3.72.83 \(\int \frac {-24+8 e^x-8 x-4 x^2}{78-12 x+13 x^2-2 x^3+e^x (-13+2 x)+(-12+2 e^x-2 x^2) \log (18-3 e^x+3 x^2)} \, dx\)

Optimal. Leaf size=21 \[ \log \left (\left (-\frac {13}{2}+x+\log \left (3 \left (6-e^x+x^2\right )\right )\right )^2\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6741, 12, 6684} \begin {gather*} 2 \log \left (-2 \log \left (3 \left (x^2-e^x+6\right )\right )-2 x+13\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 + 8*E^x - 8*x - 4*x^2)/(78 - 12*x + 13*x^2 - 2*x^3 + E^x*(-13 + 2*x) + (-12 + 2*E^x - 2*x^2)*Log[18 -
3*E^x + 3*x^2]),x]

[Out]

2*Log[13 - 2*x - 2*Log[3*(6 - E^x + x^2)]]

Rule 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (-6+2 e^x-2 x-x^2\right )}{\left (6-e^x+x^2\right ) \left (13-2 x-2 \log \left (3 \left (6-e^x+x^2\right )\right )\right )} \, dx\\ &=4 \int \frac {-6+2 e^x-2 x-x^2}{\left (6-e^x+x^2\right ) \left (13-2 x-2 \log \left (3 \left (6-e^x+x^2\right )\right )\right )} \, dx\\ &=2 \log \left (13-2 x-2 \log \left (3 \left (6-e^x+x^2\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 25, normalized size = 1.19 \begin {gather*} 2 \log \left (13-2 x-\log (9)-2 \log \left (6-e^x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 + 8*E^x - 8*x - 4*x^2)/(78 - 12*x + 13*x^2 - 2*x^3 + E^x*(-13 + 2*x) + (-12 + 2*E^x - 2*x^2)*Lo
g[18 - 3*E^x + 3*x^2]),x]

[Out]

2*Log[13 - 2*x - Log[9] - 2*Log[6 - E^x + x^2]]

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fricas [A]  time = 0.88, size = 22, normalized size = 1.05 \begin {gather*} 2 \, \log \left (2 \, x + 2 \, \log \left (3 \, x^{2} - 3 \, e^{x} + 18\right ) - 13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)-4*x^2-8*x-24)/((2*exp(x)-2*x^2-12)*log(-3*exp(x)+3*x^2+18)+(2*x-13)*exp(x)-2*x^3+13*x^2-12
*x+78),x, algorithm="fricas")

[Out]

2*log(2*x + 2*log(3*x^2 - 3*e^x + 18) - 13)

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giac [A]  time = 0.24, size = 22, normalized size = 1.05 \begin {gather*} 2 \, \log \left (2 \, x + 2 \, \log \left (3 \, x^{2} - 3 \, e^{x} + 18\right ) - 13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)-4*x^2-8*x-24)/((2*exp(x)-2*x^2-12)*log(-3*exp(x)+3*x^2+18)+(2*x-13)*exp(x)-2*x^3+13*x^2-12
*x+78),x, algorithm="giac")

[Out]

2*log(2*x + 2*log(3*x^2 - 3*e^x + 18) - 13)

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maple [A]  time = 0.07, size = 19, normalized size = 0.90

method	result	size
risch	\(2 \ln \left (x -\frac {13}{2}+\ln \left (-3 \,{\mathrm e}^{x}+3 x^{2}+18\right )\right )\)	\(19\)
norman	\(2 \ln \left (2 \ln \left (-3 \,{\mathrm e}^{x}+3 x^{2}+18\right )+2 x -13\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(x)-4*x^2-8*x-24)/((2*exp(x)-2*x^2-12)*ln(-3*exp(x)+3*x^2+18)+(2*x-13)*exp(x)-2*x^3+13*x^2-12*x+78),
x,method=_RETURNVERBOSE)

[Out]

2*ln(x-13/2+ln(-3*exp(x)+3*x^2+18))

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maxima [C]  time = 0.49, size = 21, normalized size = 1.00 \begin {gather*} 2 \, \log \left (i \, \pi + x + \log \relax (3) + \log \left (-x^{2} + e^{x} - 6\right ) - \frac {13}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)-4*x^2-8*x-24)/((2*exp(x)-2*x^2-12)*log(-3*exp(x)+3*x^2+18)+(2*x-13)*exp(x)-2*x^3+13*x^2-12
*x+78),x, algorithm="maxima")

[Out]

2*log(I*pi + x + log(3) + log(-x^2 + e^x - 6) - 13/2)

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mupad [B]  time = 0.54, size = 18, normalized size = 0.86 \begin {gather*} 2\,\ln \left (x+\ln \left (3\,x^2-3\,{\mathrm {e}}^x+18\right )-\frac {13}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - 8*exp(x) + 4*x^2 + 24)/(12*x + log(3*x^2 - 3*exp(x) + 18)*(2*x^2 - 2*exp(x) + 12) - exp(x)*(2*x - 1
3) - 13*x^2 + 2*x^3 - 78),x)

[Out]

2*log(x + log(3*x^2 - 3*exp(x) + 18) - 13/2)

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sympy [A]  time = 0.36, size = 20, normalized size = 0.95 \begin {gather*} 2 \log {\left (x + \log {\left (3 x^{2} - 3 e^{x} + 18 \right )} - \frac {13}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(x)-4*x**2-8*x-24)/((2*exp(x)-2*x**2-12)*ln(-3*exp(x)+3*x**2+18)+(2*x-13)*exp(x)-2*x**3+13*x**
2-12*x+78),x)

[Out]

2*log(x + log(3*x**2 - 3*exp(x) + 18) - 13/2)

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