3.13.50 \(\int \frac {e^{15-5 x+(-3+x+\log (3+x)) \log (\frac {1}{4} (8+\log (x)))} (-9-160 x-39 x^2+(-20 x-5 x^2) \log (x)+(3+x) \log (3+x)+(32 x+8 x^2+(4 x+x^2) \log (x)) \log (\frac {1}{4} (8+\log (x))))}{(3+x)^5 (24 x+8 x^2+(3 x+x^2) \log (x))} \, dx\)

Optimal. Leaf size=27 \[ e^{(3-x-\log (3+x)) \left (5-\log \left (2+\frac {\log (x)}{4}\right )\right )} \]

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Rubi [F]  time = 31.39, 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 {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) \left (-9-160 x-39 x^2+\left (-20 x-5 x^2\right ) \log (x)+(3+x) \log (3+x)+\left (32 x+8 x^2+\left (4 x+x^2\right ) \log (x)\right ) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{(3+x)^5 \left (24 x+8 x^2+\left (3 x+x^2\right ) \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*(-9 - 160*x - 39*x^2 + (-20*x - 5*x^2)*Log[x] +
(3 + x)*Log[3 + x] + (32*x + 8*x^2 + (4*x + x^2)*Log[x])*Log[(8 + Log[x])/4]))/((3 + x)^5*(24*x + 8*x^2 + (3*x
+ x^2)*Log[x])),x]

[Out]

-5*Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])/(3 + x)^6, x] - 5*Defer[Int][E^(15 - 5*
x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])/(3 + x)^5, x] - Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*
Log[(8 + Log[x])/4])/(x*(8 + Log[x])), x]/81 + 2*Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x
])/4])/((3 + x)^5*(8 + Log[x])), x] + Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])/((3
+ x)^4*(8 + Log[x])), x]/3 + Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])/((3 + x)^3*(8
+ Log[x])), x]/9 + Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])/((3 + x)^2*(8 + Log[x]
)), x]/27 + Defer[Int][E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])/((3 + x)*(8 + Log[x])), x]/81
+ Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[3 + x])/(x*(8 + Log[x])), x]/243 -
Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[3 + x])/((3 + x)^5*(8 + Log[x])), x]/
3 - Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[3 + x])/((3 + x)^4*(8 + Log[x])),
x]/9 - Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[3 + x])/((3 + x)^3*(8 + Log[x
])), x]/27 - Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[3 + x])/((3 + x)^2*(8 +
Log[x])), x]/81 - Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[3 + x])/((3 + x)*(8
+ Log[x])), x]/243 + Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[(8 + Log[x])/4]
)/(3 + x)^6, x] + Defer[Int][(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*Log[(8 + Log[x])/4])/(3
+ x)^5, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) \left (-9-160 x-39 x^2+\left (-20 x-5 x^2\right ) \log (x)+(3+x) \log (3+x)+\left (32 x+8 x^2+\left (4 x+x^2\right ) \log (x)\right ) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{x (3+x)^6 (8+\log (x))} \, dx\\ &=\int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) \left (-9-160 x-39 x^2-5 x (4+x) \log (x)+(3+x) \log (3+x)+x (4+x) (8+\log (x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{x (3+x)^6 (8+\log (x))} \, dx\\ &=\int \left (-\frac {160 \exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{(3+x)^6 (8+\log (x))}-\frac {9 \exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{x (3+x)^6 (8+\log (x))}-\frac {39 \exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) x}{(3+x)^6 (8+\log (x))}-\frac {5 \exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) (4+x) \log (x)}{(3+x)^6 (8+\log (x))}+\frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) \log (3+x)}{x (3+x)^5 (8+\log (x))}+\frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) (4+x) \log \left (\frac {1}{4} (8+\log (x))\right )}{(3+x)^6}\right ) \, dx\\ &=-\left (5 \int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) (4+x) \log (x)}{(3+x)^6 (8+\log (x))} \, dx\right )-9 \int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{x (3+x)^6 (8+\log (x))} \, dx-39 \int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) x}{(3+x)^6 (8+\log (x))} \, dx-160 \int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{(3+x)^6 (8+\log (x))} \, dx+\int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) \log (3+x)}{x (3+x)^5 (8+\log (x))} \, dx+\int \frac {\exp \left (15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )\right ) (4+x) \log \left (\frac {1}{4} (8+\log (x))\right )}{(3+x)^6} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.51, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{15-5 x+(-3+x+\log (3+x)) \log \left (\frac {1}{4} (8+\log (x))\right )} \left (-9-160 x-39 x^2+\left (-20 x-5 x^2\right ) \log (x)+(3+x) \log (3+x)+\left (32 x+8 x^2+\left (4 x+x^2\right ) \log (x)\right ) \log \left (\frac {1}{4} (8+\log (x))\right )\right )}{(3+x)^5 \left (24 x+8 x^2+\left (3 x+x^2\right ) \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*(-9 - 160*x - 39*x^2 + (-20*x - 5*x^2)*Log
[x] + (3 + x)*Log[3 + x] + (32*x + 8*x^2 + (4*x + x^2)*Log[x])*Log[(8 + Log[x])/4]))/((3 + x)^5*(24*x + 8*x^2
+ (3*x + x^2)*Log[x])),x]

