3.39.70 \(\int \frac {3+117 x+30 x^2-45 x^3+(24 x+6 x^2-9 x^3) \log (3 x)+(-15 x-3 x \log (3 x)) \log (5+\log (3 x))}{5 x+x \log (3 x)} \, dx\)

Optimal. Leaf size=19 \[ 3 (1-x) \left (-8+x^2+\log (5+\log (3 x))\right ) \]

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Rubi [F]  time = 0.58, 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+117 x+30 x^2-45 x^3+\left (24 x+6 x^2-9 x^3\right ) \log (3 x)+(-15 x-3 x \log (3 x)) \log (5+\log (3 x))}{5 x+x \log (3 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3 + 117*x + 30*x^2 - 45*x^3 + (24*x + 6*x^2 - 9*x^3)*Log[3*x] + (-15*x - 3*x*Log[3*x])*Log[5 + Log[3*x]])
/(5*x + x*Log[3*x]),x]

[Out]

24*x + 3*x^2 - 3*x^3 - ExpIntegralEi[5 + Log[3*x]]/E^5 + 3*Log[5 + Log[3*x]] - Defer[Subst][Defer[Int][Log[5 +
Log[x]], x], x, 3*x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3+117 x+30 x^2-45 x^3+\left (24 x+6 x^2-9 x^3\right ) \log (3 x)+(-15 x-3 x \log (3 x)) \log (5+\log (3 x))}{x (5+\log (3 x))} \, dx\\ &=\int \left (-\frac {3 \left (-1-39 x-10 x^2+15 x^3-8 x \log (3 x)-2 x^2 \log (3 x)+3 x^3 \log (3 x)\right )}{x (5+\log (3 x))}-3 \log (5+\log (3 x))\right ) \, dx\\ &=-\left (3 \int \frac {-1-39 x-10 x^2+15 x^3-8 x \log (3 x)-2 x^2 \log (3 x)+3 x^3 \log (3 x)}{x (5+\log (3 x))} \, dx\right )-3 \int \log (5+\log (3 x)) \, dx\\ &=-\left (3 \int \left (-8-2 x+3 x^2+\frac {-1+x}{x (5+\log (3 x))}\right ) \, dx\right )-\operatorname {Subst}(\int \log (5+\log (x)) \, dx,x,3 x)\\ &=24 x+3 x^2-3 x^3-3 \int \frac {-1+x}{x (5+\log (3 x))} \, dx-\operatorname {Subst}(\int \log (5+\log (x)) \, dx,x,3 x)\\ &=24 x+3 x^2-3 x^3-3 \int \left (\frac {1}{5+\log (3 x)}-\frac {1}{x (5+\log (3 x))}\right ) \, dx-\operatorname {Subst}(\int \log (5+\log (x)) \, dx,x,3 x)\\ &=24 x+3 x^2-3 x^3-3 \int \frac {1}{5+\log (3 x)} \, dx+3 \int \frac {1}{x (5+\log (3 x))} \, dx-\operatorname {Subst}(\int \log (5+\log (x)) \, dx,x,3 x)\\ &=24 x+3 x^2-3 x^3+3 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,5+\log (3 x)\right )-\operatorname {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log (3 x)\right )-\operatorname {Subst}(\int \log (5+\log (x)) \, dx,x,3 x)\\ &=24 x+3 x^2-3 x^3-\frac {\text {Ei}(5+\log (3 x))}{e^5}+3 \log (5+\log (3 x))-\operatorname {Subst}(\int \log (5+\log (x)) \, dx,x,3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 32, normalized size = 1.68 \begin {gather*} -3 \left (-8 x-x^2+x^3-\log (5+\log (3 x))+x \log (5+\log (3 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 117*x + 30*x^2 - 45*x^3 + (24*x + 6*x^2 - 9*x^3)*Log[3*x] + (-15*x - 3*x*Log[3*x])*Log[5 + Log[
3*x]])/(5*x + x*Log[3*x]),x]

[Out]

-3*(-8*x - x^2 + x^3 - Log[5 + Log[3*x]] + x*Log[5 + Log[3*x]])

