3.8.15 \(\int \frac {e^x (-72-36 x)+(e^x (-600+16 x+72 x^2)+e^x (-300+8 x+36 x^2) \log (25-9 x)) \log (2+\log (25-9 x)) \log (\log (2+\log (25-9 x)))}{(-50+18 x+(-25+9 x) \log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))} \, dx\)

Optimal. Leaf size=20 \[ \frac {4 e^x (2+x)}{\log (\log (2+\log (25-9 x)))} \]

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Rubi [F]  time = 2.63, 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 {e^x (-72-36 x)+\left (e^x \left (-600+16 x+72 x^2\right )+e^x \left (-300+8 x+36 x^2\right ) \log (25-9 x)\right ) \log (2+\log (25-9 x)) \log (\log (2+\log (25-9 x)))}{(-50+18 x+(-25+9 x) \log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-72 - 36*x) + (E^x*(-600 + 16*x + 72*x^2) + E^x*(-300 + 8*x + 36*x^2)*Log[25 - 9*x])*Log[2 + Log[25
- 9*x]]*Log[Log[2 + Log[25 - 9*x]]])/((-50 + 18*x + (-25 + 9*x)*Log[25 - 9*x])*Log[2 + Log[25 - 9*x]]*Log[Log[
2 + Log[25 - 9*x]]]^2),x]

[Out]

-4*Defer[Int][E^x/((2 + Log[25 - 9*x])*Log[2 + Log[25 - 9*x]]*Log[Log[2 + Log[25 - 9*x]]]^2), x] - 172*Defer[I
nt][E^x/((-25 + 9*x)*(2 + Log[25 - 9*x])*Log[2 + Log[25 - 9*x]]*Log[Log[2 + Log[25 - 9*x]]]^2), x] + 12*Defer[
Int][E^x/Log[Log[2 + Log[25 - 9*x]]], x] + 4*Defer[Int][(E^x*x)/Log[Log[2 + Log[25 - 9*x]]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^x \left (\frac {9 (2+x)}{(2+\log (25-9 x)) \log (2+\log (25-9 x))}-\left (-75+2 x+9 x^2\right ) \log (\log (2+\log (25-9 x)))\right )}{(25-9 x) \log ^2(\log (2+\log (25-9 x)))} \, dx\\ &=4 \int \frac {e^x \left (\frac {9 (2+x)}{(2+\log (25-9 x)) \log (2+\log (25-9 x))}-\left (-75+2 x+9 x^2\right ) \log (\log (2+\log (25-9 x)))\right )}{(25-9 x) \log ^2(\log (2+\log (25-9 x)))} \, dx\\ &=4 \int \left (-\frac {9 e^x (2+x)}{(-25+9 x) (2+\log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))}+\frac {e^x (3+x)}{\log (\log (2+\log (25-9 x)))}\right ) \, dx\\ &=4 \int \frac {e^x (3+x)}{\log (\log (2+\log (25-9 x)))} \, dx-36 \int \frac {e^x (2+x)}{(-25+9 x) (2+\log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))} \, dx\\ &=4 \int \left (\frac {3 e^x}{\log (\log (2+\log (25-9 x)))}+\frac {e^x x}{\log (\log (2+\log (25-9 x)))}\right ) \, dx-36 \int \left (\frac {e^x}{9 (2+\log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))}+\frac {43 e^x}{9 (-25+9 x) (2+\log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))}\right ) \, dx\\ &=-\left (4 \int \frac {e^x}{(2+\log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))} \, dx\right )+4 \int \frac {e^x x}{\log (\log (2+\log (25-9 x)))} \, dx+12 \int \frac {e^x}{\log (\log (2+\log (25-9 x)))} \, dx-172 \int \frac {e^x}{(-25+9 x) (2+\log (25-9 x)) \log (2+\log (25-9 x)) \log ^2(\log (2+\log (25-9 x)))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 20, normalized size = 1.00 \begin {gather*} \frac {4 e^x (2+x)}{\log (\log (2+\log (25-9 x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-72 - 36*x) + (E^x*(-600 + 16*x + 72*x^2) + E^x*(-300 + 8*x + 36*x^2)*Log[25 - 9*x])*Log[2 + L
og[25 - 9*x]]*Log[Log[2 + Log[25 - 9*x]]])/((-50 + 18*x + (-25 + 9*x)*Log[25 - 9*x])*Log[2 + Log[25 - 9*x]]*Lo
g[Log[2 + Log[25 - 9*x]]]^2),x]

[Out]

(4*E^x*(2 + x))/Log[Log[2 + Log[25 - 9*x]]]

