3.90.90 \(\int \frac {16+8 x^2+x^4+e^{5+x} (-16 x-8 x^3-x^5)+e^5 (-4 x+x^3) \log (2)+(e^x (-16 x-8 x^3-x^5)+(-4 x+x^3) \log (2)) \log (\frac {x}{4})}{e^5 (256 x+128 x^3+16 x^5)+(256 x+128 x^3+16 x^5) \log (\frac {x}{4})} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{16} \left (2-e^x-\frac {\log (2)}{\frac {4}{x}+x}+\log \left (e^5+\log \left (\frac {x}{4}\right )\right )\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 2.60, 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 {16+8 x^2+x^4+e^{5+x} \left (-16 x-8 x^3-x^5\right )+e^5 \left (-4 x+x^3\right ) \log (2)+\left (e^x \left (-16 x-8 x^3-x^5\right )+\left (-4 x+x^3\right ) \log (2)\right ) \log \left (\frac {x}{4}\right )}{e^5 \left (256 x+128 x^3+16 x^5\right )+\left (256 x+128 x^3+16 x^5\right ) \log \left (\frac {x}{4}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16 + 8*x^2 + x^4 + E^(5 + x)*(-16*x - 8*x^3 - x^5) + E^5*(-4*x + x^3)*Log[2] + (E^x*(-16*x - 8*x^3 - x^5)
+ (-4*x + x^3)*Log[2])*Log[x/4])/(E^5*(256*x + 128*x^3 + 16*x^5) + (256*x + 128*x^3 + 16*x^5)*Log[x/4]),x]

[Out]

-1/16*E^x - (x*Log[2])/(32*(4 + x^2)) + (ArcTan[x/2]*Log[2])/64 - (x*Log[16])/(128*(4 + x^2)) - (ArcTan[x/2]*L
og[16])/256 + Log[E^5 + Log[x/4]]/16 - (E^5*Log[2]*Defer[Int][x^2/((4 + x^2)^2*(E^5 + Log[x/4])), x])/16 + (E^
5*Log[2]*Defer[Int][(-4 + x^2)/((4 + x^2)^2*(E^5 + Log[x/4])), x])/16 + (E^5*Log[16]*Defer[Subst][Defer[Int][1
/((1 + 4*x^2)^2*(E^5 + Log[x])), x], x, x/4])/64

