3.83.15 \(\int \frac {-e^{3+x}+e^{3+x} x \log (e^4 x)+x \log ^2(e^4 x)}{e^3 x \log ^2(e^4 x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^{25+e}}{4}+\frac {x}{e^3}+\frac {e^x}{\log \left (e^4 x\right )} \]

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Rubi [A]  time = 0.34, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 6688, 2288} \begin {gather*} \frac {x}{e^3}+\frac {e^x (4 x+x \log (x))}{x (\log (x)+4)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/E^3 + (E^x*(4*x + x*Log[x]))/(x*(4 + Log[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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 27, normalized size = 1.00 \begin {gather*} \frac {x}{e^3}+\frac {e^x (4 x+x \log (x))}{x (4+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.05, size = 22, normalized size = 0.81 \begin {gather*} \frac {{\left (x \log \left (x e^{4}\right ) + e^{\left (x + 3\right )}\right )} e^{\left (-3\right )}}{\log \left (x e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 21, normalized size = 0.78 \begin {gather*} \frac {{\left (x \log \relax (x) + 4 \, x + e^{\left (x + 3\right )}\right )} e^{\left (-3\right )}}{\log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 16, normalized size = 0.59

method	result	size
risch	\(x \,{\mathrm e}^{-3}+\frac {{\mathrm e}^{x}}{\ln \left (x \,{\mathrm e}^{4}\right )}\)	\(16\)
default	\({\mathrm e}^{-3} \left (x +\frac {{\mathrm e}^{3} {\mathrm e}^{x}}{\ln \left (x \,{\mathrm e}^{4}\right )}\right )\)	\(20\)
norman	\(\frac {x \,{\mathrm e}^{-3} \ln \left (x \,{\mathrm e}^{4}\right )+{\mathrm e}^{x}}{\ln \left (x \,{\mathrm e}^{4}\right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-3)+exp(x)/ln(x*exp(4))

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maxima [A]  time = 0.42, size = 16, normalized size = 0.59 \begin {gather*} {\left (x + \frac {e^{\left (x + 3\right )}}{\log \relax (x) + 4}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + e^(x + 3)/(log(x) + 4))*e^(-3)

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mupad [B]  time = 5.18, size = 14, normalized size = 0.52 \begin {gather*} x\,{\mathrm {e}}^{-3}+\frac {{\mathrm {e}}^x}{\ln \relax (x)+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-3) + exp(x)/(log(x) + 4)

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sympy [A]  time = 0.26, size = 14, normalized size = 0.52 \begin {gather*} \frac {x}{e^{3}} + \frac {e^{x}}{\log {\left (x e^{4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-3) + exp(x)/log(x*exp(4))

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