3.54.48 \(\int \frac {e^5 x+(-12 x-4 x^2) \log (-3-x)+e^5 (-6-2 x) \log (-3-x) \log (\log (-3-x))}{(3 x^3+x^4) \log (-3-x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {4+\frac {e^5 \log (\log (-3-x))}{x}}{x} \]

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Rubi [F]  time = 0.63, 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^5 x+\left (-12 x-4 x^2\right ) \log (-3-x)+e^5 (-6-2 x) \log (-3-x) \log (\log (-3-x))}{\left (3 x^3+x^4\right ) \log (-3-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*x + (-12*x - 4*x^2)*Log[-3 - x] + E^5*(-6 - 2*x)*Log[-3 - x]*Log[Log[-3 - x]])/((3*x^3 + x^4)*Log[-3
- x]),x]

[Out]

4/x + (E^5*Log[Log[-3 - x]])/9 + (E^5*Defer[Int][1/(x^2*Log[-3 - x]), x])/3 - (E^5*Defer[Int][1/(x*Log[-3 - x]
), x])/9 - 2*E^5*Defer[Int][Log[Log[-3 - x]]/x^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 x+\left (-12 x-4 x^2\right ) \log (-3-x)+e^5 (-6-2 x) \log (-3-x) \log (\log (-3-x))}{x^3 (3+x) \log (-3-x)} \, dx\\ &=\int \frac {-4 x+\frac {e^5 x}{(3+x) \log (-3-x)}-2 e^5 \log (\log (-3-x))}{x^3} \, dx\\ &=\int \left (\frac {e^5-12 \log (-3-x)-4 x \log (-3-x)}{x^2 (3+x) \log (-3-x)}-\frac {2 e^5 \log (\log (-3-x))}{x^3}\right ) \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\right )+\int \frac {e^5-12 \log (-3-x)-4 x \log (-3-x)}{x^2 (3+x) \log (-3-x)} \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\right )+\int \frac {-4+\frac {e^5}{(3+x) \log (-3-x)}}{x^2} \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\right )+\int \left (-\frac {4}{x^2}+\frac {e^5}{x^2 (3+x) \log (-3-x)}\right ) \, dx\\ &=\frac {4}{x}+e^5 \int \frac {1}{x^2 (3+x) \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}+e^5 \int \left (\frac {1}{3 x^2 \log (-3-x)}-\frac {1}{9 x \log (-3-x)}+\frac {1}{9 (3+x) \log (-3-x)}\right ) \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{9} e^5 \int \frac {1}{(3+x) \log (-3-x)} \, dx+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{9} e^5 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-3-x\right )+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{9} e^5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-3-x)\right )+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ &=\frac {4}{x}+\frac {1}{9} e^5 \log (\log (-3-x))-\frac {1}{9} e^5 \int \frac {1}{x \log (-3-x)} \, dx+\frac {1}{3} e^5 \int \frac {1}{x^2 \log (-3-x)} \, dx-\left (2 e^5\right ) \int \frac {\log (\log (-3-x))}{x^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 20, normalized size = 1.00 \begin {gather*} \frac {4}{x}+\frac {e^5 \log (\log (-3-x))}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*x + (-12*x - 4*x^2)*Log[-3 - x] + E^5*(-6 - 2*x)*Log[-3 - x]*Log[Log[-3 - x]])/((3*x^3 + x^4)*L
og[-3 - x]),x]

[Out]

4/x + (E^5*Log[Log[-3 - x]])/x^2

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fricas [A]  time = 0.61, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{5} \log \left (\log \left (-x - 3\right )\right ) + 4 \, x}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(5)*log(-3-x)*log(log(-3-x))+(-4*x^2-12*x)*log(-3-x)+x*exp(5))/(x^4+3*x^3)/log(-3-x),x,
algorithm="fricas")

[Out]

(e^5*log(log(-x - 3)) + 4*x)/x^2

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giac [A]  time = 0.13, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{5} \log \left (\log \left (-x - 3\right )\right ) + 4 \, x}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(5)*log(-3-x)*log(log(-3-x))+(-4*x^2-12*x)*log(-3-x)+x*exp(5))/(x^4+3*x^3)/log(-3-x),x,
algorithm="giac")

[Out]

(e^5*log(log(-x - 3)) + 4*x)/x^2

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maple [A]  time = 0.18, size = 20, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{5} \ln \left (\ln \left (-3-x \right )\right )}{x^{2}}+\frac {4}{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-6)*exp(5)*ln(-3-x)*ln(ln(-3-x))+(-4*x^2-12*x)*ln(-3-x)+x*exp(5))/(x^4+3*x^3)/ln(-3-x),x,method=_RET
URNVERBOSE)

[Out]

exp(5)/x^2*ln(ln(-3-x))+4/x

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maxima [A]  time = 0.43, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{5} \log \left (\log \left (-x - 3\right )\right ) + 4 \, x}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(5)*log(-3-x)*log(log(-3-x))+(-4*x^2-12*x)*log(-3-x)+x*exp(5))/(x^4+3*x^3)/log(-3-x),x,
algorithm="maxima")

[Out]

(e^5*log(log(-x - 3)) + 4*x)/x^2

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mupad [B]  time = 3.84, size = 18, normalized size = 0.90 \begin {gather*} \frac {4\,x+\ln \left (\ln \left (-x-3\right )\right )\,{\mathrm {e}}^5}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*x + log(log(- x - 3))*exp(5))/x^2

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sympy [A]  time = 0.37, size = 17, normalized size = 0.85 \begin {gather*} \frac {4}{x} + \frac {e^{5} \log {\left (\log {\left (- x - 3 \right )} \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-6)*exp(5)*ln(-3-x)*ln(ln(-3-x))+(-4*x**2-12*x)*ln(-3-x)+x*exp(5))/(x**4+3*x**3)/ln(-3-x),x)

[Out]

4/x + exp(5)*log(log(-x - 3))/x**2

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