\(\int \frac {(2-x-2 x^2+x^3) (d+e x+f x^2+g x^3+h x^4)}{(4-5 x^2+x^4)^2} \, dx\) [89]

Optimal result

Integrand size = 46, antiderivative size = 106 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {d-2 e+4 f-8 g+16 h}{12 (2+x)}-\frac {1}{18} (d+e+f+g+h) \log (1-x)+\frac {1}{48} (d+2 e+4 f+8 g+16 h) \log (2-x)+\frac {1}{6} (d-e+f-g+h) \log (1+x)-\frac {1}{144} (19 d-26 e+28 f-8 g-80 h) \log (2+x) \]

[Out]

1/12*(d-2*e+4*f-8*g+16*h)/(2+x)-1/18*(d+e+f+g+h)*ln(1-x)+1/48*(d+2*e+4*f+8*g+16*h)*ln(2-x)+1/6*(d-e+f-g+h)*ln(
1+x)-1/144*(19*d-26*e+28*f-8*g-80*h)*ln(2+x)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1600, 6874} \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {d-2 e+4 f-8 g+16 h}{12 (x+2)}-\frac {1}{18} \log (1-x) (d+e+f+g+h)+\frac {1}{48} \log (2-x) (d+2 e+4 f+8 g+16 h)+\frac {1}{6} \log (x+1) (d-e+f-g+h)-\frac {1}{144} \log (x+2) (19 d-26 e+28 f-8 g-80 h) \]

[In]

Int[((2 - x - 2*x^2 + x^3)*(d + e*x + f*x^2 + g*x^3 + h*x^4))/(4 - 5*x^2 + x^4)^2,x]

[Out]

(d - 2*e + 4*f - 8*g + 16*h)/(12*(2 + x)) - ((d + e + f + g + h)*Log[1 - x])/18 + ((d + 2*e + 4*f + 8*g + 16*h
)*Log[2 - x])/48 + ((d - e + f - g + h)*Log[1 + x])/6 - ((19*d - 26*e + 28*f - 8*g - 80*h)*Log[2 + x])/144

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x+f x^2+g x^3+h x^4}{(2+x)^2 \left (2-x-2 x^2+x^3\right )} \, dx \\ & = \int \left (\frac {d+2 e+4 f+8 g+16 h}{48 (-2+x)}+\frac {-d-e-f-g-h}{18 (-1+x)}+\frac {d-e+f-g+h}{6 (1+x)}+\frac {-d+2 e-4 f+8 g-16 h}{12 (2+x)^2}+\frac {-19 d+26 e-28 f+8 g+80 h}{144 (2+x)}\right ) \, dx \\ & = \frac {d-2 e+4 f-8 g+16 h}{12 (2+x)}-\frac {1}{18} (d+e+f+g+h) \log (1-x)+\frac {1}{48} (d+2 e+4 f+8 g+16 h) \log (2-x)+\frac {1}{6} (d-e+f-g+h) \log (1+x)-\frac {1}{144} (19 d-26 e+28 f-8 g-80 h) \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \left (\frac {12 (d-2 e+4 f-8 g+16 h)}{2+x}+24 (d-e+f-g+h) \log (-1-x)-8 (d+e+f+g+h) \log (1-x)+3 (d+2 (e+2 f+4 g+8 h)) \log (2-x)+(-19 d+26 e-28 f+8 g+80 h) \log (2+x)\right ) \]

[In]

Integrate[((2 - x - 2*x^2 + x^3)*(d + e*x + f*x^2 + g*x^3 + h*x^4))/(4 - 5*x^2 + x^4)^2,x]

[Out]

((12*(d - 2*e + 4*f - 8*g + 16*h))/(2 + x) + 24*(d - e + f - g + h)*Log[-1 - x] - 8*(d + e + f + g + h)*Log[1
- x] + 3*(d + 2*(e + 2*f + 4*g + 8*h))*Log[2 - x] + (-19*d + 26*e - 28*f + 8*g + 80*h)*Log[2 + x])/144

