3.74.20 \(\int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} (x \log (5)+2 e^{5+\log ^2(x)} \log (x))}{e^5 x} \, dx\)

Optimal. Leaf size=21 \[ e^{17+e^{\log ^2(x)}+\frac {x \log (5)}{e^5}}-x \]

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 14, 2288} \begin {gather*} 5^{\frac {x}{e^5}} e^{e^{\log ^2(x)}+17}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^5*x) + E^((17*E^5 + E^(5 + Log[x]^2) + x*Log[5])/E^5)*(x*Log[5] + 2*E^(5 + Log[x]^2)*Log[x]))/(E^5*x)
,x]

[Out]

5^(x/E^5)*E^(17 + E^Log[x]^2) - 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 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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-e^5 x+e^{\frac {17 e^5+e^{5+\log ^2(x)}+x \log (5)}{e^5}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{x} \, dx}{e^5}\\ &=\frac {\int \left (-e^5+\frac {5^{\frac {x}{e^5}} e^{17+e^{\log ^2(x)}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{x}\right ) \, dx}{e^5}\\ &=-x+\frac {\int \frac {5^{\frac {x}{e^5}} e^{17+e^{\log ^2(x)}} \left (x \log (5)+2 e^{5+\log ^2(x)} \log (x)\right )}{x} \, dx}{e^5}\\ &=5^{\frac {x}{e^5}} e^{17+e^{\log ^2(x)}}-x\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 22, normalized size = 1.05 \begin {gather*} 5^{\frac {x}{e^5}} e^{17+e^{\log ^2(x)}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^5*x) + E^((17*E^5 + E^(5 + Log[x]^2) + x*Log[5])/E^5)*(x*Log[5] + 2*E^(5 + Log[x]^2)*Log[x]))/(
E^5*x),x]

[Out]

5^(x/E^5)*E^(17 + E^Log[x]^2) - x

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 24, normalized size = 1.14 \begin {gather*} -x + e^{\left ({\left (x \log \relax (5) + 17 \, e^{5} + e^{\left (\log \relax (x)^{2} + 5\right )}\right )} e^{\left (-5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)*log(x)*exp(log(x)^2)+x*log(5))*exp((exp(5)*exp(log(x)^2)+x*log(5)+17*exp(5))/exp(5))-x*ex
p(5))/x/exp(5),x, algorithm="fricas")

[Out]

-x + e^((x*log(5) + 17*e^5 + e^(log(x)^2 + 5))*e^(-5))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x e^{5} - {\left (x \log \relax (5) + 2 \, e^{\left (\log \relax (x)^{2} + 5\right )} \log \relax (x)\right )} e^{\left ({\left (x \log \relax (5) + 17 \, e^{5} + e^{\left (\log \relax (x)^{2} + 5\right )}\right )} e^{\left (-5\right )}\right )}\right )} e^{\left (-5\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)*log(x)*exp(log(x)^2)+x*log(5))*exp((exp(5)*exp(log(x)^2)+x*log(5)+17*exp(5))/exp(5))-x*ex
p(5))/x/exp(5),x, algorithm="giac")

[Out]

integrate(-(x*e^5 - (x*log(5) + 2*e^(log(x)^2 + 5)*log(x))*e^((x*log(5) + 17*e^5 + e^(log(x)^2 + 5))*e^(-5)))*
e^(-5)/x, x)

________________________________________________________________________________________

maple [A]  time = 0.11, size = 25, normalized size = 1.19

method	result	size
risch	\(-x +{\mathrm e}^{\left ({\mathrm e}^{5+\ln \relax (x )^{2}}+x \ln \relax (5)+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}\)	\(25\)
norman	\(-x +{\mathrm e}^{\left ({\mathrm e}^{5} {\mathrm e}^{\ln \relax (x )^{2}}+x \ln \relax (5)+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}\)	\(28\)
default	\({\mathrm e}^{-5} \left ({\mathrm e}^{\left ({\mathrm e}^{5+\ln \relax (x )^{2}}+x \ln \relax (5)+17 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}} {\mathrm e}^{5}-x \,{\mathrm e}^{5}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(5)*ln(x)*exp(ln(x)^2)+x*ln(5))*exp((exp(5)*exp(ln(x)^2)+x*ln(5)+17*exp(5))/exp(5))-x*exp(5))/x/exp
(5),x,method=_RETURNVERBOSE)

[Out]

-x+exp((exp(5+ln(x)^2)+x*ln(5)+17*exp(5))*exp(-5))

________________________________________________________________________________________

maxima [A]  time = 0.58, size = 25, normalized size = 1.19 \begin {gather*} -{\left (x e^{5} - e^{\left (x e^{\left (-5\right )} \log \relax (5) + e^{\left (\log \relax (x)^{2}\right )} + 22\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)*log(x)*exp(log(x)^2)+x*log(5))*exp((exp(5)*exp(log(x)^2)+x*log(5)+17*exp(5))/exp(5))-x*ex
p(5))/x/exp(5),x, algorithm="maxima")

[Out]

-(x*e^5 - e^(x*e^(-5)*log(5) + e^(log(x)^2) + 22))*e^(-5)

________________________________________________________________________________________

mupad [B]  time = 4.64, size = 19, normalized size = 0.90 \begin {gather*} 5^{x\,{\mathrm {e}}^{-5}}\,{\mathrm {e}}^{{\mathrm {e}}^{{\ln \relax (x)}^2}}\,{\mathrm {e}}^{17}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5)*(x*exp(5) - exp(exp(-5)*(17*exp(5) + x*log(5) + exp(5)*exp(log(x)^2)))*(x*log(5) + 2*exp(5)*exp(
log(x)^2)*log(x))))/x,x)

[Out]

5^(x*exp(-5))*exp(exp(log(x)^2))*exp(17) - x

________________________________________________________________________________________

sympy [A]  time = 0.47, size = 26, normalized size = 1.24 \begin {gather*} - x + e^{\frac {x \log {\relax (5 )} + e^{5} e^{\log {\relax (x )}^{2}} + 17 e^{5}}{e^{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)*ln(x)*exp(ln(x)**2)+x*ln(5))*exp((exp(5)*exp(ln(x)**2)+x*ln(5)+17*exp(5))/exp(5))-x*exp(5
))/x/exp(5),x)

[Out]

-x + exp((x*log(5) + exp(5)*exp(log(x)**2) + 17*exp(5))*exp(-5))

________________________________________________________________________________________