3.89.99 \(\int \frac {e^{-1+x} (e^{1-x} (1+x^3) \log (4)+(-6 x^3-6 x^4) \log (4))}{x^3} \, dx\) [8899]

Optimal. Leaf size=26 \[ -e^5+\left (-\frac {1}{2 x^2}+x-6 e^{-1+x} x\right ) \log (4) \]

[Out]

2*ln(2)*(x-1/2/x^2-6*x/exp(1-x))-exp(5)

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Rubi [A]
time = 0.25, antiderivative size = 35, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6820, 12, 2207, 2225} \begin {gather*} -\frac {\log (4)}{2 x^2}+x \log (4)+6 e^{x-1} \log (4)-6 e^{x-1} (x+1) \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 + x)*(E^(1 - x)*(1 + x^3)*Log[4] + (-6*x^3 - 6*x^4)*Log[4]))/x^3,x]

[Out]

6*E^(-1 + x)*Log[4] - Log[4]/(2*x^2) + x*Log[4] - 6*E^(-1 + x)*(1 + x)*Log[4]

Rule 12

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

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6820

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 20, normalized size = 0.77 \begin {gather*} \left (-\frac {1}{2 x^2}+x-6 e^{-1+x} x\right ) \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 + x)*(E^(1 - x)*(1 + x^3)*Log[4] + (-6*x^3 - 6*x^4)*Log[4]))/x^3,x]

[Out]

(-1/2*1/x^2 + x - 6*E^(-1 + x)*x)*Log[4]

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Maple [A]
time = 1.46, size = 39, normalized size = 1.50

method	result	size
risch	\(2 x \ln \left (2\right )-\frac {\ln \left (2\right )}{x^{2}}-12 x \ln \left (2\right ) {\mathrm e}^{x -1}\)	\(23\)
derivativedivides	\(-\frac {\ln \left (2\right )}{x^{2}}-2 \left (1-x \right ) \ln \left (2\right )-12 \ln \left (2\right ) {\mathrm e}^{x -1}+12 \ln \left (2\right ) {\mathrm e}^{x -1} \left (1-x \right )\)	\(39\)
default	\(-\frac {\ln \left (2\right )}{x^{2}}-2 \left (1-x \right ) \ln \left (2\right )-12 \ln \left (2\right ) {\mathrm e}^{x -1}+12 \ln \left (2\right ) {\mathrm e}^{x -1} \left (1-x \right )\)	\(39\)
norman	\(\frac {\left (-12 x^{3} \ln \left (2\right )-\ln \left (2\right ) {\mathrm e}^{1-x}+2 x^{3} \ln \left (2\right ) {\mathrm e}^{1-x}\right ) {\mathrm e}^{x -1}}{x^{2}}\)	\(44\)
meijerg	\(2 \ln \left (2\right ) {\mathrm e}^{x -2-{\mathrm e}^{-1} x} \left (1-{\mathrm e}\right )^{2} \left (\frac {{\mathrm e}^{2} \left (9 x^{2} {\mathrm e}^{-2} \left (1-{\mathrm e}\right )^{2}+12 x \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )+6\right )}{12 x^{2} \left (1-{\mathrm e}\right )^{2}}-\frac {{\mathrm e}^{2+x \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )} \left (3+3 x \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )\right )}{6 x^{2} \left (1-{\mathrm e}\right )^{2}}-\frac {\ln \left (-x \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )\right )}{2}-\frac {\expIntegral \left (1, -x \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )\right )}{2}-\frac {5}{4}+\frac {\ln \left (x \right )}{2}+\frac {i \pi }{2}+\frac {\ln \left (1-{\mathrm e}\right )}{2}-\frac {{\mathrm e}^{2}}{2 x^{2} \left (1-{\mathrm e}\right )^{2}}-\frac {{\mathrm e}}{x \left (1-{\mathrm e}\right )}\right )-\frac {2 \ln \left (2\right ) {\mathrm e}^{x -{\mathrm e}^{-1} x +1} \left (1-{\mathrm e}^{x \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )}\right )}{1-{\mathrm e}}-12 \ln \left (2\right ) {\mathrm e}^{x -{\mathrm e}^{-1} x +1} \left (1-\frac {\left (2-2 \,{\mathrm e}^{-1} x \right ) {\mathrm e}^{{\mathrm e}^{-1} x}}{2}\right )+12 \ln \left (2\right ) {\mathrm e}^{x -{\mathrm e}^{-1} x} \left (1-{\mathrm e}^{{\mathrm e}^{-1} x}\right )\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(x^3+1)*ln(2)*exp(1-x)+2*(-6*x^4-6*x^3)*ln(2))/x^3/exp(1-x),x,method=_RETURNVERBOSE)

[Out]

-ln(2)/x^2-2*(1-x)*ln(2)-12*ln(2)*exp(x-1)+12*ln(2)*exp(x-1)*(1-x)

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Maxima [A]
time = 0.26, size = 32, normalized size = 1.23 \begin {gather*} -12 \, {\left (x - 1\right )} e^{\left (x - 1\right )} \log \left (2\right ) + 2 \, x \log \left (2\right ) - 12 \, e^{\left (x - 1\right )} \log \left (2\right ) - \frac {\log \left (2\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x),x, algorithm="maxima")

[Out]

-12*(x - 1)*e^(x - 1)*log(2) + 2*x*log(2) - 12*e^(x - 1)*log(2) - log(2)/x^2

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Fricas [A]
time = 0.38, size = 28, normalized size = 1.08 \begin {gather*} -\frac {12 \, x^{3} e^{\left (x - 1\right )} \log \left (2\right ) - {\left (2 \, x^{3} - 1\right )} \log \left (2\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x),x, algorithm="fricas")

[Out]

-(12*x^3*e^(x - 1)*log(2) - (2*x^3 - 1)*log(2))/x^2

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Sympy [A]
time = 0.07, size = 24, normalized size = 0.92 \begin {gather*} - 12 x e^{x - 1} \log {\left (2 \right )} + 2 x \log {\left (2 \right )} - \frac {\log {\left (2 \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(x**3+1)*ln(2)*exp(1-x)+2*(-6*x**4-6*x**3)*ln(2))/x**3/exp(1-x),x)

[Out]

-12*x*exp(x - 1)*log(2) + 2*x*log(2) - log(2)/x**2

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Giac [A]
time = 0.43, size = 31, normalized size = 1.19 \begin {gather*} \frac {{\left (2 \, x^{3} e \log \left (2\right ) - 12 \, x^{3} e^{x} \log \left (2\right ) - e \log \left (2\right )\right )} e^{\left (-1\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(x^3+1)*log(2)*exp(1-x)+2*(-6*x^4-6*x^3)*log(2))/x^3/exp(1-x),x, algorithm="giac")

[Out]

(2*x^3*e*log(2) - 12*x^3*e^x*log(2) - e*log(2))*e^(-1)/x^2

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Mupad [B]
time = 0.11, size = 23, normalized size = 0.88 \begin {gather*} x\,{\mathrm {e}}^{-1}\,\ln \left (2\right )\,\left (2\,\mathrm {e}-12\,{\mathrm {e}}^x\right )-\frac {\ln \left (2\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 1)*(2*log(2)*(6*x^3 + 6*x^4) - 2*exp(1 - x)*log(2)*(x^3 + 1)))/x^3,x)

[Out]

x*exp(-1)*log(2)*(2*exp(1) - 12*exp(x)) - log(2)/x^2

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