3.21.77 \(\int \frac {-3 e^{5+x}-3 x+(2 x+e^{5+x} (1+x)) \log (2 x^3)}{e^{10+2 x} x^2+2 e^{5+x} x^3+x^4} \, dx\)

Optimal. Leaf size=20 \[ -\frac {\log \left (2 x^3\right )}{x \left (e^{5+x}+x\right )} \]

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Rubi [F]  time = 1.68, 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 {-3 e^{5+x}-3 x+\left (2 x+e^{5+x} (1+x)\right ) \log \left (2 x^3\right )}{e^{10+2 x} x^2+2 e^{5+x} x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*E^(5 + x) - 3*x + (2*x + E^(5 + x)*(1 + x))*Log[2*x^3])/(E^(10 + 2*x)*x^2 + 2*E^(5 + x)*x^3 + x^4),x]

[Out]

-(Log[2*x^3]*Defer[Int][(E^(5 + x) + x)^(-2), x]) + Log[2*x^3]*Defer[Int][1/(x*(E^(5 + x) + x)^2), x] - 3*Defe
r[Int][1/(x^2*(E^(5 + x) + x)), x] + Log[2*x^3]*Defer[Int][1/(x^2*(E^(5 + x) + x)), x] + Log[2*x^3]*Defer[Int]
[1/(x*(E^(5 + x) + x)), x] + 3*Defer[Int][Defer[Int][(E^(5 + x) + x)^(-2), x]/x, x] - 3*Defer[Int][Defer[Int][
1/(x*(E^(5 + x) + x)^2), x]/x, x] - 3*Defer[Int][Defer[Int][1/(x^2*(E^(5 + x) + x)), x]/x, x] - 3*Defer[Int][D
efer[Int][(E^(5 + x)*x + x^2)^(-1), x]/x, x]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 20, normalized size = 1.00 \begin {gather*} -\frac {\log \left (2 x^3\right )}{x \left (e^{5+x}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^(5 + x) - 3*x + (2*x + E^(5 + x)*(1 + x))*Log[2*x^3])/(E^(10 + 2*x)*x^2 + 2*E^(5 + x)*x^3 + x^
4),x]

[Out]

-(Log[2*x^3]/(x*(E^(5 + x) + x)))

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fricas [A]  time = 1.14, size = 20, normalized size = 1.00 \begin {gather*} -\frac {\log \left (2 \, x^{3}\right )}{x^{2} + x e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(5+x)+2*x)*log(2*x^3)-3*exp(5+x)-3*x)/(x^2*exp(5+x)^2+2*x^3*exp(5+x)+x^4),x, algorithm="f
ricas")

[Out]

-log(2*x^3)/(x^2 + x*e^(x + 5))

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giac [A]  time = 1.40, size = 21, normalized size = 1.05 \begin {gather*} -\frac {\log \relax (2) + 3 \, \log \relax (x)}{x^{2} + x e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(5+x)+2*x)*log(2*x^3)-3*exp(5+x)-3*x)/(x^2*exp(5+x)^2+2*x^3*exp(5+x)+x^4),x, algorithm="g
iac")

[Out]

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

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maple [C]  time = 0.13, size = 158, normalized size = 7.90

method	result	size
risch	\(-\frac {3 \ln \relax (x )}{x \left (x +{\mathrm e}^{5+x}\right )}-\frac {-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+2 \ln \relax (2)}{2 x \left (x +{\mathrm e}^{5+x}\right )}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(5+x)+2*x)*ln(2*x^3)-3*exp(5+x)-3*x)/(x^2*exp(5+x)^2+2*x^3*exp(5+x)+x^4),x,method=_RETURNVERBOS
E)

[Out]

-3/x/(x+exp(5+x))*ln(x)-1/2*(-I*Pi*csgn(I*x^2)*csgn(I*x)^2+2*I*Pi*csgn(I*x^2)^2*csgn(I*x)-I*Pi*csgn(I*x)*csgn(
I*x^2)*csgn(I*x^3)+I*Pi*csgn(I*x)*csgn(I*x^3)^2-I*Pi*csgn(I*x^2)^3+I*Pi*csgn(I*x^2)*csgn(I*x^3)^2-I*Pi*csgn(I*
x^3)^3+2*ln(2))/x/(x+exp(5+x))

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maxima [A]  time = 0.88, size = 21, normalized size = 1.05 \begin {gather*} -\frac {\log \relax (2) + 3 \, \log \relax (x)}{x^{2} + x e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(5+x)+2*x)*log(2*x^3)-3*exp(5+x)-3*x)/(x^2*exp(5+x)^2+2*x^3*exp(5+x)+x^4),x, algorithm="m
axima")

[Out]

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

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mupad [B]  time = 1.42, size = 19, normalized size = 0.95 \begin {gather*} -\frac {\ln \left (2\,x^3\right )}{x\,\left (x+{\mathrm {e}}^{x+5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 3*exp(x + 5) - log(2*x^3)*(2*x + exp(x + 5)*(x + 1)))/(2*x^3*exp(x + 5) + x^2*exp(2*x + 10) + x^4)
,x)

[Out]

-log(2*x^3)/(x*(x + exp(x + 5)))

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sympy [A]  time = 0.25, size = 17, normalized size = 0.85 \begin {gather*} - \frac {\log {\left (2 x^{3} \right )}}{x^{2} + x e^{x + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(5+x)+2*x)*ln(2*x**3)-3*exp(5+x)-3*x)/(x**2*exp(5+x)**2+2*x**3*exp(5+x)+x**4),x)

[Out]

-log(2*x**3)/(x**2 + x*exp(x + 5))

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