3.13.96 \(\int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} (8 x-6 x^2+x^3)+(-8 x+2 x^2+e^{1+x} (-72+18 x)) \log (x)+(36-9 x) \log ^2(x)+(e^{2+2 x} (-48-9 x)-e^{1+x} x^2+(x^2+e^{1+x} (96+18 x)) \log (x)+(-48-9 x) \log ^2(x)) \log (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)})}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+(e^{1+x} (-96-18 x)-x^2) \log (x)+(48+9 x) \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ (4-x) \log \left (-4+\frac {1}{4} x \left (-3+\frac {x}{3 \left (-e^{1+x}+\log (x)\right )}\right )\right ) \]

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Rubi [F]  time = 40.03, 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^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2 + 2*x)*(36 - 9*x) + 4*x - x^2 + E^(1 + x)*(8*x - 6*x^2 + x^3) + (-8*x + 2*x^2 + E^(1 + x)*(-72 + 18*
x))*Log[x] + (36 - 9*x)*Log[x]^2 + (E^(2 + 2*x)*(-48 - 9*x) - E^(1 + x)*x^2 + (x^2 + E^(1 + x)*(96 + 18*x))*Lo
g[x] + (-48 - 9*x)*Log[x]^2)*Log[(x^2 + E^(1 + x)*(48 + 9*x) + (-48 - 9*x)*Log[x])/(-12*E^(1 + x) + 12*Log[x])
])/(E^(1 + x)*x^2 + E^(2 + 2*x)*(48 + 9*x) + (E^(1 + x)*(-96 - 18*x) - x^2)*Log[x] + (48 + 9*x)*Log[x]^2),x]

[Out]

