\(\int \frac {(d+e x) (1+2 x+x^2)^5}{x^{11}} \, dx\) [577]

Optimal result

Integrand size = 19, antiderivative size = 138 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=-\frac {d}{10 x^{10}}-\frac {10 d+e}{9 x^9}-\frac {5 (9 d+2 e)}{8 x^8}-\frac {15 (8 d+3 e)}{7 x^7}-\frac {5 (7 d+4 e)}{x^6}-\frac {42 (6 d+5 e)}{5 x^5}-\frac {21 (5 d+6 e)}{2 x^4}-\frac {10 (4 d+7 e)}{x^3}-\frac {15 (3 d+8 e)}{2 x^2}-\frac {5 (2 d+9 e)}{x}+e x+(d+10 e) \log (x) \]

[Out]

-1/10*d/x^10+1/9*(-10*d-e)/x^9-5/8*(9*d+2*e)/x^8-15/7*(8*d+3*e)/x^7-5*(7*d+4*e)/x^6-42/5*(6*d+5*e)/x^5-21/2*(5
*d+6*e)/x^4-10*(4*d+7*e)/x^3-15/2*(3*d+8*e)/x^2-5*(2*d+9*e)/x+e*x+(d+10*e)*ln(x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {27, 77} \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=-\frac {10 d+e}{9 x^9}-\frac {5 (9 d+2 e)}{8 x^8}-\frac {15 (8 d+3 e)}{7 x^7}-\frac {5 (7 d+4 e)}{x^6}-\frac {42 (6 d+5 e)}{5 x^5}-\frac {21 (5 d+6 e)}{2 x^4}-\frac {10 (4 d+7 e)}{x^3}-\frac {15 (3 d+8 e)}{2 x^2}-\frac {5 (2 d+9 e)}{x}+(d+10 e) \log (x)-\frac {d}{10 x^{10}}+e x \]

[In]

Int[((d + e*x)*(1 + 2*x + x^2)^5)/x^11,x]

[Out]

-1/10*d/x^10 - (10*d + e)/(9*x^9) - (5*(9*d + 2*e))/(8*x^8) - (15*(8*d + 3*e))/(7*x^7) - (5*(7*d + 4*e))/x^6 -
(42*(6*d + 5*e))/(5*x^5) - (21*(5*d + 6*e))/(2*x^4) - (10*(4*d + 7*e))/x^3 - (15*(3*d + 8*e))/(2*x^2) - (5*(2
*d + 9*e))/x + e*x + (d + 10*e)*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> 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]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+x)^{10} (d+e x)}{x^{11}} \, dx \\ & = \int \left (e+\frac {d}{x^{11}}+\frac {10 d+e}{x^{10}}+\frac {5 (9 d+2 e)}{x^9}+\frac {15 (8 d+3 e)}{x^8}+\frac {30 (7 d+4 e)}{x^7}+\frac {42 (6 d+5 e)}{x^6}+\frac {42 (5 d+6 e)}{x^5}+\frac {30 (4 d+7 e)}{x^4}+\frac {15 (3 d+8 e)}{x^3}+\frac {5 (2 d+9 e)}{x^2}+\frac {d+10 e}{x}\right ) \, dx \\ & = -\frac {d}{10 x^{10}}-\frac {10 d+e}{9 x^9}-\frac {5 (9 d+2 e)}{8 x^8}-\frac {15 (8 d+3 e)}{7 x^7}-\frac {5 (7 d+4 e)}{x^6}-\frac {42 (6 d+5 e)}{5 x^5}-\frac {21 (5 d+6 e)}{2 x^4}-\frac {10 (4 d+7 e)}{x^3}-\frac {15 (3 d+8 e)}{2 x^2}-\frac {5 (2 d+9 e)}{x}+e x+(d+10 e) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=-\frac {d}{10 x^{10}}+\frac {-10 d-e}{9 x^9}-\frac {5 (9 d+2 e)}{8 x^8}-\frac {15 (8 d+3 e)}{7 x^7}-\frac {5 (7 d+4 e)}{x^6}-\frac {42 (6 d+5 e)}{5 x^5}-\frac {21 (5 d+6 e)}{2 x^4}-\frac {10 (4 d+7 e)}{x^3}-\frac {15 (3 d+8 e)}{2 x^2}-\frac {5 (2 d+9 e)}{x}+e x+(d+10 e) \log (x) \]

