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

Optimal. Leaf size=177 \[ -\frac{d-e+f-g+h-i}{36 (x+1)}+\frac{d+e+f+g+h+i}{12 (1-x)}+\frac{d+2 e+4 f+8 g+16 h+32 i}{36 (2-x)}+\frac{1}{36} \log (1-x) (2 d+5 e+8 f+11 g+14 h+17 i)-\frac{1}{432} \log (2-x) (35 d+58 e+92 f+136 g+176 h+160 i)+\frac{1}{108} \log (x+1) (2 d+e-4 f+7 g-10 h+13 i)+\frac{1}{144} \log (x+2) (d-2 e+4 f-8 g+16 h-32 i) \]

[Out]

(d + e + f + g + h + i)/(12*(1 - x)) + (d + 2*e + 4*f + 8*g + 16*h + 32*i)/(36*(2 - x)) - (d - e + f - g + h -
i)/(36*(1 + x)) + ((2*d + 5*e + 8*f + 11*g + 14*h + 17*i)*Log[1 - x])/36 - ((35*d + 58*e + 92*f + 136*g + 176
*h + 160*i)*Log[2 - x])/432 + ((2*d + e - 4*f + 7*g - 10*h + 13*i)*Log[1 + x])/108 + ((d - 2*e + 4*f - 8*g + 1
6*h - 32*i)*Log[2 + x])/144

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Rubi [A]  time = 0.343235, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {1586, 6742} \[ -\frac{d-e+f-g+h-i}{36 (x+1)}+\frac{d+e+f+g+h+i}{12 (1-x)}+\frac{d+2 e+4 f+8 g+16 h+32 i}{36 (2-x)}+\frac{1}{36} \log (1-x) (2 d+5 e+8 f+11 g+14 h+17 i)-\frac{1}{432} \log (2-x) (35 d+58 e+92 f+136 g+176 h+160 i)+\frac{1}{108} \log (x+1) (2 d+e-4 f+7 g-10 h+13 i)+\frac{1}{144} \log (x+2) (d-2 e+4 f-8 g+16 h-32 i) \]

Antiderivative was successfully verified.

[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)^2,x]

[Out]

(d + e + f + g + h + i)/(12*(1 - x)) + (d + 2*e + 4*f + 8*g + 16*h + 32*i)/(36*(2 - x)) - (d - e + f - g + h -
i)/(36*(1 + x)) + ((2*d + 5*e + 8*f + 11*g + 14*h + 17*i)*Log[1 - x])/36 - ((35*d + 58*e + 92*f + 136*g + 176
*h + 160*i)*Log[2 - x])/432 + ((2*d + e - 4*f + 7*g - 10*h + 13*i)*Log[1 + x])/108 + ((d - 2*e + 4*f - 8*g + 1
6*h - 32*i)*Log[2 + x])/144

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 6742

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

Rubi steps

\begin{align*} \int \frac{(2+x) \left (d+e x+f x^2+g x^3+h x^4+102 x^5\right )}{\left (4-5 x^2+x^4\right )^2} \, dx &=\int \frac{d+e x+f x^2+g x^3+h x^4+102 x^5}{(2+x) \left (2-x-2 x^2+x^3\right )^2} \, dx\\ &=\int \left (\frac{3264+d+2 e+4 f+8 g+16 h}{36 (-2+x)^2}+\frac{-16320-35 d-58 e-92 f-136 g-176 h}{432 (-2+x)}+\frac{102+d+e+f+g+h}{12 (-1+x)^2}+\frac{1734+2 d+5 e+8 f+11 g+14 h}{36 (-1+x)}+\frac{-102+d-e+f-g+h}{36 (1+x)^2}+\frac{1326+2 d+e-4 f+7 g-10 h}{108 (1+x)}+\frac{-3264+d-2 e+4 f-8 g+16 h}{144 (2+x)}\right ) \, dx\\ &=\frac{102+d+e+f+g+h}{12 (1-x)}+\frac{3264+d+2 e+4 f+8 g+16 h}{36 (2-x)}+\frac{102-d+e-f+g-h}{36 (1+x)}+\frac{1}{36} (1734+2 d+5 e+8 f+11 g+14 h) \log (1-x)-\frac{1}{432} (16320+35 d+58 e+92 f+136 g+176 h) \log (2-x)+\frac{1}{108} (1326+2 d+e-4 f+7 g-10 h) \log (1+x)-\frac{1}{144} (3264-d+2 e-4 f+8 g-16 h) \log (2+x)\\ \end{align*}