4*Log[16 + 3*x] - x*Log[-1/12*(x^2 + 3*E^(1 + x)*(16 + 3*x) - 3*(16 + 3*x)*Log[x])/(E^(1 + x) - Log[x])] + 4*D
efer[Int][1/(x*(E^(1 + x) - Log[x])), x] - 4*Defer[Int][Log[x]/(E^(1 + x) - Log[x]), x] - (44*Defer[Int][(48*E
^(1 + x) + 9*E^(1 + x)*x + x^2 - 48*Log[x] - 9*x*Log[x])^(-1), x])/3 - 192*Defer[Int][1/(x*(48*E^(1 + x) + 9*E
^(1 + x)*x + x^2 - 48*Log[x] - 9*x*Log[x])), x] + 4*Defer[Int][x/(48*E^(1 + x) + 9*E^(1 + x)*x + x^2 - 48*Log[
x] - 9*x*Log[x]), x] - 4*Defer[Int][x^2/(48*E^(1 + x) + 9*E^(1 + x)*x + x^2 - 48*Log[x] - 9*x*Log[x]), x] - (1
024*Defer[Int][1/((16 + 3*x)*(48*E^(1 + x) + 9*E^(1 + x)*x + x^2 - 48*Log[x] - 9*x*Log[x])), x])/3 + 36*Defer[
Int][(x*Log[x])/(48*E^(1 + x) + 9*E^(1 + x)*x + x^2 - 48*Log[x] - 9*x*Log[x]), x] - 192*Defer[Int][Log[x]/(-48
*E^(1 + x) - 9*E^(1 + x)*x - x^2 + 48*Log[x] + 9*x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {(-4+x) (-1+x \log (x))}{x \left (e^{1+x}-\log (x)\right )}+\frac {(-4+x) \left (768+288 x-5 x^2+13 x^3+3 x^4-768 x \log (x)-288 x^2 \log (x)-27 x^3 \log (x)\right )}{x (16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}+\frac {12-3 x-16 \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right )-3 x \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right )}{16+3 x}\right ) \, dx\\ &=\int \frac {(-4+x) (-1+x \log (x))}{x \left (e^{1+x}-\log (x)\right )} \, dx+\int \frac {(-4+x) \left (768+288 x-5 x^2+13 x^3+3 x^4-768 x \log (x)-288 x^2 \log (x)-27 x^3 \log (x)\right )}{x (16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )} \, dx+\int \frac {12-3 x-16 \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right )-3 x \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right )}{16+3 x} \, dx\\ &=\int \left (\frac {-1+x \log (x)}{e^{1+x}-\log (x)}-\frac {4 (-1+x \log (x))}{x \left (e^{1+x}-\log (x)\right )}\right ) \, dx+\int \left (-\frac {768+288 x-5 x^2+13 x^3+3 x^4-768 x \log (x)-288 x^2 \log (x)-27 x^3 \log (x)}{4 x \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}+\frac {7 \left (768+288 x-5 x^2+13 x^3+3 x^4-768 x \log (x)-288 x^2 \log (x)-27 x^3 \log (x)\right )}{4 (16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}\right ) \, dx+\int \left (-\frac {3 (-4+x)}{16+3 x}-\log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right )\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {768+288 x-5 x^2+13 x^3+3 x^4-768 x \log (x)-288 x^2 \log (x)-27 x^3 \log (x)}{x \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )} \, dx\right )+\frac {7}{4} \int \frac {768+288 x-5 x^2+13 x^3+3 x^4-768 x \log (x)-288 x^2 \log (x)-27 x^3 \log (x)}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )} \, dx-3 \int \frac {-4+x}{16+3 x} \, dx-4 \int \frac {-1+x \log (x)}{x \left (e^{1+x}-\log (x)\right )} \, dx+\int \frac {-1+x \log (x)}{e^{1+x}-\log (x)} \, dx-\int \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right ) \, dx\\ &=-x \log \left (-\frac {x^2+3 e^{1+x} (16+3 x)-3 (16+3 x) \log (x)}{12 \left (e^{1+x}-\log (x)\right )}\right )-\frac {1}{4} \int \left (\frac {288}{48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)}+\frac {768}{x \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}-\frac {5 x}{48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)}+\frac {13 x^2}{48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)}+\frac {3 x^3}{48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)}-\frac {288 x \log (x)}{48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)}-\frac {27 x^2 \log (x)}{48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)}+\frac {768 \log (x)}{-48 e^{1+x}-9 e^{1+x} x-x^2+48 \log (x)+9 x \log (x)}\right ) \, dx+\frac {7}{4} \int \left (\frac {768}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}+\frac {288 x}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}-\frac {5 x^2}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}+\frac {13 x^3}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}+\frac {3 x^4}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}-\frac {768 x \log (x)}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}-\frac {288 x^2 \log (x)}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}-\frac {27 x^3 \log (x)}{(16+3 x) \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )}\right ) \, dx-3 \int \left (\frac {1}{3}-\frac {28}{3 (16+3 x)}\right ) \, dx-4 \int \left (-\frac {1}{x \left (e^{1+x}-\log (x)\right )}+\frac {\log (x)}{e^{1+x}-\log (x)}\right ) \, dx+\int \left (-\frac {1}{e^{1+x}-\log (x)}+\frac {x \log (x)}{e^{1+x}-\log (x)}\right ) \, dx+\int \frac {x \left (9 e^{2+2 x}+x-e^{1+x} (-2+x) x-2 \left (9 e^{1+x}+x\right ) \log (x)+9 \log ^2(x)\right )}{\left (e^{1+x}-\log (x)\right ) \left (x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 5.62, size = 89, normalized size = 2.70 \begin {gather*} -4 \log \left (e^{1+x}-\log (x)\right )+4 \log \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )-x \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + 2*x)*(36 - 9*x) + 4*x - x^2 + E^(1 + x)*(8*x - 6*x^2 + x^3) + (-8*x + 2*x^2 + E^(1 + x)*(-72
+ 18*x))*Log[x] + (36 - 9*x)*Log[x]^2 + (E^(2 + 2*x)*(-48 - 9*x) - E^(1 + x)*x^2 + (x^2 + E^(1 + x)*(96 + 18*
x))*Log[x] + (-48 - 9*x)*Log[x]^2)*Log[(x^2 + E^(1 + x)*(48 + 9*x) + (-48 - 9*x)*Log[x])/(-12*E^(1 + x) + 12*L
og[x])])/(E^(1 + x)*x^2 + E^(2 + 2*x)*(48 + 9*x) + (E^(1 + x)*(-96 - 18*x) - x^2)*Log[x] + (48 + 9*x)*Log[x]^2
),x]

