3.1.99 \(\int \frac {(2+x) (d+e x+f x^2)}{(4-5 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=122 \[ -\frac {d-e+f}{36 (x+1)}+\frac {d+e+f}{12 (1-x)}+\frac {d+2 e+4 f}{36 (2-x)}+\frac {1}{36} \log (1-x) (2 d+5 e+8 f)-\frac {1}{432} \log (2-x) (35 d+58 e+92 f)+\frac {1}{108} \log (x+1) (2 d+e-4 f)+\frac {1}{144} \log (x+2) (d-2 e+4 f) \]

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Rubi [A]  time = 0.22, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1586, 6742} \begin {gather*} -\frac {d-e+f}{36 (x+1)}+\frac {d+e+f}{12 (1-x)}+\frac {d+2 e+4 f}{36 (2-x)}+\frac {1}{36} \log (1-x) (2 d+5 e+8 f)-\frac {1}{432} \log (2-x) (35 d+58 e+92 f)+\frac {1}{108} \log (x+1) (2 d+e-4 f)+\frac {1}{144} \log (x+2) (d-2 e+4 f) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e + f)/(12*(1 - x)) + (d + 2*e + 4*f)/(36*(2 - x)) - (d - e + f)/(36*(1 + x)) + ((2*d + 5*e + 8*f)*Log[1
- x])/36 - ((35*d + 58*e + 92*f)*Log[2 - x])/432 + ((2*d + e - 4*f)*Log[1 + x])/108 + ((d - 2*e + 4*f)*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\right )}{\left (4-5 x^2+x^4\right )^2} \, dx &=\int \frac {d+e x+f x^2}{(2+x) \left (2-x-2 x^2+x^3\right )^2} \, dx\\ &=\int \left (\frac {d+2 e+4 f}{36 (-2+x)^2}+\frac {-35 d-58 e-92 f}{432 (-2+x)}+\frac {d+e+f}{12 (-1+x)^2}+\frac {2 d+5 e+8 f}{36 (-1+x)}+\frac {d-e+f}{36 (1+x)^2}+\frac {2 d+e-4 f}{108 (1+x)}+\frac {d-2 e+4 f}{144 (2+x)}\right ) \, dx\\ &=\frac {d+e+f}{12 (1-x)}+\frac {d+2 e+4 f}{36 (2-x)}-\frac {d-e+f}{36 (1+x)}+\frac {1}{36} (2 d+5 e+8 f) \log (1-x)-\frac {1}{432} (35 d+58 e+92 f) \log (2-x)+\frac {1}{108} (2 d+e-4 f) \log (1+x)+\frac {1}{144} (d-2 e+4 f) \log (2+x)\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 121, normalized size = 0.99 \begin {gather*} \frac {1}{432} \left (\frac {12 \left (d \left (-5 x^2+6 x+5\right )+e \left (10-4 x^2\right )+2 f \left (-4 x^2+3 x+4\right )\right )}{x^3-2 x^2-x+2}+12 \log (1-x) (2 d+5 e+8 f)-\log (2-x) (35 d+58 e+92 f)+4 \log (x+1) (2 d+e-4 f)+3 \log (x+2) (d-2 e+4 f)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((12*(d*(5 + 6*x - 5*x^2) + e*(10 - 4*x^2) + 2*f*(4 + 3*x - 4*x^2)))/(2 - x - 2*x^2 + x^3) + 12*(2*d + 5*e + 8
*f)*Log[1 - x] - (35*d + 58*e + 92*f)*Log[2 - x] + 4*(2*d + e - 4*f)*Log[1 + x] + 3*(d - 2*e + 4*f)*Log[2 + x]
)/432

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(2+x) \left (d+e x+f x^2\right )}{\left (4-5 x^2+x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((2 + x)*(d + e*x + f*x^2))/(4 - 5*x^2 + x^4)^2,x]

[Out]

IntegrateAlgebraic[((2 + x)*(d + e*x + f*x^2))/(4 - 5*x^2 + x^4)^2, x]

