\(\int \frac {(2+x) (d+e x+f x^2+g x^3+h x^4+i x^5)}{4-5 x^2+x^4} \, dx\) [84]

Optimal result

Integrand size = 41, antiderivative size = 96 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=(g+2 h+5 i) x+\frac {1}{2} (h+2 i) x^2+\frac {i x^3}{3}-\frac {1}{2} (d+e+f+g+h+i) \log (1-x)+\frac {1}{3} (d+2 e+4 f+8 g+16 h+32 i) \log (2-x)+\frac {1}{6} (d-e+f-g+h-i) \log (1+x) \]

[Out]

(g+2*h+5*i)*x+1/2*(h+2*i)*x^2+1/3*i*x^3-1/2*(d+e+f+g+h+i)*ln(1-x)+1/3*(d+2*e+4*f+8*g+16*h+32*i)*ln(2-x)+1/6*(d
-e+f-g+h-i)*ln(1+x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {1600, 2099} \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=-\frac {1}{2} \log (1-x) (d+e+f+g+h+i)+\frac {1}{3} \log (2-x) (d+2 e+4 f+8 g+16 h+32 i)+\frac {1}{6} \log (x+1) (d-e+f-g+h-i)+x (g+2 h+5 i)+\frac {1}{2} x^2 (h+2 i)+\frac {i x^3}{3} \]

[In]

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

[Out]

(g + 2*h + 5*i)*x + ((h + 2*i)*x^2)/2 + (i*x^3)/3 - ((d + e + f + g + h + i)*Log[1 - x])/2 + ((d + 2*e + 4*f +
8*g + 16*h + 32*i)*Log[2 - x])/3 + ((d - e + f - g + h - i)*Log[1 + x])/6

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 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{2-x-2 x^2+x^3} \, dx \\ & = \int \left (g \left (1+\frac {2 h+5 i}{g}\right )+\frac {d+2 e+4 f+8 g+16 h+32 i}{3 (-2+x)}+\frac {-d-e-f-g-h-i}{2 (-1+x)}+(h+2 i) x+i x^2+\frac {d-e+f-g+h-i}{6 (1+x)}\right ) \, dx \\ & = (g+2 h+5 i) x+\frac {1}{2} (h+2 i) x^2+\frac {i x^3}{3}-\frac {1}{2} (d+e+f+g+h+i) \log (1-x)+\frac {1}{3} (d+2 e+4 f+8 g+16 h+32 i) \log (2-x)+\frac {1}{6} (d-e+f-g+h-i) \log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{6} \left (6 (g+2 h+5 i) x+3 (h+2 i) x^2+2 i x^3-3 (d+e+f+g+h+i) \log (1-x)+2 (d+2 e+4 (f+2 g+4 h+8 i)) \log (2-x)+(d-e+f-g+h-i) \log (1+x)\right ) \]

[In]

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

[Out]

(6*(g + 2*h + 5*i)*x + 3*(h + 2*i)*x^2 + 2*i*x^3 - 3*(d + e + f + g + h + i)*Log[1 - x] + 2*(d + 2*e + 4*(f +
2*g + 4*h + 8*i))*Log[2 - x] + (d - e + f - g + h - i)*Log[1 + x])/6

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.03

method	result	size
norman	\(\left (\frac {h}{2}+i \right ) x^{2}+\left (g +2 h +5 i \right ) x +\frac {i \,x^{3}}{3}+\left (-\frac {d}{2}-\frac {e}{2}-\frac {f}{2}-\frac {g}{2}-\frac {h}{2}-\frac {i}{2}\right ) \ln \left (x -1\right )+\left (\frac {d}{3}+\frac {2 e}{3}+\frac {4 f}{3}+\frac {8 g}{3}+\frac {16 h}{3}+\frac {32 i}{3}\right ) \ln \left (x -2\right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right ) \ln \left (x +1\right )\)	\(99\)
default	\(\frac {i \,x^{3}}{3}+\frac {h \,x^{2}}{2}+i \,x^{2}+g x +2 h x +5 i x +\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{2}-\frac {e}{2}-\frac {f}{2}-\frac {g}{2}-\frac {h}{2}-\frac {i}{2}\right ) \ln \left (x -1\right )+\left (\frac {d}{3}+\frac {2 e}{3}+\frac {4 f}{3}+\frac {8 g}{3}+\frac {16 h}{3}+\frac {32 i}{3}\right ) \ln \left (x -2\right )\)	\(102\)
parallelrisch	\(g x +i \,x^{2}+\frac {\ln \left (x -2\right ) d}{3}+\frac {2 \ln \left (x -2\right ) e}{3}-\frac {\ln \left (x -1\right ) d}{2}-\frac {\ln \left (x -1\right ) e}{2}+\frac {h \,x^{2}}{2}+\frac {i \,x^{3}}{3}-\frac {\ln \left (x +1\right ) i}{6}+\frac {\ln \left (x +1\right ) f}{6}+2 h x +\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}-\frac {\ln \left (x +1\right ) g}{6}+\frac {32 \ln \left (x -2\right ) i}{3}-\frac {\ln \left (x -1\right ) i}{2}+\frac {8 \ln \left (x -2\right ) g}{3}-\frac {\ln \left (x -1\right ) g}{2}+\frac {4 \ln \left (x -2\right ) f}{3}-\frac {\ln \left (x -1\right ) f}{2}+\frac {16 \ln \left (x -2\right ) h}{3}-\frac {\ln \left (x -1\right ) h}{2}+5 i x +\frac {\ln \left (x +1\right ) h}{6}\)	\(156\)
risch	\(\frac {i \,x^{3}}{3}+\frac {h \,x^{2}}{2}+i \,x^{2}+g x +2 h x +5 i x +\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {\ln \left (x +1\right ) f}{6}-\frac {\ln \left (x +1\right ) g}{6}+\frac {\ln \left (x +1\right ) h}{6}-\frac {\ln \left (x +1\right ) i}{6}-\frac {\ln \left (1-x \right ) d}{2}-\frac {\ln \left (1-x \right ) e}{2}-\frac {\ln \left (1-x \right ) f}{2}-\frac {\ln \left (1-x \right ) g}{2}-\frac {\ln \left (1-x \right ) h}{2}-\frac {\ln \left (1-x \right ) i}{2}+\frac {\ln \left (2-x \right ) d}{3}+\frac {2 \ln \left (2-x \right ) e}{3}+\frac {4 \ln \left (2-x \right ) f}{3}+\frac {8 \ln \left (2-x \right ) g}{3}+\frac {16 \ln \left (2-x \right ) h}{3}+\frac {32 \ln \left (2-x \right ) i}{3}\)	\(180\)