[Out]

-4*Log[E^(1 + x) - Log[x]] + 4*Log[48*E^(1 + x) + 9*E^(1 + x)*x + x^2 - 48*Log[x] - 9*x*Log[x]] - x*Log[(x^2 +
E^(1 + x)*(48 + 9*x) - 3*(16 + 3*x)*Log[x])/(12*(-E^(1 + x) + Log[x]))]

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fricas [A]  time = 0.69, size = 43, normalized size = 1.30 \begin {gather*} -{\left (x - 4\right )} \log \left (-\frac {x^{2} + 3 \, {\left (3 \, x + 16\right )} e^{\left (x + 1\right )} - 3 \, {\left (3 \, x + 16\right )} \log \relax (x)}{12 \, {\left (e^{\left (x + 1\right )} - \log \relax (x)\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-48)*log(x)^2+((18*x+96)*exp(x+1)+x^2)*log(x)+(-9*x-48)*exp(x+1)^2-x^2*exp(x+1))*log(((-9*x-4
8)*log(x)+(9*x+48)*exp(x+1)+x^2)/(12*log(x)-12*exp(x+1)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(x+1)+2*x^2-8*x)*lo
g(x)+(-9*x+36)*exp(x+1)^2+(x^3-6*x^2+8*x)*exp(x+1)-x^2+4*x)/((9*x+48)*log(x)^2+((-18*x-96)*exp(x+1)-x^2)*log(x
)+(9*x+48)*exp(x+1)^2+x^2*exp(x+1)),x, algorithm="fricas")

[Out]

-(x - 4)*log(-1/12*(x^2 + 3*(3*x + 16)*e^(x + 1) - 3*(3*x + 16)*log(x))/(e^(x + 1) - log(x)))

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giac [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((((-9*x-48)*log(x)^2+((18*x+96)*exp(x+1)+x^2)*log(x)+(-9*x-48)*exp(x+1)^2-x^2*exp(x+1))*log(((-9*x-4
8)*log(x)+(9*x+48)*exp(x+1)+x^2)/(12*log(x)-12*exp(x+1)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(x+1)+2*x^2-8*x)*lo
g(x)+(-9*x+36)*exp(x+1)^2+(x^3-6*x^2+8*x)*exp(x+1)-x^2+4*x)/((9*x+48)*log(x)^2+((-18*x-96)*exp(x+1)-x^2)*log(x
)+(9*x+48)*exp(x+1)^2+x^2*exp(x+1)),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.26, size = 467, normalized size = 14.15

method	result	size
risch	\(-x \ln \left (x^{2}+\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x +48 \,{\mathrm e}^{x +1}-48 \ln \relax (x )\right )+x \ln \left (-\ln \relax (x )+{\mathrm e}^{x +1}\right )+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )-{\mathrm e}^{x +1}}\right ) \mathrm {csgn}\left (i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )}{\ln \relax (x )-{\mathrm e}^{x +1}}\right )}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )}{\ln \relax (x )-{\mathrm e}^{x +1}}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )}{\ln \relax (x )-{\mathrm e}^{x +1}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )-{\mathrm e}^{x +1}}\right ) \mathrm {csgn}\left (\frac {i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )}{\ln \relax (x )-{\mathrm e}^{x +1}}\right )^{2}}{2}+i \pi x \mathrm {csgn}\left (\frac {i \left (-x^{2}-\left (9 \,{\mathrm e}^{x +1}-9 \ln \relax (x )\right ) x -48 \,{\mathrm e}^{x +1}+48 \ln \relax (x )\right )}{\ln \relax (x )-{\mathrm e}^{x +1}}\right )^{2}-i \pi x +2 x \ln \relax (2)+x \ln \relax (3)+4 \ln \left (3 x +16\right )+4 \ln \left (\ln \relax (x )-\frac {x^{2}+9 x \,{\mathrm e}^{x +1}+48 \,{\mathrm e}^{x +1}}{3 \left (3 x +16\right )}\right )-4 \ln \left (\ln \relax (x )-{\mathrm e}^{x +1}\right )\)	\(467\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-9*x-48)*ln(x)^2+((18*x+96)*exp(x+1)+x^2)*ln(x)+(-9*x-48)*exp(x+1)^2-x^2*exp(x+1))*ln(((-9*x-48)*ln(x)+
(9*x+48)*exp(x+1)+x^2)/(12*ln(x)-12*exp(x+1)))+(-9*x+36)*ln(x)^2+((18*x-72)*exp(x+1)+2*x^2-8*x)*ln(x)+(-9*x+36
)*exp(x+1)^2+(x^3-6*x^2+8*x)*exp(x+1)-x^2+4*x)/((9*x+48)*ln(x)^2+((-18*x-96)*exp(x+1)-x^2)*ln(x)+(9*x+48)*exp(
x+1)^2+x^2*exp(x+1)),x,method=_RETURNVERBOSE)

