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

Optimal. Leaf size=68 \[ \log (x+2) (d-2 e+4 f-8 g+16 h)+x (e-2 f+4 g-8 h)+\frac{1}{2} x^2 (f-2 g+4 h)+\frac{1}{3} x^3 (g-2 h)+\frac{h x^4}{4} \]

[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]

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Rubi [A]  time = 0.117464, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {1586, 1850} \[ \log (x+2) (d-2 e+4 f-8 g+16 h)+x (e-2 f+4 g-8 h)+\frac{1}{2} x^2 (f-2 g+4 h)+\frac{1}{3} x^3 (g-2 h)+\frac{h x^4}{4} \]

Antiderivative was successfully verified.

[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 1586

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 1850

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*} \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 &=\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]  time = 0.0231874, size = 68, normalized size = 1. \[ \log (x+2) (d-2 e+4 f-8 g+16 h)+x (e-2 f+4 g-8 h)+\frac{1}{2} x^2 (f-2 g+4 h)+\frac{1}{3} x^3 (g-2 h)+\frac{h x^4}{4} \]

Antiderivative was successfully verified.

[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]

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Maple [A]  time = 0.003, size = 87, normalized size = 1.3 \begin{align*}{\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}+ex-2\,fx+4\,gx-8\,hx+\ln \left ( 2+x \right ) d-2\,\ln \left ( 2+x \right ) e+4\,\ln \left ( 2+x \right ) f-8\,\ln \left ( 2+x \right ) g+16\,\ln \left ( 2+x \right ) h \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[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+ln(2+x)*d-2*ln(2+x)*e+4*ln(2+x)*f-
8*ln(2+x)*g+16*ln(2+x)*h

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Maxima [A]  time = 1.03224, size = 84, normalized size = 1.24 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.51703, size = 169, normalized size = 2.49 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.341934, size = 63, normalized size = 0.93 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.07238, size = 100, normalized size = 1.47 \begin{align*} \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} - 2 \, f x + 4 \, g x - 8 \, h x + x e +{\left (d + 4 \, f - 8 \, g + 16 \, h - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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