3.89.24 \(\int \frac {12+3 e^{-1+x}+6 x+6 x \log (3)+15 \log (x)}{e^{-1+x}-x+x^2+x^2 \log (3)+5 x \log (x)} \, dx\)

Optimal. Leaf size=30 \[ 3 \log \left (x-x \left (x+\frac {e^{-1+x}+x^2 \log (3)}{x}+5 \log (x)\right )\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 2, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 6684} \begin {gather*} 3 \log \left (e x^2+e x^2 \log (3)-e x+e^x+5 e x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 3*E^(-1 + x) + 6*x + 6*x*Log[3] + 15*Log[x])/(E^(-1 + x) - x + x^2 + x^2*Log[3] + 5*x*Log[x]),x]

[Out]

3*Log[E^x - E*x + E*x^2 + E*x^2*Log[3] + 5*E*x*Log[x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12+3 e^{-1+x}+x (6+6 \log (3))+15 \log (x)}{e^{-1+x}-x+x^2+x^2 \log (3)+5 x \log (x)} \, dx\\ &=\int \frac {12+3 e^{-1+x}+x (6+6 \log (3))+15 \log (x)}{e^{-1+x}-x+x^2 (1+\log (3))+5 x \log (x)} \, dx\\ &=3 \log \left (e^x-e x+e x^2+e x^2 \log (3)+5 e x \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {12+3 e^{-1+x}+6 x+6 x \log (3)+15 \log (x)}{e^{-1+x}-x+x^2+x^2 \log (3)+5 x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(12 + 3*E^(-1 + x) + 6*x + 6*x*Log[3] + 15*Log[x])/(E^(-1 + x) - x + x^2 + x^2*Log[3] + 5*x*Log[x]),
x]

[Out]

Integrate[(12 + 3*E^(-1 + x) + 6*x + 6*x*Log[3] + 15*Log[x])/(E^(-1 + x) - x + x^2 + x^2*Log[3] + 5*x*Log[x]),
x]

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fricas [A]  time = 0.56, size = 34, normalized size = 1.13 \begin {gather*} 3 \, \log \relax (x) + 3 \, \log \left (\frac {x^{2} \log \relax (3) + x^{2} + 5 \, x \log \relax (x) - x + e^{\left (x - 1\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*log(x)+3*exp(x-1)+6*x*log(3)+6*x+12)/(5*x*log(x)+exp(x-1)+x^2*log(3)+x^2-x),x, algorithm="fricas
")

[Out]

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

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giac [A]  time = 0.12, size = 32, normalized size = 1.07 \begin {gather*} 3 \, \log \left (x^{2} e \log \relax (3) + x^{2} e + 5 \, x e \log \relax (x) - x e + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*log(x)+3*exp(x-1)+6*x*log(3)+6*x+12)/(5*x*log(x)+exp(x-1)+x^2*log(3)+x^2-x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 26, normalized size = 0.87

method	result	size
default	\(3 \ln \left (5 x \ln \relax (x )+{\mathrm e}^{x -1}+x^{2} \ln \relax (3)+x^{2}-x \right )\)	\(26\)
norman	\(3 \ln \left (5 x \ln \relax (x )+{\mathrm e}^{x -1}+x^{2} \ln \relax (3)+x^{2}-x \right )\)	\(26\)
risch	\(3 \ln \relax (x )+3 \ln \left (\ln \relax (x )+\frac {x^{2} \ln \relax (3)+x^{2}-x +{\mathrm e}^{x -1}}{5 x}\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*ln(x)+3*exp(x-1)+6*x*ln(3)+6*x+12)/(5*x*ln(x)+exp(x-1)+x^2*ln(3)+x^2-x),x,method=_RETURNVERBOSE)

[Out]

3*ln(5*x*ln(x)+exp(x-1)+x^2*ln(3)+x^2-x)

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maxima [A]  time = 0.37, size = 25, normalized size = 0.83 \begin {gather*} 3 \, \log \left (x^{2} \log \relax (3) + x^{2} + 5 \, x \log \relax (x) - x + e^{\left (x - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*log(x)+3*exp(x-1)+6*x*log(3)+6*x+12)/(5*x*log(x)+exp(x-1)+x^2*log(3)+x^2-x),x, algorithm="maxima
")

[Out]

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

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mupad [B]  time = 5.57, size = 25, normalized size = 0.83 \begin {gather*} 3\,\ln \left ({\mathrm {e}}^{x-1}-x+x^2\,\ln \relax (3)+5\,x\,\ln \relax (x)+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + 3*exp(x - 1) + 15*log(x) + 6*x*log(3) + 12)/(exp(x - 1) - x + x^2*log(3) + 5*x*log(x) + x^2),x)

[Out]

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

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sympy [A]  time = 0.31, size = 26, normalized size = 0.87 \begin {gather*} 3 \log {\left (x^{2} + x^{2} \log {\relax (3 )} + 5 x \log {\relax (x )} - x + e^{x - 1} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*ln(x)+3*exp(x-1)+6*x*ln(3)+6*x+12)/(5*x*ln(x)+exp(x-1)+x**2*ln(3)+x**2-x),x)

[Out]

3*log(x**2 + x**2*log(3) + 5*x*log(x) - x + exp(x - 1))

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