Integrand size = 27, antiderivative size = 132 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=-\frac {a^6 A}{6 x^6}-\frac {a^5 (6 A b+a B)}{5 x^5}-\frac {3 a^4 b (5 A b+2 a B)}{4 x^4}-\frac {5 a^3 b^2 (4 A b+3 a B)}{3 x^3}-\frac {5 a^2 b^3 (3 A b+4 a B)}{2 x^2}-\frac {3 a b^4 (2 A b+5 a B)}{x}+b^6 B x+b^5 (A b+6 a B) \log (x) \] Output:
-1/6*a^6*A/x^6-1/5*a^5*(6*A*b+B*a)/x^5-3/4*a^4*b*(5*A*b+2*B*a)/x^4-5/3*a^3 *b^2*(4*A*b+3*B*a)/x^3-5/2*a^2*b^3*(3*A*b+4*B*a)/x^2-3*a*b^4*(2*A*b+5*B*a) /x+b^6*B*x+b^5*(A*b+6*B*a)*ln(x)
Time = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=-\frac {360 a A b^5 x^5-60 b^6 B x^7+450 a^2 b^4 x^4 (A+2 B x)+200 a^3 b^3 x^3 (2 A+3 B x)+75 a^4 b^2 x^2 (3 A+4 B x)+18 a^5 b x (4 A+5 B x)+2 a^6 (5 A+6 B x)}{60 x^6}+b^5 (A b+6 a B) \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^7,x]
Output:
-1/60*(360*a*A*b^5*x^5 - 60*b^6*B*x^7 + 450*a^2*b^4*x^4*(A + 2*B*x) + 200* a^3*b^3*x^3*(2*A + 3*B*x) + 75*a^4*b^2*x^2*(3*A + 4*B*x) + 18*a^5*b*x*(4*A + 5*B*x) + 2*a^6*(5*A + 6*B*x))/x^6 + b^5*(A*b + 6*a*B)*Log[x]
Time = 0.49 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x)}{x^7} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^7}dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^7}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^6 A}{x^7}+\frac {a^5 (a B+6 A b)}{x^6}+\frac {3 a^4 b (2 a B+5 A b)}{x^5}+\frac {5 a^3 b^2 (3 a B+4 A b)}{x^4}+\frac {5 a^2 b^3 (4 a B+3 A b)}{x^3}+\frac {b^5 (6 a B+A b)}{x}+\frac {3 a b^4 (5 a B+2 A b)}{x^2}+b^6 B\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^6 A}{6 x^6}-\frac {a^5 (a B+6 A b)}{5 x^5}-\frac {3 a^4 b (2 a B+5 A b)}{4 x^4}-\frac {5 a^3 b^2 (3 a B+4 A b)}{3 x^3}-\frac {5 a^2 b^3 (4 a B+3 A b)}{2 x^2}+b^5 \log (x) (6 a B+A b)-\frac {3 a b^4 (5 a B+2 A b)}{x}+b^6 B x\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^7,x]
Output:
-1/6*(a^6*A)/x^6 - (a^5*(6*A*b + a*B))/(5*x^5) - (3*a^4*b*(5*A*b + 2*a*B)) /(4*x^4) - (5*a^3*b^2*(4*A*b + 3*a*B))/(3*x^3) - (5*a^2*b^3*(3*A*b + 4*a*B ))/(2*x^2) - (3*a*b^4*(2*A*b + 5*a*B))/x + b^6*B*x + b^5*(A*b + 6*a*B)*Log [x]
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 = 0.94 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {a^{6} A}{6 x^{6}}-\frac {a^{5} \left (6 A b +B a \right )}{5 x^{5}}-\frac {3 a^{4} b \left (5 A b +2 B a \right )}{4 x^{4}}-\frac {5 a^{3} b^{2} \left (4 A b +3 B a \right )}{3 x^{3}}-\frac {5 a^{2} b^{3} \left (3 A b +4 B a \right )}{2 x^{2}}-\frac {3 a \,b^{4} \left (2 A b +5 B a \right )}{x}+b^{6} B x +b^{5} \left (A b +6 B a \right ) \ln \left (x \right )\) | \(123\) |
risch | \(b^{6} B x +\frac {\left (-6 A a \,b^{5}-15 B \,a^{2} b^{4}\right ) x^{5}+\left (-\frac {15}{2} A \,a^{2} b^{4}-10 B \,a^{3} b^{3}\right ) x^{4}+\left (-\frac {20}{3} A \,a^{3} b^{3}-5 B \,a^{4} b^{2}\right ) x^{3}+\left (-\frac {15}{4} A \,a^{4} b^{2}-\frac {3}{2} B \,a^{5} b \right ) x^{2}+\left (-\frac {6}{5} A \,a^{5} b -\frac {1}{5} B \,a^{6}\right ) x -\frac {A \,a^{6}}{6}}{x^{6}}+A \ln \left (x \right ) b^{6}+6 B \ln \left (x \right ) a \,b^{5}\) | \(139\) |
