3.74.91 \(\int \frac {64^x \log ^{-1+x}(25+50 x+25 x^2+e^{e^x} (-10-20 x-10 x^2)+e^{2 e^x} (1+2 x+x^2)) (-10 x+e^{e^x} (2 x+e^x (2 x+2 x^2))+(-5-5 x+e^{e^x} (1+x)) \log (25+50 x+25 x^2+e^{e^x} (-10-20 x-10 x^2)+e^{2 e^x} (1+2 x+x^2)) \log (64 \log (25+50 x+25 x^2+e^{e^x} (-10-20 x-10 x^2)+e^{2 e^x} (1+2 x+x^2))))}{-5-5 x+e^{e^x} (1+x)} \, dx\)

Optimal. Leaf size=26 \[ 2+64^x \log ^x\left (\left (5-e^{e^x}\right )^2 (1+x)^2\right ) \]

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Rubi [F]  time = 5.79, 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 {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(64^x*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]^(-1 + x)*(-10*x +
E^E^x*(2*x + E^x*(2*x + 2*x^2)) + (-5 - 5*x + E^E^x*(1 + x))*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x
^2) + E^(2*E^x)*(1 + 2*x + x^2)]*Log[64*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 +
2*x + x^2)]]))/(-5 - 5*x + E^E^x*(1 + x)),x]

[Out]

Defer[Int][2^(1 + 6*x)*Log[(-5 + E^E^x)^2*(1 + x)^2]^(-1 + x), x] + Defer[Int][(2^(1 + 6*x)*Log[(-5 + E^E^x)^2
*(1 + x)^2]^(-1 + x))/(-1 - x), x] + Defer[Int][(2^(1 + 6*x)*E^(E^x + x)*x*Log[(-5 + E^E^x)^2*(1 + x)^2]^(-1 +
x))/(-5 + E^E^x), x] + Defer[Int][64^x*Log[(-5 + E^E^x)^2*(1 + x)^2]^x*Log[64*Log[(-5 + E^E^x)^2*(1 + x)^2]],
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (10 x-2 e^{e^x} x \left (1+e^x (1+x)\right )-\left (-5+e^{e^x}\right ) (1+x) \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{\left (5-e^{e^x}\right ) (1+x)} \, dx\\ &=\int \left (\frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}}+\frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (2 x+\log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )+x \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{1+x}\right ) \, dx\\ &=\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (2 x+\log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )+x \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{1+x} \, dx\\ &=\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \frac {64^x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \left (2 x+(1+x) \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right )}{1+x} \, dx\\ &=\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \left (\frac {2^{1+6 x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{1+x}+64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right )\right ) \, dx\\ &=\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \frac {2^{1+6 x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{1+x} \, dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right ) \, dx\\ &=\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int \left (2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )+\frac {2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-1-x}\right ) \, dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right ) \, dx\\ &=\int 2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \, dx+\int \frac {2^{1+6 x} \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-1-x} \, dx+\int \frac {2^{1+6 x} e^{e^x+x} x \log ^{-1+x}\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )}{-5+e^{e^x}} \, dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.98, size = 22, normalized size = 0.85 \begin {gather*} 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (1+x)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64^x*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]^(-1 + x)*(-1
0*x + E^E^x*(2*x + E^x*(2*x + 2*x^2)) + (-5 - 5*x + E^E^x*(1 + x))*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x
- 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]*Log[64*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)
*(1 + 2*x + x^2)]]))/(-5 - 5*x + E^E^x*(1 + x)),x]

[Out]

64^x*Log[(-5 + E^E^x)^2*(1 + x)^2]^x

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fricas [B]  time = 0.90, size = 42, normalized size = 1.62 \begin {gather*} \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right )^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(exp(x))-5*x-5)*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+2
5)*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x
)*exp(exp(x))-10*x)*exp(x*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)))
/((x+1)*exp(exp(x))-5*x-5)/log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x, algo
rithm="fricas")

[Out]

(64*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x + 25))^x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left ({\left (x + 1\right )} e^{\left (e^{x}\right )} - 5 \, x - 5\right )} \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right ) \log \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right ) + 2 \, {\left ({\left (x^{2} + x\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} - 10 \, x\right )} \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right )^{x}}{{\left ({\left (x + 1\right )} e^{\left (e^{x}\right )} - 5 \, x - 5\right )} \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(exp(x))-5*x-5)*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+2
5)*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x
)*exp(exp(x))-10*x)*exp(x*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)))
/((x+1)*exp(exp(x))-5*x-5)/log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x, algo
rithm="giac")