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03

method	result	size
default	\(\left (\frac {5 h}{9}+\frac {g}{18}-\frac {7 f}{36}+\frac {13 e}{72}-\frac {19 d}{144}\right ) \ln \left (x +2\right )-\frac {-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}+\frac {2 g}{3}-\frac {4 h}{3}}{x +2}+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{18}-\frac {e}{18}-\frac {f}{18}-\frac {g}{18}-\frac {h}{18}\right ) \ln \left (x -1\right )+\left (\frac {d}{48}+\frac {e}{24}+\frac {f}{12}+\frac {g}{6}+\frac {h}{3}\right ) \ln \left (x -2\right )\)	\(109\)
norman	\(\frac {\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}+\frac {2 g}{3}-\frac {4 h}{3}\right ) x +\left (\frac {d}{12}-\frac {e}{6}+\frac {f}{3}-\frac {2 g}{3}+\frac {4 h}{3}\right ) x^{3}+\left (-\frac {8 h}{3}+\frac {4 g}{3}-\frac {2 f}{3}+\frac {e}{3}-\frac {d}{6}\right ) x^{2}+\frac {8 h}{3}-\frac {4 g}{3}+\frac {2 f}{3}-\frac {e}{3}+\frac {d}{6}}{x^{4}-5 x^{2}+4}+\left (-\frac {d}{18}-\frac {e}{18}-\frac {f}{18}-\frac {g}{18}-\frac {h}{18}\right ) \ln \left (x -1\right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}\right ) \ln \left (x +1\right )+\left (\frac {d}{48}+\frac {e}{24}+\frac {f}{12}+\frac {g}{6}+\frac {h}{3}\right ) \ln \left (x -2\right )+\left (\frac {5 h}{9}+\frac {g}{18}-\frac {7 f}{36}+\frac {13 e}{72}-\frac {19 d}{144}\right ) \ln \left (x +2\right )\)	\(173\)
risch	\(-\frac {\ln \left (x -1\right ) d}{18}-\frac {\ln \left (x -1\right ) e}{18}+\frac {\ln \left (x +1\right ) f}{6}+\frac {\ln \left (2-x \right ) f}{12}+\frac {5 \ln \left (-x -2\right ) h}{9}+\frac {d}{12 x +24}+\frac {\ln \left (2-x \right ) d}{48}+\frac {\ln \left (2-x \right ) e}{24}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}-\frac {\ln \left (x +1\right ) g}{6}+\frac {\ln \left (2-x \right ) g}{6}-\frac {\ln \left (x -1\right ) g}{18}-\frac {19 \ln \left (-x -2\right ) d}{144}+\frac {13 \ln \left (-x -2\right ) e}{72}-\frac {e}{6 \left (x +2\right )}+\frac {f}{3 x +6}+\frac {4 h}{3 \left (x +2\right )}-\frac {\ln \left (x -1\right ) f}{18}-\frac {\ln \left (x -1\right ) h}{18}-\frac {2 g}{3 \left (x +2\right )}+\frac {\ln \left (-x -2\right ) g}{18}+\frac {\ln \left (2-x \right ) h}{3}+\frac {\ln \left (x +1\right ) h}{6}-\frac {7 \ln \left (-x -2\right ) f}{36}\)	\(202\)
parallelrisch	\(\frac {48 f -96 g +12 d +192 h -24 e +48 \ln \left (x -2\right ) x h -8 \ln \left (x -1\right ) x h +24 \ln \left (x +1\right ) x h +80 \ln \left (x +2\right ) x h +6 \ln \left (x -2\right ) d +12 \ln \left (x -2\right ) e -16 \ln \left (x -1\right ) d -16 \ln \left (x -1\right ) e +24 \ln \left (x -2\right ) x g -8 \ln \left (x -1\right ) x g -24 \ln \left (x +1\right ) x g +8 \ln \left (x +2\right ) x g -56 \ln \left (x +2\right ) f +48 \ln \left (x +1\right ) f +26 \ln \left (x +2\right ) x e +6 \ln \left (x -2\right ) x e -8 \ln \left (x -1\right ) x d -8 \ln \left (x -1\right ) x e +24 \ln \left (x +1\right ) x d -24 \ln \left (x +1\right ) x e -19 \ln \left (x +2\right ) x d -38 \ln \left (x +2\right ) d +12 \ln \left (x -2\right ) x f -8 \ln \left (x -1\right ) x f +24 \ln \left (x +1\right ) x f -28 \ln \left (x +2\right ) x f +52 \ln \left (x +2\right ) e +48 \ln \left (x +1\right ) d -48 \ln \left (x +1\right ) e +3 \ln \left (x -2\right ) x d -48 \ln \left (x +1\right ) g +16 \ln \left (x +2\right ) g +48 \ln \left (x -2\right ) g -16 \ln \left (x -1\right ) g +24 \ln \left (x -2\right ) f -16 \ln \left (x -1\right ) f +96 \ln \left (x -2\right ) h -16 \ln \left (x -1\right ) h +48 \ln \left (x +1\right ) h +160 \ln \left (x +2\right ) h}{144 x +288}\)	\(324\)

[In]

int((x^3-2*x^2-x+2)*(h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x,method=_RETURNVERBOSE)

[Out]

(5/9*h+1/18*g-7/36*f+13/72*e-19/144*d)*ln(x+2)-(-1/12*d+1/6*e-1/3*f+2/3*g-4/3*h)/(x+2)+(1/6*d-1/6*e+1/6*f-1/6*
g+1/6*h)*ln(x+1)+(-1/18*d-1/18*e-1/18*f-1/18*g-1/18*h)*ln(x-1)+(1/48*d+1/24*e+1/12*f+1/6*g+1/3*h)*ln(x-2)