[In]

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

[Out]

(1/2*h+i)*x^2+(g+2*h+5*i)*x+1/3*i*x^3+(-1/2*d-1/2*e-1/2*f-1/2*g-1/2*h-1/2*i)*ln(x-1)+(1/3*d+2/3*e+4/3*f+8/3*g+
16/3*h+32/3*i)*ln(x-2)+(1/6*d-1/6*e+1/6*f-1/6*g+1/6*h-1/6*i)*ln(x+1)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{3} \, i x^{3} + \frac {1}{2} \, {\left (h + 2 \, i\right )} x^{2} + {\left (g + 2 \, h + 5 \, i\right )} x + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac {1}{2} \, {\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac {1}{3} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \]

[In]

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

[Out]

1/3*i*x^3 + 1/2*(h + 2*i)*x^2 + (g + 2*h + 5*i)*x + 1/6*(d - e + f - g + h - i)*log(x + 1) - 1/2*(d + e + f +
g + h + i)*log(x - 1) + 1/3*(d + 2*e + 4*f + 8*g + 16*h + 32*i)*log(x - 2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{3} \, i x^{3} + \frac {1}{2} \, {\left (h + 2 \, i\right )} x^{2} + {\left (g + 2 \, h + 5 \, i\right )} x + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac {1}{2} \, {\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac {1}{3} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \]

[In]

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

[Out]

1/3*i*x^3 + 1/2*(h + 2*i)*x^2 + (g + 2*h + 5*i)*x + 1/6*(d - e + f - g + h - i)*log(x + 1) - 1/2*(d + e + f +
g + h + i)*log(x - 1) + 1/3*(d + 2*e + 4*f + 8*g + 16*h + 32*i)*log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{3} \, i x^{3} + \frac {1}{2} \, h x^{2} + i x^{2} + g x + 2 \, h x + 5 \, i x + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{2} \, {\left (d + e + f + g + h + i\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{3} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left ({\left | x - 2 \right |}\right ) \]

[In]

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

[Out]

1/3*i*x^3 + 1/2*h*x^2 + i*x^2 + g*x + 2*h*x + 5*i*x + 1/6*(d - e + f - g + h - i)*log(abs(x + 1)) - 1/2*(d + e
+ f + g + h + i)*log(abs(x - 1)) + 1/3*(d + 2*e + 4*f + 8*g + 16*h + 32*i)*log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 7.93 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.03 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=x\,\left (g+2\,h+5\,i\right )+\frac {i\,x^3}{3}-\ln \left (x-1\right )\,\left (\frac {d}{2}+\frac {e}{2}+\frac {f}{2}+\frac {g}{2}+\frac {h}{2}+\frac {i}{2}\right )+\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right )+\ln \left (x-2\right )\,\left (\frac {d}{3}+\frac {2\,e}{3}+\frac {4\,f}{3}+\frac {8\,g}{3}+\frac {16\,h}{3}+\frac {32\,i}{3}\right )+x^2\,\left (\frac {h}{2}+i\right ) \]

[In]

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

[Out]

x*(g + 2*h + 5*i) + (i*x^3)/3 - log(x - 1)*(d/2 + e/2 + f/2 + g/2 + h/2 + i/2) + log(x + 1)*(d/6 - e/6 + f/6 -
g/6 + h/6 - i/6) + log(x - 2)*(d/3 + (2*e)/3 + (4*f)/3 + (8*g)/3 + (16*h)/3 + (32*i)/3) + x^2*(h/2 + i)