\(\int \frac {5 e^{5-x}+(5+e^{5-x} (-5-5 x)) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx\) [6473]

Optimal result

Integrand size = 46, antiderivative size = 21 \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5 \log (x) \left (e^{5-x}+\log (5 x)\right )}{2 x} \]

[Out]

5/2*(ln(5*x)+exp(5-x))*ln(x)/x

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 14, 2326, 2341, 2340, 2413} \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5 \log (5 x) \log (x)}{2 x}+\frac {5 e^{5-x} \log (x)}{2 x} \]

[In]

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

[Out]

(5*E^(5 - x)*Log[x])/(2*x) + (5*Log[x]*Log[5*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 2326

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 2340

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

Rule 2341

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

Rule 2413

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{x^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {5 e^{5-x} (-1+\log (x)+x \log (x))}{x^2}-\frac {5 (-\log (x)-\log (5 x)+\log (x) \log (5 x))}{x^2}\right ) \, dx \\ & = -\left (\frac {5}{2} \int \frac {e^{5-x} (-1+\log (x)+x \log (x))}{x^2} \, dx\right )-\frac {5}{2} \int \frac {-\log (x)-\log (5 x)+\log (x) \log (5 x)}{x^2} \, dx \\ & = \frac {5 e^{5-x} \log (x)}{2 x}-\frac {5}{2} \int \left (-\frac {\log (x)}{x^2}+\frac {(-1+\log (x)) \log (5 x)}{x^2}\right ) \, dx \\ & = \frac {5 e^{5-x} \log (x)}{2 x}+\frac {5}{2} \int \frac {\log (x)}{x^2} \, dx-\frac {5}{2} \int \frac {(-1+\log (x)) \log (5 x)}{x^2} \, dx \\ & = -\frac {5}{2 x}-\frac {5 \log (x)}{2 x}+\frac {5 e^{5-x} \log (x)}{2 x}+\frac {5 \log (x) \log (5 x)}{2 x}-\frac {5}{2} \int \frac {\log (x)}{x^2} \, dx \\ & = \frac {5 e^{5-x} \log (x)}{2 x}+\frac {5 \log (x) \log (5 x)}{2 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5 e^{-x} \log (x) \left (e^5+e^x \log (5 x)\right )}{2 x} \]

[In]

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

[Out]

(5*Log[x]*(E^5 + E^x*Log[5*x]))/(2*E^x*x)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\(\frac {10 \ln \left (x \right ) \ln \left (5 x \right )+10 \ln \left (x \right ) {\mathrm e}^{5-x}}{4 x}\)	\(25\)
risch	\(\frac {5 \ln \left (x \right )^{2}}{2 x}+\frac {5 \left (2 \ln \left (5\right )+2 \,{\mathrm e}^{5-x}\right ) \ln \left (x \right )}{4 x}\)	\(31\)
default	\(-\frac {5 \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{2}+\frac {5 \ln \left (x \right )^{2}}{2 x}+\frac {5 \ln \left (x \right )}{2 x}+\frac {5 \ln \left (x \right ) {\mathrm e}^{5-x}}{2 x}-\frac {5 \ln \left (5 x \right )}{2 x}\)	\(57\)
parts	\(-\frac {5 \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{2}+\frac {5 \ln \left (x \right )^{2}}{2 x}+\frac {5 \ln \left (x \right )}{2 x}+\frac {5 \ln \left (x \right ) {\mathrm e}^{5-x}}{2 x}-\frac {5 \ln \left (5 x \right )}{2 x}\)	\(57\)

[In]

int(1/2*((-5*ln(x)+5)*ln(5*x)+((-5*x-5)*exp(5-x)+5)*ln(x)+5*exp(5-x))/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4/x*(10*ln(x)*ln(5*x)+10*ln(x)*exp(5-x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5 \, {\left ({\left (e^{\left (-x + 5\right )} + \log \left (5\right )\right )} \log \left (x\right ) + \log \left (x\right )^{2}\right )}}{2 \, x} \]

[In]

integrate(1/2*((-5*log(x)+5)*log(5*x)+((-5*x-5)*exp(5-x)+5)*log(x)+5*exp(5-x))/x^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5 e^{5 - x} \log {\left (x \right )}}{2 x} + \frac {5 \log {\left (x \right )}^{2}}{2 x} + \frac {5 \log {\left (5 \right )} \log {\left (x \right )}}{2 x} \]

[In]

integrate(1/2*((-5*ln(x)+5)*ln(5*x)+((-5*x-5)*exp(5-x)+5)*ln(x)+5*exp(5-x))/x**2,x)

[Out]

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

Maxima [F]

\[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\int { -\frac {5 \, {\left ({\left (\log \left (x\right ) - 1\right )} \log \left (5 \, x\right ) + {\left ({\left (x + 1\right )} e^{\left (-x + 5\right )} - 1\right )} \log \left (x\right ) - e^{\left (-x + 5\right )}\right )}}{2 \, x^{2}} \,d x } \]

[In]

integrate(1/2*((-5*log(x)+5)*log(5*x)+((-5*x-5)*exp(5-x)+5)*log(x)+5*exp(5-x))/x^2,x, algorithm="maxima")

[Out]

-5/2*e^5*gamma(-1, x) + 5/2*((log(5) + 1)*log(x) + e^(-x + 5)*log(x) + log(x)^2 + 1)/x - 5/2*log(x)/x - 5/2/x
- 5/2*integrate(e^(-x + 5)/x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5 \, {\left (e^{\left (-x + 5\right )} \log \left (x\right ) + \log \left (5\right ) \log \left (x\right ) + \log \left (x\right )^{2}\right )}}{2 \, x} \]

[In]

integrate(1/2*((-5*log(x)+5)*log(5*x)+((-5*x-5)*exp(5-x)+5)*log(x)+5*exp(5-x))/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.61 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {5 e^{5-x}+\left (5+e^{5-x} (-5-5 x)\right ) \log (x)+(5-5 \log (x)) \log (5 x)}{2 x^2} \, dx=\frac {5\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left ({\mathrm {e}}^5+{\mathrm {e}}^x\,\ln \left (x\right )+{\mathrm {e}}^x\,\ln \left (5\right )\right )}{2\,x} \]

[In]

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

[Out]

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