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

Optimal. Leaf size=19 $$(4+e^{19x})\log (5-x-\log (\log (x)))$$

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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} $$\begin{array}{c}\frac{e^{19x}(x^2(-\log (x))\log (-x-\log (\log (x))+5)+5x\log (x)\log (-x-\log (\log (x))+5)-x\log (x)\log (\log (x))\log (-x-\log (\log (x))+5))}{x\log (x)(-x-\log (\log (x))+5)}+4\log (-x-\log (\log (x))+5)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-4-e^{19x}-x\log (x)(4+e^{19x}+19e^{19x}(-5+x+\log (\log (x)))\log (5-x-\log (\log (x))))}{x\log (x)(5-x-\log (\log (x)))}dx\\ & =\int (\frac{4(1+x\log (x))}{x\log (x)(-5+x+\log (\log (x)))}+\frac{e^{19x}(1+x\log (x)-95x\log (x)\log (5-x-\log (\log (x)))+19x^2\log (x)\log (5-x-\log (\log (x)))+19x\log (x)\log (\log (x))\log (5-x-\log (\log (x))))}{x\log (x)(-5+x+\log (\log (x)))})dx\\ & =4\int \frac{1+x\log (x)}{x\log (x)(-5+x+\log (\log (x)))}dx+\int \frac{e^{19x}(1+x\log (x)-95x\log (x)\log (5-x-\log (\log (x)))+19x^2\log (x)\log (5-x-\log (\log (x)))+19x\log (x)\log (\log (x))\log (5-x-\log (\log (x))))}{x\log (x)(-5+x+\log (\log (x)))}dx\\ & =4\log (5-x-\log (\log (x)))+\frac{e^{19x}(5x\log (x)\log (5-x-\log (\log (x)))-x^2\log (x)\log (5-x-\log (\log (x)))-x\log (x)\log (\log (x))\log (5-x-\log (\log (x))))}{x\log (x)(5-x-\log (\log (x)))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.31, size = 19, normalized size = 1.00 $$\begin{array}{c}(4+e^{19x})\log (5-x-\log (\log (x)))\end{array}$$

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 $$\begin{array}{c}(e^{\left(19x\right)}+4)\log (-x-\log (\log (x))+5)\end{array}$$

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 $$\begin{array}{c}{e}^{\left(19x\right)}\log (-x-\log (\log (x))+5)+4\log (x+\log (\log (x))-5)\end{array}$$

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	${\mathrm{e}}^{19x}\ln (-\ln (\ln (x))+5-x)+4\ln (\ln (\ln (x))+x-5)$	$27$

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 $$\begin{array}{c}(e^{\left(19x\right)}+4)\log (-x-\log (\log (x))+5)\end{array}$$

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 $$\begin{array}{c}-\int \frac{4{\mathrm{e}}^{5x}+{\mathrm{e}}^{24x}-\ln (5-\ln (\ln (x))-x)({\mathrm{e}}^{24x}\ln (x)(95x-19x^2)-19x\ln (\ln (x)){\mathrm{e}}^{24x}\ln (x))+\ln (x)(4x{\mathrm{e}}^{5x}+x{\mathrm{e}}^{24x})}{{\mathrm{e}}^{5x}\ln (x)(5x-x^2)-x\ln (\ln (x)){\mathrm{e}}^{5x}\ln (x)}dx\end{array}$$

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 $$\begin{array}{c}{e}^{19x}\log (-x-\log (\log (x))+5)+4\log (x+\log (\log (x))-5)\end{array}$$

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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