3.34.30 \(\int \frac {(e^x (1+4 x+4 \log ^2(3))+\log (x)) \log ^2(e^x (1+4 x+4 \log ^2(3))+\log (x))+e^{\frac {x}{\log (e^x (1+4 x+4 \log ^2(3))+\log (x))}} (-1+e^x (-5 x-4 x^2-4 x \log ^2(3))+(e^x (1+4 x+4 \log ^2(3))+\log (x)) \log (e^x (1+4 x+4 \log ^2(3))+\log (x)))}{(e^x (1+4 x+4 \log ^2(3))+\log (x)) \log ^2(e^x (1+4 x+4 \log ^2(3))+\log (x))} \, dx\)

Optimal. Leaf size=26 \[ e^{\frac {x}{\log \left (e^x \left (1+4 \left (x+\log ^2(3)\right )\right )+\log (x)\right )}}+x \]

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Rubi [F]  time = 30.21, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2 + E^(x/Log[E^x*(1 + 4*x
+ 4*Log[3]^2) + Log[x]])*(-1 + E^x*(-5*x - 4*x^2 - 4*x*Log[3]^2) + (E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E
^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]))/((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) +
Log[x]]^2),x]

[Out]

x - Defer[Int][E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2, x] -
Defer[Int][(E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*x)/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2, x]
+ (1 + 4*Log[3]^2)*Defer[Int][E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])/((1 + 4*x + 4*Log[3]^2)*Log[E^x*(
1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x] + Defer[Int][E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])/((-4*E^x*x
- E^x*(1 + 4*Log[3]^2) - Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x] + (1 + 4*Log[3]^2)*Defer[Int]
[E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])/((-1 - 4*x - 4*Log[3]^2)*(4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log
[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x] + (1 + 4*Log[3]^2)*Defer[Int][E^(x/Log[E^x*(1 + 4*x + 4*L
og[3]^2) + Log[x]])/((1 + 4*x + 4*Log[3]^2)*(4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log
[3]^2) + Log[x]]^2), x] - ((1 + 4*Log[3]^2)*Defer[Int][(E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*Log[x])
/((4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x])/4 + ((5 + 4*Log[3
]^2)*Defer[Int][(E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*Log[x])/((4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log
[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x])/4 + Defer[Int][(E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Lo
g[x]])*x*Log[x])/((4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x] +
((1 + 4*Log[3]^2)^2*Defer[Int][(E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*Log[x])/((1 + 4*x + 4*Log[3]^2)
*(4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x])/4 - ((1 + 4*Log[3]
^2)*(5 + 4*Log[3]^2)*Defer[Int][(E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*Log[x])/((1 + 4*x + 4*Log[3]^2
)*(4*E^x*x + E^x*(1 + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2), x])/4 + Defer[Int][E^
(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1-e^x x \left (5+4 x+4 \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}\right ) \, dx\\ &=x+\int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1-e^x x \left (5+4 x+4 \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx\\ &=x+\int \left (\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1-4 x-4 \log ^2(3)+4 x^2 \log (x)+5 x \left (1+\frac {4 \log ^2(3)}{5}\right ) \log (x)\right )}{\left (1+4 x+4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}+\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-4 x^2-5 x \left (1+\frac {4 \log ^2(3)}{5}\right )+4 x \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+\left (1+4 \log ^2(3)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (1+4 x+4 \log ^2(3)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}\right ) \, dx\\ &=x+\int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1-4 x-4 \log ^2(3)+4 x^2 \log (x)+5 x \left (1+\frac {4 \log ^2(3)}{5}\right ) \log (x)\right )}{\left (1+4 x+4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx+\int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-4 x^2-5 x \left (1+\frac {4 \log ^2(3)}{5}\right )+4 x \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+\left (1+4 \log ^2(3)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (1+4 x+4 \log ^2(3)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx\\ &=x+\int \left (\frac {4 e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x}{\left (-1-4 x-4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}+\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (1+4 \log ^2(3)\right )}{\left (-1-4 x-4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}+\frac {4 e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x^2 \log (x)}{\left (1+4 x+4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}+\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x \left (5+4 \log ^2(3)\right ) \log (x)}{\left (1+4 x+4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}\right ) \, dx+\int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-x \left (5+4 x+4 \log ^2(3)\right )+\left (1+4 x+4 \log ^2(3)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (1+4 x+4 \log ^2(3)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx\\ &=x+4 \int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x}{\left (-1-4 x-4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx+4 \int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x^2 \log (x)}{\left (1+4 x+4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx+\left (1+4 \log ^2(3)\right ) \int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}}}{\left (-1-4 x-4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx+\left (5+4 \log ^2(3)\right ) \int \frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x \log (x)}{\left (1+4 x+4 \log ^2(3)\right ) \left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx+\int \left (-\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} x \left (5+4 x+4 \log ^2(3)\right )}{\left (1+4 x+4 \log ^2(3)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}+\frac {e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}}}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 27, normalized size = 1.04 \begin {gather*} e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2 + E^(x/Log[E^x*(1
+ 4*x + 4*Log[3]^2) + Log[x]])*(-1 + E^x*(-5*x - 4*x^2 - 4*x*Log[3]^2) + (E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])
*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]))/((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]
^2) + Log[x]]^2),x]

