3.32.53 \(\int \frac {-2 x^4+2 e^{1+x} x^4+(12 x^3-4 e^{1+x} x^3+4 x^4) \log (-3+e^{1+x}-x)+(e^{1+2 x}+e^x (-3-x)) \log ^3(-3+e^{1+x}-x)}{(-3+e^{1+x}-x) \log ^3(-3+e^{1+x}-x)} \, dx\)

Optimal. Leaf size=22 \[ e^x-\frac {x^4}{\log ^2\left (-3+e^{1+x}-x\right )} \]

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Rubi [F]  time = 2.01, 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 {-2 x^4+2 e^{1+x} x^4+\left (12 x^3-4 e^{1+x} x^3+4 x^4\right ) \log \left (-3+e^{1+x}-x\right )+\left (e^{1+2 x}+e^x (-3-x)\right ) \log ^3\left (-3+e^{1+x}-x\right )}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x^4 + 2*E^(1 + x)*x^4 + (12*x^3 - 4*E^(1 + x)*x^3 + 4*x^4)*Log[-3 + E^(1 + x) - x] + (E^(1 + 2*x) + E^
x*(-3 - x))*Log[-3 + E^(1 + x) - x]^3)/((-3 + E^(1 + x) - x)*Log[-3 + E^(1 + x) - x]^3),x]

[Out]

E^x + 2*Defer[Int][x^4/Log[-3 + E^(1 + x) - x]^3, x] + 4*Defer[Int][x^4/((-3 + E^(1 + x) - x)*Log[-3 + E^(1 +
x) - x]^3), x] + 2*Defer[Int][x^5/((-3 + E^(1 + x) - x)*Log[-3 + E^(1 + x) - x]^3), x] - 4*Defer[Int][x^3/Log[
-3 + E^(1 + x) - x]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x+\frac {2 \left (-1+e^{1+x}\right ) x^4}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )}-\frac {4 x^3}{\log ^2\left (-3+e^{1+x}-x\right )}\right ) \, dx\\ &=2 \int \frac {\left (-1+e^{1+x}\right ) x^4}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )} \, dx-4 \int \frac {x^3}{\log ^2\left (-3+e^{1+x}-x\right )} \, dx+\int e^x \, dx\\ &=e^x+2 \int \left (\frac {x^4}{\log ^3\left (-3+e^{1+x}-x\right )}+\frac {x^4 (2+x)}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )}\right ) \, dx-4 \int \frac {x^3}{\log ^2\left (-3+e^{1+x}-x\right )} \, dx\\ &=e^x+2 \int \frac {x^4}{\log ^3\left (-3+e^{1+x}-x\right )} \, dx+2 \int \frac {x^4 (2+x)}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )} \, dx-4 \int \frac {x^3}{\log ^2\left (-3+e^{1+x}-x\right )} \, dx\\ &=e^x+2 \int \left (\frac {2 x^4}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )}+\frac {x^5}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )}\right ) \, dx+2 \int \frac {x^4}{\log ^3\left (-3+e^{1+x}-x\right )} \, dx-4 \int \frac {x^3}{\log ^2\left (-3+e^{1+x}-x\right )} \, dx\\ &=e^x+2 \int \frac {x^4}{\log ^3\left (-3+e^{1+x}-x\right )} \, dx+2 \int \frac {x^5}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )} \, dx+4 \int \frac {x^4}{\left (-3+e^{1+x}-x\right ) \log ^3\left (-3+e^{1+x}-x\right )} \, dx-4 \int \frac {x^3}{\log ^2\left (-3+e^{1+x}-x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 22, normalized size = 1.00 \begin {gather*} e^x-\frac {x^4}{\log ^2\left (-3+e^{1+x}-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^4 + 2*E^(1 + x)*x^4 + (12*x^3 - 4*E^(1 + x)*x^3 + 4*x^4)*Log[-3 + E^(1 + x) - x] + (E^(1 + 2*x
) + E^x*(-3 - x))*Log[-3 + E^(1 + x) - x]^3)/((-3 + E^(1 + x) - x)*Log[-3 + E^(1 + x) - x]^3),x]

[Out]

E^x - x^4/Log[-3 + E^(1 + x) - x]^2

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fricas [B]  time = 0.52, size = 41, normalized size = 1.86 \begin {gather*} -\frac {{\left (x^{4} e - e^{\left (x + 1\right )} \log \left (-x + e^{\left (x + 1\right )} - 3\right )^{2}\right )} e^{\left (-1\right )}}{\log \left (-x + e^{\left (x + 1\right )} - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(x+1)+(-3-x)*exp(x))*log(exp(x+1)-3-x)^3+(-4*x^3*exp(x+1)+4*x^4+12*x^3)*log(exp(x+1)-3-x
)+2*x^4*exp(x+1)-2*x^4)/(exp(x+1)-3-x)/log(exp(x+1)-3-x)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.35, size = 34, normalized size = 1.55 \begin {gather*} -\frac {x^{4} - e^{x} \log \left (-x + e^{\left (x + 1\right )} - 3\right )^{2}}{\log \left (-x + e^{\left (x + 1\right )} - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(x+1)+(-3-x)*exp(x))*log(exp(x+1)-3-x)^3+(-4*x^3*exp(x+1)+4*x^4+12*x^3)*log(exp(x+1)-3-x
)+2*x^4*exp(x+1)-2*x^4)/(exp(x+1)-3-x)/log(exp(x+1)-3-x)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 21, normalized size = 0.95