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fricas [A]  time = 0.72, size = 19, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (x + 2\right )} e^{x}}{\log \left (\log \left (\log \left (-9 \, x + 25\right ) + 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((36*x^2+8*x-300)*exp(x)*log(-9*x+25)+(72*x^2+16*x-600)*exp(x))*log(log(-9*x+25)+2)*log(log(log(-9*
x+25)+2))+(-36*x-72)*exp(x))/((9*x-25)*log(-9*x+25)+18*x-50)/log(log(-9*x+25)+2)/log(log(log(-9*x+25)+2))^2,x,
algorithm="fricas")

[Out]

4*(x + 2)*e^x/log(log(log(-9*x + 25) + 2))

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giac [A]  time = 0.78, size = 23, normalized size = 1.15 \begin {gather*} \frac {4 \, {\left (x e^{x} + 2 \, e^{x}\right )}}{\log \left (\log \left (\log \left (-9 \, x + 25\right ) + 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((36*x^2+8*x-300)*exp(x)*log(-9*x+25)+(72*x^2+16*x-600)*exp(x))*log(log(-9*x+25)+2)*log(log(log(-9*
x+25)+2))+(-36*x-72)*exp(x))/((9*x-25)*log(-9*x+25)+18*x-50)/log(log(-9*x+25)+2)/log(log(log(-9*x+25)+2))^2,x,
algorithm="giac")

[Out]

4*(x*e^x + 2*e^x)/log(log(log(-9*x + 25) + 2))

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maple [A]  time = 0.04, size = 20, normalized size = 1.00

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{x} \left (2+x \right )}{\ln \left (\ln \left (\ln \left (-9 x +25\right )+2\right )\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((36*x^2+8*x-300)*exp(x)*ln(-9*x+25)+(72*x^2+16*x-600)*exp(x))*ln(ln(-9*x+25)+2)*ln(ln(ln(-9*x+25)+2))+(-
36*x-72)*exp(x))/((9*x-25)*ln(-9*x+25)+18*x-50)/ln(ln(-9*x+25)+2)/ln(ln(ln(-9*x+25)+2))^2,x,method=_RETURNVERB
OSE)

[Out]

4*exp(x)*(2+x)/ln(ln(ln(-9*x+25)+2))

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maxima [A]  time = 0.84, size = 19, normalized size = 0.95 \begin {gather*} \frac {4 \, {\left (x + 2\right )} e^{x}}{\log \left (\log \left (\log \left (-9 \, x + 25\right ) + 2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((36*x^2+8*x-300)*exp(x)*log(-9*x+25)+(72*x^2+16*x-600)*exp(x))*log(log(-9*x+25)+2)*log(log(log(-9*
x+25)+2))+(-36*x-72)*exp(x))/((9*x-25)*log(-9*x+25)+18*x-50)/log(log(-9*x+25)+2)/log(log(log(-9*x+25)+2))^2,x,
algorithm="maxima")

[Out]

4*(x + 2)*e^x/log(log(log(-9*x + 25) + 2))

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mupad [B]  time = 1.22, size = 19, normalized size = 0.95 \begin {gather*} \frac {4\,{\mathrm {e}}^x\,\left (x+2\right )}{\ln \left (\ln \left (\ln \left (25-9\,x\right )+2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(36*x + 72) - log(log(25 - 9*x) + 2)*log(log(log(25 - 9*x) + 2))*(exp(x)*(16*x + 72*x^2 - 600) +
exp(x)*log(25 - 9*x)*(8*x + 36*x^2 - 300)))/(log(log(25 - 9*x) + 2)*log(log(log(25 - 9*x) + 2))^2*(18*x + log(
25 - 9*x)*(9*x - 25) - 50)),x)

[Out]

(4*exp(x)*(x + 2))/log(log(log(25 - 9*x) + 2))

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sympy [A]  time = 0.75, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (4 x + 8\right ) e^{x}}{\log {\left (\log {\left (\log {\left (25 - 9 x \right )} + 2 \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((36*x**2+8*x-300)*exp(x)*ln(-9*x+25)+(72*x**2+16*x-600)*exp(x))*ln(ln(-9*x+25)+2)*ln(ln(ln(-9*x+25
)+2))+(-36*x-72)*exp(x))/((9*x-25)*ln(-9*x+25)+18*x-50)/ln(ln(-9*x+25)+2)/ln(ln(ln(-9*x+25)+2))**2,x)

[Out]

(4*x + 8)*exp(x)/log(log(log(25 - 9*x) + 2))

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