[In]

Integrate[((d + e*x)*(1 + 2*x + x^2)^5)/x^11,x]

[Out]

-1/10*d/x^10 + (-10*d - e)/(9*x^9) - (5*(9*d + 2*e))/(8*x^8) - (15*(8*d + 3*e))/(7*x^7) - (5*(7*d + 4*e))/x^6
- (42*(6*d + 5*e))/(5*x^5) - (21*(5*d + 6*e))/(2*x^4) - (10*(4*d + 7*e))/x^3 - (15*(3*d + 8*e))/(2*x^2) - (5*(
2*d + 9*e))/x + e*x + (d + 10*e)*Log[x]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86

method	result	size
risch	\(e x +\frac {\left (-10 d -45 e \right ) x^{9}+\left (-\frac {45 d}{2}-60 e \right ) x^{8}+\left (-40 d -70 e \right ) x^{7}+\left (-\frac {105 d}{2}-63 e \right ) x^{6}+\left (-\frac {252 d}{5}-42 e \right ) x^{5}+\left (-35 d -20 e \right ) x^{4}+\left (-\frac {120 d}{7}-\frac {45 e}{7}\right ) x^{3}+\left (-\frac {45 d}{8}-\frac {5 e}{4}\right ) x^{2}+\left (-\frac {10 d}{9}-\frac {e}{9}\right ) x -\frac {d}{10}}{x^{10}}+d \ln \left (x \right )+10 e \ln \left (x \right )\)	\(119\)
norman	\(\frac {e \,x^{11}+\left (-40 d -70 e \right ) x^{7}+\left (-35 d -20 e \right ) x^{4}+\left (-10 d -45 e \right ) x^{9}+\left (-\frac {252 d}{5}-42 e \right ) x^{5}+\left (-\frac {120 d}{7}-\frac {45 e}{7}\right ) x^{3}+\left (-\frac {105 d}{2}-63 e \right ) x^{6}+\left (-\frac {45 d}{2}-60 e \right ) x^{8}+\left (-\frac {45 d}{8}-\frac {5 e}{4}\right ) x^{2}+\left (-\frac {10 d}{9}-\frac {e}{9}\right ) x -\frac {d}{10}}{x^{10}}+\left (d +10 e \right ) \ln \left (x \right )\)	\(120\)
default	\(e x -\frac {210 d +120 e}{6 x^{6}}-\frac {210 d +252 e}{4 x^{4}}-\frac {d}{10 x^{10}}-\frac {252 d +210 e}{5 x^{5}}+\left (d +10 e \right ) \ln \left (x \right )-\frac {45 d +120 e}{2 x^{2}}-\frac {10 d +e}{9 x^{9}}-\frac {120 d +45 e}{7 x^{7}}-\frac {10 d +45 e}{x}-\frac {120 d +210 e}{3 x^{3}}-\frac {45 d +10 e}{8 x^{8}}\)	\(125\)
parallelrisch	\(\frac {2520 \ln \left (x \right ) x^{10} d +25200 \ln \left (x \right ) x^{10} e +2520 e \,x^{11}-25200 d \,x^{9}-113400 e \,x^{9}-56700 d \,x^{8}-151200 e \,x^{8}-100800 d \,x^{7}-176400 e \,x^{7}-132300 d \,x^{6}-158760 e \,x^{6}-127008 d \,x^{5}-105840 e \,x^{5}-88200 d \,x^{4}-50400 e \,x^{4}-43200 d \,x^{3}-16200 e \,x^{3}-14175 d \,x^{2}-3150 e \,x^{2}-2800 d x -280 e x -252 d}{2520 x^{10}}\)	\(136\)

