x + c*x^2),x]
Output:
a*A*d*x + ((A*b*d + a*B*d + a*A*e)*x^2)/2 + ((b*B*d + A*c*d + A*b*e + a*B* e)*x^3)/3 + ((B*c*d + b*B*e + A*c*e)*x^4)/4 + (B*c*e*x^5)/5
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {B c e \,x^{5}}{5}+\frac {\left (\left (A e +B d \right ) c +B b e \right ) x^{4}}{4}+\frac {\left (A c d +b \left (A e +B d \right )+B a e \right ) x^{3}}{3}+\frac {\left (A b d +a \left (A e +B d \right )\right ) x^{2}}{2}+A a d x\) | \(76\) |
norman | \(\frac {B c e \,x^{5}}{5}+\left (\frac {1}{4} A c e +\frac {1}{4} B b e +\frac {1}{4} B c d \right ) x^{4}+\left (\frac {1}{3} A b e +\frac {1}{3} A c d +\frac {1}{3} B a e +\frac {1}{3} B b d \right ) x^{3}+\left (\frac {1}{2} a A e +\frac {1}{2} A b d +\frac {1}{2} B a d \right ) x^{2}+A a d x\) | \(80\) |
orering | \(\frac {x \left (12 B c e \,x^{4}+15 A c e \,x^{3}+15 B b e \,x^{3}+15 B c d \,x^{3}+20 A b e \,x^{2}+20 A c d \,x^{2}+20 B a e \,x^{2}+20 B b d \,x^{2}+30 A a e x +30 A b d x +30 B a d x +60 A a d \right )}{60}\) | \(92\) |
gosper | \(\frac {1}{5} B c e \,x^{5}+\frac {1}{4} x^{4} A c e +\frac {1}{4} x^{4} B b e +\frac {1}{4} x^{4} B c d +\frac {1}{3} x^{3} A b e +\frac {1}{3} x^{3} A c d +\frac {1}{3} x^{3} B a e +\frac {1}{3} B b d \,x^{3}+\frac {1}{2} x^{2} a A e +\frac {1}{2} A b d \,x^{2}+\frac {1}{2} x^{2} B a d +A a d x\) | \(95\) |
risch | \(\frac {1}{5} B c e \,x^{5}+\frac {1}{4} x^{4} A c e +\frac {1}{4} x^{4} B b e +\frac {1}{4} x^{4} B c d +\frac {1}{3} x^{3} A b e +\frac {1}{3} x^{3} A c d +\frac {1}{3} x^{3} B a e +\frac {1}{3} B b d \,x^{3}+\frac {1}{2} x^{2} a A e +\frac {1}{2} A b d \,x^{2}+\frac {1}{2} x^{2} B a d +A a d x\) | \(95\) |
parallelrisch | \(\frac {1}{5} B c e \,x^{5}+\frac {1}{4} x^{4} A c e +\frac {1}{4} x^{4} B b e +\frac {1}{4} x^{4} B c d +\frac {1}{3} x^{3} A b e +\frac {1}{3} x^{3} A c d +\frac {1}{3} x^{3} B a e +\frac {1}{3} B b d \,x^{3}+\frac {1}{2} x^{2} a A e +\frac {1}{2} A b d \,x^{2}+\frac {1}{2} x^{2} B a d +A a d x\) | \(95\) |
Input:
int((B*x+A)*(e*x+d)*(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
1/5*B*c*e*x^5+1/4*((A*e+B*d)*c+B*b*e)*x^4+1/3*(A*c*d+b*(A*e+B*d)+B*a*e)*x^ 3+1/2*(A*b*d+a*(A*e+B*d))*x^2+A*a*d*x
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=\frac {1}{5} \, B c e x^{5} + \frac {1}{4} \, {\left (B c d + {\left (B b + A c\right )} e\right )} x^{4} + A a d x + \frac {1}{3} \, {\left ({\left (B b + A c\right )} d + {\left (B a + A b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a e + {\left (B a + A b\right )} d\right )} x^{2} \] Input:
integrate((B*x+A)*(e*x+d)*(c*x^2+b*x+a),x, algorithm="fricas")
Output:
1/5*B*c*e*x^5 + 1/4*(B*c*d + (B*b + A*c)*e)*x^4 + A*a*d*x + 1/3*((B*b + A* c)*d + (B*a + A*b)*e)*x^3 + 1/2*(A*a*e + (B*a + A*b)*d)*x^2
Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=A a d x + \frac {B c e x^{5}}{5} + x^{4} \left (\frac {A c e}{4} + \frac {B b e}{4} + \frac {B c d}{4}\right ) + x^{3} \left (\frac {A b e}{3} + \frac {A c d}{3} + \frac {B a e}{3} + \frac {B b d}{3}\right ) + x^{2} \left (\frac {A a e}{2} + \frac {A b d}{2} + \frac {B a d}{2}\right ) \] Input:
integrate((B*x+A)*(e*x+d)*(c*x**2+b*x+a),x)
Output:
A*a*d*x + B*c*e*x**5/5 + x**4*(A*c*e/4 + B*b*e/4 + B*c*d/4) + x**3*(A*b*e/ 3 + A*c*d/3 + B*a*e/3 + B*b*d/3) + x**2*(A*a*e/2 + A*b*d/2 + B*a*d/2)
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=\frac {1}{5} \, B c e x^{5} + \frac {1}{4} \, {\left (B c d + {\left (B b + A c\right )} e\right )} x^{4} + A a d x + \frac {1}{3} \, {\left ({\left (B b + A c\right )} d + {\left (B a + A b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a e + {\left (B a + A b\right )} d\right )} x^{2} \] Input:
integrate((B*x+A)*(e*x+d)*(c*x^2+b*x+a),x, algorithm="maxima")
Output:
1/5*B*c*e*x^5 + 1/4*(B*c*d + (B*b + A*c)*e)*x^4 + A*a*d*x + 1/3*((B*b + A* c)*d + (B*a + A*b)*e)*x^3 + 1/2*(A*a*e + (B*a + A*b)*d)*x^2
Time = 0.13 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=\frac {1}{5} \, B c e x^{5} + \frac {1}{4} \, B c d x^{4} + \frac {1}{4} \, B b e x^{4} + \frac {1}{4} \, A c e x^{4} + \frac {1}{3} \, B b d x^{3} + \frac {1}{3} \, A c d x^{3} + \frac {1}{3} \, B a e x^{3} + \frac {1}{3} \, A b e x^{3} + \frac {1}{2} \, B a d x^{2} + \frac {1}{2} \, A b d x^{2} + \frac {1}{2} \, A a e x^{2} + A a d x \] Input:
integrate((B*x+A)*(e*x+d)*(c*x^2+b*x+a),x, algorithm="giac")
Output:
1/5*B*c*e*x^5 + 1/4*B*c*d*x^4 + 1/4*B*b*e*x^4 + 1/4*A*c*e*x^4 + 1/3*B*b*d* x^3 + 1/3*A*c*d*x^3 + 1/3*B*a*e*x^3 + 1/3*A*b*e*x^3 + 1/2*B*a*d*x^2 + 1/2* A*b*d*x^2 + 1/2*A*a*e*x^2 + A*a*d*x
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=\frac {B\,c\,e\,x^5}{5}+\left (\frac {A\,c\,e}{4}+\frac {B\,b\,e}{4}+\frac {B\,c\,d}{4}\right )\,x^4+\left (\frac {A\,b\,e}{3}+\frac {A\,c\,d}{3}+\frac {B\,a\,e}{3}+\frac {B\,b\,d}{3}\right )\,x^3+\left (\frac {A\,a\,e}{2}+\frac {A\,b\,d}{2}+\frac {B\,a\,d}{2}\right )\,x^2+A\,a\,d\,x \] Input:
int((A + B*x)*(d + e*x)*(a + b*x + c*x^2),x)
Output:
x^3*((A*b*e)/3 + (A*c*d)/3 + (B*a*e)/3 + (B*b*d)/3) + x^2*((A*a*e)/2 + (A* b*d)/2 + (B*a*d)/2) + x^4*((A*c*e)/4 + (B*b*e)/4 + (B*c*d)/4) + (B*c*e*x^5 )/5 + A*a*d*x
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01 \[ \int (A+B x) (d+e x) \left (a+b x+c x^2\right ) \, dx=\frac {x \left (12 b c e \,x^{4}+15 a c e \,x^{3}+15 b^{2} e \,x^{3}+15 b c d \,x^{3}+40 a b e \,x^{2}+20 a c d \,x^{2}+20 b^{2} d \,x^{2}+30 a^{2} e x +60 a b d x +60 a^{2} d \right )}{60} \] Input:
int((B*x+A)*(e*x+d)*(c*x^2+b*x+a),x)
Output:
(x*(60*a**2*d + 30*a**2*e*x + 60*a*b*d*x + 40*a*b*e*x**2 + 20*a*c*d*x**2 + 15*a*c*e*x**3 + 20*b**2*d*x**2 + 15*b**2*e*x**3 + 15*b*c*d*x**3 + 12*b*c* e*x**4))/60