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

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

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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} $$\begin{array}{c}x-e^{\frac{1}{2\log (5x)}}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{2}\int \frac{e^{\frac{1}{2\log (5x)}}+2x{\log }^2(5x)}{x{\log }^2(5x)}dx\\ & =\frac{1}{2}\int (2+\frac{e^{\frac{1}{2\log (5x)}}}{x{\log }^2(5x)})dx\\ & =x+\frac{1}{2}\int \frac{e^{\frac{1}{2\log (5x)}}}{x{\log }^2(5x)}dx\\ & =x+\frac{1}{2}\mathrm{Subst}(\int \frac{e^{\frac{1}{2}/x}}{x^2}dx,x,\log (5x))\\ & =-e^{\frac{1}{2\log (5x)}}+x\end{array}\end{array}$$

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

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

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

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))

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

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)))

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

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

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))

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

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))

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

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)))

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

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)))

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