3.32.58 \(\int \frac {54-18 x+3 x^2+e^{x/3} x^3+18 x \log (2)}{3 x^3} \, dx\)

Optimal. Leaf size=25 \[ e^{x/3}-\frac {(-3+x-x \log (2))^2}{x^2}+\log (x) \]

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6, 12, 14, 2194} \begin {gather*} -\frac {9}{x^2}+e^{x/3}+\log (x)+\frac {6 (1-\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(54 - 18*x + 3*x^2 + E^(x/3)*x^3 + 18*x*Log[2])/(3*x^3),x]

[Out]

E^(x/3) - 9/x^2 + (6*(1 - Log[2]))/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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {54+3 x^2+e^{x/3} x^3+x (-18+18 \log (2))}{3 x^3} \, dx\\ &=\frac {1}{3} \int \frac {54+3 x^2+e^{x/3} x^3+x (-18+18 \log (2))}{x^3} \, dx\\ &=\frac {1}{3} \int \left (e^{x/3}+\frac {3 \left (18+x^2-6 x (1-\log (2))\right )}{x^3}\right ) \, dx\\ &=\frac {1}{3} \int e^{x/3} \, dx+\int \frac {18+x^2-6 x (1-\log (2))}{x^3} \, dx\\ &=e^{x/3}+\int \left (\frac {18}{x^3}+\frac {1}{x}+\frac {6 (-1+\log (2))}{x^2}\right ) \, dx\\ &=e^{x/3}-\frac {9}{x^2}+\frac {6 (1-\log (2))}{x}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 0.96 \begin {gather*} e^{x/3}-\frac {9}{x^2}-\frac {6 (-1+\log (2))}{x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(54 - 18*x + 3*x^2 + E^(x/3)*x^3 + 18*x*Log[2])/(3*x^3),x]

[Out]

E^(x/3) - 9/x^2 - (6*(-1 + Log[2]))/x + Log[x]

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fricas [A]  time = 0.51, size = 28, normalized size = 1.12 \begin {gather*} \frac {x^{2} e^{\left (\frac {1}{3} \, x\right )} + x^{2} \log \relax (x) - 6 \, x \log \relax (2) + 6 \, x - 9}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(x^3*exp(1/3*x)+18*x*log(2)+3*x^2-18*x+54)/x^3,x, algorithm="fricas")

[Out]

(x^2*e^(1/3*x) + x^2*log(x) - 6*x*log(2) + 6*x - 9)/x^2

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giac [A]  time = 0.33, size = 28, normalized size = 1.12 \begin {gather*} \frac {x^{2} e^{\left (\frac {1}{3} \, x\right )} + x^{2} \log \relax (x) - 6 \, x \log \relax (2) + 6 \, x - 9}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(x^3*exp(1/3*x)+18*x*log(2)+3*x^2-18*x+54)/x^3,x, algorithm="giac")

[Out]

(x^2*e^(1/3*x) + x^2*log(x) - 6*x*log(2) + 6*x - 9)/x^2

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maple [A]  time = 0.06, size = 23, normalized size = 0.92

method	result	size
risch	\(\frac {\left (-18 \ln \relax (2)+18\right ) x -27}{3 x^{2}}+\ln \relax (x )+{\mathrm e}^{\frac {x}{3}}\)	\(23\)
norman	\(\frac {-9+x^{2} {\mathrm e}^{\frac {x}{3}}+\left (-6 \ln \relax (2)+6\right ) x}{x^{2}}+\ln \relax (x )\)	\(26\)
derivativedivides	\(\ln \left (\frac {x}{3}\right )-\frac {9}{x^{2}}+\frac {6}{x}-\frac {6 \ln \relax (2)}{x}+{\mathrm e}^{\frac {x}{3}}\)	\(27\)
default	\(\ln \left (\frac {x}{3}\right )-\frac {9}{x^{2}}+\frac {6}{x}-\frac {6 \ln \relax (2)}{x}+{\mathrm e}^{\frac {x}{3}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(x^3*exp(1/3*x)+18*x*ln(2)+3*x^2-18*x+54)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/3*((-18*ln(2)+18)*x-27)/x^2+ln(x)+exp(1/3*x)

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maxima [A]  time = 0.41, size = 24, normalized size = 0.96 \begin {gather*} -\frac {6 \, \log \relax (2)}{x} + \frac {6}{x} - \frac {9}{x^{2}} + e^{\left (\frac {1}{3} \, x\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(x^3*exp(1/3*x)+18*x*log(2)+3*x^2-18*x+54)/x^3,x, algorithm="maxima")

[Out]

-6*log(2)/x + 6/x - 9/x^2 + e^(1/3*x) + log(x)

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mupad [B]  time = 1.87, size = 20, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{x/3}+\ln \relax (x)-\frac {x\,\left (\ln \left (64\right )-6\right )+9}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x*log(2) - 6*x + (x^3*exp(x/3))/3 + x^2 + 18)/x^3,x)

[Out]

exp(x/3) + log(x) - (x*(log(64) - 6) + 9)/x^2

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sympy [A]  time = 0.19, size = 20, normalized size = 0.80 \begin {gather*} e^{\frac {x}{3}} + \log {\relax (x )} + \frac {x \left (6 - 6 \log {\relax (2 )}\right ) - 9}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(x**3*exp(1/3*x)+18*x*ln(2)+3*x**2-18*x+54)/x**3,x)

[Out]

exp(x/3) + log(x) + (x*(6 - 6*log(2)) - 9)/x**2

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