3.85.60 \(\int \frac {e^{x/4} (4-8 x)+e^{x/4} (-7 x-2 x^2) \log (\frac {1}{5} x \log (5))}{20 x} \, dx\) [8460]

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

[Out]

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

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Rubi [A]
time = 0.09, 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, 2225, 2209, 2634, 2207, 2230} \begin {gather*} \frac {1}{5} e^{x/4} \log \left (\frac {1}{5} x \log (5)\right )-\frac {2}{5} e^{x/4} x \log \left (\frac {1}{5} x \log (5)\right ) \end {gather*}

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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2209

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

Rule 2225

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 2230

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] &&  !TrueQ[$UseGamma]

Rule 2634

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

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Mathematica [A]
time = 0.13, size = 24, normalized size = 1.00 \begin {gather*} -\frac {1}{5} e^{x/4} (-1+2 x) \log \left (\frac {1}{5} x \log (5)\right ) \end {gather*}

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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Maple [A]
time = 0.18, size = 18, normalized size = 0.75

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

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 {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.40, size = 17, normalized size = 0.71 \begin {gather*} -\frac {1}{5} \, {\left (2 \, x - 1\right )} e^{\left (\frac {1}{4} \, x\right )} \log \left (\frac {1}{5} \, x \log \left (5\right )\right ) \end {gather*}

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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Sympy [A]
time = 0.13, size = 26, normalized size = 1.08 \begin {gather*} \frac {\left (- 2 x \log {\left (\frac {x \log {\left (5 \right )}}{5} \right )} + \log {\left (\frac {x \log {\left (5 \right )}}{5} \right )}\right ) e^{\frac {x}{4}}}{5} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).
time = 0.41, size = 41, normalized size = 1.71 \begin {gather*} \frac {2}{5} \, x e^{\left (\frac {1}{4} \, x\right )} \log \left (5\right ) - \frac {2}{5} \, x e^{\left (\frac {1}{4} \, x\right )} \log \left (x \log \left (5\right )\right ) - \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )} \log \left (5\right ) + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )} \log \left (x \log \left (5\right )\right ) \end {gather*}

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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Mupad [B]
time = 5.41, size = 21, normalized size = 0.88 \begin {gather*} -\frac {{\mathrm {e}}^{x/4}\,\left (2\,x-1\right )\,\left (\ln \left (\ln \left (5\right )\right )-\ln \left (5\right )+\ln \left (x\right )\right )}{5} \end {gather*}

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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