Integrand size = 19, antiderivative size = 139 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=-\frac {d}{x}+5 (9 d+2 e) x+\frac {15}{2} (8 d+3 e) x^2+10 (7 d+4 e) x^3+\frac {21}{2} (6 d+5 e) x^4+\frac {42}{5} (5 d+6 e) x^5+5 (4 d+7 e) x^6+\frac {15}{7} (3 d+8 e) x^7+\frac {5}{8} (2 d+9 e) x^8+\frac {1}{9} (d+10 e) x^9+\frac {e x^{10}}{10}+(10 d+e) \log (x) \] Output:
-d/x+5*(9*d+2*e)*x+15/2*(8*d+3*e)*x^2+10*(7*d+4*e)*x^3+21/2*(6*d+5*e)*x^4+ 42/5*(5*d+6*e)*x^5+5*(4*d+7*e)*x^6+15/7*(3*d+8*e)*x^7+5/8*(2*d+9*e)*x^8+1/ 9*(d+10*e)*x^9+1/10*e*x^10+(10*d+e)*ln(x)
Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=-\frac {d}{x}+5 (9 d+2 e) x+\frac {15}{2} (8 d+3 e) x^2+10 (7 d+4 e) x^3+\frac {21}{2} (6 d+5 e) x^4+\frac {42}{5} (5 d+6 e) x^5+5 (4 d+7 e) x^6+\frac {15}{7} (3 d+8 e) x^7+\frac {5}{8} (2 d+9 e) x^8+\frac {1}{9} (d+10 e) x^9+\frac {e x^{10}}{10}+(10 d+e) \log (x) \] Input:
Integrate[((d + e*x)*(1 + 2*x + x^2)^5)/x^2,x]
Output:
-(d/x) + 5*(9*d + 2*e)*x + (15*(8*d + 3*e)*x^2)/2 + 10*(7*d + 4*e)*x^3 + ( 21*(6*d + 5*e)*x^4)/2 + (42*(5*d + 6*e)*x^5)/5 + 5*(4*d + 7*e)*x^6 + (15*( 3*d + 8*e)*x^7)/7 + (5*(2*d + 9*e)*x^8)/8 + ((d + 10*e)*x^9)/9 + (e*x^10)/ 10 + (10*d + e)*Log[x]
Time = 0.51 (sec) , antiderivative size = 139, 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.
\(\displaystyle \int \frac {\left (x^2+2 x+1\right )^5 (d+e x)}{x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \int \frac {(x+1)^{10} (d+e x)}{x^2}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (x^8 (d+10 e)+5 x^7 (2 d+9 e)+15 x^6 (3 d+8 e)+30 x^5 (4 d+7 e)+42 x^4 (5 d+6 e)+42 x^3 (6 d+5 e)+30 x^2 (7 d+4 e)+15 x (8 d+3 e)+\frac {10 d+e}{x}+5 (9 d+2 e)+\frac {d}{x^2}+e x^9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} x^9 (d+10 e)+\frac {5}{8} x^8 (2 d+9 e)+\frac {15}{7} x^7 (3 d+8 e)+5 x^6 (4 d+7 e)+\frac {42}{5} x^5 (5 d+6 e)+\frac {21}{2} x^4 (6 d+5 e)+10 x^3 (7 d+4 e)+\frac {15}{2} x^2 (8 d+3 e)+5 x (9 d+2 e)+(10 d+e) \log (x)-\frac {d}{x}+\frac {e x^{10}}{10}\) |
Input:
Int[((d + e*x)*(1 + 2*x + x^2)^5)/x^2,x]
Output:
-(d/x) + 5*(9*d + 2*e)*x + (15*(8*d + 3*e)*x^2)/2 + 10*(7*d + 4*e)*x^3 + ( 21*(6*d + 5*e)*x^4)/2 + (42*(5*d + 6*e)*x^5)/5 + 5*(4*d + 7*e)*x^6 + (15*( 3*d + 8*e)*x^7)/7 + (5*(2*d + 9*e)*x^8)/8 + ((d + 10*e)*x^9)/9 + (e*x^10)/ 10 + (10*d + e)*Log[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 = 0.79 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {\left (20 d +35 e \right ) x^{7}+\left (42 d +\frac {252 e}{5}\right ) x^{6}+\left (45 d +10 e \right ) x^{2}+\left (60 d +\frac {45 e}{2}\right ) x^{3}+\left (63 d +\frac {105 e}{2}\right ) x^{5}+\left (70 d +40 e \right ) x^{4}+\left (\frac {d}{9}+\frac {10 e}{9}\right ) x^{10}+\left (\frac {5 d}{4}+\frac {45 e}{8}\right ) x^{9}+\left (\frac {45 d}{7}+\frac {120 e}{7}\right ) x^{8}-d +\frac {e \,x^{11}}{10}}{x}+\left (10 d +e \right ) \ln \left (x \right )\) | \(123\) |
default | \(\frac {e \,x^{10}}{10}+\frac {d \,x^{9}}{9}+\frac {10 e \,x^{9}}{9}+\frac {5 d \,x^{8}}{4}+\frac {45 e \,x^{8}}{8}+\frac {45 d \,x^{7}}{7}+\frac {120 e \,x^{7}}{7}+20 d \,x^{6}+35 e \,x^{6}+42 d \,x^{5}+\frac {252 x^{5} e}{5}+63 d \,x^{4}+\frac {105 x^{4} e}{2}+70 d \,x^{3}+40 x^{3} e +60 d \,x^{2}+\frac {45 e \,x^{2}}{2}+45 d x +10 e x +\left (10 d +e \right ) \ln \left (x \right )-\frac {d}{x}\) | \(126\) |
risch | \(\frac {e \,x^{10}}{10}+\frac {d \,x^{9}}{9}+\frac {10 e \,x^{9}}{9}+\frac {5 d \,x^{8}}{4}+\frac {45 e \,x^{8}}{8}+\frac {45 d \,x^{7}}{7}+\frac {120 e \,x^{7}}{7}+20 d \,x^{6}+35 e \,x^{6}+42 d \,x^{5}+\frac {252 x^{5} e}{5}+63 d \,x^{4}+\frac {105 x^{4} e}{2}+70 d \,x^{3}+40 x^{3} e +60 d \,x^{2}+\frac {45 e \,x^{2}}{2}+45 d x +10 e x +10 d \ln \left (x \right )+e \ln \left (x \right )-\frac {d}{x}\) | \(127\) |
parallelrisch | \(\frac {252 e \,x^{11}+280 d \,x^{10}+2800 e \,x^{10}+3150 d \,x^{9}+14175 e \,x^{9}+16200 d \,x^{8}+43200 e \,x^{8}+50400 d \,x^{7}+88200 e \,x^{7}+105840 d \,x^{6}+127008 e \,x^{6}+158760 d \,x^{5}+132300 x^{5} e +176400 d \,x^{4}+100800 x^{4} e +151200 d \,x^{3}+56700 x^{3} e +25200 \ln \left (x \right ) x d +2520 \ln \left (x \right ) x e +113400 d \,x^{2}+25200 e \,x^{2}-2520 d}{2520 x}\) | \(136\) |
Input:
int((e*x+d)*(x^2+2*x+1)^5/x^2,x,method=_RETURNVERBOSE)
Output:
((20*d+35*e)*x^7+(42*d+252/5*e)*x^6+(45*d+10*e)*x^2+(60*d+45/2*e)*x^3+(63* d+105/2*e)*x^5+(70*d+40*e)*x^4+(1/9*d+10/9*e)*x^10+(5/4*d+45/8*e)*x^9+(45/ 7*d+120/7*e)*x^8-d+1/10*e*x^11)/x+(10*d+e)*ln(x)
Time = 0.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=\frac {252 \, e x^{11} + 280 \, {\left (d + 10 \, e\right )} x^{10} + 1575 \, {\left (2 \, d + 9 \, e\right )} x^{9} + 5400 \, {\left (3 \, d + 8 \, e\right )} x^{8} + 12600 \, {\left (4 \, d + 7 \, e\right )} x^{7} + 21168 \, {\left (5 \, d + 6 \, e\right )} x^{6} + 26460 \, {\left (6 \, d + 5 \, e\right )} x^{5} + 25200 \, {\left (7 \, d + 4 \, e\right )} x^{4} + 18900 \, {\left (8 \, d + 3 \, e\right )} x^{3} + 12600 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 2520 \, {\left (10 \, d + e\right )} x \log \left (x\right ) - 2520 \, d}{2520 \, x} \] Input:
