Integrand size = 25, antiderivative size = 61 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=-\frac {a (A b-a B) (a+b x)^5}{5 b^3}+\frac {(A b-2 a B) (a+b x)^6}{6 b^3}+\frac {B (a+b x)^7}{7 b^3} \] Output:
-1/5*a*(A*b-B*a)*(b*x+a)^5/b^3+1/6*(A*b-2*B*a)*(b*x+a)^6/b^3+1/7*B*(b*x+a) ^7/b^3
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{210} x^2 \left (35 a^4 (3 A+2 B x)+70 a^3 b x (4 A+3 B x)+63 a^2 b^2 x^2 (5 A+4 B x)+28 a b^3 x^3 (6 A+5 B x)+5 b^4 x^4 (7 A+6 B x)\right ) \] Input:
Integrate[x*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(x^2*(35*a^4*(3*A + 2*B*x) + 70*a^3*b*x*(4*A + 3*B*x) + 63*a^2*b^2*x^2*(5* A + 4*B*x) + 28*a*b^3*x^3*(6*A + 5*B*x) + 5*b^4*x^4*(7*A + 6*B*x)))/210
Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1184, 27, 85, 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.
\(\displaystyle \int x \left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x) \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^4 x (a+b x)^4 (A+B x)dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x (a+b x)^4 (A+B x)dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {(a+b x)^5 (A b-2 a B)}{b^2}+\frac {a (a+b x)^4 (a B-A b)}{b^2}+\frac {B (a+b x)^6}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x)^6 (A b-2 a B)}{6 b^3}-\frac {a (a+b x)^5 (A b-a B)}{5 b^3}+\frac {B (a+b x)^7}{7 b^3}\) |
Input:
Int[x*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
-1/5*(a*(A*b - a*B)*(a + b*x)^5)/b^3 + ((A*b - 2*a*B)*(a + b*x)^6)/(6*b^3) + (B*(a + b*x)^7)/(7*b^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.92 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.59
method | result | size |
norman | \(\frac {B \,b^{4} x^{7}}{7}+\left (\frac {1}{6} A \,b^{4}+\frac {2}{3} B a \,b^{3}\right ) x^{6}+\left (\frac {4}{5} A a \,b^{3}+\frac {6}{5} B \,a^{2} b^{2}\right ) x^{5}+\left (\frac {3}{2} a^{2} A \,b^{2}+B \,a^{3} b \right ) x^{4}+\left (\frac {4}{3} A \,a^{3} b +\frac {1}{3} a^{4} B \right ) x^{3}+\frac {a^{4} A \,x^{2}}{2}\) | \(97\) |
gosper | \(\frac {x^{2} \left (30 b^{4} B \,x^{5}+35 A \,b^{4} x^{4}+140 B a \,b^{3} x^{4}+168 A a \,b^{3} x^{3}+252 B \,a^{2} b^{2} x^{3}+315 A \,a^{2} b^{2} x^{2}+210 B \,a^{3} b \,x^{2}+280 A \,a^{3} b x +70 a^{4} B x +105 a^{4} A \right )}{210}\) | \(100\) |
default | \(\frac {B \,b^{4} x^{7}}{7}+\frac {\left (A \,b^{4}+4 B a \,b^{3}\right ) x^{6}}{6}+\frac {\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) x^{5}}{5}+\frac {\left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) x^{4}}{4}+\frac {\left (4 A \,a^{3} b +a^{4} B \right ) x^{3}}{3}+\frac {a^{4} A \,x^{2}}{2}\) | \(100\) |
risch | \(\frac {1}{7} B \,b^{4} x^{7}+\frac {1}{6} A \,b^{4} x^{6}+\frac {2}{3} B a \,b^{3} x^{6}+\frac {4}{5} A a \,b^{3} x^{5}+\frac {6}{5} a^{2} B \,b^{2} x^{5}+\frac {3}{2} a^{2} A \,b^{2} x^{4}+x^{4} B \,a^{3} b +\frac {4}{3} x^{3} A \,a^{3} b +\frac {1}{3} x^{3} a^{4} B +\frac {1}{2} a^{4} A \,x^{2}\) | \(101\) |
parallelrisch | \(\frac {1}{7} B \,b^{4} x^{7}+\frac {1}{6} A \,b^{4} x^{6}+\frac {2}{3} B a \,b^{3} x^{6}+\frac {4}{5} A a \,b^{3} x^{5}+\frac {6}{5} a^{2} B \,b^{2} x^{5}+\frac {3}{2} a^{2} A \,b^{2} x^{4}+x^{4} B \,a^{3} b +\frac {4}{3} x^{3} A \,a^{3} b +\frac {1}{3} x^{3} a^{4} B +\frac {1}{2} a^{4} A \,x^{2}\) | \(101\) |
orering | \(\frac {x^{2} \left (30 b^{4} B \,x^{5}+35 A \,b^{4} x^{4}+140 B a \,b^{3} x^{4}+168 A a \,b^{3} x^{3}+252 B \,a^{2} b^{2} x^{3}+315 A \,a^{2} b^{2} x^{2}+210 B \,a^{3} b \,x^{2}+280 A \,a^{3} b x +70 a^{4} B x +105 a^{4} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{210 \left (b x +a \right )^{4}}\) | \(125\) |
Input:
int(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
1/7*B*b^4*x^7+(1/6*A*b^4+2/3*B*a*b^3)*x^6+(4/5*A*a*b^3+6/5*B*a^2*b^2)*x^5+ (3/2*a^2*A*b^2+B*a^3*b)*x^4+(4/3*A*a^3*b+1/3*a^4*B)*x^3+1/2*a^4*A*x^2
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.62 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{7} \, B b^{4} x^{7} + \frac {1}{2} \, A a^{4} x^{2} + \frac {1}{6} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{6} + \frac {2}{5} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{5} + \frac {1}{2} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{3} \] Input:
integrate(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
1/7*B*b^4*x^7 + 1/2*A*a^4*x^2 + 1/6*(4*B*a*b^3 + A*b^4)*x^6 + 2/5*(3*B*a^2 *b^2 + 2*A*a*b^3)*x^5 + 1/2*(2*B*a^3*b + 3*A*a^2*b^2)*x^4 + 1/3*(B*a^4 + 4 *A*a^3*b)*x^3
Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (53) = 106\).
Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.75 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {A a^{4} x^{2}}{2} + \frac {B b^{4} x^{7}}{7} + x^{6} \left (\frac {A b^{4}}{6} + \frac {2 B a b^{3}}{3}\right ) + x^{5} \cdot \left (\frac {4 A a b^{3}}{5} + \frac {6 B a^{2} b^{2}}{5}\right ) + x^{4} \cdot \left (\frac {3 A a^{2} b^{2}}{2} + B a^{3} b\right ) + x^{3} \cdot \left (\frac {4 A a^{3} b}{3} + \frac {B a^{4}}{3}\right ) \] Input:
integrate(x*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
A*a**4*x**2/2 + B*b**4*x**7/7 + x**6*(A*b**4/6 + 2*B*a*b**3/3) + x**5*(4*A *a*b**3/5 + 6*B*a**2*b**2/5) + x**4*(3*A*a**2*b**2/2 + B*a**3*b) + x**3*(4 *A*a**3*b/3 + B*a**4/3)
Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.62 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{7} \, B b^{4} x^{7} + \frac {1}{2} \, A a^{4} x^{2} + \frac {1}{6} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{6} + \frac {2}{5} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{5} + \frac {1}{2} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{3} \] Input:
integrate(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/7*B*b^4*x^7 + 1/2*A*a^4*x^2 + 1/6*(4*B*a*b^3 + A*b^4)*x^6 + 2/5*(3*B*a^2 *b^2 + 2*A*a*b^3)*x^5 + 1/2*(2*B*a^3*b + 3*A*a^2*b^2)*x^4 + 1/3*(B*a^4 + 4 *A*a^3*b)*x^3
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.64 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{7} \, B b^{4} x^{7} + \frac {2}{3} \, B a b^{3} x^{6} + \frac {1}{6} \, A b^{4} x^{6} + \frac {6}{5} \, B a^{2} b^{2} x^{5} + \frac {4}{5} \, A a b^{3} x^{5} + B a^{3} b x^{4} + \frac {3}{2} \, A a^{2} b^{2} x^{4} + \frac {1}{3} \, B a^{4} x^{3} + \frac {4}{3} \, A a^{3} b x^{3} + \frac {1}{2} \, A a^{4} x^{2} \] Input:
integrate(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
1/7*B*b^4*x^7 + 2/3*B*a*b^3*x^6 + 1/6*A*b^4*x^6 + 6/5*B*a^2*b^2*x^5 + 4/5* A*a*b^3*x^5 + B*a^3*b*x^4 + 3/2*A*a^2*b^2*x^4 + 1/3*B*a^4*x^3 + 4/3*A*a^3* b*x^3 + 1/2*A*a^4*x^2
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.49 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=x^3\,\left (\frac {B\,a^4}{3}+\frac {4\,A\,b\,a^3}{3}\right )+x^6\,\left (\frac {A\,b^4}{6}+\frac {2\,B\,a\,b^3}{3}\right )+\frac {A\,a^4\,x^2}{2}+\frac {B\,b^4\,x^7}{7}+\frac {a^2\,b\,x^4\,\left (3\,A\,b+2\,B\,a\right )}{2}+\frac {2\,a\,b^2\,x^5\,\left (2\,A\,b+3\,B\,a\right )}{5} \] Input:
int(x*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
x^3*((B*a^4)/3 + (4*A*a^3*b)/3) + x^6*((A*b^4)/6 + (2*B*a*b^3)/3) + (A*a^4 *x^2)/2 + (B*b^4*x^7)/7 + (a^2*b*x^4*(3*A*b + 2*B*a))/2 + (2*a*b^2*x^5*(2* A*b + 3*B*a))/5
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int x (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {x^{2} \left (6 b^{5} x^{5}+35 a \,b^{4} x^{4}+84 a^{2} b^{3} x^{3}+105 a^{3} b^{2} x^{2}+70 a^{4} b x +21 a^{5}\right )}{42} \] Input:
int(x*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
(x**2*(21*a**5 + 70*a**4*b*x + 105*a**3*b**2*x**2 + 84*a**2*b**3*x**3 + 35 *a*b**4*x**4 + 6*b**5*x**5))/42