[Out]

Integrate[(E^(15 - 5*x + (-3 + x + Log[3 + x])*Log[(8 + Log[x])/4])*(-9 - 160*x - 39*x^2 + (-20*x - 5*x^2)*Log
[x] + (3 + x)*Log[3 + x] + (32*x + 8*x^2 + (4*x + x^2)*Log[x])*Log[(8 + Log[x])/4]))/((3 + x)^5*(24*x + 8*x^2
+ (3*x + x^2)*Log[x])), x]

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fricas [A]  time = 0.76, size = 27, normalized size = 1.00 \begin {gather*} e^{\left ({\left (x + \log \left (x + 3\right ) - 3\right )} \log \left (\frac {1}{4} \, \log \relax (x) + 2\right ) - 5 \, x - 5 \, \log \left (x + 3\right ) + 15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+4*x)*log(x)+8*x^2+32*x)*log(1/4*log(x)+2)+(3+x)*log(3+x)+(-5*x^2-20*x)*log(x)-39*x^2-160*x-9)
*exp((log(3+x)+x-3)*log(1/4*log(x)+2)-5*log(3+x)+15-5*x)/((x^2+3*x)*log(x)+8*x^2+24*x),x, algorithm="fricas")

[Out]

e^((x + log(x + 3) - 3)*log(1/4*log(x) + 2) - 5*x - 5*log(x + 3) + 15)

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giac [B]  time = 2.97, size = 42, normalized size = 1.56 \begin {gather*} e^{\left (x \log \left (\frac {1}{4} \, \log \relax (x) + 2\right ) + \log \left (x + 3\right ) \log \left (\frac {1}{4} \, \log \relax (x) + 2\right ) - 5 \, x - 5 \, \log \left (x + 3\right ) - 3 \, \log \left (\frac {1}{4} \, \log \relax (x) + 2\right ) + 15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+4*x)*log(x)+8*x^2+32*x)*log(1/4*log(x)+2)+(3+x)*log(3+x)+(-5*x^2-20*x)*log(x)-39*x^2-160*x-9)
*exp((log(3+x)+x-3)*log(1/4*log(x)+2)-5*log(3+x)+15-5*x)/((x^2+3*x)*log(x)+8*x^2+24*x),x, algorithm="giac")

[Out]

e^(x*log(1/4*log(x) + 2) + log(x + 3)*log(1/4*log(x) + 2) - 5*x - 5*log(x + 3) - 3*log(1/4*log(x) + 2) + 15)

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maple [A]  time = 0.03, size = 27, normalized size = 1.00

method	result	size
risch	\(\frac {\left (\frac {\ln \relax (x )}{4}+2\right )^{\ln \left (3+x \right )+x -3} {\mathrm e}^{15-5 x}}{\left (3+x \right )^{5}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2+4*x)*ln(x)+8*x^2+32*x)*ln(1/4*ln(x)+2)+(3+x)*ln(3+x)+(-5*x^2-20*x)*ln(x)-39*x^2-160*x-9)*exp((ln(3+
x)+x-3)*ln(1/4*ln(x)+2)-5*ln(3+x)+15-5*x)/((x^2+3*x)*ln(x)+8*x^2+24*x),x,method=_RETURNVERBOSE)

[Out]

(1/4*ln(x)+2)^(ln(3+x)+x-3)/(3+x)^5*exp(15-5*x)

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maxima [B]  time = 0.79, size = 149, normalized size = 5.52 \begin {gather*} \frac {64 \, e^{\left (-2 \, x \log \relax (2) - 2 \, \log \relax (2) \log \left (x + 3\right ) + x \log \left (\log \relax (x) + 8\right ) + \log \left (x + 3\right ) \log \left (\log \relax (x) + 8\right ) - 5 \, x + 15\right )}}{512 \, x^{5} + 7680 \, x^{4} + {\left (x^{5} + 15 \, x^{4} + 90 \, x^{3} + 270 \, x^{2} + 405 \, x + 243\right )} \log \relax (x)^{3} + 46080 \, x^{3} + 24 \, {\left (x^{5} + 15 \, x^{4} + 90 \, x^{3} + 270 \, x^{2} + 405 \, x + 243\right )} \log \relax (x)^{2} + 138240 \, x^{2} + 192 \, {\left (x^{5} + 15 \, x^{4} + 90 \, x^{3} + 270 \, x^{2} + 405 \, x + 243\right )} \log \relax (x) + 207360 \, x + 124416} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2+4*x)*log(x)+8*x^2+32*x)*log(1/4*log(x)+2)+(3+x)*log(3+x)+(-5*x^2-20*x)*log(x)-39*x^2-160*x-9)
*exp((log(3+x)+x-3)*log(1/4*log(x)+2)-5*log(3+x)+15-5*x)/((x^2+3*x)*log(x)+8*x^2+24*x),x, algorithm="maxima")