integrate((e*x+d)*(x^2+2*x+1)^5/x^2,x, algorithm="fricas")
Output:
1/2520*(252*e*x^11 + 280*(d + 10*e)*x^10 + 1575*(2*d + 9*e)*x^9 + 5400*(3* d + 8*e)*x^8 + 12600*(4*d + 7*e)*x^7 + 21168*(5*d + 6*e)*x^6 + 26460*(6*d + 5*e)*x^5 + 25200*(7*d + 4*e)*x^4 + 18900*(8*d + 3*e)*x^3 + 12600*(9*d + 2*e)*x^2 + 2520*(10*d + e)*x*log(x) - 2520*d)/x
Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=- \frac {d}{x} + \frac {e x^{10}}{10} + x^{9} \left (\frac {d}{9} + \frac {10 e}{9}\right ) + x^{8} \cdot \left (\frac {5 d}{4} + \frac {45 e}{8}\right ) + x^{7} \cdot \left (\frac {45 d}{7} + \frac {120 e}{7}\right ) + x^{6} \cdot \left (20 d + 35 e\right ) + x^{5} \cdot \left (42 d + \frac {252 e}{5}\right ) + x^{4} \cdot \left (63 d + \frac {105 e}{2}\right ) + x^{3} \cdot \left (70 d + 40 e\right ) + x^{2} \cdot \left (60 d + \frac {45 e}{2}\right ) + x \left (45 d + 10 e\right ) + \left (10 d + e\right ) \log {\left (x \right )} \] Input:
integrate((e*x+d)*(x**2+2*x+1)**5/x**2,x)
Output:
-d/x + e*x**10/10 + x**9*(d/9 + 10*e/9) + x**8*(5*d/4 + 45*e/8) + x**7*(45 *d/7 + 120*e/7) + x**6*(20*d + 35*e) + x**5*(42*d + 252*e/5) + x**4*(63*d + 105*e/2) + x**3*(70*d + 40*e) + x**2*(60*d + 45*e/2) + x*(45*d + 10*e) + (10*d + e)*log(x)
Time = 0.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=\frac {1}{10} \, e x^{10} + \frac {1}{9} \, {\left (d + 10 \, e\right )} x^{9} + \frac {5}{8} \, {\left (2 \, d + 9 \, e\right )} x^{8} + \frac {15}{7} \, {\left (3 \, d + 8 \, e\right )} x^{7} + 5 \, {\left (4 \, d + 7 \, e\right )} x^{6} + \frac {42}{5} \, {\left (5 \, d + 6 \, e\right )} x^{5} + \frac {21}{2} \, {\left (6 \, d + 5 \, e\right )} x^{4} + 10 \, {\left (7 \, d + 4 \, e\right )} x^{3} + \frac {15}{2} \, {\left (8 \, d + 3 \, e\right )} x^{2} + 5 \, {\left (9 \, d + 2 \, e\right )} x + {\left (10 \, d + e\right )} \log \left (x\right ) - \frac {d}{x} \] Input:
integrate((e*x+d)*(x^2+2*x+1)^5/x^2,x, algorithm="maxima")
Output:
1/10*e*x^10 + 1/9*(d + 10*e)*x^9 + 5/8*(2*d + 9*e)*x^8 + 15/7*(3*d + 8*e)* x^7 + 5*(4*d + 7*e)*x^6 + 42/5*(5*d + 6*e)*x^5 + 21/2*(6*d + 5*e)*x^4 + 10 *(7*d + 4*e)*x^3 + 15/2*(8*d + 3*e)*x^2 + 5*(9*d + 2*e)*x + (10*d + e)*log (x) - d/x
Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=\frac {1}{10} \, e x^{10} + \frac {1}{9} \, d x^{9} + \frac {10}{9} \, e x^{9} + \frac {5}{4} \, d x^{8} + \frac {45}{8} \, e x^{8} + \frac {45}{7} \, d x^{7} + \frac {120}{7} \, e x^{7} + 20 \, d x^{6} + 35 \, e x^{6} + 42 \, d x^{5} + \frac {252}{5} \, e x^{5} + 63 \, d x^{4} + \frac {105}{2} \, e x^{4} + 70 \, d x^{3} + 40 \, e x^{3} + 60 \, d x^{2} + \frac {45}{2} \, e x^{2} + 45 \, d x + 10 \, e x + {\left (10 \, d + e\right )} \log \left ({\left | x \right |}\right ) - \frac {d}{x} \] Input:
integrate((e*x+d)*(x^2+2*x+1)^5/x^2,x, algorithm="giac")
Output:
1/10*e*x^10 + 1/9*d*x^9 + 10/9*e*x^9 + 5/4*d*x^8 + 45/8*e*x^8 + 45/7*d*x^7 + 120/7*e*x^7 + 20*d*x^6 + 35*e*x^6 + 42*d*x^5 + 252/5*e*x^5 + 63*d*x^4 + 105/2*e*x^4 + 70*d*x^3 + 40*e*x^3 + 60*d*x^2 + 45/2*e*x^2 + 45*d*x + 10*e *x + (10*d + e)*log(abs(x)) - d/x
Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=x^9\,\left (\frac {d}{9}+\frac {10\,e}{9}\right )+x^6\,\left (20\,d+35\,e\right )+x^8\,\left (\frac {5\,d}{4}+\frac {45\,e}{8}\right )+x^2\,\left (60\,d+\frac {45\,e}{2}\right )+x^3\,\left (70\,d+40\,e\right )+x^4\,\left (63\,d+\frac {105\,e}{2}\right )+x^7\,\left (\frac {45\,d}{7}+\frac {120\,e}{7}\right )+x^5\,\left (42\,d+\frac {252\,e}{5}\right )-\frac {d}{x}+\frac {e\,x^{10}}{10}+x\,\left (45\,d+10\,e\right )+\ln \left (x\right )\,\left (10\,d+e\right ) \] Input:
int(((d + e*x)*(2*x + x^2 + 1)^5)/x^2,x)
Output:
x^9*(d/9 + (10*e)/9) + x^6*(20*d + 35*e) + x^8*((5*d)/4 + (45*e)/8) + x^2* (60*d + (45*e)/2) + x^3*(70*d + 40*e) + x^4*(63*d + (105*e)/2) + x^7*((45* d)/7 + (120*e)/7) + x^5*(42*d + (252*e)/5) - d/x + (e*x^10)/10 + x*(45*d + 10*e) + log(x)*(10*d + e)
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^2} \, dx=\frac {25200 \,\mathrm {log}\left (x \right ) d x +2520 \,\mathrm {log}\left (x \right ) e x +280 d \,x^{10}+3150 d \,x^{9}+16200 d \,x^{8}+50400 d \,x^{7}+105840 d \,x^{6}+158760 d \,x^{5}+176400 d \,x^{4}+151200 d \,x^{3}+113400 d \,x^{2}-2520 d +252 e \,x^{11}+2800 e \,x^{10}+14175 e \,x^{9}+43200 e \,x^{8}+88200 e \,x^{7}+127008 e \,x^{6}+132300 e \,x^{5}+100800 e \,x^{4}+56700 e \,x^{3}+25200 e \,x^{2}}{2520 x} \] Input:
int((e*x+d)*(x^2+2*x+1)^5/x^2,x)
Output:
(25200*log(x)*d*x + 2520*log(x)*e*x + 280*d*x**10 + 3150*d*x**9 + 16200*d* x**8 + 50400*d*x**7 + 105840*d*x**6 + 158760*d*x**5 + 176400*d*x**4 + 1512 00*d*x**3 + 113400*d*x**2 - 2520*d + 252*e*x**11 + 2800*e*x**10 + 14175*e* x**9 + 43200*e*x**8 + 88200*e*x**7 + 127008*e*x**6 + 132300*e*x**5 + 10080 0*e*x**4 + 56700*e*x**3 + 25200*e*x**2)/(2520*x)