3.89.68 \(\int \frac {4 e^{5 x}+e^{24 x}+(4 e^{5 x} x+e^{24 x} x) \log (x)+(e^{24 x} (-95 x+19 x^2) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))) \log (5-x-\log (\log (x)))}{e^{5 x} (-5 x+x^2) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=19 \[ \left (4+e^{19 x}\right ) \log (5-x-\log (\log (x))) \]

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Rubi [B]  time = 1.37, antiderivative size = 93, normalized size of antiderivative = 4.89, number of steps used = 5, number of rules used = 4, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6688, 6742, 6684, 2288} \begin {gather*} \frac {e^{19 x} \left (x^2 (-\log (x)) \log (-x-\log (\log (x))+5)+5 x \log (x) \log (-x-\log (\log (x))+5)-x \log (x) \log (\log (x)) \log (-x-\log (\log (x))+5)\right )}{x \log (x) (-x-\log (\log (x))+5)}+4 \log (-x-\log (\log (x))+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^(5*x) + E^(24*x) + (4*E^(5*x)*x + E^(24*x)*x)*Log[x] + (E^(24*x)*(-95*x + 19*x^2)*Log[x] + 19*E^(24*x
)*x*Log[x]*Log[Log[x]])*Log[5 - x - Log[Log[x]]])/(E^(5*x)*(-5*x + x^2)*Log[x] + E^(5*x)*x*Log[x]*Log[Log[x]])
,x]

[Out]

4*Log[5 - x - Log[Log[x]]] + (E^(19*x)*(5*x*Log[x]*Log[5 - x - Log[Log[x]]] - x^2*Log[x]*Log[5 - x - Log[Log[x
]]] - x*Log[x]*Log[Log[x]]*Log[5 - x - Log[Log[x]]]))/(x*Log[x]*(5 - x - Log[Log[x]]))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

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 {-4-e^{19 x}-x \log (x) \left (4+e^{19 x}+19 e^{19 x} (-5+x+\log (\log (x))) \log (5-x-\log (\log (x)))\right )}{x \log (x) (5-x-\log (\log (x)))} \, dx\\ &=\int \left (\frac {4 (1+x \log (x))}{x \log (x) (-5+x+\log (\log (x)))}+\frac {e^{19 x} \left (1+x \log (x)-95 x \log (x) \log (5-x-\log (\log (x)))+19 x^2 \log (x) \log (5-x-\log (\log (x)))+19 x \log (x) \log (\log (x)) \log (5-x-\log (\log (x)))\right )}{x \log (x) (-5+x+\log (\log (x)))}\right ) \, dx\\ &=4 \int \frac {1+x \log (x)}{x \log (x) (-5+x+\log (\log (x)))} \, dx+\int \frac {e^{19 x} \left (1+x \log (x)-95 x \log (x) \log (5-x-\log (\log (x)))+19 x^2 \log (x) \log (5-x-\log (\log (x)))+19 x \log (x) \log (\log (x)) \log (5-x-\log (\log (x)))\right )}{x \log (x) (-5+x+\log (\log (x)))} \, dx\\ &=4 \log (5-x-\log (\log (x)))+\frac {e^{19 x} \left (5 x \log (x) \log (5-x-\log (\log (x)))-x^2 \log (x) \log (5-x-\log (\log (x)))-x \log (x) \log (\log (x)) \log (5-x-\log (\log (x)))\right )}{x \log (x) (5-x-\log (\log (x)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 19, normalized size = 1.00 \begin {gather*} \left (4+e^{19 x}\right ) \log (5-x-\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(5*x) + E^(24*x) + (4*E^(5*x)*x + E^(24*x)*x)*Log[x] + (E^(24*x)*(-95*x + 19*x^2)*Log[x] + 19*E
^(24*x)*x*Log[x]*Log[Log[x]])*Log[5 - x - Log[Log[x]]])/(E^(5*x)*(-5*x + x^2)*Log[x] + E^(5*x)*x*Log[x]*Log[Lo
g[x]]),x]

[Out]

(4 + E^(19*x))*Log[5 - x - Log[Log[x]]]

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fricas [A]  time = 0.51, size = 18, normalized size = 0.95 \begin {gather*} {\left (e^{\left (19 \, x\right )} + 4\right )} \log \left (-x - \log \left (\log \relax (x)\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log(x))*log(-log(log(x))+5-x)+(x*exp(24*
x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*exp(5*x))/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x, alg
orithm="fricas")