[Out]

64*e^(-2*x*log(2) - 2*log(2)*log(x + 3) + x*log(log(x) + 8) + log(x + 3)*log(log(x) + 8) - 5*x + 15)/(512*x^5
+ 7680*x^4 + (x^5 + 15*x^4 + 90*x^3 + 270*x^2 + 405*x + 243)*log(x)^3 + 46080*x^3 + 24*(x^5 + 15*x^4 + 90*x^3
+ 270*x^2 + 405*x + 243)*log(x)^2 + 138240*x^2 + 192*(x^5 + 15*x^4 + 90*x^3 + 270*x^2 + 405*x + 243)*log(x) +
207360*x + 124416)

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mupad [B]  time = 1.31, size = 182, normalized size = 6.74 \begin {gather*} \frac {64\,{\mathrm {e}}^{15-5\,x}\,{\left (\frac {\ln \relax (x)}{4}+2\right )}^{x+\ln \left (x+3\right )}}{x^5\,{\ln \relax (x)}^3+24\,x^5\,{\ln \relax (x)}^2+192\,x^5\,\ln \relax (x)+512\,x^5+15\,x^4\,{\ln \relax (x)}^3+360\,x^4\,{\ln \relax (x)}^2+2880\,x^4\,\ln \relax (x)+7680\,x^4+90\,x^3\,{\ln \relax (x)}^3+2160\,x^3\,{\ln \relax (x)}^2+17280\,x^3\,\ln \relax (x)+46080\,x^3+270\,x^2\,{\ln \relax (x)}^3+6480\,x^2\,{\ln \relax (x)}^2+51840\,x^2\,\ln \relax (x)+138240\,x^2+405\,x\,{\ln \relax (x)}^3+9720\,x\,{\ln \relax (x)}^2+77760\,x\,\ln \relax (x)+207360\,x+243\,{\ln \relax (x)}^3+5832\,{\ln \relax (x)}^2+46656\,\ln \relax (x)+124416} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(log(x)/4 + 2)*(x + log(x + 3) - 3) - 5*log(x + 3) - 5*x + 15)*(160*x - log(log(x)/4 + 2)*(32*x +
log(x)*(4*x + x^2) + 8*x^2) - log(x + 3)*(x + 3) + log(x)*(20*x + 5*x^2) + 39*x^2 + 9))/(24*x + log(x)*(3*x +
x^2) + 8*x^2),x)

[Out]

(64*exp(15 - 5*x)*(log(x)/4 + 2)^(x + log(x + 3)))/(207360*x + 46656*log(x) + 9720*x*log(x)^2 + 51840*x^2*log(
x) + 405*x*log(x)^3 + 17280*x^3*log(x) + 2880*x^4*log(x) + 192*x^5*log(x) + 5832*log(x)^2 + 243*log(x)^3 + 648
0*x^2*log(x)^2 + 270*x^2*log(x)^3 + 2160*x^3*log(x)^2 + 90*x^3*log(x)^3 + 360*x^4*log(x)^2 + 15*x^4*log(x)^3 +
24*x^5*log(x)^2 + x^5*log(x)^3 + 77760*x*log(x) + 138240*x^2 + 46080*x^3 + 7680*x^4 + 512*x^5 + 124416)

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sympy [B]  time = 1.93, size = 46, normalized size = 1.70 \begin {gather*} \frac {e^{- 5 x + \left (x + \log {\left (x + 3 \right )} - 3\right ) \log {\left (\frac {\log {\relax (x )}}{4} + 2 \right )} + 15}}{x^{5} + 15 x^{4} + 90 x^{3} + 270 x^{2} + 405 x + 243} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2+4*x)*ln(x)+8*x**2+32*x)*ln(1/4*ln(x)+2)+(3+x)*ln(3+x)+(-5*x**2-20*x)*ln(x)-39*x**2-160*x-9)*
exp((ln(3+x)+x-3)*ln(1/4*ln(x)+2)-5*ln(3+x)+15-5*x)/((x**2+3*x)*ln(x)+8*x**2+24*x),x)

[Out]

exp(-5*x + (x + log(x + 3) - 3)*log(log(x)/4 + 2) + 15)/(x**5 + 15*x**4 + 90*x**3 + 270*x**2 + 405*x + 243)

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