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4+x^2\right )^2-e^{5+x} x \left (4+x^2\right )^2+e^5 x \left (-4+x^2\right ) \log (2)-x \left (e^x \left (4+x^2\right )^2-x^2 \log (2)+\log (16)\right ) \log \left (\frac {x}{4}\right )}{16 x \left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx\\ &=\frac {1}{16} \int \frac {\left (4+x^2\right )^2-e^{5+x} x \left (4+x^2\right )^2+e^5 x \left (-4+x^2\right ) \log (2)-x \left (e^x \left (4+x^2\right )^2-x^2 \log (2)+\log (16)\right ) \log \left (\frac {x}{4}\right )}{x \left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx\\ &=\frac {1}{16} \int \left (-e^x+\frac {1}{x \left (e^5+\log \left (\frac {x}{4}\right )\right )}+\frac {e^5 (-2+x) (2+x) \log (2)}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}+\frac {x^2 \log (2) \log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}-\frac {\log (16) \log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}\right ) \, dx\\ &=-\frac {\int e^x \, dx}{16}+\frac {1}{16} \int \frac {1}{x \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \log (2) \int \frac {x^2 \log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {(-2+x) (2+x)}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{16} \log (16) \int \frac {\log \left (\frac {x}{4}\right )}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx\\ &=-\frac {e^x}{16}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^5+\log \left (\frac {x}{4}\right )\right )+\frac {1}{16} \log (2) \int \left (\frac {x^2}{\left (4+x^2\right )^2}-\frac {e^5 x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )}\right ) \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{4} \log (16) \operatorname {Subst}\left (\int \frac {\log (x)}{\left (4+16 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right )\\ &=-\frac {e^x}{16}+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )+\frac {1}{16} \log (2) \int \frac {x^2}{\left (4+x^2\right )^2} \, dx-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{4} \log (16) \operatorname {Subst}\left (\int \left (\frac {1}{16 \left (1+4 x^2\right )^2}-\frac {e^5}{16 \left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )}\right ) \, dx,x,\frac {x}{4}\right )\\ &=-\frac {e^x}{16}-\frac {x \log (2)}{32 \left (4+x^2\right )}+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )+\frac {1}{32} \log (2) \int \frac {1}{4+x^2} \, dx-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{64} \log (16) \operatorname {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2} \, dx,x,\frac {x}{4}\right )+\frac {1}{64} \left (e^5 \log (16)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right )\\ &=-\frac {e^x}{16}-\frac {x \log (2)}{32 \left (4+x^2\right )}+\frac {1}{64} \tan ^{-1}\left (\frac {x}{2}\right ) \log (2)-\frac {x \log (16)}{128 \left (4+x^2\right )}+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx-\frac {1}{128} \log (16) \operatorname {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\frac {x}{4}\right )+\frac {1}{64} \left (e^5 \log (16)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right )\\ &=-\frac {e^x}{16}-\frac {x \log (2)}{32 \left (4+x^2\right )}+\frac {1}{64} \tan ^{-1}\left (\frac {x}{2}\right ) \log (2)-\frac {x \log (16)}{128 \left (4+x^2\right )}-\frac {1}{256} \tan ^{-1}\left (\frac {x}{2}\right ) \log (16)+\frac {1}{16} \log \left (e^5+\log \left (\frac {x}{4}\right )\right )-\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{16} \left (e^5 \log (2)\right ) \int \frac {-4+x^2}{\left (4+x^2\right )^2 \left (e^5+\log \left (\frac {x}{4}\right )\right )} \, dx+\frac {1}{64} \left (e^5 \log (16)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+4 x^2\right )^2 \left (e^5+\log (x)\right )} \, dx,x,\frac {x}{4}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 4.15, size = 36, normalized size = 1.03 \begin {gather*} \frac {1}{16} \left (1-e^x-\frac {x \log (256)}{8 \left (4+x^2\right )}+\log \left (e^5+\log \left (\frac {x}{4}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 8*x^2 + x^4 + E^(5 + x)*(-16*x - 8*x^3 - x^5) + E^5*(-4*x + x^3)*Log[2] + (E^x*(-16*x - 8*x^3
- x^5) + (-4*x + x^3)*Log[2])*Log[x/4])/(E^5*(256*x + 128*x^3 + 16*x^5) + (256*x + 128*x^3 + 16*x^5)*Log[x/4])
,x]

[Out]

(1 - E^x - (x*Log[256])/(8*(4 + x^2)) + Log[E^5 + Log[x/4]])/16

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 45, normalized size = 1.29 \begin {gather*} -\frac {{\left (x e^{5} \log \relax (2) - {\left (x^{2} + 4\right )} e^{5} \log \left (e^{5} + \log \left (\frac {1}{4} \, x\right )\right ) + {\left (x^{2} + 4\right )} e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{16 \, {\left (x^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8*x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*ex
p(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128*x^3+256*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x, algorithm="fric
as")

[Out]

-1/16*(x*e^5*log(2) - (x^2 + 4)*e^5*log(e^5 + log(1/4*x)) + (x^2 + 4)*e^(x + 5))*e^(-5)/(x^2 + 4)

________________________________________________________________________________________

giac [A]  time = 0.17, size = 51, normalized size = 1.46 \begin {gather*} -\frac {x^{2} e^{x} - x^{2} \log \left (e^{5} - 2 \, \log \relax (2) + \log \relax (x)\right ) + x \log \relax (2) + 4 \, e^{x} - 4 \, \log \left (e^{5} - 2 \, \log \relax (2) + \log \relax (x)\right )}{16 \, {\left (x^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8*x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*ex
p(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128*x^3+256*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x, algorithm="giac
")

