3.27.66 \(\int \frac {-27+117 x+9 x^2-15 x^3+e^8 (-9+33 x-10 x^2-3 x^3+x^4)+(-3 x+9 x^2+3 x^3) \log (-x^3+3 x^4+x^5)}{e^8 (-9 x+33 x^2-10 x^3-3 x^4+x^5)} \, dx\)

Optimal. Leaf size=32 \[ \log (x)+\frac {\log \left (x^2 \left (-x+x^2 (3+x)\right )\right )}{e^8 \left (1-\frac {x}{3}\right )} \]

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Rubi [A]  time = 0.44, antiderivative size = 31, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 6, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {12, 6688, 6742, 632, 31, 2525} \begin {gather*} \frac {3 \log \left (-x^3 \left (-x^2-3 x+1\right )\right )}{e^8 (3-x)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-27 + 117*x + 9*x^2 - 15*x^3 + E^8*(-9 + 33*x - 10*x^2 - 3*x^3 + x^4) + (-3*x + 9*x^2 + 3*x^3)*Log[-x^3 +
 3*x^4 + x^5])/(E^8*(-9*x + 33*x^2 - 10*x^3 - 3*x^4 + x^5)),x]

[Out]

Log[x] + (3*Log[-(x^3*(1 - 3*x - x^2))])/(E^8*(3 - x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
 1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-27+117 x+9 x^2-15 x^3+e^8 \left (-9+33 x-10 x^2-3 x^3+x^4\right )+\left (-3 x+9 x^2+3 x^3\right ) \log \left (-x^3+3 x^4+x^5\right )}{-9 x+33 x^2-10 x^3-3 x^4+x^5} \, dx}{e^8}\\ &=\frac {\int \left (\frac {e^8}{x}+\frac {9-36 x-15 x^2}{x \left (3-10 x+x^3\right )}+\frac {3 \log \left (x^3 \left (-1+3 x+x^2\right )\right )}{(-3+x)^2}\right ) \, dx}{e^8}\\ &=\log (x)+\frac {\int \frac {9-36 x-15 x^2}{x \left (3-10 x+x^3\right )} \, dx}{e^8}+\frac {3 \int \frac {\log \left (x^3 \left (-1+3 x+x^2\right )\right )}{(-3+x)^2} \, dx}{e^8}\\ &=\log (x)+\frac {3 \log \left (-x^3 \left (1-3 x-x^2\right )\right )}{e^8 (3-x)}+\frac {\int \left (-\frac {78}{17 (-3+x)}+\frac {3}{x}+\frac {3 (20+9 x)}{17 \left (-1+3 x+x^2\right )}\right ) \, dx}{e^8}+\frac {3 \int \frac {-3+12 x+5 x^2}{x \left (3-10 x+x^3\right )} \, dx}{e^8}\\ &=-\frac {78 \log (3-x)}{17 e^8}+\log (x)+\frac {3 \log (x)}{e^8}+\frac {3 \log \left (-x^3 \left (1-3 x-x^2\right )\right )}{e^8 (3-x)}+\frac {3 \int \frac {20+9 x}{-1+3 x+x^2} \, dx}{17 e^8}+\frac {3 \int \left (\frac {26}{17 (-3+x)}-\frac {1}{x}+\frac {-20-9 x}{17 \left (-1+3 x+x^2\right )}\right ) \, dx}{e^8}\\ &=\log (x)+\frac {3 \log \left (-x^3 \left (1-3 x-x^2\right )\right )}{e^8 (3-x)}+\frac {3 \int \frac {-20-9 x}{-1+3 x+x^2} \, dx}{17 e^8}+\frac {\left (3 \left (9-\sqrt {13}\right )\right ) \int \frac {1}{\frac {3}{2}+\frac {\sqrt {13}}{2}+x} \, dx}{34 e^8}+\frac {\left (3 \left (9+\sqrt {13}\right )\right ) \int \frac {1}{\frac {3}{2}-\frac {\sqrt {13}}{2}+x} \, dx}{34 e^8}\\ &=\log (x)+\frac {3 \left (9+\sqrt {13}\right ) \log \left (3-\sqrt {13}+2 x\right )}{34 e^8}+\frac {3 \left (9-\sqrt {13}\right ) \log \left (3+\sqrt {13}+2 x\right )}{34 e^8}+\frac {3 \log \left (-x^3 \left (1-3 x-x^2\right )\right )}{e^8 (3-x)}-\frac {\left (3 \left (9-\sqrt {13}\right )\right ) \int \frac {1}{\frac {3}{2}+\frac {\sqrt {13}}{2}+x} \, dx}{34 e^8}-\frac {\left (3 \left (9+\sqrt {13}\right )\right ) \int \frac {1}{\frac {3}{2}-\frac {\sqrt {13}}{2}+x} \, dx}{34 e^8}\\ &=\log (x)+\frac {3 \log \left (-x^3 \left (1-3 x-x^2\right )\right )}{e^8 (3-x)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.15, size = 78, normalized size = 2.44 \begin {gather*} \frac {3 \left (9+\sqrt {13}\right ) \log \left (-3+\sqrt {13}-2 x\right )}{34 e^8}+\log (x)-\frac {3 \left (9+\sqrt {13}\right ) \log \left (3-\sqrt {13}+2 x\right )}{34 e^8}-\frac {3 \log \left (x^3 \left (-1+3 x+x^2\right )\right )}{e^8 (-3+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-27 + 117*x + 9*x^2 - 15*x^3 + E^8*(-9 + 33*x - 10*x^2 - 3*x^3 + x^4) + (-3*x + 9*x^2 + 3*x^3)*Log[
-x^3 + 3*x^4 + x^5])/(E^8*(-9*x + 33*x^2 - 10*x^3 - 3*x^4 + x^5)),x]