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fricas [B]  time = 1.32, size = 267, normalized size = 2.19 \begin {gather*} -\frac {12 \, {\left (5 \, d + 4 \, e + 8 \, f\right )} x^{2} - 72 \, {\left (d + f\right )} x - 3 \, {\left ({\left (d - 2 \, e + 4 \, f\right )} x^{3} - 2 \, {\left (d - 2 \, e + 4 \, f\right )} x^{2} - {\left (d - 2 \, e + 4 \, f\right )} x + 2 \, d - 4 \, e + 8 \, f\right )} \log \left (x + 2\right ) - 4 \, {\left ({\left (2 \, d + e - 4 \, f\right )} x^{3} - 2 \, {\left (2 \, d + e - 4 \, f\right )} x^{2} - {\left (2 \, d + e - 4 \, f\right )} x + 4 \, d + 2 \, e - 8 \, f\right )} \log \left (x + 1\right ) - 12 \, {\left ({\left (2 \, d + 5 \, e + 8 \, f\right )} x^{3} - 2 \, {\left (2 \, d + 5 \, e + 8 \, f\right )} x^{2} - {\left (2 \, d + 5 \, e + 8 \, f\right )} x + 4 \, d + 10 \, e + 16 \, f\right )} \log \left (x - 1\right ) + {\left ({\left (35 \, d + 58 \, e + 92 \, f\right )} x^{3} - 2 \, {\left (35 \, d + 58 \, e + 92 \, f\right )} x^{2} - {\left (35 \, d + 58 \, e + 92 \, f\right )} x + 70 \, d + 116 \, e + 184 \, f\right )} \log \left (x - 2\right ) - 60 \, d - 120 \, e - 96 \, f}{432 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/432*(12*(5*d + 4*e + 8*f)*x^2 - 72*(d + f)*x - 3*((d - 2*e + 4*f)*x^3 - 2*(d - 2*e + 4*f)*x^2 - (d - 2*e +
4*f)*x + 2*d - 4*e + 8*f)*log(x + 2) - 4*((2*d + e - 4*f)*x^3 - 2*(2*d + e - 4*f)*x^2 - (2*d + e - 4*f)*x + 4*
d + 2*e - 8*f)*log(x + 1) - 12*((2*d + 5*e + 8*f)*x^3 - 2*(2*d + 5*e + 8*f)*x^2 - (2*d + 5*e + 8*f)*x + 4*d +
10*e + 16*f)*log(x - 1) + ((35*d + 58*e + 92*f)*x^3 - 2*(35*d + 58*e + 92*f)*x^2 - (35*d + 58*e + 92*f)*x + 70
*d + 116*e + 184*f)*log(x - 2) - 60*d - 120*e - 96*f)/(x^3 - 2*x^2 - x + 2)

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giac [A]  time = 0.33, size = 118, normalized size = 0.97 \begin {gather*} \frac {1}{144} \, {\left (d + 4 \, f - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{108} \, {\left (2 \, d - 4 \, f + e\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{36} \, {\left (2 \, d + 8 \, f + 5 \, e\right )} \log \left ({\left | x - 1 \right |}\right ) - \frac {1}{432} \, {\left (35 \, d + 92 \, f + 58 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) - \frac {{\left (5 \, d + 8 \, f + 4 \, e\right )} x^{2} - 6 \, {\left (d + f\right )} x - 5 \, d - 8 \, f - 10 \, e}{36 \, {\left (x + 1\right )} {\left (x - 1\right )} {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(d + 4*f - 2*e)*log(abs(x + 2)) + 1/108*(2*d - 4*f + e)*log(abs(x + 1)) + 1/36*(2*d + 8*f + 5*e)*log(abs
(x - 1)) - 1/432*(35*d + 92*f + 58*e)*log(abs(x - 2)) - 1/36*((5*d + 8*f + 4*e)*x^2 - 6*(d + f)*x - 5*d - 8*f
- 10*e)/((x + 1)*(x - 1)*(x - 2))