[Out]

-x*ln(x^2+(9*exp(x+1)-9*ln(x))*x+48*exp(x+1)-48*ln(x))+x*ln(-ln(x)+exp(x+1))+1/2*I*Pi*x*csgn(I/(ln(x)-exp(x+1)
))*csgn(I*(-x^2-(9*exp(x+1)-9*ln(x))*x-48*exp(x+1)+48*ln(x)))*csgn(I/(ln(x)-exp(x+1))*(-x^2-(9*exp(x+1)-9*ln(x
))*x-48*exp(x+1)+48*ln(x)))-1/2*I*Pi*x*csgn(I/(ln(x)-exp(x+1))*(-x^2-(9*exp(x+1)-9*ln(x))*x-48*exp(x+1)+48*ln(
x)))^3+1/2*I*Pi*x*csgn(I*(-x^2-(9*exp(x+1)-9*ln(x))*x-48*exp(x+1)+48*ln(x)))*csgn(I/(ln(x)-exp(x+1))*(-x^2-(9*
exp(x+1)-9*ln(x))*x-48*exp(x+1)+48*ln(x)))^2+1/2*I*Pi*x*csgn(I/(ln(x)-exp(x+1)))*csgn(I/(ln(x)-exp(x+1))*(-x^2
-(9*exp(x+1)-9*ln(x))*x-48*exp(x+1)+48*ln(x)))^2+I*Pi*x*csgn(I/(ln(x)-exp(x+1))*(-x^2-(9*exp(x+1)-9*ln(x))*x-4
8*exp(x+1)+48*ln(x)))^2-I*Pi*x+2*x*ln(2)+x*ln(3)+4*ln(3*x+16)+4*ln(ln(x)-1/3*(x^2+9*x*exp(x+1)+48*exp(x+1))/(3
*x+16))-4*ln(ln(x)-exp(x+1))

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maxima [B]  time = 0.81, size = 120, normalized size = 3.64 \begin {gather*} x {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} - x \log \left (x^{2} + 3 \, {\left (3 \, x e + 16 \, e\right )} e^{x} - 3 \, {\left (3 \, x + 16\right )} \log \relax (x)\right ) + x \log \left (-e^{\left (x + 1\right )} + \log \relax (x)\right ) - 4 \, \log \left ({\left (e^{\left (x + 1\right )} - \log \relax (x)\right )} e^{\left (-1\right )}\right ) + 4 \, \log \left (3 \, x + 16\right ) + 4 \, \log \left (\frac {x^{2} + 3 \, {\left (3 \, x e + 16 \, e\right )} e^{x} - 3 \, {\left (3 \, x + 16\right )} \log \relax (x)}{3 \, {\left (3 \, x e + 16 \, e\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-48)*log(x)^2+((18*x+96)*exp(x+1)+x^2)*log(x)+(-9*x-48)*exp(x+1)^2-x^2*exp(x+1))*log(((-9*x-4
8)*log(x)+(9*x+48)*exp(x+1)+x^2)/(12*log(x)-12*exp(x+1)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(x+1)+2*x^2-8*x)*lo
g(x)+(-9*x+36)*exp(x+1)^2+(x^3-6*x^2+8*x)*exp(x+1)-x^2+4*x)/((9*x+48)*log(x)^2+((-18*x-96)*exp(x+1)-x^2)*log(x
)+(9*x+48)*exp(x+1)^2+x^2*exp(x+1)),x, algorithm="maxima")

