\(\int \frac {(-32 x+10 e^5 x-8 e^{15} x) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx\) [8408]

Optimal result

Integrand size = 36, antiderivative size = 20 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=\left (-5+\frac {16}{e^5}+4 e^{10}+\frac {\log (5)}{x}\right )^2 \]

[Out]

(4*exp(5)^2+ln(5)/x+16/exp(5)-5)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 192, 37} \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=\frac {\left (\left (16-5 e^5+4 e^{15}\right ) x+e^5 \log (5)\right )^2}{e^{10} x^2} \]

[In]

Int[((-32*x + 10*E^5*x - 8*E^15*x)*Log[5] - 2*E^5*Log[5]^2)/(E^5*x^3),x]

[Out]

((16 - 5*E^5 + 4*E^15)*x + E^5*Log[5])^2/(E^10*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 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 192

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{x^3} \, dx}{e^5} \\ & = \frac {\int \frac {-2 \left (16-5 e^5+4 e^{15}\right ) x \log (5)-2 e^5 \log ^2(5)}{x^3} \, dx}{e^5} \\ & = \frac {\left (\left (16-5 e^5+4 e^{15}\right ) x+e^5 \log (5)\right )^2}{e^{10} x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=-\frac {2 \log (5) \left (\frac {-16+5 e^5-4 e^{15}}{x}-\frac {e^5 \log (5)}{2 x^2}\right )}{e^5} \]

[In]

Integrate[((-32*x + 10*E^5*x - 8*E^15*x)*Log[5] - 2*E^5*Log[5]^2)/(E^5*x^3),x]

[Out]

(-2*Log[5]*((-16 + 5*E^5 - 4*E^15)/x - (E^5*Log[5])/(2*x^2)))/E^5

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55

method	result	size
norman	\(\frac {\ln \left (5\right )^{2}+2 \left (4 \,{\mathrm e}^{15}-5 \,{\mathrm e}^{5}+16\right ) \ln \left (5\right ) {\mathrm e}^{-5} x}{x^{2}}\)	\(31\)
gosper	\(\frac {\ln \left (5\right ) \left (8 x \,{\mathrm e}^{15}+{\mathrm e}^{5} \ln \left (5\right )-10 x \,{\mathrm e}^{5}+32 x \right ) {\mathrm e}^{-5}}{x^{2}}\)	\(32\)
default	\(2 \,{\mathrm e}^{-5} \ln \left (5\right ) \left (-\frac {-4 \,{\mathrm e}^{15}+5 \,{\mathrm e}^{5}-16}{x}+\frac {{\mathrm e}^{5} \ln \left (5\right )}{2 x^{2}}\right )\)	\(34\)
risch	\(\frac {{\mathrm e}^{-5} \left (\left (8 \ln \left (5\right ) {\mathrm e}^{15}-10 \,{\mathrm e}^{5} \ln \left (5\right )+32 \ln \left (5\right )\right ) x +{\mathrm e}^{5} \ln \left (5\right )^{2}\right )}{x^{2}}\)	\(38\)
parallelrisch	\(\frac {{\mathrm e}^{-5} \left ({\mathrm e}^{5} \ln \left (5\right )^{2}-10 x \,{\mathrm e}^{5} \ln \left (5\right )+8 \ln \left (5\right ) {\mathrm e}^{15} x +32 x \ln \left (5\right )\right )}{x^{2}}\)	\(38\)

[In]

int((-2*exp(5)*ln(5)^2+(-8*x*exp(5)^3+10*x*exp(5)-32*x)*ln(5))/x^3/exp(5),x,method=_RETURNVERBOSE)

[Out]

(ln(5)^2+2*(4*exp(5)^3-5*exp(5)+16)*ln(5)/exp(5)*x)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=\frac {{\left (e^{5} \log \left (5\right )^{2} + 2 \, {\left (4 \, x e^{15} - 5 \, x e^{5} + 16 \, x\right )} \log \left (5\right )\right )} e^{\left (-5\right )}}{x^{2}} \]

[In]

integrate((-2*exp(5)*log(5)^2+(-8*x*exp(5)^3+10*x*exp(5)-32*x)*log(5))/x^3/exp(5),x, algorithm="fricas")

[Out]

(e^5*log(5)^2 + 2*(4*x*e^15 - 5*x*e^5 + 16*x)*log(5))*e^(-5)/x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=- \frac {x \left (- 8 e^{15} \log {\left (5 \right )} - 32 \log {\left (5 \right )} + 10 e^{5} \log {\left (5 \right )}\right ) - e^{5} \log {\left (5 \right )}^{2}}{x^{2} e^{5}} \]

[In]

integrate((-2*exp(5)*ln(5)**2+(-8*x*exp(5)**3+10*x*exp(5)-32*x)*ln(5))/x**3/exp(5),x)

[Out]

-(x*(-8*exp(15)*log(5) - 32*log(5) + 10*exp(5)*log(5)) - exp(5)*log(5)**2)*exp(-5)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=\frac {{\left (2 \, x {\left (4 \, e^{15} - 5 \, e^{5} + 16\right )} \log \left (5\right ) + e^{5} \log \left (5\right )^{2}\right )} e^{\left (-5\right )}}{x^{2}} \]

[In]

integrate((-2*exp(5)*log(5)^2+(-8*x*exp(5)^3+10*x*exp(5)-32*x)*log(5))/x^3/exp(5),x, algorithm="maxima")

[Out]

(2*x*(4*e^15 - 5*e^5 + 16)*log(5) + e^5*log(5)^2)*e^(-5)/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=\frac {{\left (8 \, x e^{15} \log \left (5\right ) - 10 \, x e^{5} \log \left (5\right ) + e^{5} \log \left (5\right )^{2} + 32 \, x \log \left (5\right )\right )} e^{\left (-5\right )}}{x^{2}} \]

[In]

integrate((-2*exp(5)*log(5)^2+(-8*x*exp(5)^3+10*x*exp(5)-32*x)*log(5))/x^3/exp(5),x, algorithm="giac")

[Out]

(8*x*e^15*log(5) - 10*x*e^5*log(5) + e^5*log(5)^2 + 32*x*log(5))*e^(-5)/x^2

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-32 x+10 e^5 x-8 e^{15} x\right ) \log (5)-2 e^5 \log ^2(5)}{e^5 x^3} \, dx=\frac {{\ln \left (5\right )}^2+x\,{\mathrm {e}}^{-5}\,\left (32\,\ln \left (5\right )-10\,{\mathrm {e}}^5\,\ln \left (5\right )+8\,{\mathrm {e}}^{15}\,\ln \left (5\right )\right )}{x^2} \]

[In]

int(-(exp(-5)*(log(5)*(32*x - 10*x*exp(5) + 8*x*exp(15)) + 2*exp(5)*log(5)^2))/x^3,x)

[Out]

(log(5)^2 + x*exp(-5)*(32*log(5) - 10*exp(5)*log(5) + 8*exp(15)*log(5)))/x^2