norman | \(\frac {\left (-\frac {15}{2} A \,a^{2} b^{4}-10 B \,a^{3} b^{3}\right ) x^{4}+\left (-\frac {20}{3} A \,a^{3} b^{3}-5 B \,a^{4} b^{2}\right ) x^{3}+\left (-\frac {15}{4} A \,a^{4} b^{2}-\frac {3}{2} B \,a^{5} b \right ) x^{2}+\left (-\frac {6}{5} A \,a^{5} b -\frac {1}{5} B \,a^{6}\right ) x +\left (-6 A a \,b^{5}-15 B \,a^{2} b^{4}\right ) x^{5}+b^{6} B \,x^{7}-\frac {A \,a^{6}}{6}}{x^{6}}+\left (A \,b^{6}+6 B a \,b^{5}\right ) \ln \left (x \right )\) | \(141\) |
parallelrisch | \(\frac {60 A \ln \left (x \right ) x^{6} b^{6}+360 B \ln \left (x \right ) x^{6} a \,b^{5}+60 b^{6} B \,x^{7}-360 A a \,b^{5} x^{5}-900 B \,a^{2} b^{4} x^{5}-450 A \,a^{2} b^{4} x^{4}-600 B \,a^{3} b^{3} x^{4}-400 A \,a^{3} b^{3} x^{3}-300 B \,a^{4} b^{2} x^{3}-225 A \,a^{4} b^{2} x^{2}-90 B \,a^{5} b \,x^{2}-72 A \,a^{5} b x -12 B \,a^{6} x -10 A \,a^{6}}{60 x^{6}}\) | \(152\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^7,x,method=_RETURNVERBOSE)
Output:
-1/6*a^6*A/x^6-1/5*a^5*(6*A*b+B*a)/x^5-3/4*a^4*b*(5*A*b+2*B*a)/x^4-5/3*a^3 *b^2*(4*A*b+3*B*a)/x^3-5/2*a^2*b^3*(3*A*b+4*B*a)/x^2-3*a*b^4*(2*A*b+5*B*a) /x+b^6*B*x+b^5*(A*b+6*B*a)*ln(x)
Time = 0.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=\frac {60 \, B b^{6} x^{7} + 60 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} \log \left (x\right ) - 10 \, A a^{6} - 180 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} - 150 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} - 100 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} - 45 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 12 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{60 \, x^{6}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^7,x, algorithm="fricas")
Output:
1/60*(60*B*b^6*x^7 + 60*(6*B*a*b^5 + A*b^6)*x^6*log(x) - 10*A*a^6 - 180*(5 *B*a^2*b^4 + 2*A*a*b^5)*x^5 - 150*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 - 100*(3 *B*a^4*b^2 + 4*A*a^3*b^3)*x^3 - 45*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 12*(B*a ^6 + 6*A*a^5*b)*x)/x^6
Time = 1.75 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=B b^{6} x + b^{5} \left (A b + 6 B a\right ) \log {\left (x \right )} + \frac {- 10 A a^{6} + x^{5} \left (- 360 A a b^{5} - 900 B a^{2} b^{4}\right ) + x^{4} \left (- 450 A a^{2} b^{4} - 600 B a^{3} b^{3}\right ) + x^{3} \left (- 400 A a^{3} b^{3} - 300 B a^{4} b^{2}\right ) + x^{2} \left (- 225 A a^{4} b^{2} - 90 B a^{5} b\right ) + x \left (- 72 A a^{5} b - 12 B a^{6}\right )}{60 x^{6}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3/x**7,x)
Output:
B*b**6*x + b**5*(A*b + 6*B*a)*log(x) + (-10*A*a**6 + x**5*(-360*A*a*b**5 - 900*B*a**2*b**4) + x**4*(-450*A*a**2*b**4 - 600*B*a**3*b**3) + x**3*(-400 *A*a**3*b**3 - 300*B*a**4*b**2) + x**2*(-225*A*a**4*b**2 - 90*B*a**5*b) + x*(-72*A*a**5*b - 12*B*a**6))/(60*x**6)
Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=B b^{6} x + {\left (6 \, B a b^{5} + A b^{6}\right )} \log \left (x\right ) - \frac {10 \, A a^{6} + 180 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 150 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 100 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 45 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 12 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{60 \, x^{6}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^7,x, algorithm="maxima")
Output:
B*b^6*x + (6*B*a*b^5 + A*b^6)*log(x) - 1/60*(10*A*a^6 + 180*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 150*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 100*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 45*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 12*(B*a^6 + 6*A*a^5 *b)*x)/x^6
Time = 0.19 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=B b^{6} x + {\left (6 \, B a b^{5} + A b^{6}\right )} \log \left ({\left | x \right |}\right ) - \frac {10 \, A a^{6} + 180 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 150 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 100 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 45 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 12 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x}{60 \, x^{6}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^7,x, algorithm="giac")
Output:
B*b^6*x + (6*B*a*b^5 + A*b^6)*log(abs(x)) - 1/60*(10*A*a^6 + 180*(5*B*a^2* b^4 + 2*A*a*b^5)*x^5 + 150*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 100*(3*B*a^4* b^2 + 4*A*a^3*b^3)*x^3 + 45*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 12*(B*a^6 + 6* A*a^5*b)*x)/x^6
Time = 10.63 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=\ln \left (x\right )\,\left (A\,b^6+6\,B\,a\,b^5\right )-\frac {x\,\left (\frac {B\,a^6}{5}+\frac {6\,A\,b\,a^5}{5}\right )+\frac {A\,a^6}{6}+x^2\,\left (\frac {3\,B\,a^5\,b}{2}+\frac {15\,A\,a^4\,b^2}{4}\right )+x^5\,\left (15\,B\,a^2\,b^4+6\,A\,a\,b^5\right )+x^3\,\left (5\,B\,a^4\,b^2+\frac {20\,A\,a^3\,b^3}{3}\right )+x^4\,\left (10\,B\,a^3\,b^3+\frac {15\,A\,a^2\,b^4}{2}\right )}{x^6}+B\,b^6\,x \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3)/x^7,x)
Output:
log(x)*(A*b^6 + 6*B*a*b^5) - (x*((B*a^6)/5 + (6*A*a^5*b)/5) + (A*a^6)/6 + x^2*((15*A*a^4*b^2)/4 + (3*B*a^5*b)/2) + x^5*(15*B*a^2*b^4 + 6*A*a*b^5) + x^3*((20*A*a^3*b^3)/3 + 5*B*a^4*b^2) + x^4*((15*A*a^2*b^4)/2 + 10*B*a^3*b^ 3))/x^6 + B*b^6*x
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^7} \, dx=\frac {420 \,\mathrm {log}\left (x \right ) a \,b^{6} x^{6}-10 a^{7}-84 a^{6} b x -315 a^{5} b^{2} x^{2}-700 a^{4} b^{3} x^{3}-1050 a^{3} b^{4} x^{4}-1260 a^{2} b^{5} x^{5}+60 b^{7} x^{7}}{60 x^{6}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3/x^7,x)
Output:
(420*log(x)*a*b**6*x**6 - 10*a**7 - 84*a**6*b*x - 315*a**5*b**2*x**2 - 700 *a**4*b**3*x**3 - 1050*a**3*b**4*x**4 - 1260*a**2*b**5*x**5 + 60*b**7*x**7 )/(60*x**6)