[Out]

E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]) + x

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fricas [A]  time = 0.58, size = 25, normalized size = 0.96 \begin {gather*} x + e^{\left (\frac {x}{\log \left ({\left (4 \, \log \relax (3)^{2} + 4 \, x + 1\right )} e^{x} + \log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+1)*exp(x))+(-4*x*log(3)^2-4*x^2-5*x)
*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^2+4*x+1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3
)^2+4*x+1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3)^2+4*x+1)*exp(x))^2,x, algorithm=
"fricas")

[Out]

x + e^(x/log((4*log(3)^2 + 4*x + 1)*e^x + log(x)))

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giac [A]  time = 7.21, size = 26, normalized size = 1.00 \begin {gather*} x + e^{\left (\frac {x}{\log \left (4 \, e^{x} \log \relax (3)^{2} + 4 \, x e^{x} + e^{x} + \log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+1)*exp(x))+(-4*x*log(3)^2-4*x^2-5*x)
*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^2+4*x+1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3
)^2+4*x+1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3)^2+4*x+1)*exp(x))^2,x, algorithm=
"giac")

[Out]

x + e^(x/log(4*e^x*log(3)^2 + 4*x*e^x + e^x + log(x)))

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maple [A]  time = 0.05, size = 26, normalized size = 1.00

method	result	size
risch	\(x +{\mathrm e}^{\frac {x}{\ln \left (\ln \relax (x )+\left (4 \ln \relax (3)^{2}+4 x +1\right ) {\mathrm e}^{x}\right )}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((ln(x)+(4*ln(3)^2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))+(-4*x*ln(3)^2-4*x^2-5*x)*exp(x)-1)*e
xp(x/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x)))+(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))^2
)/(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))^2,x,method=_RETURNVERBOSE)

[Out]

x+exp(x/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+1)*exp(x))+(-4*x*log(3)^2-4*x^2-5*x)
*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^2+4*x+1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3
)^2+4*x+1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3)^2+4*x+1)*exp(x))^2,x, algorithm=
"maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 2.38, size = 26, normalized size = 1.00 \begin {gather*} x+{\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^x+\ln \relax (x)+4\,{\mathrm {e}}^x\,{\ln \relax (3)}^2+4\,x\,{\mathrm {e}}^x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x/log(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1)))*(exp(x)*(5*x + 4*x*log(3)^2 + 4*x^2) - log(log(x) + e
xp(x)*(4*x + 4*log(3)^2 + 1))*(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1)) + 1) - log(log(x) + exp(x)*(4*x + 4*log
(3)^2 + 1))^2*(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1)))/(log(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1))^2*(log(x)
+ exp(x)*(4*x + 4*log(3)^2 + 1))),x)

[Out]

x + exp(x/log(exp(x) + log(x) + 4*exp(x)*log(3)^2 + 4*x*exp(x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((ln(x)+(4*ln(3)**2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)**2+4*x+1)*exp(x))+(-4*x*ln(3)**2-4*x**2-5*x)*e
xp(x)-1)*exp(x/ln(ln(x)+(4*ln(3)**2+4*x+1)*exp(x)))+(ln(x)+(4*ln(3)**2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)**2+4*x
+1)*exp(x))**2)/(ln(x)+(4*ln(3)**2+4*x+1)*exp(x))/ln(ln(x)+(4*ln(3)**2+4*x+1)*exp(x))**2,x)

[Out]

Timed out

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