Fricas [A] (verification not implemented)

none

Time = 3.26 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {{\left ({\left (19 \, d - 26 \, e + 28 \, f - 8 \, g - 80 \, h\right )} x + 38 \, d - 52 \, e + 56 \, f - 16 \, g - 160 \, h\right )} \log \left (x + 2\right ) - 24 \, {\left ({\left (d - e + f - g + h\right )} x + 2 \, d - 2 \, e + 2 \, f - 2 \, g + 2 \, h\right )} \log \left (x + 1\right ) + 8 \, {\left ({\left (d + e + f + g + h\right )} x + 2 \, d + 2 \, e + 2 \, f + 2 \, g + 2 \, h\right )} \log \left (x - 1\right ) - 3 \, {\left ({\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h\right )} x + 2 \, d + 4 \, e + 8 \, f + 16 \, g + 32 \, h\right )} \log \left (x - 2\right ) - 12 \, d + 24 \, e - 48 \, f + 96 \, g - 192 \, h}{144 \, {\left (x + 2\right )}} \]

[In]

integrate((x^3-2*x^2-x+2)*(h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="fricas")

[Out]

-1/144*(((19*d - 26*e + 28*f - 8*g - 80*h)*x + 38*d - 52*e + 56*f - 16*g - 160*h)*log(x + 2) - 24*((d - e + f
- g + h)*x + 2*d - 2*e + 2*f - 2*g + 2*h)*log(x + 1) + 8*((d + e + f + g + h)*x + 2*d + 2*e + 2*f + 2*g + 2*h)
*log(x - 1) - 3*((d + 2*e + 4*f + 8*g + 16*h)*x + 2*d + 4*e + 8*f + 16*g + 32*h)*log(x - 2) - 12*d + 24*e - 48
*f + 96*g - 192*h)/(x + 2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((x**3-2*x**2-x+2)*(h*x**4+g*x**3+f*x**2+e*x+d)/(x**4-5*x**2+4)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {1}{144} \, {\left (19 \, d - 26 \, e + 28 \, f - 8 \, g - 80 \, h\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f - g + h\right )} \log \left (x + 1\right ) - \frac {1}{18} \, {\left (d + e + f + g + h\right )} \log \left (x - 1\right ) + \frac {1}{48} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h\right )} \log \left (x - 2\right ) + \frac {d - 2 \, e + 4 \, f - 8 \, g + 16 \, h}{12 \, {\left (x + 2\right )}} \]

[In]

integrate((x^3-2*x^2-x+2)*(h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="maxima")

[Out]

-1/144*(19*d - 26*e + 28*f - 8*g - 80*h)*log(x + 2) + 1/6*(d - e + f - g + h)*log(x + 1) - 1/18*(d + e + f + g
+ h)*log(x - 1) + 1/48*(d + 2*e + 4*f + 8*g + 16*h)*log(x - 2) + 1/12*(d - 2*e + 4*f - 8*g + 16*h)/(x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {1}{144} \, {\left (19 \, d - 26 \, e + 28 \, f - 8 \, g - 80 \, h\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, {\left (d - e + f - g + h\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{18} \, {\left (d + e + f + g + h\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{48} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h\right )} \log \left ({\left | x - 2 \right |}\right ) + \frac {d - 2 \, e + 4 \, f - 8 \, g + 16 \, h}{12 \, {\left (x + 2\right )}} \]

[In]

integrate((x^3-2*x^2-x+2)*(h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="giac")

[Out]

-1/144*(19*d - 26*e + 28*f - 8*g - 80*h)*log(abs(x + 2)) + 1/6*(d - e + f - g + h)*log(abs(x + 1)) - 1/18*(d +
e + f + g + h)*log(abs(x - 1)) + 1/48*(d + 2*e + 4*f + 8*g + 16*h)*log(abs(x - 2)) + 1/12*(d - 2*e + 4*f - 8*
g + 16*h)/(x + 2)

Mupad [B] (verification not implemented)

Time = 8.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {\frac {d}{12}-\frac {e}{6}+\frac {f}{3}-\frac {2\,g}{3}+\frac {4\,h}{3}}{x+2}+\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}\right )-\ln \left (x-1\right )\,\left (\frac {d}{18}+\frac {e}{18}+\frac {f}{18}+\frac {g}{18}+\frac {h}{18}\right )+\ln \left (x-2\right )\,\left (\frac {d}{48}+\frac {e}{24}+\frac {f}{12}+\frac {g}{6}+\frac {h}{3}\right )+\ln \left (x+2\right )\,\left (\frac {13\,e}{72}-\frac {19\,d}{144}-\frac {7\,f}{36}+\frac {g}{18}+\frac {5\,h}{9}\right ) \]

[In]

int(-((x + 2*x^2 - x^3 - 2)*(d + e*x + f*x^2 + g*x^3 + h*x^4))/(x^4 - 5*x^2 + 4)^2,x)

[Out]

(d/12 - e/6 + f/3 - (2*g)/3 + (4*h)/3)/(x + 2) + log(x + 1)*(d/6 - e/6 + f/6 - g/6 + h/6) - log(x - 1)*(d/18 +
e/18 + f/18 + g/18 + h/18) + log(x - 2)*(d/48 + e/24 + f/12 + g/6 + h/3) + log(x + 2)*((13*e)/72 - (19*d)/144
- (7*f)/36 + g/18 + (5*h)/9)