3.36.23 \(\int \frac {15 x+e^3 (1+10 x-5 x^2)+(3 x+e^3 (2 x-x^2)) \log (x)+(5 e^3 x+e^3 x \log (x)) \log (5+\log (x))}{15 x+e^3 (15 x-5 x^2)+(3 x+e^3 (3 x-x^2)) \log (x)+(5 e^3 x+e^3 x \log (x)) \log (5+\log (x))} \, dx\)

Optimal. Leaf size=22 \[ x+\log \left (\frac {1}{5} \left (3+\frac {3}{e^3}-x+\log (5+\log (x))\right )\right ) \]

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Rubi [A]  time = 0.82, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 3, integrand size = 118, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6741, 6742, 6684} \begin {gather*} x+\log \left (-e^3 x+e^3 \log (\log (x)+5)+3 \left (1+e^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*x + E^3*(1 + 10*x - 5*x^2) + (3*x + E^3*(2*x - x^2))*Log[x] + (5*E^3*x + E^3*x*Log[x])*Log[5 + Log[x]]
)/(15*x + E^3*(15*x - 5*x^2) + (3*x + E^3*(3*x - x^2))*Log[x] + (5*E^3*x + E^3*x*Log[x])*Log[5 + Log[x]]),x]

[Out]

x + Log[3*(1 + E^3) - E^3*x + E^3*Log[5 + Log[x]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 x+e^3 \left (1+10 x-5 x^2\right )+\left (3 x+e^3 \left (2 x-x^2\right )\right ) \log (x)+\left (5 e^3 x+e^3 x \log (x)\right ) \log (5+\log (x))}{x (5+\log (x)) \left (3 \left (1+e^3\right )-e^3 x+e^3 \log (5+\log (x))\right )} \, dx\\ &=\int \left (1+\frac {e^3 (1-5 x-x \log (x))}{x (5+\log (x)) \left (3 \left (1+e^3\right )-e^3 x+e^3 \log (5+\log (x))\right )}\right ) \, dx\\ &=x+e^3 \int \frac {1-5 x-x \log (x)}{x (5+\log (x)) \left (3 \left (1+e^3\right )-e^3 x+e^3 \log (5+\log (x))\right )} \, dx\\ &=x+\log \left (3 \left (1+e^3\right )-e^3 x+e^3 \log (5+\log (x))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 25, normalized size = 1.14 \begin {gather*} x+\log \left (3+3 e^3-e^3 x+e^3 \log (5+\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*x + E^3*(1 + 10*x - 5*x^2) + (3*x + E^3*(2*x - x^2))*Log[x] + (5*E^3*x + E^3*x*Log[x])*Log[5 + L
og[x]])/(15*x + E^3*(15*x - 5*x^2) + (3*x + E^3*(3*x - x^2))*Log[x] + (5*E^3*x + E^3*x*Log[x])*Log[5 + Log[x]]
),x]

[Out]

x + Log[3 + 3*E^3 - E^3*x + E^3*Log[5 + Log[x]]]

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fricas [A]  time = 0.78, size = 20, normalized size = 0.91 \begin {gather*} x + \log \left (-{\left (x - 3\right )} e^{3} + e^{3} \log \left (\log \relax (x) + 5\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)*log(x)+5*x*exp(3))*log(5+log(x))+((-x^2+2*x)*exp(3)+3*x)*log(x)+(-5*x^2+10*x+1)*exp(3)+15
*x)/((x*exp(3)*log(x)+5*x*exp(3))*log(5+log(x))+((-x^2+3*x)*exp(3)+3*x)*log(x)+(-5*x^2+15*x)*exp(3)+15*x),x, a
lgorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 22, normalized size = 1.00 \begin {gather*} x + \log \left (-x e^{3} + e^{3} \log \left (\log \relax (x) + 5\right ) + 3 \, e^{3} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)*log(x)+5*x*exp(3))*log(5+log(x))+((-x^2+2*x)*exp(3)+3*x)*log(x)+(-5*x^2+10*x+1)*exp(3)+15
*x)/((x*exp(3)*log(x)+5*x*exp(3))*log(5+log(x))+((-x^2+3*x)*exp(3)+3*x)*log(x)+(-5*x^2+15*x)*exp(3)+15*x),x, a
lgorithm="giac")