[In]

int((e*x+d)*(x^2+2*x+1)^5/x^11,x,method=_RETURNVERBOSE)

[Out]

e*x+((-10*d-45*e)*x^9+(-45/2*d-60*e)*x^8+(-40*d-70*e)*x^7+(-105/2*d-63*e)*x^6+(-252/5*d-42*e)*x^5+(-35*d-20*e)
*x^4+(-120/7*d-45/7*e)*x^3+(-45/8*d-5/4*e)*x^2+(-10/9*d-1/9*e)*x-1/10*d)/x^10+d*ln(x)+10*e*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=\frac {2520 \, e x^{11} + 2520 \, {\left (d + 10 \, e\right )} x^{10} \log \left (x\right ) - 12600 \, {\left (2 \, d + 9 \, e\right )} x^{9} - 18900 \, {\left (3 \, d + 8 \, e\right )} x^{8} - 25200 \, {\left (4 \, d + 7 \, e\right )} x^{7} - 26460 \, {\left (5 \, d + 6 \, e\right )} x^{6} - 21168 \, {\left (6 \, d + 5 \, e\right )} x^{5} - 12600 \, {\left (7 \, d + 4 \, e\right )} x^{4} - 5400 \, {\left (8 \, d + 3 \, e\right )} x^{3} - 1575 \, {\left (9 \, d + 2 \, e\right )} x^{2} - 280 \, {\left (10 \, d + e\right )} x - 252 \, d}{2520 \, x^{10}} \]

[In]

integrate((e*x+d)*(x^2+2*x+1)^5/x^11,x, algorithm="fricas")

[Out]

1/2520*(2520*e*x^11 + 2520*(d + 10*e)*x^10*log(x) - 12600*(2*d + 9*e)*x^9 - 18900*(3*d + 8*e)*x^8 - 25200*(4*d
+ 7*e)*x^7 - 26460*(5*d + 6*e)*x^6 - 21168*(6*d + 5*e)*x^5 - 12600*(7*d + 4*e)*x^4 - 5400*(8*d + 3*e)*x^3 - 1
575*(9*d + 2*e)*x^2 - 280*(10*d + e)*x - 252*d)/x^10

Sympy [A] (verification not implemented)

Time = 4.68 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=e x + \left (d + 10 e\right ) \log {\left (x \right )} + \frac {- 252 d + x^{9} \left (- 25200 d - 113400 e\right ) + x^{8} \left (- 56700 d - 151200 e\right ) + x^{7} \left (- 100800 d - 176400 e\right ) + x^{6} \left (- 132300 d - 158760 e\right ) + x^{5} \left (- 127008 d - 105840 e\right ) + x^{4} \left (- 88200 d - 50400 e\right ) + x^{3} \left (- 43200 d - 16200 e\right ) + x^{2} \left (- 14175 d - 3150 e\right ) + x \left (- 2800 d - 280 e\right )}{2520 x^{10}} \]

[In]

integrate((e*x+d)*(x**2+2*x+1)**5/x**11,x)

[Out]

e*x + (d + 10*e)*log(x) + (-252*d + x**9*(-25200*d - 113400*e) + x**8*(-56700*d - 151200*e) + x**7*(-100800*d
- 176400*e) + x**6*(-132300*d - 158760*e) + x**5*(-127008*d - 105840*e) + x**4*(-88200*d - 50400*e) + x**3*(-4
3200*d - 16200*e) + x**2*(-14175*d - 3150*e) + x*(-2800*d - 280*e))/(2520*x**10)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=e x + {\left (d + 10 \, e\right )} \log \left (x\right ) - \frac {12600 \, {\left (2 \, d + 9 \, e\right )} x^{9} + 18900 \, {\left (3 \, d + 8 \, e\right )} x^{8} + 25200 \, {\left (4 \, d + 7 \, e\right )} x^{7} + 26460 \, {\left (5 \, d + 6 \, e\right )} x^{6} + 21168 \, {\left (6 \, d + 5 \, e\right )} x^{5} + 12600 \, {\left (7 \, d + 4 \, e\right )} x^{4} + 5400 \, {\left (8 \, d + 3 \, e\right )} x^{3} + 1575 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 280 \, {\left (10 \, d + e\right )} x + 252 \, d}{2520 \, x^{10}} \]