[Out]

(3*(9 + Sqrt[13])*Log[-3 + Sqrt[13] - 2*x])/(34*E^8) + Log[x] - (3*(9 + Sqrt[13])*Log[3 - Sqrt[13] + 2*x])/(34
*E^8) - (3*Log[x^3*(-1 + 3*x + x^2)])/(E^8*(-3 + x))

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fricas [A]  time = 0.68, size = 34, normalized size = 1.06 \begin {gather*} \frac {{\left ({\left (x - 3\right )} e^{8} \log \relax (x) - 3 \, \log \left (x^{5} + 3 \, x^{4} - x^{3}\right )\right )} e^{\left (-8\right )}}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+9*x^2-3*x)*log(x^5+3*x^4-x^3)+(x^4-3*x^3-10*x^2+33*x-9)*exp(8)-15*x^3+9*x^2+117*x-27)/(x^5-3
*x^4-10*x^3+33*x^2-9*x)/exp(8),x, algorithm="fricas")

[Out]

((x - 3)*e^8*log(x) - 3*log(x^5 + 3*x^4 - x^3))*e^(-8)/(x - 3)

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giac [A]  time = 0.36, size = 38, normalized size = 1.19 \begin {gather*} \frac {{\left (x e^{8} \log \relax (x) - 3 \, e^{8} \log \relax (x) - 3 \, \log \left (x^{5} + 3 \, x^{4} - x^{3}\right )\right )} e^{\left (-8\right )}}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+9*x^2-3*x)*log(x^5+3*x^4-x^3)+(x^4-3*x^3-10*x^2+33*x-9)*exp(8)-15*x^3+9*x^2+117*x-27)/(x^5-3
*x^4-10*x^3+33*x^2-9*x)/exp(8),x, algorithm="giac")

[Out]

(x*e^8*log(x) - 3*e^8*log(x) - 3*log(x^5 + 3*x^4 - x^3))*e^(-8)/(x - 3)

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maple [A]  time = 0.14, size = 28, normalized size = 0.88




method result size



risch \(-\frac {3 \,{\mathrm e}^{-8} \ln \left (x^{5}+3 x^{4}-x^{3}\right )}{x -3}+\ln \relax (x )\) \(28\)
norman \(-\frac {3 \,{\mathrm e}^{-8} \ln \left (x^{5}+3 x^{4}-x^{3}\right )}{x -3}+\ln \relax (x )\) \(30\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3+9*x^2-3*x)*ln(x^5+3*x^4-x^3)+(x^4-3*x^3-10*x^2+33*x-9)*exp(8)-15*x^3+9*x^2+117*x-27)/(x^5-3*x^4-10
*x^3+33*x^2-9*x)/exp(8),x,method=_RETURNVERBOSE)

[Out]

-3*exp(-8)*ln(x^5+3*x^4-x^3)/(x-3)+ln(x)

