\(\int \frac {(2-x-2 x^2+x^3) (d+e x+f x^2+g x^3+h x^4)}{4-5 x^2+x^4} \, dx\) [71]

Optimal result

Integrand size = 46, antiderivative size = 68 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=(e-2 f+4 g-8 h) x+\frac {1}{2} (f-2 g+4 h) x^2+\frac {1}{3} (g-2 h) x^3+\frac {h x^4}{4}+(d-2 e+4 f-8 g+16 h) \log (2+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, 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, 1864} \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\log (x+2) (d-2 e+4 f-8 g+16 h)+x (e-2 f+4 g-8 h)+\frac {1}{2} x^2 (f-2 g+4 h)+\frac {1}{3} x^3 (g-2 h)+\frac {h x^4}{4} \]

[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),x]

[Out]

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

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 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=(e-2 f+4 g-8 h) x+\frac {1}{2} (f-2 g+4 h) x^2+\frac {1}{3} (g-2 h) x^3+\frac {h x^4}{4}+(d-2 e+4 f-8 g+16 h) \log (2+x) \]

[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),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96

method	result	size
norman	\(\left (\frac {g}{3}-\frac {2 h}{3}\right ) x^{3}+\left (\frac {f}{2}-g +2 h \right ) x^{2}+\left (e -2 f +4 g -8 h \right ) x +\frac {h \,x^{4}}{4}+\left (d -2 e +4 f -8 g +16 h \right ) \ln \left (x +2\right )\)	\(65\)
default	\(\frac {h \,x^{4}}{4}+\frac {g \,x^{3}}{3}-\frac {2 h \,x^{3}}{3}+\frac {f \,x^{2}}{2}-g \,x^{2}+2 h \,x^{2}+e x -2 f x +4 g x -8 h x +\left (d -2 e +4 f -8 g +16 h \right ) \ln \left (x +2\right )\)	\(72\)
risch	\(\frac {h \,x^{4}}{4}+\frac {g \,x^{3}}{3}-\frac {2 h \,x^{3}}{3}+\frac {f \,x^{2}}{2}-g \,x^{2}+2 h \,x^{2}+e x -2 f x +4 g x -8 h x +\ln \left (x +2\right ) d -2 \ln \left (x +2\right ) e +4 \ln \left (x +2\right ) f -8 \ln \left (x +2\right ) g +16 \ln \left (x +2\right ) h\)	\(87\)
parallelrisch	\(\frac {h \,x^{4}}{4}+\frac {g \,x^{3}}{3}-\frac {2 h \,x^{3}}{3}+\frac {f \,x^{2}}{2}-g \,x^{2}+2 h \,x^{2}+e x -2 f x +4 g x -8 h x +\ln \left (x +2\right ) d -2 \ln \left (x +2\right ) e +4 \ln \left (x +2\right ) f -8 \ln \left (x +2\right ) g +16 \ln \left (x +2\right ) h\)	\(87\)

[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),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {1}{4} \, h x^{4} + \frac {1}{3} \, {\left (g - 2 \, h\right )} x^{3} + \frac {1}{2} \, {\left (f - 2 \, g + 4 \, h\right )} x^{2} + {\left (e - 2 \, f + 4 \, g - 8 \, h\right )} x + {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left (x + 2\right ) \]

[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),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {h x^{4}}{4} + x^{3} \left (\frac {g}{3} - \frac {2 h}{3}\right ) + x^{2} \left (\frac {f}{2} - g + 2 h\right ) + x \left (e - 2 f + 4 g - 8 h\right ) + \left (d - 2 e + 4 f - 8 g + 16 h\right ) \log {\left (x + 2 \right )} \]

[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),x)

[Out]

h*x**4/4 + x**3*(g/3 - 2*h/3) + x**2*(f/2 - g + 2*h) + x*(e - 2*f + 4*g - 8*h) + (d - 2*e + 4*f - 8*g + 16*h)*
log(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {1}{4} \, h x^{4} + \frac {1}{3} \, {\left (g - 2 \, h\right )} x^{3} + \frac {1}{2} \, {\left (f - 2 \, g + 4 \, h\right )} x^{2} + {\left (e - 2 \, f + 4 \, g - 8 \, h\right )} x + {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left (x + 2\right ) \]

[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),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {1}{4} \, h x^{4} + \frac {1}{3} \, g x^{3} - \frac {2}{3} \, h x^{3} + \frac {1}{2} \, f x^{2} - g x^{2} + 2 \, h x^{2} + e x - 2 \, f x + 4 \, g x - 8 \, h x + {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left ({\left | x + 2 \right |}\right ) \]

[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),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=x^3\,\left (\frac {g}{3}-\frac {2\,h}{3}\right )+\ln \left (x+2\right )\,\left (d-2\,e+4\,f-8\,g+16\,h\right )+\frac {h\,x^4}{4}+x^2\,\left (\frac {f}{2}-g+2\,h\right )+x\,\left (e-2\,f+4\,g-8\,h\right ) \]

[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),x)

[Out]

x^3*(g/3 - (2*h)/3) + log(x + 2)*(d - 2*e + 4*f - 8*g + 16*h) + (h*x^4)/4 + x^2*(f/2 - g + 2*h) + x*(e - 2*f +
4*g - 8*h)