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fricas [A]  time = 0.86, size = 26, normalized size = 1.37 \begin {gather*} -3 \, x^{3} + 3 \, x^{2} - 3 \, {\left (x - 1\right )} \log \left (\log \left (3 \, x\right ) + 5\right ) + 24 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x*log(3*x)-15*x)*log(5+log(3*x))+(-9*x^3+6*x^2+24*x)*log(3*x)-45*x^3+30*x^2+117*x+3)/(x*log(3*x
)+5*x),x, algorithm="fricas")

[Out]

-3*x^3 + 3*x^2 - 3*(x - 1)*log(log(3*x) + 5) + 24*x

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giac [A]  time = 1.13, size = 33, normalized size = 1.74 \begin {gather*} -3 \, x^{3} + 3 \, x^{2} - 3 \, x \log \left (\log \left (3 \, x\right ) + 5\right ) + 24 \, x + 3 \, \log \left (\log \left (3 \, x\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x*log(3*x)-15*x)*log(5+log(3*x))+(-9*x^3+6*x^2+24*x)*log(3*x)-45*x^3+30*x^2+117*x+3)/(x*log(3*x
)+5*x),x, algorithm="giac")

[Out]

-3*x^3 + 3*x^2 - 3*x*log(log(3*x) + 5) + 24*x + 3*log(log(3*x) + 5)

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maple [A]  time = 0.04, size = 34, normalized size = 1.79

method	result	size
risch	\(-3 \ln \left (5+\ln \left (3 x \right )\right ) x -3 x^{3}+3 x^{2}+24 x +3 \ln \left (5+\ln \left (3 x \right )\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x*ln(3*x)-15*x)*ln(5+ln(3*x))+(-9*x^3+6*x^2+24*x)*ln(3*x)-45*x^3+30*x^2+117*x+3)/(x*ln(3*x)+5*x),x,me
thod=_RETURNVERBOSE)

[Out]

-3*ln(5+ln(3*x))*x-3*x^3+3*x^2+24*x+3*ln(5+ln(3*x))

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maxima [A]  time = 0.46, size = 33, normalized size = 1.74 \begin {gather*} -3 \, x^{3} + 3 \, x^{2} - 3 \, x \log \left (\log \relax (3) + \log \relax (x) + 5\right ) + 24 \, x + 3 \, \log \left (\log \relax (3) + \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x*log(3*x)-15*x)*log(5+log(3*x))+(-9*x^3+6*x^2+24*x)*log(3*x)-45*x^3+30*x^2+117*x+3)/(x*log(3*x
)+5*x),x, algorithm="maxima")

[Out]

-3*x^3 + 3*x^2 - 3*x*log(log(3) + log(x) + 5) + 24*x + 3*log(log(3) + log(x) + 5)

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mupad [B]  time = 2.37, size = 33, normalized size = 1.74 \begin {gather*} 24\,x+3\,\ln \left (\ln \left (3\,x\right )+5\right )+3\,x^2-3\,x^3-3\,x\,\ln \left (\ln \left (3\,x\right )+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((117*x + log(3*x)*(24*x + 6*x^2 - 9*x^3) - log(log(3*x) + 5)*(15*x + 3*x*log(3*x)) + 30*x^2 - 45*x^3 + 3)/
(5*x + x*log(3*x)),x)

[Out]

24*x + 3*log(log(3*x) + 5) + 3*x^2 - 3*x^3 - 3*x*log(log(3*x) + 5)

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sympy [A]  time = 0.38, size = 34, normalized size = 1.79 \begin {gather*} - 3 x^{3} + 3 x^{2} - 3 x \log {\left (\log {\left (3 x \right )} + 5 \right )} + 24 x + 3 \log {\left (\log {\left (3 x \right )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x*ln(3*x)-15*x)*ln(5+ln(3*x))+(-9*x**3+6*x**2+24*x)*ln(3*x)-45*x**3+30*x**2+117*x+3)/(x*ln(3*x)
+5*x),x)

[Out]

-3*x**3 + 3*x**2 - 3*x*log(log(3*x) + 5) + 24*x + 3*log(log(3*x) + 5)

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