[Out]

(e^(19*x) + 4)*log(-x - log(log(x)) + 5)

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giac [A]  time = 0.25, size = 26, normalized size = 1.37 \begin {gather*} e^{\left (19 \, x\right )} \log \left (-x - \log \left (\log \relax (x)\right ) + 5\right ) + 4 \, \log \left (x + \log \left (\log \relax (x)\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log(x))*log(-log(log(x))+5-x)+(x*exp(24*
x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*exp(5*x))/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x, alg
orithm="giac")

[Out]

e^(19*x)*log(-x - log(log(x)) + 5) + 4*log(x + log(log(x)) - 5)

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maple [A]  time = 0.06, size = 27, normalized size = 1.42

method	result	size
risch	\({\mathrm e}^{19 x} \ln \left (-\ln \left (\ln \relax (x )\right )+5-x \right )+4 \ln \left (\ln \left (\ln \relax (x )\right )+x -5\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((19*x*exp(24*x)*ln(x)*ln(ln(x))+(19*x^2-95*x)*exp(24*x)*ln(x))*ln(-ln(ln(x))+5-x)+(x*exp(24*x)+4*x*exp(5*
x))*ln(x)+exp(24*x)+4*exp(5*x))/(x*exp(5*x)*ln(x)*ln(ln(x))+(x^2-5*x)*exp(5*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

exp(19*x)*ln(-ln(ln(x))+5-x)+4*ln(ln(ln(x))+x-5)

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maxima [A]  time = 0.42, size = 18, normalized size = 0.95 \begin {gather*} {\left (e^{\left (19 \, x\right )} + 4\right )} \log \left (-x - \log \left (\log \relax (x)\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log(x))*log(-log(log(x))+5-x)+(x*exp(24*
x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*exp(5*x))/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x, alg
orithm="maxima")

[Out]

(e^(19*x) + 4)*log(-x - log(log(x)) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {4\,{\mathrm {e}}^{5\,x}+{\mathrm {e}}^{24\,x}-\ln \left (5-\ln \left (\ln \relax (x)\right )-x\right )\,\left ({\mathrm {e}}^{24\,x}\,\ln \relax (x)\,\left (95\,x-19\,x^2\right )-19\,x\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{24\,x}\,\ln \relax (x)\right )+\ln \relax (x)\,\left (4\,x\,{\mathrm {e}}^{5\,x}+x\,{\mathrm {e}}^{24\,x}\right )}{{\mathrm {e}}^{5\,x}\,\ln \relax (x)\,\left (5\,x-x^2\right )-x\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{5\,x}\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(5*x) + exp(24*x) - log(5 - log(log(x)) - x)*(exp(24*x)*log(x)*(95*x - 19*x^2) - 19*x*log(log(x))*e
xp(24*x)*log(x)) + log(x)*(4*x*exp(5*x) + x*exp(24*x)))/(exp(5*x)*log(x)*(5*x - x^2) - x*log(log(x))*exp(5*x)*
log(x)),x)

[Out]

-int((4*exp(5*x) + exp(24*x) - log(5 - log(log(x)) - x)*(exp(24*x)*log(x)*(95*x - 19*x^2) - 19*x*log(log(x))*e
xp(24*x)*log(x)) + log(x)*(4*x*exp(5*x) + x*exp(24*x)))/(exp(5*x)*log(x)*(5*x - x^2) - x*log(log(x))*exp(5*x)*
log(x)), x)

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sympy [A]  time = 13.95, size = 26, normalized size = 1.37 \begin {gather*} e^{19 x} \log {\left (- x - \log {\left (\log {\relax (x )} \right )} + 5 \right )} + 4 \log {\left (x + \log {\left (\log {\relax (x )} \right )} - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((19*x*exp(24*x)*ln(x)*ln(ln(x))+(19*x**2-95*x)*exp(24*x)*ln(x))*ln(-ln(ln(x))+5-x)+(x*exp(24*x)+4*x
*exp(5*x))*ln(x)+exp(24*x)+4*exp(5*x))/(x*exp(5*x)*ln(x)*ln(ln(x))+(x**2-5*x)*exp(5*x)*ln(x)),x)

[Out]

exp(19*x)*log(-x - log(log(x)) + 5) + 4*log(x + log(log(x)) - 5)

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