3.15.29 \(\int \frac {29+8 e^x x+(-4-e^x x) \log (x)}{-8 x+x \log (x)} \, dx\)

Optimal. Leaf size=27 \[ -e^x-\log (x)+3 \log \left (\frac {1}{x \left (-4+\frac {\log (x)}{2}\right )}\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 19, normalized size of antiderivative = 0.70, number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2561, 6742, 2194, 2365, 43} \begin {gather*} -e^x-4 \log (x)-3 \log (8-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(29 + 8*E^x*x + (-4 - E^x*x)*Log[x])/(-8*x + x*Log[x]),x]

[Out]

-E^x - 4*Log[x] - 3*Log[8 - Log[x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2365

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(c_.)*(x_)^(n_.)]*(e_.))^(q_.))/(x_), x_Symbol]
:> Dist[1/n, Subst[Int[(a + b*x)^p*(d + e*x)^q, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x]

Rule 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {29+8 e^x x+\left (-4-e^x x\right ) \log (x)}{x (-8+\log (x))} \, dx\\ &=\int \left (-e^x+\frac {29-4 \log (x)}{x (-8+\log (x))}\right ) \, dx\\ &=-\int e^x \, dx+\int \frac {29-4 \log (x)}{x (-8+\log (x))} \, dx\\ &=-e^x+\operatorname {Subst}\left (\int \frac {29-4 x}{-8+x} \, dx,x,\log (x)\right )\\ &=-e^x+\operatorname {Subst}\left (\int \left (-4-\frac {3}{-8+x}\right ) \, dx,x,\log (x)\right )\\ &=-e^x-4 \log (x)-3 \log (8-\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 19, normalized size = 0.70 \begin {gather*} -e^x-4 \log (x)-3 \log (8-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(29 + 8*E^x*x + (-4 - E^x*x)*Log[x])/(-8*x + x*Log[x]),x]

[Out]

-E^x - 4*Log[x] - 3*Log[8 - Log[x]]

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fricas [A]  time = 1.30, size = 16, normalized size = 0.59 \begin {gather*} -e^{x} - 4 \, \log \relax (x) - 3 \, \log \left (\log \relax (x) - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-4)*log(x)+8*exp(x)*x+29)/(x*log(x)-8*x),x, algorithm="fricas")

[Out]

-e^x - 4*log(x) - 3*log(log(x) - 8)

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giac [A]  time = 0.28, size = 16, normalized size = 0.59 \begin {gather*} -e^{x} - 4 \, \log \relax (x) - 3 \, \log \left (\log \relax (x) - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-4)*log(x)+8*exp(x)*x+29)/(x*log(x)-8*x),x, algorithm="giac")

[Out]

-e^x - 4*log(x) - 3*log(log(x) - 8)

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maple [A]  time = 0.04, size = 17, normalized size = 0.63

method	result	size
default	\(-4 \ln \relax (x )-3 \ln \left (\ln \relax (x )-8\right )-{\mathrm e}^{x}\)	\(17\)
norman	\(-4 \ln \relax (x )-3 \ln \left (\ln \relax (x )-8\right )-{\mathrm e}^{x}\)	\(17\)
risch	\(-4 \ln \relax (x )-3 \ln \left (\ln \relax (x )-8\right )-{\mathrm e}^{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x-4)*ln(x)+8*exp(x)*x+29)/(x*ln(x)-8*x),x,method=_RETURNVERBOSE)

[Out]

-4*ln(x)-3*ln(ln(x)-8)-exp(x)

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maxima [A]  time = 0.71, size = 16, normalized size = 0.59 \begin {gather*} -e^{x} - 4 \, \log \relax (x) - 3 \, \log \left (\log \relax (x) - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-4)*log(x)+8*exp(x)*x+29)/(x*log(x)-8*x),x, algorithm="maxima")

[Out]

-e^x - 4*log(x) - 3*log(log(x) - 8)

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mupad [B]  time = 1.03, size = 16, normalized size = 0.59 \begin {gather*} -3\,\ln \left (\ln \relax (x)-8\right )-{\mathrm {e}}^x-4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x*exp(x) - log(x)*(x*exp(x) + 4) + 29)/(8*x - x*log(x)),x)

[Out]

- 3*log(log(x) - 8) - exp(x) - 4*log(x)

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sympy [A]  time = 0.33, size = 17, normalized size = 0.63 \begin {gather*} - e^{x} - 4 \log {\relax (x )} - 3 \log {\left (\log {\relax (x )} - 8 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-4)*ln(x)+8*exp(x)*x+29)/(x*ln(x)-8*x),x)

[Out]

-exp(x) - 4*log(x) - 3*log(log(x) - 8)

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