Integrand size = 19, antiderivative size = 138 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=-\frac {d}{3 x^3}-\frac {10 d+e}{2 x^2}-\frac {5 (9 d+2 e)}{x}+30 (7 d+4 e) x+21 (6 d+5 e) x^2+14 (5 d+6 e) x^3+\frac {15}{2} (4 d+7 e) x^4+3 (3 d+8 e) x^5+\frac {5}{6} (2 d+9 e) x^6+\frac {1}{7} (d+10 e) x^7+\frac {e x^8}{8}+15 (8 d+3 e) \log (x) \]
-1/3*d/x^3+1/2*(-10*d-e)/x^2-5*(9*d+2*e)/x+30*(7*d+4*e)*x+21*(6*d+5*e)*x^2 +14*(5*d+6*e)*x^3+15/2*(4*d+7*e)*x^4+3*(3*d+8*e)*x^5+5/6*(2*d+9*e)*x^6+1/7 *(d+10*e)*x^7+1/8*e*x^8+15*(8*d+3*e)*ln(x)
Time = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=-\frac {d}{3 x^3}+\frac {-10 d-e}{2 x^2}-\frac {5 (9 d+2 e)}{x}+30 (7 d+4 e) x+21 (6 d+5 e) x^2+14 (5 d+6 e) x^3+\frac {15}{2} (4 d+7 e) x^4+3 (3 d+8 e) x^5+\frac {5}{6} (2 d+9 e) x^6+\frac {1}{7} (d+10 e) x^7+\frac {e x^8}{8}+15 (8 d+3 e) \log (x) \]
-1/3*d/x^3 + (-10*d - e)/(2*x^2) - (5*(9*d + 2*e))/x + 30*(7*d + 4*e)*x + 21*(6*d + 5*e)*x^2 + 14*(5*d + 6*e)*x^3 + (15*(4*d + 7*e)*x^4)/2 + 3*(3*d + 8*e)*x^5 + (5*(2*d + 9*e)*x^6)/6 + ((d + 10*e)*x^7)/7 + (e*x^8)/8 + 15*( 8*d + 3*e)*Log[x]
Time = 0.30 (sec) , antiderivative size = 138, 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.
\(\displaystyle \int \frac {\left (x^2+2 x+1\right )^5 (d+e x)}{x^4} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \int \frac {(x+1)^{10} (d+e x)}{x^4}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (x^6 (d+10 e)+5 x^5 (2 d+9 e)+15 x^4 (3 d+8 e)+30 x^3 (4 d+7 e)+\frac {10 d+e}{x^3}+42 x^2 (5 d+6 e)+\frac {5 (9 d+2 e)}{x^2}+42 x (6 d+5 e)+\frac {15 (8 d+3 e)}{x}+30 (7 d+4 e)+\frac {d}{x^4}+e x^7\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{7} x^7 (d+10 e)+\frac {5}{6} x^6 (2 d+9 e)+3 x^5 (3 d+8 e)+\frac {15}{2} x^4 (4 d+7 e)+14 x^3 (5 d+6 e)+21 x^2 (6 d+5 e)-\frac {10 d+e}{2 x^2}+30 x (7 d+4 e)-\frac {5 (9 d+2 e)}{x}+15 (8 d+3 e) \log (x)-\frac {d}{3 x^3}+\frac {e x^8}{8}\) |
-1/3*d/x^3 - (10*d + e)/(2*x^2) - (5*(9*d + 2*e))/x + 30*(7*d + 4*e)*x + 2 1*(6*d + 5*e)*x^2 + 14*(5*d + 6*e)*x^3 + (15*(4*d + 7*e)*x^4)/2 + 3*(3*d + 8*e)*x^5 + (5*(2*d + 9*e)*x^6)/6 + ((d + 10*e)*x^7)/7 + (e*x^8)/8 + 15*(8 *d + 3*e)*Log[x]
3.6.70.3.1 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 = 0.89
method | result | size |
norman | \(\frac {\left (-45 d -10 e \right ) x^{2}+\left (-5 d -\frac {e}{2}\right ) x +\left (9 d +24 e \right ) x^{8}+\left (30 d +\frac {105 e}{2}\right ) x^{7}+\left (70 d +84 e \right ) x^{6}+\left (126 d +105 e \right ) x^{5}+\left (210 d +120 e \right ) x^{4}+\left (\frac {d}{7}+\frac {10 e}{7}\right ) x^{10}+\left (\frac {5 d}{3}+\frac {15 e}{2}\right ) x^{9}-\frac {d}{3}+\frac {e \,x^{11}}{8}}{x^{3}}+\left (120 d +45 e \right ) \ln \left (x \right )\) | \(123\) |