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maple [A]  time = 0.02, size = 158, normalized size = 1.30 \begin {gather*} \frac {d \ln \left (x +2\right )}{144}-\frac {35 d \ln \left (x -2\right )}{432}+\frac {d \ln \left (x -1\right )}{18}+\frac {d \ln \left (x +1\right )}{54}-\frac {e \ln \left (x +2\right )}{72}-\frac {29 e \ln \left (x -2\right )}{216}+\frac {5 e \ln \left (x -1\right )}{36}+\frac {e \ln \left (x +1\right )}{108}+\frac {f \ln \left (x +2\right )}{36}-\frac {23 f \ln \left (x -2\right )}{108}+\frac {2 f \ln \left (x -1\right )}{9}-\frac {f \ln \left (x +1\right )}{27}-\frac {d}{36 \left (x -2\right )}-\frac {d}{36 \left (x +1\right )}-\frac {d}{12 \left (x -1\right )}-\frac {e}{18 \left (x -2\right )}+\frac {e}{36 x +36}-\frac {e}{12 \left (x -1\right )}-\frac {f}{9 \left (x -2\right )}-\frac {f}{36 \left (x +1\right )}-\frac {f}{12 \left (x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35/432*d*ln(x-2)-29/216*e*ln(x-2)-23/108*f*ln(x-2)-1/36/(x-2)*d-1/18/(x-2)*e-1/9/(x-2)*f-1/36/(x+1)*d+1/36/(x
+1)*e-1/36/(x+1)*f+1/54*d*ln(x+1)+1/108*e*ln(x+1)-1/27*f*ln(x+1)-1/12/(x-1)*d-1/12/(x-1)*e-1/12/(x-1)*f+1/18*d
*ln(x-1)+5/36*e*ln(x-1)+2/9*f*ln(x-1)+1/144*d*ln(x+2)-1/72*e*ln(x+2)+1/36*f*ln(x+2)

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maxima [A]  time = 0.44, size = 108, normalized size = 0.89 \begin {gather*} \frac {1}{144} \, {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) + \frac {1}{108} \, {\left (2 \, d + e - 4 \, f\right )} \log \left (x + 1\right ) + \frac {1}{36} \, {\left (2 \, d + 5 \, e + 8 \, f\right )} \log \left (x - 1\right ) - \frac {1}{432} \, {\left (35 \, d + 58 \, e + 92 \, f\right )} \log \left (x - 2\right ) - \frac {{\left (5 \, d + 4 \, e + 8 \, f\right )} x^{2} - 6 \, {\left (d + f\right )} x - 5 \, d - 10 \, e - 8 \, f}{36 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(d - 2*e + 4*f)*log(x + 2) + 1/108*(2*d + e - 4*f)*log(x + 1) + 1/36*(2*d + 5*e + 8*f)*log(x - 1) - 1/43
2*(35*d + 58*e + 92*f)*log(x - 2) - 1/36*((5*d + 4*e + 8*f)*x^2 - 6*(d + f)*x - 5*d - 10*e - 8*f)/(x^3 - 2*x^2
- x + 2)

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mupad [B]  time = 0.13, size = 113, normalized size = 0.93 \begin {gather*} \ln \left (x-1\right )\,\left (\frac {d}{18}+\frac {5\,e}{36}+\frac {2\,f}{9}\right )+\ln \left (x+1\right )\,\left (\frac {d}{54}+\frac {e}{108}-\frac {f}{27}\right )+\ln \left (x+2\right )\,\left (\frac {d}{144}-\frac {e}{72}+\frac {f}{36}\right )-\ln \left (x-2\right )\,\left (\frac {35\,d}{432}+\frac {29\,e}{216}+\frac {23\,f}{108}\right )-\frac {\left (-\frac {5\,d}{36}-\frac {e}{9}-\frac {2\,f}{9}\right )\,x^2+\left (\frac {d}{6}+\frac {f}{6}\right )\,x+\frac {5\,d}{36}+\frac {5\,e}{18}+\frac {2\,f}{9}}{-x^3+2\,x^2+x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1)*(d/18 + (5*e)/36 + (2*f)/9) + log(x + 1)*(d/54 + e/108 - f/27) + log(x + 2)*(d/144 - e/72 + f/36) -
log(x - 2)*((35*d)/432 + (29*e)/216 + (23*f)/108) - ((5*d)/36 + (5*e)/18 + (2*f)/9 + x*(d/6 + f/6) - x^2*((5*
d)/36 + e/9 + (2*f)/9))/(x + 2*x^2 - x^3 - 2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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