Mathematica [A]  time = 0.119405, size = 195, normalized size = 1.1 \[ \frac{-5 d x^2+6 d x+5 d-4 e x^2+10 e-8 f x^2+6 f x+8 f-10 g x^2+16 g-20 h x^2+6 h x+20 h-34 i x^2+40 i}{36 \left (x^3-2 x^2-x+2\right )}+\frac{1}{36} \log (1-x) (2 d+5 e+8 f+11 g+14 h+17 i)+\frac{1}{432} \log (2-x) (-35 d-58 e-92 f-136 g-176 h-160 i)+\frac{1}{108} \log (x+1) (2 d+e-4 f+7 g-10 h+13 i)+\frac{1}{144} \log (x+2) (d-2 e+4 f-8 g+16 h-32 i) \]

Antiderivative was successfully verified.

[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)^2,x]

[Out]

(5*d + 10*e + 8*f + 16*g + 20*h + 40*i + 6*d*x + 6*f*x + 6*h*x - 5*d*x^2 - 4*e*x^2 - 8*f*x^2 - 10*g*x^2 - 20*h
*x^2 - 34*i*x^2)/(36*(2 - x - 2*x^2 + x^3)) + ((2*d + 5*e + 8*f + 11*g + 14*h + 17*i)*Log[1 - x])/36 + ((-35*d
- 58*e - 92*f - 136*g - 176*h - 160*i)*Log[2 - x])/432 + ((2*d + e - 4*f + 7*g - 10*h + 13*i)*Log[1 + x])/108
+ ((d - 2*e + 4*f - 8*g + 16*h - 32*i)*Log[2 + x])/144

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Maple [A]  time = 0.016, size = 314, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( 2+x \right ) d}{144}}-{\frac{\ln \left ( 2+x \right ) e}{72}}+{\frac{\ln \left ( 1+x \right ) d}{54}}+{\frac{\ln \left ( 1+x \right ) e}{108}}-{\frac{35\,\ln \left ( x-2 \right ) d}{432}}-{\frac{29\,\ln \left ( x-2 \right ) e}{216}}+{\frac{\ln \left ( x-1 \right ) d}{18}}+{\frac{5\,\ln \left ( x-1 \right ) e}{36}}-{\frac{8\,i}{9\,x-18}}-{\frac{i}{12\,x-12}}+{\frac{i}{36+36\,x}}-{\frac{4\,h}{9\,x-18}}-{\frac{h}{12\,x-12}}-{\frac{h}{36+36\,x}}-{\frac{d}{36+36\,x}}+{\frac{e}{36+36\,x}}-{\frac{2\,g}{9\,x-18}}-{\frac{d}{36\,x-72}}-{\frac{e}{18\,x-36}}-{\frac{g}{12\,x-12}}-{\frac{d}{12\,x-12}}-{\frac{e}{12\,x-12}}+{\frac{g}{36+36\,x}}-{\frac{f}{36+36\,x}}-{\frac{f}{9\,x-18}}-{\frac{f}{12\,x-12}}-{\frac{10\,\ln \left ( x-2 \right ) i}{27}}+{\frac{17\,\ln \left ( x-1 \right ) i}{36}}-{\frac{2\,\ln \left ( 2+x \right ) i}{9}}+{\frac{13\,\ln \left ( 1+x \right ) i}{108}}-{\frac{\ln \left ( 2+x \right ) g}{18}}+{\frac{7\,\ln \left ( 1+x \right ) g}{108}}-{\frac{17\,\ln \left ( x-2 \right ) g}{54}}+{\frac{11\,\ln \left ( x-1 \right ) g}{36}}+{\frac{\ln \left ( 2+x \right ) h}{9}}-{\frac{5\,\ln \left ( 1+x \right ) h}{54}}-{\frac{11\,\ln \left ( x-2 \right ) h}{27}}+{\frac{7\,\ln \left ( x-1 \right ) h}{18}}-{\frac{23\,\ln \left ( x-2 \right ) f}{108}}+{\frac{2\,\ln \left ( x-1 \right ) f}{9}}+{\frac{\ln \left ( 2+x \right ) f}{36}}-{\frac{\ln \left ( 1+x \right ) f}{27}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*ln(2+x)*d-1/72*ln(2+x)*e+1/54*ln(1+x)*d+1/108*ln(1+x)*e-35/432*ln(x-2)*d-29/216*ln(x-2)*e+1/18*ln(x-1)*d
+5/36*ln(x-1)*e-8/9/(x-2)*i-1/12/(x-1)*i+1/36/(1+x)*i-4/9/(x-2)*h-1/12/(x-1)*h-1/36/(1+x)*h-1/36/(1+x)*d+1/36/
(1+x)*e-2/9/(x-2)*g-1/36/(x-2)*d-1/18/(x-2)*e-1/12/(x-1)*g-1/12/(x-1)*d-1/12/(x-1)*e+1/36/(1+x)*g-1/36/(1+x)*f
-1/9/(x-2)*f-1/12/(x-1)*f-10/27*ln(x-2)*i+17/36*ln(x-1)*i-2/9*ln(2+x)*i+13/108*ln(1+x)*i-1/18*ln(2+x)*g+7/108*
ln(1+x)*g-17/54*ln(x-2)*g+11/36*ln(x-1)*g+1/9*ln(2+x)*h-5/54*ln(1+x)*h-11/27*ln(x-2)*h+7/18*ln(x-1)*h-23/108*l
n(x-2)*f+2/9*ln(x-1)*f+1/36*ln(2+x)*f-1/27*ln(1+x)*f