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maxima [B]  time = 0.71, size = 308, normalized size = 9.62 \begin {gather*} -\frac {1}{7514} \, {\left ({\left (1161 \, \sqrt {13} \log \left (\frac {2 \, x - \sqrt {13} + 3}{2 \, x + \sqrt {13} + 3}\right ) - \frac {1326}{x - 3} + 4329 \, \log \left (x^{2} + 3 \, x - 1\right ) - 1144 \, \log \left (x - 3\right ) - 7514 \, \log \relax (x)\right )} e^{8} + {\left (261 \, \sqrt {13} \log \left (\frac {2 \, x - \sqrt {13} + 3}{2 \, x + \sqrt {13} + 3}\right ) + \frac {11934}{x - 3} - 949 \, \log \left (x^{2} + 3 \, x - 1\right ) - 5616 \, \log \left (x - 3\right )\right )} e^{8} + 3 \, {\left (83 \, \sqrt {13} \log \left (\frac {2 \, x - \sqrt {13} + 3}{2 \, x + \sqrt {13} + 3}\right ) - \frac {3978}{x - 3} - 273 \, \log \left (x^{2} + 3 \, x - 1\right ) + 546 \, \log \left (x - 3\right )\right )} e^{8} - 33 \, {\left (47 \, \sqrt {13} \log \left (\frac {2 \, x - \sqrt {13} + 3}{2 \, x + \sqrt {13} + 3}\right ) - \frac {442}{x - 3} + 117 \, \log \left (x^{2} + 3 \, x - 1\right ) - 234 \, \log \left (x - 3\right )\right )} e^{8} - 20 \, {\left (6 \, \sqrt {13} \log \left (\frac {2 \, x - \sqrt {13} + 3}{2 \, x + \sqrt {13} + 3}\right ) + \frac {663}{x - 3} - 65 \, \log \left (x^{2} + 3 \, x - 1\right ) + 130 \, \log \left (x - 3\right )\right )} e^{8} + \frac {22542 \, {\left (\log \left (x^{2} + 3 \, x - 1\right ) + 3 \, \log \relax (x)\right )}}{x - 3}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3+9*x^2-3*x)*log(x^5+3*x^4-x^3)+(x^4-3*x^3-10*x^2+33*x-9)*exp(8)-15*x^3+9*x^2+117*x-27)/(x^5-3
*x^4-10*x^3+33*x^2-9*x)/exp(8),x, algorithm="maxima")

[Out]

-1/7514*((1161*sqrt(13)*log((2*x - sqrt(13) + 3)/(2*x + sqrt(13) + 3)) - 1326/(x - 3) + 4329*log(x^2 + 3*x - 1
) - 1144*log(x - 3) - 7514*log(x))*e^8 + (261*sqrt(13)*log((2*x - sqrt(13) + 3)/(2*x + sqrt(13) + 3)) + 11934/
(x - 3) - 949*log(x^2 + 3*x - 1) - 5616*log(x - 3))*e^8 + 3*(83*sqrt(13)*log((2*x - sqrt(13) + 3)/(2*x + sqrt(
13) + 3)) - 3978/(x - 3) - 273*log(x^2 + 3*x - 1) + 546*log(x - 3))*e^8 - 33*(47*sqrt(13)*log((2*x - sqrt(13)
+ 3)/(2*x + sqrt(13) + 3)) - 442/(x - 3) + 117*log(x^2 + 3*x - 1) - 234*log(x - 3))*e^8 - 20*(6*sqrt(13)*log((
2*x - sqrt(13) + 3)/(2*x + sqrt(13) + 3)) + 663/(x - 3) - 65*log(x^2 + 3*x - 1) + 130*log(x - 3))*e^8 + 22542*
(log(x^2 + 3*x - 1) + 3*log(x))/(x - 3))*e^(-8)

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mupad [B]  time = 1.75, size = 27, normalized size = 0.84 \begin {gather*} \ln \relax (x)-\frac {3\,\ln \left (x^5+3\,x^4-x^3\right )\,{\mathrm {e}}^{-8}}{x-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(117*x + log(3*x^4 - x^3 + x^5)*(9*x^2 - 3*x + 3*x^3) - exp(8)*(10*x^2 - 33*x + 3*x^3 - x^4 + 9)
 + 9*x^2 - 15*x^3 - 27))/(9*x - 33*x^2 + 10*x^3 + 3*x^4 - x^5),x)

[Out]

log(x) - (3*log(3*x^4 - x^3 + x^5)*exp(-8))/(x - 3)

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sympy [A]  time = 0.24, size = 27, normalized size = 0.84 \begin {gather*} \log {\relax (x )} - \frac {3 \log {\left (x^{5} + 3 x^{4} - x^{3} \right )}}{x e^{8} - 3 e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3+9*x**2-3*x)*ln(x**5+3*x**4-x**3)+(x**4-3*x**3-10*x**2+33*x-9)*exp(8)-15*x**3+9*x**2+117*x-2
7)/(x**5-3*x**4-10*x**3+33*x**2-9*x)/exp(8),x)

[Out]

log(x) - 3*log(x**5 + 3*x**4 - x**3)/(x*exp(8) - 3*exp(8))

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