3.85.60 $\int \frac{e^{x/4}(4-8x)+e^{x/4}(-7x-2x^2)\log (\frac{1}{5}x\log (5))}{20x}dx$

Optimal. Leaf size=24 $$\frac{1}{5}{e}^{x/4}(1-2x)\log (\frac{1}{5}x\log (5))$$

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Rubi [A]  time = 0.13, antiderivative size = 40, normalized size of antiderivative = 1.67, number of steps used = 14, number of rules used = 7, integrand size = 46, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.152, Rules used = {12, 14, 2194, 2178, 2554, 2176, 2199} $$\begin{array}{c}\frac{1}{5}{e}^{x/4}\log (\frac{1}{5}x\log (5))-\frac{2}{5}{e}^{x/4}x\log (\frac{1}{5}x\log (5))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(x/4)*(4 - 8*x) + E^(x/4)*(-7*x - 2*x^2)*Log[(x*Log[5])/5])/(20*x),x]

[Out]

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

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{20}\int \frac{e^{x/4}(4-8x)+e^{x/4}(-7x-2x^2)\log (\frac{1}{5}x\log (5))}{x}dx\\ & =\frac{1}{20}\int (-8e^{x/4}+\frac{4e^{x/4}}{x}-7e^{x/4}\log (\frac{1}{5}x\log (5))-2e^{x/4}x\log (\frac{1}{5}x\log (5)))dx\\ & =-(\frac{1}{10}\int e^{x/4}x\log (\frac{1}{5}x\log (5))dx)+\frac{1}{5}\int \frac{e^{x/4}}{x}dx-\frac{7}{20}\int e^{x/4}\log (\frac{1}{5}x\log (5))dx-\frac{2}{5}\int e^{x/4}dx\\ & =-\frac{8e^{x/4}}{5}+\frac{\text{Ei}\left(\frac{x}{4}\right)}{5}+\frac{1}{5}{e}^{x/4}\log (\frac{1}{5}x\log (5))-\frac{2}{5}{e}^{x/4}x\log (\frac{1}{5}x\log (5))+\frac{1}{10}\int \frac{4e^{x/4}(-4+x)}{x}dx+\frac{7}{20}\int \frac{4e^{x/4}}{x}dx\\ & =-\frac{8e^{x/4}}{5}+\frac{\text{Ei}\left(\frac{x}{4}\right)}{5}+\frac{1}{5}{e}^{x/4}\log (\frac{1}{5}x\log (5))-\frac{2}{5}{e}^{x/4}x\log (\frac{1}{5}x\log (5))+\frac{2}{5}\int \frac{e^{x/4}(-4+x)}{x}dx+\frac{7}{5}\int \frac{e^{x/4}}{x}dx\\ & =-\frac{8e^{x/4}}{5}+\frac{8\text{Ei}\left(\frac{x}{4}\right)}{5}+\frac{1}{5}{e}^{x/4}\log (\frac{1}{5}x\log (5))-\frac{2}{5}{e}^{x/4}x\log (\frac{1}{5}x\log (5))+\frac{2}{5}\int (e^{x/4}-\frac{4e^{x/4}}{x})dx\\ & =-\frac{8e^{x/4}}{5}+\frac{8\text{Ei}\left(\frac{x}{4}\right)}{5}+\frac{1}{5}{e}^{x/4}\log (\frac{1}{5}x\log (5))-\frac{2}{5}{e}^{x/4}x\log (\frac{1}{5}x\log (5))+\frac{2}{5}\int e^{x/4}dx-\frac{8}{5}\int \frac{e^{x/4}}{x}dx\\ & =\frac{1}{5}{e}^{x/4}\log (\frac{1}{5}x\log (5))-\frac{2}{5}{e}^{x/4}x\log (\frac{1}{5}x\log (5))\end{array}\end{array}$$

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Mathematica [A]  time = 0.17, size = 24, normalized size = 1.00 $$\begin{array}{c}-\frac{1}{5}{e}^{x/4}(-1+2x)\log (\frac{1}{5}x\log (5))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/4)*(4 - 8*x) + E^(x/4)*(-7*x - 2*x^2)*Log[(x*Log[5])/5])/(20*x),x]

[Out]

-1/5*(E^(x/4)*(-1 + 2*x)*Log[(x*Log[5])/5])