[Out]

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

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maple [A]  time = 0.28, size = 23, normalized size = 1.05

method	result	size
norman	\(x +\ln \left (x \,{\mathrm e}^{3}-{\mathrm e}^{3} \ln \left (5+\ln \relax (x )\right )-3 \,{\mathrm e}^{3}-3\right )\)	\(23\)
risch	\(x +\ln \left (\ln \left (5+\ln \relax (x )\right )-\left (x \,{\mathrm e}^{3}-3 \,{\mathrm e}^{3}-3\right ) {\mathrm e}^{-3}\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(3)*ln(x)+5*x*exp(3))*ln(5+ln(x))+((-x^2+2*x)*exp(3)+3*x)*ln(x)+(-5*x^2+10*x+1)*exp(3)+15*x)/((x*ex
p(3)*ln(x)+5*x*exp(3))*ln(5+ln(x))+((-x^2+3*x)*exp(3)+3*x)*ln(x)+(-5*x^2+15*x)*exp(3)+15*x),x,method=_RETURNVE
RBOSE)

[Out]

x+ln(x*exp(3)-exp(3)*ln(5+ln(x))-3*exp(3)-3)

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maxima [A]  time = 0.51, size = 26, normalized size = 1.18 \begin {gather*} x + \log \left (-{\left (x e^{3} - e^{3} \log \left (\log \relax (x) + 5\right ) - 3 \, e^{3} - 3\right )} e^{\left (-3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)*log(x)+5*x*exp(3))*log(5+log(x))+((-x^2+2*x)*exp(3)+3*x)*log(x)+(-5*x^2+10*x+1)*exp(3)+15
*x)/((x*exp(3)*log(x)+5*x*exp(3))*log(5+log(x))+((-x^2+3*x)*exp(3)+3*x)*log(x)+(-5*x^2+15*x)*exp(3)+15*x),x, a
lgorithm="maxima")

[Out]

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

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mupad [B]  time = 2.83, size = 17, normalized size = 0.77 \begin {gather*} x+\ln \left (3\,{\mathrm {e}}^{-3}-x+\ln \left (\ln \relax (x)+5\right )+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x + exp(3)*(10*x - 5*x^2 + 1) + log(x)*(3*x + exp(3)*(2*x - x^2)) + log(log(x) + 5)*(5*x*exp(3) + x*ex
p(3)*log(x)))/(15*x + exp(3)*(15*x - 5*x^2) + log(x)*(3*x + exp(3)*(3*x - x^2)) + log(log(x) + 5)*(5*x*exp(3)
+ x*exp(3)*log(x))),x)

[Out]

x + log(3*exp(-3) - x + log(log(x) + 5) + 3)

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sympy [A]  time = 0.43, size = 24, normalized size = 1.09 \begin {gather*} x + \log {\left (\frac {- x e^{3} + 3 + 3 e^{3}}{e^{3}} + \log {\left (\log {\relax (x )} + 5 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(3)*ln(x)+5*x*exp(3))*ln(5+ln(x))+((-x**2+2*x)*exp(3)+3*x)*ln(x)+(-5*x**2+10*x+1)*exp(3)+15*x
)/((x*exp(3)*ln(x)+5*x*exp(3))*ln(5+ln(x))+((-x**2+3*x)*exp(3)+3*x)*ln(x)+(-5*x**2+15*x)*exp(3)+15*x),x)

[Out]

x + log((-x*exp(3) + 3 + 3*exp(3))*exp(-3) + log(log(x) + 5))

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