3.92.17 \(\int \frac {3-x+e^x (-3-5 x-x^2+x^3) \log (16)+(-x+e^x (1+x^2) \log (16)) \log (x+e^x (-1-x^2) \log (16)) \log (\log (x+e^x (-1-x^2) \log (16)))}{(-x+e^x (1+x^2) \log (16)) \log (x+e^x (-1-x^2) \log (16))} \, dx\)

Optimal. Leaf size=23 \[ (-3+x) \log \left (\log \left (x-\left (e^x+e^x x^2\right ) \log (16)\right )\right ) \]

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Rubi [F]  time = 5.44, 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 {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3 - x + E^x*(-3 - 5*x - x^2 + x^3)*Log[16] + (-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]
*Log[Log[x + E^x*(-1 - x^2)*Log[16]]])/((-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]),x]

[Out]

-Defer[Int][Log[x - E^x*(1 + x^2)*Log[16]]^(-1), x] + (3 - I)*Defer[Int][1/((I - x)*Log[x - E^x*(1 + x^2)*Log[
16]]), x] + Defer[Int][x/Log[x - E^x*(1 + x^2)*Log[16]], x] - (3 + I)*Defer[Int][1/((I + x)*Log[x - E^x*(1 + x
^2)*Log[16]]), x] - 3*Defer[Int][1/((-x + E^x*Log[16] + E^x*x^2*Log[16])*Log[x - E^x*(1 + x^2)*Log[16]]), x] +
(1 + 3*I)*Defer[Int][1/((I - x)*(-x + E^x*Log[16] + E^x*x^2*Log[16])*Log[x - E^x*(1 + x^2)*Log[16]]), x] - 2*
Defer[Int][x/((-x + E^x*Log[16] + E^x*x^2*Log[16])*Log[x - E^x*(1 + x^2)*Log[16]]), x] + Defer[Int][x^2/((-x +
E^x*Log[16] + E^x*x^2*Log[16])*Log[x - E^x*(1 + x^2)*Log[16]]), x] - (1 - 3*I)*Defer[Int][1/((I + x)*(-x + E^
x*Log[16] + E^x*x^2*Log[16])*Log[x - E^x*(1 + x^2)*Log[16]]), x] + Defer[Int][Log[Log[x - E^x*(1 + x^2)*Log[16
]]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {(-3+x) \left (-1+e^x (1+x)^2 \log (16)\right )}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right )\right ) \, dx\\ &=\int \frac {(-3+x) \left (-1+e^x (1+x)^2 \log (16)\right )}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=\int \left (\frac {(-3+x) (1+x)^2}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {3-4 x-2 x^2-2 x^3+x^4}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=\int \frac {(-3+x) (1+x)^2}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {3-4 x-2 x^2-2 x^3+x^4}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=\int \left (-\frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}-\frac {2 (1+3 x)}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+\int \left (-\frac {3}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}-\frac {2 x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}-\frac {2 (-3+x)}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \frac {1+3 x}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx\right )-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-2 \int \frac {-3+x}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \left (\frac {1}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {3 x}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx\right )-2 \int \left (-\frac {3}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {x}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \frac {1}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx\right )-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-2 \int \frac {x}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-6 \int \frac {x}{\left (1+x^2\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+6 \int \frac {1}{\left (1+x^2\right ) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (2 \int \left (\frac {i}{2 (i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {i}{2 (i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx\right )-2 \int \left (-\frac {1}{2 (i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {1}{2 (i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-6 \int \left (-\frac {1}{2 (i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {1}{2 (i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx+6 \int \left (\frac {i}{2 (i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}+\frac {i}{2 (i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )}\right ) \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ &=-\left (i \int \frac {1}{(i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx\right )-i \int \frac {1}{(i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+3 i \int \frac {1}{(i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+3 i \int \frac {1}{(i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-2 \int \frac {x}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+3 \int \frac {1}{(i-x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{(i+x) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-3 \int \frac {1}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x}{\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {1}{(i-x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \frac {x^2}{\left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx-\int \frac {1}{(i+x) \left (-x+e^x \log (16)+e^x x^2 \log (16)\right ) \log \left (x-e^x \left (1+x^2\right ) \log (16)\right )} \, dx+\int \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 37, normalized size = 1.61 \begin {gather*} -3 \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right )+x \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - x + E^x*(-3 - 5*x - x^2 + x^3)*Log[16] + (-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Lo
g[16]]*Log[Log[x + E^x*(-1 - x^2)*Log[16]]])/((-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]),x]

[Out]

-3*Log[Log[x - E^x*(1 + x^2)*Log[16]]] + x*Log[Log[x - E^x*(1 + x^2)*Log[16]]]

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fricas [A]  time = 0.79, size = 19, normalized size = 0.83 \begin {gather*} {\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} + 1\right )} e^{x} \log \relax (2) + x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.53, size = 43, normalized size = 1.87 \begin {gather*} x \log \left (\log \left (-4 \, x^{2} e^{x} \log \relax (2) - 4 \, e^{x} \log \relax (2) + x\right )\right ) - 3 \, \log \left (\log \left (-4 \, x^{2} e^{x} \log \relax (2) - 4 \, e^{x} \log \relax (2) + x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 40, normalized size = 1.74

method	result	size
risch	\(\ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \relax (2) {\mathrm e}^{x}+x \right )\right ) x -3 \ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \relax (2) {\mathrm e}^{x}+x \right )\right )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 21, normalized size = 0.91 \begin {gather*} {\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} \log \relax (2) + \log \relax (2)\right )} e^{x} + x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.87, size = 19, normalized size = 0.83 \begin {gather*} \ln \left (\ln \left (x-4\,{\mathrm {e}}^x\,\ln \relax (2)\,\left (x^2+1\right )\right )\right )\,\left (x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x - 4*exp(x)*log(2)*(x^2 + 1)))*(x - 3)

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sympy [A]  time = 3.69, size = 44, normalized size = 1.91 \begin {gather*} \left (x - 1\right ) \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\relax (2 )} \right )} \right )} - 2 \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\relax (2 )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 1)*log(log(x + (-4*x**2 - 4)*exp(x)*log(2))) - 2*log(log(x + (-4*x**2 - 4)*exp(x)*log(2)))

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