default | \(\frac {e \,x^{8}}{8}+\frac {d \,x^{7}}{7}+\frac {10 e \,x^{7}}{7}+\frac {5 d \,x^{6}}{3}+\frac {15 e \,x^{6}}{2}+9 d \,x^{5}+24 e \,x^{5}+30 d \,x^{4}+\frac {105 e \,x^{4}}{2}+70 d \,x^{3}+84 e \,x^{3}+126 d \,x^{2}+105 e \,x^{2}+210 d x +120 e x +\left (120 d +45 e \right ) \ln \left (x \right )-\frac {10 d +e}{2 x^{2}}-\frac {45 d +10 e}{x}-\frac {d}{3 x^{3}}\) | \(126\) |
risch | \(\frac {e \,x^{8}}{8}+\frac {d \,x^{7}}{7}+\frac {10 e \,x^{7}}{7}+\frac {5 d \,x^{6}}{3}+\frac {15 e \,x^{6}}{2}+9 d \,x^{5}+24 e \,x^{5}+30 d \,x^{4}+\frac {105 e \,x^{4}}{2}+70 d \,x^{3}+84 e \,x^{3}+126 d \,x^{2}+105 e \,x^{2}+210 d x +120 e x +\frac {\left (-45 d -10 e \right ) x^{2}+\left (-5 d -\frac {e}{2}\right ) x -\frac {d}{3}}{x^{3}}+120 d \ln \left (x \right )+45 e \ln \left (x \right )\) | \(126\) |
parallelrisch | \(\frac {21 e \,x^{11}+24 d \,x^{10}+240 e \,x^{10}+280 d \,x^{9}+1260 e \,x^{9}+1512 d \,x^{8}+4032 e \,x^{8}+5040 d \,x^{7}+8820 e \,x^{7}+11760 d \,x^{6}+14112 e \,x^{6}+21168 d \,x^{5}+17640 e \,x^{5}+20160 \ln \left (x \right ) x^{3} d +7560 \ln \left (x \right ) x^{3} e +35280 d \,x^{4}+20160 e \,x^{4}-7560 d \,x^{2}-1680 e \,x^{2}-840 d x -84 e x -56 d}{168 x^{3}}\) | \(136\) |
((-45*d-10*e)*x^2+(-5*d-1/2*e)*x+(9*d+24*e)*x^8+(30*d+105/2*e)*x^7+(70*d+8 4*e)*x^6+(126*d+105*e)*x^5+(210*d+120*e)*x^4+(1/7*d+10/7*e)*x^10+(5/3*d+15 /2*e)*x^9-1/3*d+1/8*e*x^11)/x^3+(120*d+45*e)*ln(x)
Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=\frac {21 \, e x^{11} + 24 \, {\left (d + 10 \, e\right )} x^{10} + 140 \, {\left (2 \, d + 9 \, e\right )} x^{9} + 504 \, {\left (3 \, d + 8 \, e\right )} x^{8} + 1260 \, {\left (4 \, d + 7 \, e\right )} x^{7} + 2352 \, {\left (5 \, d + 6 \, e\right )} x^{6} + 3528 \, {\left (6 \, d + 5 \, e\right )} x^{5} + 5040 \, {\left (7 \, d + 4 \, e\right )} x^{4} + 2520 \, {\left (8 \, d + 3 \, e\right )} x^{3} \log \left (x\right ) - 840 \, {\left (9 \, d + 2 \, e\right )} x^{2} - 84 \, {\left (10 \, d + e\right )} x - 56 \, d}{168 \, x^{3}} \]
1/168*(21*e*x^11 + 24*(d + 10*e)*x^10 + 140*(2*d + 9*e)*x^9 + 504*(3*d + 8 *e)*x^8 + 1260*(4*d + 7*e)*x^7 + 2352*(5*d + 6*e)*x^6 + 3528*(6*d + 5*e)*x ^5 + 5040*(7*d + 4*e)*x^4 + 2520*(8*d + 3*e)*x^3*log(x) - 840*(9*d + 2*e)* x^2 - 84*(10*d + e)*x - 56*d)/x^3
Time = 0.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=\frac {e x^{8}}{8} + x^{7} \left (\frac {d}{7} + \frac {10 e}{7}\right ) + x^{6} \cdot \left (\frac {5 d}{3} + \frac {15 e}{2}\right ) + x^{5} \cdot \left (9 d + 24 e\right ) + x^{4} \cdot \left (30 d + \frac {105 e}{2}\right ) + x^{3} \cdot \left (70 d + 84 e\right ) + x^{2} \cdot \left (126 d + 105 e\right ) + x \left (210 d + 120 e\right ) + 15 \cdot \left (8 d + 3 e\right ) \log {\left (x \right )} + \frac {- 2 d + x^{2} \left (- 270 d - 60 e\right ) + x \left (- 30 d - 3 e\right )}{6 x^{3}} \]