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fricas [A]  time = 0.83, size = 17, normalized size = 0.71 $$\begin{array}{c}-\frac{1}{5}(2x-1)e^{\left(\frac{1}{4}x\right)}\log (\frac{1}{5}x\log (5))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x^2-7*x)*exp(1/4*x)*log(1/5*x*log(5))+(-8*x+4)*exp(1/4*x))/x,x, algorithm="fricas")

[Out]

-1/5*(2*x - 1)*e^(1/4*x)*log(1/5*x*log(5))

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giac [B]  time = 0.18, size = 41, normalized size = 1.71 $$\begin{array}{c}\frac{2}{5}x{e}^{\left(\frac{1}{4}x\right)}\log (5)-\frac{2}{5}x{e}^{\left(\frac{1}{4}x\right)}\log (x\log (5))-\frac{1}{5}{e}^{\left(\frac{1}{4}x\right)}\log (5)+\frac{1}{5}{e}^{\left(\frac{1}{4}x\right)}\log (x\log (5))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x^2-7*x)*exp(1/4*x)*log(1/5*x*log(5))+(-8*x+4)*exp(1/4*x))/x,x, algorithm="giac")

[Out]

2/5*x*e^(1/4*x)*log(5) - 2/5*x*e^(1/4*x)*log(x*log(5)) - 1/5*e^(1/4*x)*log(5) + 1/5*e^(1/4*x)*log(x*log(5))

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maple [A]  time = 0.04, size = 18, normalized size = 0.75

method	result	size
risch	$-\frac{{\mathrm{e}}^{\frac{x}{4}}(2x-1)\ln \left(\frac{x\ln (5)}{5}\right)}{5}$	$18$
norman	$\frac{{\mathrm{e}}^{\frac{x}{4}}\ln \left(\frac{x\ln (5)}{5}\right)}{5}-\frac{2x{\mathrm{e}}^{\frac{x}{4}}\ln \left(\frac{x\ln (5)}{5}\right)}{5}$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/20*((-2*x^2-7*x)*exp(1/4*x)*ln(1/5*x*ln(5))+(-8*x+4)*exp(1/4*x))/x,x,method=_RETURNVERBOSE)

[Out]

-1/5*exp(1/4*x)*(2*x-1)*ln(1/5*x*ln(5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}-\frac{2}{5}(x-4)e^{\left(\frac{1}{4}x\right)}\log (x)-\frac{7}{5}{e}^{\left(\frac{1}{4}x\right)}\log (\frac{1}{5}x\log (5))+\frac{8}{5}\mathrm{E}\mathrm{i}\left(\frac{1}{4}x\right)-\frac{8}{5}{e}^{\left(\frac{1}{4}x\right)}+\frac{1}{10}\int \frac{(x^2(\log (5)-\log (\log (5)))+4x-16)e^{\left(\frac{1}{4}x\right)}}{x}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x^2-7*x)*exp(1/4*x)*log(1/5*x*log(5))+(-8*x+4)*exp(1/4*x))/x,x, algorithm="maxima")

[Out]

-2/5*(x - 4)*e^(1/4*x)*log(x) - 7/5*e^(1/4*x)*log(1/5*x*log(5)) + 8/5*Ei(1/4*x) - 8/5*e^(1/4*x) + 1/10*integra
te((x^2*(log(5) - log(log(5))) + 4*x - 16)*e^(1/4*x)/x, x)

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mupad [B]  time = 5.41, size = 21, normalized size = 0.88 $$\begin{array}{c}-\frac{{\mathrm{e}}^{x/4}(2x-1)(\ln (\ln (5))-\ln (5)+\ln (x))}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x/4)*(8*x - 4))/20 + (exp(x/4)*log((x*log(5))/5)*(7*x + 2*x^2))/20)/x,x)

[Out]

-(exp(x/4)*(2*x - 1)*(log(log(5)) - log(5) + log(x)))/5

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sympy [A]  time = 0.32, size = 26, normalized size = 1.08 $$\begin{array}{c}\frac{(-2x\log \left(\frac{x\log (5)}{5}\right)+\log \left(\frac{x\log (5)}{5}\right))e^{\frac{x}{4}}}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((-2*x**2-7*x)*exp(1/4*x)*ln(1/5*x*ln(5))+(-8*x+4)*exp(1/4*x))/x,x)

[Out]

(-2*x*log(x*log(5)/5) + log(x*log(5)/5))*exp(x/4)/5

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