[Out]

integrate((((x + 1)*e^(e^x) - 5*x - 5)*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 5
0*x + 25)*log(64*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x + 25)) + 2*((x^2 +
x)*e^x + x)*e^(e^x) - 10*x)*(64*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x +
25))^x/(((x + 1)*e^(e^x) - 5*x - 5)*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x
+ 25)), x)

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maple [C]  time = 0.34, size = 170, normalized size = 6.54

method	result	size
risch	\(\left (128 \ln \left (x +1\right )+128 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )-32 i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\mathrm {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )\right )^{2}-32 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left (x +1\right )^{2}\right )+\mathrm {csgn}\left (i \left (x +1\right )\right )\right )^{2}-32 i \pi \,\mathrm {csgn}\left (i \left (x +1\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left (x +1\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\mathrm {csgn}\left (i \left (x +1\right )^{2}\right )\right ) \left (-\mathrm {csgn}\left (i \left (x +1\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\mathrm {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )\right )\right )^{x}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(exp(x))-5*x-5)*ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)*ln(6
4*ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x)*exp(exp
(x))-10*x)*exp(x*ln(64*ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)))/((x+1)*exp
(exp(x))-5*x-5)/ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x,method=_RETURNVER
BOSE)

[Out]

(128*ln(x+1)+128*ln(exp(exp(x))-5)-32*I*Pi*csgn(I*(exp(exp(x))-5)^2)*(-csgn(I*(exp(exp(x))-5)^2)+csgn(I*(exp(e
xp(x))-5)))^2-32*I*Pi*csgn(I*(x+1)^2)*(-csgn(I*(x+1)^2)+csgn(I*(x+1)))^2-32*I*Pi*csgn(I*(x+1)^2*(exp(exp(x))-5
)^2)*(-csgn(I*(x+1)^2*(exp(exp(x))-5)^2)+csgn(I*(x+1)^2))*(-csgn(I*(x+1)^2*(exp(exp(x))-5)^2)+csgn(I*(exp(exp(
x))-5)^2)))^x

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maxima [A]  time = 0.81, size = 21, normalized size = 0.81 \begin {gather*} e^{\left (7 \, x \log \relax (2) + x \log \left (\log \left (x + 1\right ) + \log \left (e^{\left (e^{x}\right )} - 5\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(exp(x))-5*x-5)*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+2
5)*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x
)*exp(exp(x))-10*x)*exp(x*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)))
/((x+1)*exp(exp(x))-5*x-5)/log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x, algo
rithm="maxima")

[Out]

e^(7*x*log(2) + x*log(log(x + 1) + log(e^(e^x) - 5)))

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mupad [B]  time = 4.96, size = 60, normalized size = 2.31 \begin {gather*} 2^{6\,x}\,{\ln \left (50\,x-10\,{\mathrm {e}}^{{\mathrm {e}}^x}+{\mathrm {e}}^{2\,{\mathrm {e}}^x}-20\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+2\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-10\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}+25\,x^2+x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+25\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*log(64*log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x))*(20*x + 10*x^2 + 10) + 25*x^2 + 25)))
*(10*x - exp(exp(x))*(2*x + exp(x)*(2*x + 2*x^2)) + log(64*log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(
x))*(20*x + 10*x^2 + 10) + 25*x^2 + 25))*log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x))*(20*x + 10*x^2
+ 10) + 25*x^2 + 25)*(5*x - exp(exp(x))*(x + 1) + 5)))/(log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x)
)*(20*x + 10*x^2 + 10) + 25*x^2 + 25)*(5*x - exp(exp(x))*(x + 1) + 5)),x)

[Out]

2^(6*x)*log(50*x - 10*exp(exp(x)) + exp(2*exp(x)) - 20*x*exp(exp(x)) + 2*x*exp(2*exp(x)) - 10*x^2*exp(exp(x))
+ 25*x^2 + x^2*exp(2*exp(x)) + 25)^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((((x+1)*exp(exp(x))-5*x-5)*ln((x**2+2*x+1)*exp(exp(x))**2+(-10*x**2-20*x-10)*exp(exp(x))+25*x**2+50*
x+25)*ln(64*ln((x**2+2*x+1)*exp(exp(x))**2+(-10*x**2-20*x-10)*exp(exp(x))+25*x**2+50*x+25))+((2*x**2+2*x)*exp(
x)+2*x)*exp(exp(x))-10*x)*exp(x*ln(64*ln((x**2+2*x+1)*exp(exp(x))**2+(-10*x**2-20*x-10)*exp(exp(x))+25*x**2+50
*x+25)))/((x+1)*exp(exp(x))-5*x-5)/ln((x**2+2*x+1)*exp(exp(x))**2+(-10*x**2-20*x-10)*exp(exp(x))+25*x**2+50*x+
25),x)

[Out]

Timed out

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