3.39.90 \(\int \frac {4-2 e^4+x-4 e^x x-2 \log (2 e^{-4 e^x+2 x} x^4)}{x^3} \, dx\) [3890]

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

[Out]

(ln(2*x^4*exp(x)^2/exp(4*exp(x)))+x+exp(4))/x^2

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(26)=52\).
time = 0.16, antiderivative size = 65, normalized size of antiderivative = 2.50, number of steps used = 14, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {14, 2208, 2209, 37, 2631, 12} \begin {gather*} \frac {(x+2)^2}{2 x^2}-\frac {\left (x+2 \left (2-e^4\right )\right )^2}{4 \left (2-e^4\right ) x^2}+\frac {\log \left (2 e^{-2 \left (2 e^x-x\right )} x^4\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 2*E^4 + x - 4*E^x*x - 2*Log[2*E^(-4*E^x + 2*x)*x^4])/x^3,x]

[Out]

(2 + x)^2/(2*x^2) - (2*(2 - E^4) + x)^2/(4*(2 - E^4)*x^2) + Log[(2*x^4)/E^(2*(2*E^x - 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2208

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 32, normalized size = 1.23 \begin {gather*} \frac {e^4}{x^2}+\frac {1}{x}+\frac {\log \left (2 e^{-4 e^x+2 x} x^4\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 2*E^4 + x - 4*E^x*x - 2*Log[2*E^(-4*E^x + 2*x)*x^4])/x^3,x]

[Out]

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

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Maple [A]
time = 0.51, size = 42, normalized size = 1.62

method	result	size
default	\(\frac {\ln \left (2 x^{4} {\mathrm e}^{2 x} {\mathrm e}^{-4 \,{\mathrm e}^{x}}\right )}{x^{2}}+\frac {1}{x}+\frac {2}{x^{2}}+\frac {2 \,{\mathrm e}^{4}-4}{2 x^{2}}\)	\(42\)
risch	\(-\frac {\ln \left ({\mathrm e}^{4 \,{\mathrm e}^{x}}\right )}{x^{2}}+\frac {2 x +i \pi \mathrm {csgn}\left (i x^{3}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \mathrm {csgn}\left (i x^{3}\right )^{2} \mathrm {csgn}\left (i x^{2}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}+2 i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-4 \,{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i x^{4} {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i x^{4} {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{4}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+4 \ln \left ({\mathrm e}^{x}\right )+2 \,{\mathrm e}^{4}+2 \ln \left (2\right )+8 \ln \left (x \right )-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )^{3}-i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-i \pi \mathrm {csgn}\left (i x^{4} {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )^{3}-i \pi \mathrm {csgn}\left (i x^{4}\right )^{3}-i \pi \,\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i x^{4} {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-4 \,{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x -4 \,{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )}{2 x^{2}}\)	\(524\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x^2*ln(2*x^4*exp(x)^2/exp(4*exp(x)))+1/x+2/x^2+1/2*(2*exp(4)-4)/x^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x + log(2) + 4*log(x) + 2)/x^2 - 1/x + e^4/x^2 - 2/x^2 - 4*gamma(-1, -x) + 8*integrate(e^x/x^3, x)

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Fricas [A]
time = 0.38, size = 23, normalized size = 0.88 \begin {gather*} \frac {x + e^{4} + \log \left (2 \, x^{4} e^{\left (2 \, x - 4 \, e^{x}\right )}\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.21, size = 31, normalized size = 1.19 \begin {gather*} - \frac {- x - e^{4}}{x^{2}} + \frac {\log {\left (2 x^{4} e^{2 x} e^{- 4 e^{x}} \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-x - exp(4))/x**2 + log(2*x**4*exp(2*x)*exp(-4*exp(x)))/x**2

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Giac [A]
time = 0.42, size = 20, normalized size = 0.77 \begin {gather*} \frac {3 \, x + e^{4} - 4 \, e^{x} + \log \left (2 \, x^{4}\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x + e^4 - 4*e^x + log(2*x^4))/x^2

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Mupad [B]
time = 2.30, size = 20, normalized size = 0.77 \begin {gather*} \frac {3\,x+{\mathrm {e}}^4+\ln \left (2\,x^4\right )-4\,{\mathrm {e}}^x}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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