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Maxima [A]  time = 0.985017, size = 220, normalized size = 1.24 \begin{align*} \frac{1}{144} \,{\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) + \frac{1}{108} \,{\left (2 \, d + e - 4 \, f + 7 \, g - 10 \, h + 13 \, i\right )} \log \left (x + 1\right ) + \frac{1}{36} \,{\left (2 \, d + 5 \, e + 8 \, f + 11 \, g + 14 \, h + 17 \, i\right )} \log \left (x - 1\right ) - \frac{1}{432} \,{\left (35 \, d + 58 \, e + 92 \, f + 136 \, g + 176 \, h + 160 \, i\right )} \log \left (x - 2\right ) - \frac{{\left (5 \, d + 4 \, e + 8 \, f + 10 \, g + 20 \, h + 34 \, i\right )} x^{2} - 6 \,{\left (d + f + h\right )} x - 5 \, d - 10 \, e - 8 \, f - 16 \, g - 20 \, h - 40 \, i}{36 \,{\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)^2,x, algorithm="maxima")

[Out]

1/144*(d - 2*e + 4*f - 8*g + 16*h - 32*i)*log(x + 2) + 1/108*(2*d + e - 4*f + 7*g - 10*h + 13*i)*log(x + 1) +
1/36*(2*d + 5*e + 8*f + 11*g + 14*h + 17*i)*log(x - 1) - 1/432*(35*d + 58*e + 92*f + 136*g + 176*h + 160*i)*lo
g(x - 2) - 1/36*((5*d + 4*e + 8*f + 10*g + 20*h + 34*i)*x^2 - 6*(d + f + h)*x - 5*d - 10*e - 8*f - 16*g - 20*h
- 40*i)/(x^3 - 2*x^2 - x + 2)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.08367, size = 234, normalized size = 1.32 \begin{align*} \frac{1}{144} \,{\left (d + 4 \, f - 8 \, g + 16 \, h - 32 \, i - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac{1}{108} \,{\left (2 \, d - 4 \, f + 7 \, g - 10 \, h + 13 \, i + e\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac{1}{36} \,{\left (2 \, d + 8 \, f + 11 \, g + 14 \, h + 17 \, i + 5 \, e\right )} \log \left ({\left | x - 1 \right |}\right ) - \frac{1}{432} \,{\left (35 \, d + 92 \, f + 136 \, g + 176 \, h + 160 \, i + 58 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) - \frac{{\left (5 \, d + 8 \, f + 10 \, g + 20 \, h + 34 \, i + 4 \, e\right )} x^{2} - 6 \,{\left (d + f + h\right )} x - 5 \, d - 8 \, f - 16 \, g - 20 \, h - 40 \, i - 10 \, e}{36 \,{\left (x + 1\right )}{\left (x - 1\right )}{\left (x - 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)^2,x, algorithm="giac")

[Out]

1/144*(d + 4*f - 8*g + 16*h - 32*i - 2*e)*log(abs(x + 2)) + 1/108*(2*d - 4*f + 7*g - 10*h + 13*i + e)*log(abs(
x + 1)) + 1/36*(2*d + 8*f + 11*g + 14*h + 17*i + 5*e)*log(abs(x - 1)) - 1/432*(35*d + 92*f + 136*g + 176*h + 1
60*i + 58*e)*log(abs(x - 2)) - 1/36*((5*d + 8*f + 10*g + 20*h + 34*i + 4*e)*x^2 - 6*(d + f + h)*x - 5*d - 8*f
- 16*g - 20*h - 40*i - 10*e)/((x + 1)*(x - 1)*(x - 2))