method	result	size
risch	\({\mathrm e}^{x}-\frac {x^{4}}{\ln \left ({\mathrm e}^{x +1}-3-x \right )^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*exp(x+1)+(-3-x)*exp(x))*ln(exp(x+1)-3-x)^3+(-4*x^3*exp(x+1)+4*x^4+12*x^3)*ln(exp(x+1)-3-x)+2*x^4*
exp(x+1)-2*x^4)/(exp(x+1)-3-x)/ln(exp(x+1)-3-x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 34, normalized size = 1.55 \begin {gather*} -\frac {x^{4} - e^{x} \log \left (-x + e^{\left (x + 1\right )} - 3\right )^{2}}{\log \left (-x + e^{\left (x + 1\right )} - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(x+1)+(-3-x)*exp(x))*log(exp(x+1)-3-x)^3+(-4*x^3*exp(x+1)+4*x^4+12*x^3)*log(exp(x+1)-3-x
)+2*x^4*exp(x+1)-2*x^4)/(exp(x+1)-3-x)/log(exp(x+1)-3-x)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.64, size = 274, normalized size = 12.45 \begin {gather*} {\mathrm {e}}^x-\frac {x^4+\frac {2\,x^3\,\ln \left (\mathrm {e}\,{\mathrm {e}}^x-x-3\right )\,\left (x-{\mathrm {e}}^{x+1}+3\right )}{{\mathrm {e}}^{x+1}-1}}{{\ln \left (\mathrm {e}\,{\mathrm {e}}^x-x-3\right )}^2}+\frac {\frac {2\,x^3\,\left (x-{\mathrm {e}}^{x+1}+3\right )}{{\mathrm {e}}^{x+1}-1}-\frac {2\,x^2\,\ln \left (\mathrm {e}\,{\mathrm {e}}^x-x-3\right )\,\left (x-{\mathrm {e}}^{x+1}+3\right )\,\left (4\,x-12\,{\mathrm {e}}^{x+1}+3\,{\mathrm {e}}^{2\,x+2}-2\,x\,{\mathrm {e}}^{x+1}+x^2\,{\mathrm {e}}^{x+1}+9\right )}{{\left ({\mathrm {e}}^{x+1}-1\right )}^3}}{\ln \left (\mathrm {e}\,{\mathrm {e}}^x-x-3\right )}-6\,x^2-\frac {2\,{\mathrm {e}}^{-1}\,\left (-x^4+5\,x^3+12\,x^2\right )}{{\mathrm {e}}^{-1}-{\mathrm {e}}^x}-\frac {2\,{\mathrm {e}}^{-2}\,\left (-x^5+x^4+12\,x^3+12\,x^2\right )}{{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^{x-1}+{\mathrm {e}}^{-2}}+\frac {2\,{\mathrm {e}}^{-3}\,\left (x^5+4\,x^4+4\,x^3\right )}{{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^{x-2}-{\mathrm {e}}^{-3}-3\,{\mathrm {e}}^{2\,x-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^4*exp(x + 1) + log(exp(x + 1) - x - 3)*(12*x^3 - 4*x^3*exp(x + 1) + 4*x^4) - log(exp(x + 1) - x - 3)
^3*(exp(x)*(x + 3) - exp(x + 1)*exp(x)) - 2*x^4)/(log(exp(x + 1) - x - 3)^3*(x - exp(x + 1) + 3)),x)

[Out]

exp(x) - (x^4 + (2*x^3*log(exp(1)*exp(x) - x - 3)*(x - exp(x + 1) + 3))/(exp(x + 1) - 1))/log(exp(1)*exp(x) -
x - 3)^2 + ((2*x^3*(x - exp(x + 1) + 3))/(exp(x + 1) - 1) - (2*x^2*log(exp(1)*exp(x) - x - 3)*(x - exp(x + 1)
+ 3)*(4*x - 12*exp(x + 1) + 3*exp(2*x + 2) - 2*x*exp(x + 1) + x^2*exp(x + 1) + 9))/(exp(x + 1) - 1)^3)/log(exp
(1)*exp(x) - x - 3) - 6*x^2 - (2*exp(-1)*(12*x^2 + 5*x^3 - x^4))/(exp(-1) - exp(x)) - (2*exp(-2)*(12*x^2 + 12*
x^3 + x^4 - x^5))/(exp(2*x) - 2*exp(x - 1) + exp(-2)) + (2*exp(-3)*(4*x^3 + 4*x^4 + x^5))/(exp(3*x) + 3*exp(x
- 2) - exp(-3) - 3*exp(2*x - 1))

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sympy [A]  time = 0.21, size = 19, normalized size = 0.86 \begin {gather*} - \frac {x^{4}}{\log {\left (- x + e e^{x} - 3 \right )}^{2}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*exp(x+1)+(-3-x)*exp(x))*ln(exp(x+1)-3-x)**3+(-4*x**3*exp(x+1)+4*x**4+12*x**3)*ln(exp(x+1)-3
-x)+2*x**4*exp(x+1)-2*x**4)/(exp(x+1)-3-x)/ln(exp(x+1)-3-x)**3,x)

[Out]

-x**4/log(-x + E*exp(x) - 3)**2 + exp(x)

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