e*x**8/8 + x**7*(d/7 + 10*e/7) + x**6*(5*d/3 + 15*e/2) + x**5*(9*d + 24*e) + x**4*(30*d + 105*e/2) + x**3*(70*d + 84*e) + x**2*(126*d + 105*e) + x*( 210*d + 120*e) + 15*(8*d + 3*e)*log(x) + (-2*d + x**2*(-270*d - 60*e) + x* (-30*d - 3*e))/(6*x**3)
Time = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=\frac {1}{8} \, e x^{8} + \frac {1}{7} \, {\left (d + 10 \, e\right )} x^{7} + \frac {5}{6} \, {\left (2 \, d + 9 \, e\right )} x^{6} + 3 \, {\left (3 \, d + 8 \, e\right )} x^{5} + \frac {15}{2} \, {\left (4 \, d + 7 \, e\right )} x^{4} + 14 \, {\left (5 \, d + 6 \, e\right )} x^{3} + 21 \, {\left (6 \, d + 5 \, e\right )} x^{2} + 30 \, {\left (7 \, d + 4 \, e\right )} x + 15 \, {\left (8 \, d + 3 \, e\right )} \log \left (x\right ) - \frac {30 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 3 \, {\left (10 \, d + e\right )} x + 2 \, d}{6 \, x^{3}} \]
1/8*e*x^8 + 1/7*(d + 10*e)*x^7 + 5/6*(2*d + 9*e)*x^6 + 3*(3*d + 8*e)*x^5 + 15/2*(4*d + 7*e)*x^4 + 14*(5*d + 6*e)*x^3 + 21*(6*d + 5*e)*x^2 + 30*(7*d + 4*e)*x + 15*(8*d + 3*e)*log(x) - 1/6*(30*(9*d + 2*e)*x^2 + 3*(10*d + e)* x + 2*d)/x^3
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.93 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=\frac {1}{8} \, e x^{8} + \frac {1}{7} \, d x^{7} + \frac {10}{7} \, e x^{7} + \frac {5}{3} \, d x^{6} + \frac {15}{2} \, e x^{6} + 9 \, d x^{5} + 24 \, e x^{5} + 30 \, d x^{4} + \frac {105}{2} \, e x^{4} + 70 \, d x^{3} + 84 \, e x^{3} + 126 \, d x^{2} + 105 \, e x^{2} + 210 \, d x + 120 \, e x + 15 \, {\left (8 \, d + 3 \, e\right )} \log \left ({\left | x \right |}\right ) - \frac {30 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 3 \, {\left (10 \, d + e\right )} x + 2 \, d}{6 \, x^{3}} \]
1/8*e*x^8 + 1/7*d*x^7 + 10/7*e*x^7 + 5/3*d*x^6 + 15/2*e*x^6 + 9*d*x^5 + 24 *e*x^5 + 30*d*x^4 + 105/2*e*x^4 + 70*d*x^3 + 84*e*x^3 + 126*d*x^2 + 105*e* x^2 + 210*d*x + 120*e*x + 15*(8*d + 3*e)*log(abs(x)) - 1/6*(30*(9*d + 2*e) *x^2 + 3*(10*d + e)*x + 2*d)/x^3
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^4} \, dx=x^6\,\left (\frac {5\,d}{3}+\frac {15\,e}{2}\right )+x^7\,\left (\frac {d}{7}+\frac {10\,e}{7}\right )+x^5\,\left (9\,d+24\,e\right )+x^4\,\left (30\,d+\frac {105\,e}{2}\right )+x^3\,\left (70\,d+84\,e\right )+x^2\,\left (126\,d+105\,e\right )+\ln \left (x\right )\,\left (120\,d+45\,e\right )-\frac {\left (45\,d+10\,e\right )\,x^2+\left (5\,d+\frac {e}{2}\right )\,x+\frac {d}{3}}{x^3}+\frac {e\,x^8}{8}+x\,\left (210\,d+120\,e\right ) \]