3.68.39 \(\int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 (-36+72 x-60 x^2+24 x^3-4 x^4)+e^5 (18-36 x+30 x^2-12 x^3+2 x^4) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx\)

Optimal. Leaf size=35 \[ \frac {4}{-4+x+\frac {3+x}{x}}+\left (2-\frac {1}{3} e^5 (2-\log (2))\right ) \log \left (x^2\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 33, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 4, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2074, 638, 618, 204} \begin {gather*} \frac {4 x}{x^2-3 x+3}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(108 - 180*x + 180*x^2 - 84*x^3 + 12*x^4 + E^5*(-36 + 72*x - 60*x^2 + 24*x^3 - 4*x^4) + E^5*(18 - 36*x + 3
0*x^2 - 12*x^3 + 2*x^4)*Log[2])/(27*x - 54*x^2 + 45*x^3 - 18*x^4 + 3*x^5),x]

[Out]

(4*x)/(3 - 3*x + x^2) + (2*(6 - E^5*(2 - Log[2]))*Log[x])/3

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2074

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 {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {12 (-2+x)}{\left (3-3 x+x^2\right )^2}-\frac {4}{3-3 x+x^2}+\frac {2 \left (6-e^5 (2-\log (2))\right )}{3 x}\right ) \, dx\\ &=\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)-4 \int \frac {1}{3-3 x+x^2} \, dx-12 \int \frac {-2+x}{\left (3-3 x+x^2\right )^2} \, dx\\ &=\frac {4 x}{3-3 x+x^2}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)+4 \int \frac {1}{3-3 x+x^2} \, dx+8 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-3+2 x\right )\\ &=\frac {4 x}{3-3 x+x^2}+\frac {8 \tan ^{-1}\left (\frac {3-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)-8 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-3+2 x\right )\\ &=\frac {4 x}{3-3 x+x^2}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.89 \begin {gather*} \frac {2}{3} \left (\frac {6 x}{3-3 x+x^2}+\left (6+e^5 (-2+\log (2))\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(108 - 180*x + 180*x^2 - 84*x^3 + 12*x^4 + E^5*(-36 + 72*x - 60*x^2 + 24*x^3 - 4*x^4) + E^5*(18 - 36
*x + 30*x^2 - 12*x^3 + 2*x^4)*Log[2])/(27*x - 54*x^2 + 45*x^3 - 18*x^4 + 3*x^5),x]

[Out]

(2*((6*x)/(3 - 3*x + x^2) + (6 + E^5*(-2 + Log[2]))*Log[x]))/3

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fricas [A]  time = 0.58, size = 54, normalized size = 1.54 \begin {gather*} \frac {2 \, {\left ({\left ({\left (x^{2} - 3 \, x + 3\right )} e^{5} \log \relax (2) + 6 \, x^{2} - 2 \, {\left (x^{2} - 3 \, x + 3\right )} e^{5} - 18 \, x + 18\right )} \log \relax (x) + 6 \, x\right )}}{3 \, {\left (x^{2} - 3 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x^2+72*x-36)*exp(5)+12*x^4-84*x^3+180
*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-54*x^2+27*x),x, algorithm="fricas")

[Out]

2/3*(((x^2 - 3*x + 3)*e^5*log(2) + 6*x^2 - 2*(x^2 - 3*x + 3)*e^5 - 18*x + 18)*log(x) + 6*x)/(x^2 - 3*x + 3)

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giac [A]  time = 0.34, size = 30, normalized size = 0.86 \begin {gather*} \frac {2}{3} \, {\left (e^{5} \log \relax (2) - 2 \, e^{5} + 6\right )} \log \left ({\left | x \right |}\right ) + \frac {4 \, x}{x^{2} - 3 \, x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x^2+72*x-36)*exp(5)+12*x^4-84*x^3+180
*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-54*x^2+27*x),x, algorithm="giac")

[Out]

2/3*(e^5*log(2) - 2*e^5 + 6)*log(abs(x)) + 4*x/(x^2 - 3*x + 3)

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maple [A]  time = 0.08, size = 30, normalized size = 0.86

method	result	size
default	\(\frac {2 \left ({\mathrm e}^{5} \ln \relax (2)-2 \,{\mathrm e}^{5}+6\right ) \ln \relax (x )}{3}+\frac {4 x}{x^{2}-3 x +3}\)	\(30\)
norman	\(\frac {4 x}{x^{2}-3 x +3}+\left (\frac {2 \,{\mathrm e}^{5} \ln \relax (2)}{3}-\frac {4 \,{\mathrm e}^{5}}{3}+4\right ) \ln \relax (x )\)	\(30\)
risch	\(\frac {4 x}{x^{2}-3 x +3}+\frac {2 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )}{3}-\frac {4 \,{\mathrm e}^{5} \ln \relax (x )}{3}+4 \ln \relax (x )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*ln(2)+(-4*x^4+24*x^3-60*x^2+72*x-36)*exp(5)+12*x^4-84*x^3+180*x^2-18
0*x+108)/(3*x^5-18*x^4+45*x^3-54*x^2+27*x),x,method=_RETURNVERBOSE)

[Out]

2/3*(exp(5)*ln(2)-2*exp(5)+6)*ln(x)+4*x/(x^2-3*x+3)

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maxima [A]  time = 0.37, size = 29, normalized size = 0.83 \begin {gather*} \frac {2}{3} \, {\left (e^{5} \log \relax (2) - 2 \, e^{5} + 6\right )} \log \relax (x) + \frac {4 \, x}{x^{2} - 3 \, x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x^2+72*x-36)*exp(5)+12*x^4-84*x^3+180
*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-54*x^2+27*x),x, algorithm="maxima")

[Out]

2/3*(e^5*log(2) - 2*e^5 + 6)*log(x) + 4*x/(x^2 - 3*x + 3)

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mupad [B]  time = 4.08, size = 31, normalized size = 0.89 \begin {gather*} \ln \relax (x)\,\left (\frac {2\,{\mathrm {e}}^5\,\ln \relax (2)}{3}-\frac {4\,{\mathrm {e}}^5}{3}+4\right )+\frac {12\,x}{3\,x^2-9\,x+9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((180*x^2 - exp(5)*(60*x^2 - 72*x - 24*x^3 + 4*x^4 + 36) - 180*x - 84*x^3 + 12*x^4 + exp(5)*log(2)*(30*x^2
- 36*x - 12*x^3 + 2*x^4 + 18) + 108)/(27*x - 54*x^2 + 45*x^3 - 18*x^4 + 3*x^5),x)

[Out]

log(x)*((2*exp(5)*log(2))/3 - (4*exp(5))/3 + 4) + (12*x)/(3*x^2 - 9*x + 9)

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sympy [A]  time = 0.80, size = 31, normalized size = 0.89 \begin {gather*} \frac {4 x}{x^{2} - 3 x + 3} + \frac {2 \left (- 2 e^{5} + 6 + e^{5} \log {\relax (2 )}\right ) \log {\relax (x )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4-12*x**3+30*x**2-36*x+18)*exp(5)*ln(2)+(-4*x**4+24*x**3-60*x**2+72*x-36)*exp(5)+12*x**4-84*x
**3+180*x**2-180*x+108)/(3*x**5-18*x**4+45*x**3-54*x**2+27*x),x)

[Out]

4*x/(x**2 - 3*x + 3) + 2*(-2*exp(5) + 6 + exp(5)*log(2))*log(x)/3

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