[Out]

-1/16*(x^2*e^x - x^2*log(e^5 - 2*log(2) + log(x)) + x*log(2) + 4*e^x - 4*log(e^5 - 2*log(2) + log(x)))/(x^2 +
4)

________________________________________________________________________________________

maple [A]  time = 0.37, size = 36, normalized size = 1.03

method	result	size
risch	\(-\frac {{\mathrm e}^{x} x^{2}+x \ln \relax (2)+4 \,{\mathrm e}^{x}}{16 \left (x^{2}+4\right )}+\frac {\ln \left ({\mathrm e}^{5}+\ln \left (\frac {x}{4}\right )\right )}{16}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*ln(2))*ln(1/4*x)+(-x^5-8*x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*exp(5)*ln(
2)+x^4+8*x^2+16)/((16*x^5+128*x^3+256*x)*ln(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x,method=_RETURNVERBOSE)

[Out]

-1/16*(exp(x)*x^2+x*ln(2)+4*exp(x))/(x^2+4)+1/16*ln(exp(5)+ln(1/4*x))

________________________________________________________________________________________

maxima [A]  time = 0.48, size = 35, normalized size = 1.00 \begin {gather*} -\frac {{\left (x^{2} + 4\right )} e^{x} + x \log \relax (2)}{16 \, {\left (x^{2} + 4\right )}} + \frac {1}{16} \, \log \left (e^{5} - 2 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^5-8*x^3-16*x)*exp(x)+(x^3-4*x)*log(2))*log(1/4*x)+(-x^5-8*x^3-16*x)*exp(5)*exp(x)+(x^3-4*x)*ex
p(5)*log(2)+x^4+8*x^2+16)/((16*x^5+128*x^3+256*x)*log(1/4*x)+(16*x^5+128*x^3+256*x)*exp(5)),x, algorithm="maxi
ma")

[Out]

-1/16*((x^2 + 4)*e^x + x*log(2))/(x^2 + 4) + 1/16*log(e^5 - 2*log(2) + log(x))

________________________________________________________________________________________

mupad [B]  time = 8.23, size = 29, normalized size = 0.83 \begin {gather*} \frac {\ln \left (\ln \left (\frac {x}{4}\right )+{\mathrm {e}}^5\right )}{16}-\frac {{\mathrm {e}}^x}{16}-\frac {x\,\ln \relax (2)}{16\,x^2+64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x/4)*(log(2)*(4*x - x^3) + exp(x)*(16*x + 8*x^3 + x^5)) - 8*x^2 - x^4 + exp(5)*log(2)*(4*x - x^3) +
exp(5)*exp(x)*(16*x + 8*x^3 + x^5) - 16)/(log(x/4)*(256*x + 128*x^3 + 16*x^5) + exp(5)*(256*x + 128*x^3 + 16*x
^5)),x)

[Out]

log(log(x/4) + exp(5))/16 - exp(x)/16 - (x*log(2))/(16*x^2 + 64)

________________________________________________________________________________________

sympy [A]  time = 0.65, size = 27, normalized size = 0.77 \begin {gather*} - \frac {x \log {\relax (2 )}}{16 x^{2} + 64} - \frac {e^{x}}{16} + \frac {\log {\left (\log {\left (\frac {x}{4} \right )} + e^{5} \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**5-8*x**3-16*x)*exp(x)+(x**3-4*x)*ln(2))*ln(1/4*x)+(-x**5-8*x**3-16*x)*exp(5)*exp(x)+(x**3-4*x
)*exp(5)*ln(2)+x**4+8*x**2+16)/((16*x**5+128*x**3+256*x)*ln(1/4*x)+(16*x**5+128*x**3+256*x)*exp(5)),x)

[Out]

-x*log(2)/(16*x**2 + 64) - exp(x)/16 + log(log(x/4) + exp(5))/16

________________________________________________________________________________________