Integrand size = 19, antiderivative size = 140 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=-\frac {d}{5 x^5}-\frac {10 d+e}{4 x^4}-\frac {5 (9 d+2 e)}{3 x^3}-\frac {15 (8 d+3 e)}{2 x^2}-\frac {30 (7 d+4 e)}{x}+42 (5 d+6 e) x+15 (4 d+7 e) x^2+5 (3 d+8 e) x^3+\frac {5}{4} (2 d+9 e) x^4+\frac {1}{5} (d+10 e) x^5+\frac {e x^6}{6}+42 (6 d+5 e) \log (x) \]
-1/5*d/x^5+1/4*(-10*d-e)/x^4-5/3*(9*d+2*e)/x^3-15/2*(8*d+3*e)/x^2-30*(7*d+ 4*e)/x+42*(5*d+6*e)*x+15*(4*d+7*e)*x^2+5*(3*d+8*e)*x^3+5/4*(2*d+9*e)*x^4+1 /5*(d+10*e)*x^5+1/6*e*x^6+42*(6*d+5*e)*ln(x)
Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=-\frac {d}{5 x^5}+\frac {-10 d-e}{4 x^4}-\frac {5 (9 d+2 e)}{3 x^3}-\frac {15 (8 d+3 e)}{2 x^2}-\frac {30 (7 d+4 e)}{x}+42 (5 d+6 e) x+15 (4 d+7 e) x^2+5 (3 d+8 e) x^3+\frac {5}{4} (2 d+9 e) x^4+\frac {1}{5} (d+10 e) x^5+\frac {e x^6}{6}+42 (6 d+5 e) \log (x) \]
-1/5*d/x^5 + (-10*d - e)/(4*x^4) - (5*(9*d + 2*e))/(3*x^3) - (15*(8*d + 3* e))/(2*x^2) - (30*(7*d + 4*e))/x + 42*(5*d + 6*e)*x + 15*(4*d + 7*e)*x^2 + 5*(3*d + 8*e)*x^3 + (5*(2*d + 9*e)*x^4)/4 + ((d + 10*e)*x^5)/5 + (e*x^6)/ 6 + 42*(6*d + 5*e)*Log[x]
Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1184, 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.
\(\displaystyle \int \frac {\left (x^2+2 x+1\right )^5 (d+e x)}{x^6} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \int \frac {(x+1)^{10} (d+e x)}{x^6}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {10 d+e}{x^5}+x^4 (d+10 e)+\frac {5 (9 d+2 e)}{x^4}+5 x^3 (2 d+9 e)+\frac {15 (8 d+3 e)}{x^3}+15 x^2 (3 d+8 e)+\frac {30 (7 d+4 e)}{x^2}+30 x (4 d+7 e)+\frac {42 (6 d+5 e)}{x}+42 (5 d+6 e)+\frac {d}{x^6}+e x^5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 (d+10 e)+\frac {5}{4} x^4 (2 d+9 e)-\frac {10 d+e}{4 x^4}+5 x^3 (3 d+8 e)-\frac {5 (9 d+2 e)}{3 x^3}+15 x^2 (4 d+7 e)-\frac {15 (8 d+3 e)}{2 x^2}+42 x (5 d+6 e)-\frac {30 (7 d+4 e)}{x}+42 (6 d+5 e) \log (x)-\frac {d}{5 x^5}+\frac {e x^6}{6}\) |
-1/5*d/x^5 - (10*d + e)/(4*x^4) - (5*(9*d + 2*e))/(3*x^3) - (15*(8*d + 3*e ))/(2*x^2) - (30*(7*d + 4*e))/x + 42*(5*d + 6*e)*x + 15*(4*d + 7*e)*x^2 + 5*(3*d + 8*e)*x^3 + (5*(2*d + 9*e)*x^4)/4 + ((d + 10*e)*x^5)/5 + (e*x^6)/6 + 42*(6*d + 5*e)*Log[x]
3.6.72.3.1 Defintions of rubi rules used
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 = 0.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {\left (-210 d -120 e \right ) x^{4}+\left (-60 d -\frac {45 e}{2}\right ) x^{3}+\left (-15 d -\frac {10 e}{3}\right ) x^{2}+\left (15 d +40 e \right ) x^{8}+\left (60 d +105 e \right ) x^{7}+\left (210 d +252 e \right ) x^{6}+\left (-\frac {5 d}{2}-\frac {e}{4}\right ) x +\left (\frac {d}{5}+2 e \right ) x^{10}+\left (\frac {5 d}{2}+\frac {45 e}{4}\right ) x^{9}-\frac {d}{5}+\frac {e \,x^{11}}{6}}{x^{5}}+\left (252 d +210 e \right ) \ln \left (x \right )\) | \(123\) |
