3.17.90 \(\int \frac {e^{\frac {1}{2 \log (5 x)}}+2 x \log ^2(5 x)}{2 x \log ^2(5 x)} \, dx\)

Optimal. Leaf size=17 \[ -4-e^{\frac {1}{2 \log (5 x)}}+x \]

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 6742, 2209} \begin {gather*} x-e^{\frac {1}{2 \log (5 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1/(2*Log[5*x])) + 2*x*Log[5*x]^2)/(2*x*Log[5*x]^2),x]

[Out]

-E^(1/(2*Log[5*x])) + 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 2209

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^{\frac {1}{2 \log (5 x)}}+2 x \log ^2(5 x)}{x \log ^2(5 x)} \, dx\\ &=\frac {1}{2} \int \left (2+\frac {e^{\frac {1}{2 \log (5 x)}}}{x \log ^2(5 x)}\right ) \, dx\\ &=x+\frac {1}{2} \int \frac {e^{\frac {1}{2 \log (5 x)}}}{x \log ^2(5 x)} \, dx\\ &=x+\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{\left .\frac {1}{2}\right /x}}{x^2} \, dx,x,\log (5 x)\right )\\ &=-e^{\frac {1}{2 \log (5 x)}}+x\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 16, normalized size = 0.94 \begin {gather*} -e^{\frac {1}{2 \log (5 x)}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1/(2*Log[5*x])) + 2*x*Log[5*x]^2)/(2*x*Log[5*x]^2),x]

[Out]

-E^(1/(2*Log[5*x])) + x

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 13, normalized size = 0.76 \begin {gather*} x - e^{\left (\frac {1}{2 \, \log \left (5 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - e^(1/2/log(5*x))

________________________________________________________________________________________

giac [A]  time = 0.24, size = 14, normalized size = 0.82 \begin {gather*} x - e^{\left (\frac {1}{2 \, {\left (\log \relax (5) + \log \relax (x)\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - e^(1/2/(log(5) + log(x)))

________________________________________________________________________________________

maple [A]  time = 0.02, size = 14, normalized size = 0.82

method	result	size
default	\(x -{\mathrm e}^{\frac {1}{2 \ln \left (5 x \right )}}\)	\(14\)
risch	\(x -{\mathrm e}^{\frac {1}{2 \ln \left (5 x \right )}}\)	\(14\)
norman	\(\frac {x \ln \left (5 x \right )-\ln \left (5 x \right ) {\mathrm e}^{\frac {1}{2 \ln \left (5 x \right )}}}{\ln \left (5 x \right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.85, size = 13, normalized size = 0.76 \begin {gather*} x - e^{\left (\frac {1}{2 \, \log \left (5 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - e^(1/2/log(5*x))

________________________________________________________________________________________

mupad [B]  time = 1.13, size = 13, normalized size = 0.76 \begin {gather*} x-{\mathrm {e}}^{\frac {1}{2\,\ln \left (5\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/(2*log(5*x)))/2 + x*log(5*x)^2)/(x*log(5*x)^2),x)

[Out]

x - exp(1/(2*log(5*x)))

________________________________________________________________________________________

sympy [A]  time = 0.29, size = 10, normalized size = 0.59 \begin {gather*} x - e^{\frac {1}{2 \log {\left (5 x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - exp(1/(2*log(5*x)))

________________________________________________________________________________________