\(\int \frac {4 e^{5+\frac {2}{\log (x)}}+(-x+36 x^2) \log ^2(x)}{2 x \log ^2(x)} \, dx\) [1958]

Optimal result

Integrand size = 38, antiderivative size = 30 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=2+e^5 \left (e^5-e^{\frac {2}{\log (x)}}\right )-\frac {x}{2}+9 x^2 \]

[Out]

2+9*x^2+(exp(5)-exp(2/ln(x)))*exp(5)-1/2*x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {12, 6820, 2240} \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=9 x^2-\frac {x}{2}-e^{\frac {2}{\log (x)}+5} \]

[In]

Int[(4*E^(5 + 2/Log[x]) + (-x + 36*x^2)*Log[x]^2)/(2*x*Log[x]^2),x]

[Out]

-E^(5 + 2/Log[x]) - x/2 + 9*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 2240

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

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \left (-1+36 x+\frac {4 e^{5+\frac {2}{\log (x)}}}{x \log ^2(x)}\right ) \, dx \\ & = -\frac {x}{2}+9 x^2+2 \int \frac {e^{5+\frac {2}{\log (x)}}}{x \log ^2(x)} \, dx \\ & = -\frac {x}{2}+9 x^2+2 \text {Subst}\left (\int \frac {e^{5+\frac {2}{x}}}{x^2} \, dx,x,\log (x)\right ) \\ & = -e^{5+\frac {2}{\log (x)}}-\frac {x}{2}+9 x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=-e^{5+\frac {2}{\log (x)}}-\frac {x}{2}+9 x^2 \]

[In]

Integrate[(4*E^(5 + 2/Log[x]) + (-x + 36*x^2)*Log[x]^2)/(2*x*Log[x]^2),x]

[Out]

-E^(5 + 2/Log[x]) - x/2 + 9*x^2

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70

method	result	size
default	\(9 x^{2}-\frac {x}{2}-{\mathrm e}^{5} {\mathrm e}^{\frac {2}{\ln \left (x \right )}}\)	\(21\)
parallelrisch	\(9 x^{2}-\frac {x}{2}-{\mathrm e}^{5} {\mathrm e}^{\frac {2}{\ln \left (x \right )}}\)	\(21\)
parts	\(9 x^{2}-\frac {x}{2}-{\mathrm e}^{5} {\mathrm e}^{\frac {2}{\ln \left (x \right )}}\)	\(21\)
risch	\(9 x^{2}-\frac {x}{2}-{\mathrm e}^{\frac {5 \ln \left (x \right )+2}{\ln \left (x \right )}}\)	\(24\)
norman	\(\frac {-\frac {x \ln \left (x \right )}{2}+9 x^{2} \ln \left (x \right )-\ln \left (x \right ) {\mathrm e}^{5} {\mathrm e}^{\frac {2}{\ln \left (x \right )}}}{\ln \left (x \right )}\)	\(32\)

[In]

int(1/2*(4*exp(5)*exp(2/ln(x))+(36*x^2-x)*ln(x)^2)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

9*x^2-1/2*x-exp(5)*exp(2/ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=9 \, x^{2} - \frac {1}{2} \, x - e^{\left (\frac {5 \, \log \left (x\right ) + 2}{\log \left (x\right )}\right )} \]

[In]

integrate(1/2*(4*exp(5)*exp(2/log(x))+(36*x^2-x)*log(x)^2)/x/log(x)^2,x, algorithm="fricas")

[Out]

9*x^2 - 1/2*x - e^((5*log(x) + 2)/log(x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=9 x^{2} - \frac {x}{2} - e^{5} e^{\frac {2}{\log {\left (x \right )}}} \]

[In]

integrate(1/2*(4*exp(5)*exp(2/ln(x))+(36*x**2-x)*ln(x)**2)/x/ln(x)**2,x)

[Out]

9*x**2 - x/2 - exp(5)*exp(2/log(x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=9 \, x^{2} - \frac {1}{2} \, x - e^{\left (\frac {2}{\log \left (x\right )} + 5\right )} \]

[In]

integrate(1/2*(4*exp(5)*exp(2/log(x))+(36*x^2-x)*log(x)^2)/x/log(x)^2,x, algorithm="maxima")

[Out]

9*x^2 - 1/2*x - e^(2/log(x) + 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=9 \, x^{2} - \frac {1}{2} \, x - e^{\left (\frac {2}{\log \left (x\right )} + 5\right )} \]

[In]

integrate(1/2*(4*exp(5)*exp(2/log(x))+(36*x^2-x)*log(x)^2)/x/log(x)^2,x, algorithm="giac")

[Out]

9*x^2 - 1/2*x - e^(2/log(x) + 5)

Mupad [B] (verification not implemented)

Time = 8.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {4 e^{5+\frac {2}{\log (x)}}+\left (-x+36 x^2\right ) \log ^2(x)}{2 x \log ^2(x)} \, dx=9\,x^2-{\mathrm {e}}^{\frac {2}{\ln \left (x\right )}}\,{\mathrm {e}}^5-\frac {x}{2} \]

[In]

int(-((log(x)^2*(x - 36*x^2))/2 - 2*exp(2/log(x))*exp(5))/(x*log(x)^2),x)

[Out]

9*x^2 - exp(2/log(x))*exp(5) - x/2