risch | \(\frac {e \,x^{6}}{6}+\frac {d \,x^{5}}{5}+2 e \,x^{5}+\frac {5 d \,x^{4}}{2}+\frac {45 e \,x^{4}}{4}+15 d \,x^{3}+40 e \,x^{3}+60 d \,x^{2}+105 e \,x^{2}+210 d x +252 e x +\frac {\left (-210 d -120 e \right ) x^{4}+\left (-60 d -\frac {45 e}{2}\right ) x^{3}+\left (-15 d -\frac {10 e}{3}\right ) x^{2}+\left (-\frac {5 d}{2}-\frac {e}{4}\right ) x -\frac {d}{5}}{x^{5}}+252 d \ln \left (x \right )+210 e \ln \left (x \right )\) | \(124\) |
default | \(\frac {e \,x^{6}}{6}+\frac {d \,x^{5}}{5}+2 e \,x^{5}+\frac {5 d \,x^{4}}{2}+\frac {45 e \,x^{4}}{4}+15 d \,x^{3}+40 e \,x^{3}+60 d \,x^{2}+105 e \,x^{2}+210 d x +252 e x -\frac {10 d +e}{4 x^{4}}-\frac {d}{5 x^{5}}+\left (252 d +210 e \right ) \ln \left (x \right )-\frac {120 d +45 e}{2 x^{2}}-\frac {210 d +120 e}{x}-\frac {45 d +10 e}{3 x^{3}}\) | \(126\) |
parallelrisch | \(\frac {10 e \,x^{11}+12 d \,x^{10}+120 e \,x^{10}+150 d \,x^{9}+675 e \,x^{9}+900 d \,x^{8}+2400 e \,x^{8}+3600 d \,x^{7}+6300 e \,x^{7}+15120 \ln \left (x \right ) x^{5} d +12600 \ln \left (x \right ) x^{5} e +12600 d \,x^{6}+15120 e \,x^{6}-12600 d \,x^{4}-7200 e \,x^{4}-3600 d \,x^{3}-1350 e \,x^{3}-900 d \,x^{2}-200 e \,x^{2}-150 d x -15 e x -12 d}{60 x^{5}}\) | \(136\) |
((-210*d-120*e)*x^4+(-60*d-45/2*e)*x^3+(-15*d-10/3*e)*x^2+(15*d+40*e)*x^8+ (60*d+105*e)*x^7+(210*d+252*e)*x^6+(-5/2*d-1/4*e)*x+(1/5*d+2*e)*x^10+(5/2* d+45/4*e)*x^9-1/5*d+1/6*e*x^11)/x^5+(252*d+210*e)*ln(x)
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=\frac {10 \, e x^{11} + 12 \, {\left (d + 10 \, e\right )} x^{10} + 75 \, {\left (2 \, d + 9 \, e\right )} x^{9} + 300 \, {\left (3 \, d + 8 \, e\right )} x^{8} + 900 \, {\left (4 \, d + 7 \, e\right )} x^{7} + 2520 \, {\left (5 \, d + 6 \, e\right )} x^{6} + 2520 \, {\left (6 \, d + 5 \, e\right )} x^{5} \log \left (x\right ) - 1800 \, {\left (7 \, d + 4 \, e\right )} x^{4} - 450 \, {\left (8 \, d + 3 \, e\right )} x^{3} - 100 \, {\left (9 \, d + 2 \, e\right )} x^{2} - 15 \, {\left (10 \, d + e\right )} x - 12 \, d}{60 \, x^{5}} \]
1/60*(10*e*x^11 + 12*(d + 10*e)*x^10 + 75*(2*d + 9*e)*x^9 + 300*(3*d + 8*e )*x^8 + 900*(4*d + 7*e)*x^7 + 2520*(5*d + 6*e)*x^6 + 2520*(6*d + 5*e)*x^5* log(x) - 1800*(7*d + 4*e)*x^4 - 450*(8*d + 3*e)*x^3 - 100*(9*d + 2*e)*x^2 - 15*(10*d + e)*x - 12*d)/x^5
Time = 0.86 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=\frac {e x^{6}}{6} + x^{5} \left (\frac {d}{5} + 2 e\right ) + x^{4} \cdot \left (\frac {5 d}{2} + \frac {45 e}{4}\right ) + x^{3} \cdot \left (15 d + 40 e\right ) + x^{2} \cdot \left (60 d + 105 e\right ) + x \left (210 d + 252 e\right ) + 42 \cdot \left (6 d + 5 e\right ) \log {\left (x \right )} + \frac {- 12 d + x^{4} \left (- 12600 d - 7200 e\right ) + x^{3} \left (- 3600 d - 1350 e\right ) + x^{2} \left (- 900 d - 200 e\right ) + x \left (- 150 d - 15 e\right )}{60 x^{5}} \]