[Out]

x*(log(3) + 2*log(2)) - x*log(x^2 + 3*(3*x*e + 16*e)*e^x - 3*(3*x + 16)*log(x)) + x*log(-e^(x + 1) + log(x)) -
4*log((e^(x + 1) - log(x))*e^(-1)) + 4*log(3*x + 16) + 4*log(1/3*(x^2 + 3*(3*x*e + 16*e)*e^x - 3*(3*x + 16)*l
og(x))/(3*x*e + 16*e))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{2\,x+2}\,\left (9\,x-36\right )-4\,x+\ln \left (-\frac {{\mathrm {e}}^{x+1}\,\left (9\,x+48\right )-\ln \relax (x)\,\left (9\,x+48\right )+x^2}{12\,{\mathrm {e}}^{x+1}-12\,\ln \relax (x)}\right )\,\left ({\mathrm {e}}^{2\,x+2}\,\left (9\,x+48\right )-\ln \relax (x)\,\left ({\mathrm {e}}^{x+1}\,\left (18\,x+96\right )+x^2\right )+x^2\,{\mathrm {e}}^{x+1}+{\ln \relax (x)}^2\,\left (9\,x+48\right )\right )-\ln \relax (x)\,\left ({\mathrm {e}}^{x+1}\,\left (18\,x-72\right )-8\,x+2\,x^2\right )+x^2+{\ln \relax (x)}^2\,\left (9\,x-36\right )-{\mathrm {e}}^{x+1}\,\left (x^3-6\,x^2+8\,x\right )}{{\mathrm {e}}^{2\,x+2}\,\left (9\,x+48\right )-\ln \relax (x)\,\left ({\mathrm {e}}^{x+1}\,\left (18\,x+96\right )+x^2\right )+x^2\,{\mathrm {e}}^{x+1}+{\ln \relax (x)}^2\,\left (9\,x+48\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x + 2)*(9*x - 36) - 4*x + log(-(exp(x + 1)*(9*x + 48) - log(x)*(9*x + 48) + x^2)/(12*exp(x + 1) -
12*log(x)))*(exp(2*x + 2)*(9*x + 48) - log(x)*(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x
+ 48)) - log(x)*(exp(x + 1)*(18*x - 72) - 8*x + 2*x^2) + x^2 + log(x)^2*(9*x - 36) - exp(x + 1)*(8*x - 6*x^2 +
x^3))/(exp(2*x + 2)*(9*x + 48) - log(x)*(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x + 48)
),x)

[Out]

int(-(exp(2*x + 2)*(9*x - 36) - 4*x + log(-(exp(x + 1)*(9*x + 48) - log(x)*(9*x + 48) + x^2)/(12*exp(x + 1) -
12*log(x)))*(exp(2*x + 2)*(9*x + 48) - log(x)*(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x
+ 48)) - log(x)*(exp(x + 1)*(18*x - 72) - 8*x + 2*x^2) + x^2 + log(x)^2*(9*x - 36) - exp(x + 1)*(8*x - 6*x^2 +
x^3))/(exp(2*x + 2)*(9*x + 48) - log(x)*(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x + 48)
), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-9*x-48)*ln(x)**2+((18*x+96)*exp(x+1)+x**2)*ln(x)+(-9*x-48)*exp(x+1)**2-x**2*exp(x+1))*ln(((-9*x-
48)*ln(x)+(9*x+48)*exp(x+1)+x**2)/(12*ln(x)-12*exp(x+1)))+(-9*x+36)*ln(x)**2+((18*x-72)*exp(x+1)+2*x**2-8*x)*l
n(x)+(-9*x+36)*exp(x+1)**2+(x**3-6*x**2+8*x)*exp(x+1)-x**2+4*x)/((9*x+48)*ln(x)**2+((-18*x-96)*exp(x+1)-x**2)*
ln(x)+(9*x+48)*exp(x+1)**2+x**2*exp(x+1)),x)

[Out]

Exception raised: PolynomialError

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