[In]

integrate((e*x+d)*(x^2+2*x+1)^5/x^11,x, algorithm="maxima")

[Out]

e*x + (d + 10*e)*log(x) - 1/2520*(12600*(2*d + 9*e)*x^9 + 18900*(3*d + 8*e)*x^8 + 25200*(4*d + 7*e)*x^7 + 2646
0*(5*d + 6*e)*x^6 + 21168*(6*d + 5*e)*x^5 + 12600*(7*d + 4*e)*x^4 + 5400*(8*d + 3*e)*x^3 + 1575*(9*d + 2*e)*x^
2 + 280*(10*d + e)*x + 252*d)/x^10

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=e x + {\left (d + 10 \, e\right )} \log \left ({\left | x \right |}\right ) - \frac {12600 \, {\left (2 \, d + 9 \, e\right )} x^{9} + 18900 \, {\left (3 \, d + 8 \, e\right )} x^{8} + 25200 \, {\left (4 \, d + 7 \, e\right )} x^{7} + 26460 \, {\left (5 \, d + 6 \, e\right )} x^{6} + 21168 \, {\left (6 \, d + 5 \, e\right )} x^{5} + 12600 \, {\left (7 \, d + 4 \, e\right )} x^{4} + 5400 \, {\left (8 \, d + 3 \, e\right )} x^{3} + 1575 \, {\left (9 \, d + 2 \, e\right )} x^{2} + 280 \, {\left (10 \, d + e\right )} x + 252 \, d}{2520 \, x^{10}} \]

[In]

integrate((e*x+d)*(x^2+2*x+1)^5/x^11,x, algorithm="giac")

[Out]

e*x + (d + 10*e)*log(abs(x)) - 1/2520*(12600*(2*d + 9*e)*x^9 + 18900*(3*d + 8*e)*x^8 + 25200*(4*d + 7*e)*x^7 +
26460*(5*d + 6*e)*x^6 + 21168*(6*d + 5*e)*x^5 + 12600*(7*d + 4*e)*x^4 + 5400*(8*d + 3*e)*x^3 + 1575*(9*d + 2*
e)*x^2 + 280*(10*d + e)*x + 252*d)/x^10

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x) \left (1+2 x+x^2\right )^5}{x^{11}} \, dx=e\,x-\frac {\left (10\,d+45\,e\right )\,x^9+\left (\frac {45\,d}{2}+60\,e\right )\,x^8+\left (40\,d+70\,e\right )\,x^7+\left (\frac {105\,d}{2}+63\,e\right )\,x^6+\left (\frac {252\,d}{5}+42\,e\right )\,x^5+\left (35\,d+20\,e\right )\,x^4+\left (\frac {120\,d}{7}+\frac {45\,e}{7}\right )\,x^3+\left (\frac {45\,d}{8}+\frac {5\,e}{4}\right )\,x^2+\left (\frac {10\,d}{9}+\frac {e}{9}\right )\,x+\frac {d}{10}}{x^{10}}+\ln \left (x\right )\,\left (d+10\,e\right ) \]

[In]

int(((d + e*x)*(2*x + x^2 + 1)^5)/x^11,x)

[Out]

e*x - (d/10 + x^4*(35*d + 20*e) + x^2*((45*d)/8 + (5*e)/4) + x^9*(10*d + 45*e) + x^8*((45*d)/2 + 60*e) + x^7*(
40*d + 70*e) + x^6*((105*d)/2 + 63*e) + x^3*((120*d)/7 + (45*e)/7) + x^5*((252*d)/5 + 42*e) + x*((10*d)/9 + e/
9))/x^10 + log(x)*(d + 10*e)