e*x**6/6 + x**5*(d/5 + 2*e) + x**4*(5*d/2 + 45*e/4) + x**3*(15*d + 40*e) + x**2*(60*d + 105*e) + x*(210*d + 252*e) + 42*(6*d + 5*e)*log(x) + (-12*d + x**4*(-12600*d - 7200*e) + x**3*(-3600*d - 1350*e) + x**2*(-900*d - 200* e) + x*(-150*d - 15*e))/(60*x**5)
Time = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=\frac {1}{6} \, e x^{6} + \frac {1}{5} \, {\left (d + 10 \, e\right )} x^{5} + \frac {5}{4} \, {\left (2 \, d + 9 \, e\right )} x^{4} + 5 \, {\left (3 \, d + 8 \, e\right )} x^{3} + 15 \, {\left (4 \, d + 7 \, e\right )} x^{2} + 42 \, {\left (5 \, d + 6 \, e\right )} x + 42 \, {\left (6 \, d + 5 \, e\right )} \log \left (x\right ) - \frac {1800 \, {\left (7 \, d + 4 \, e\right )} x^{4} + 450 \, {\left (8 \, d + 3 \, e\right )} x^{3} + 100 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 15 \, {\left (10 \, d + e\right )} x + 12 \, d}{60 \, x^{5}} \]
1/6*e*x^6 + 1/5*(d + 10*e)*x^5 + 5/4*(2*d + 9*e)*x^4 + 5*(3*d + 8*e)*x^3 + 15*(4*d + 7*e)*x^2 + 42*(5*d + 6*e)*x + 42*(6*d + 5*e)*log(x) - 1/60*(180 0*(7*d + 4*e)*x^4 + 450*(8*d + 3*e)*x^3 + 100*(9*d + 2*e)*x^2 + 15*(10*d + e)*x + 12*d)/x^5
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=\frac {1}{6} \, e x^{6} + \frac {1}{5} \, d x^{5} + 2 \, e x^{5} + \frac {5}{2} \, d x^{4} + \frac {45}{4} \, e x^{4} + 15 \, d x^{3} + 40 \, e x^{3} + 60 \, d x^{2} + 105 \, e x^{2} + 210 \, d x + 252 \, e x + 42 \, {\left (6 \, d + 5 \, e\right )} \log \left ({\left | x \right |}\right ) - \frac {1800 \, {\left (7 \, d + 4 \, e\right )} x^{4} + 450 \, {\left (8 \, d + 3 \, e\right )} x^{3} + 100 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 15 \, {\left (10 \, d + e\right )} x + 12 \, d}{60 \, x^{5}} \]
1/6*e*x^6 + 1/5*d*x^5 + 2*e*x^5 + 5/2*d*x^4 + 45/4*e*x^4 + 15*d*x^3 + 40*e *x^3 + 60*d*x^2 + 105*e*x^2 + 210*d*x + 252*e*x + 42*(6*d + 5*e)*log(abs(x )) - 1/60*(1800*(7*d + 4*e)*x^4 + 450*(8*d + 3*e)*x^3 + 100*(9*d + 2*e)*x^ 2 + 15*(10*d + e)*x + 12*d)/x^5
Time = 9.98 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^6} \, dx=x^5\,\left (\frac {d}{5}+2\,e\right )+x^3\,\left (15\,d+40\,e\right )+x^4\,\left (\frac {5\,d}{2}+\frac {45\,e}{4}\right )+x^2\,\left (60\,d+105\,e\right )+\ln \left (x\right )\,\left (252\,d+210\,e\right )-\frac {\left (210\,d+120\,e\right )\,x^4+\left (60\,d+\frac {45\,e}{2}\right )\,x^3+\left (15\,d+\frac {10\,e}{3}\right )\,x^2+\left (\frac {5\,d}{2}+\frac {e}{4}\right )\,x+\frac {d}{5}}{x^5}+\frac {e\,x^6}{6}+x\,\left (210\,d+252\,e\right ) \]