3.5.71 \(\int \frac {1+e^4 (3 x^3+2 x^4+3 x^5)+2 \log (\frac {5}{3 x})}{e^4 x^3} \, dx\)

Optimal. Leaf size=24 \[ x \left (3+x+x^2-\frac {\log \left (\frac {5}{3 x}\right )}{e^4 x^3}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14, 2304} \begin {gather*} x^3+x^2-\frac {\log \left (\frac {5}{3 x}\right )}{e^4 x^2}+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.08 \begin {gather*} 3 x+x^2+x^3-\frac {\log \left (\frac {5}{3 x}\right )}{e^4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^4*(3*x^3 + 2*x^4 + 3*x^5) + 2*Log[5/(3*x)])/(E^4*x^3),x]

[Out]

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

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fricas [A]  time = 0.79, size = 30, normalized size = 1.25 \begin {gather*} \frac {{\left ({\left (x^{5} + x^{4} + 3 \, x^{3}\right )} e^{4} - \log \left (\frac {5}{3 \, x}\right )\right )} e^{\left (-4\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(5/3/x)+(3*x^5+2*x^4+3*x^3)*exp(4)+1)/x^3/exp(4),x, algorithm="fricas")

[Out]

((x^5 + x^4 + 3*x^3)*e^4 - log(5/3/x))*e^(-4)/x^2

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giac [A]  time = 0.33, size = 33, normalized size = 1.38 \begin {gather*} x^{3} {\left (\frac {e^{4}}{x} + \frac {3 \, e^{4}}{x^{2}} - \frac {\log \left (\frac {5}{3 \, x}\right )}{x^{5}} + e^{4}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(5/3/x)+(3*x^5+2*x^4+3*x^3)*exp(4)+1)/x^3/exp(4),x, algorithm="giac")

[Out]

x^3*(e^4/x + 3*e^4/x^2 - log(5/3/x)/x^5 + e^4)*e^(-4)

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maple [A]  time = 0.08, size = 23, normalized size = 0.96

method	result	size
risch	\(-\frac {{\mathrm e}^{-4} \ln \left (\frac {5}{3 x}\right )}{x^{2}}+x \left (x^{2}+x +3\right )\)	\(23\)
norman	\(\frac {x^{4}+x^{5}+3 x^{3}-{\mathrm e}^{-4} \ln \left (\frac {5}{3 x}\right )}{x^{2}}\)	\(29\)
default	\({\mathrm e}^{-4} \left (-\frac {\ln \left (\frac {5}{3 x}\right )}{x^{2}}+3 x \,{\mathrm e}^{4}+x^{2} {\mathrm e}^{4}+x^{3} {\mathrm e}^{4}\right )\)	\(35\)
derivativedivides	\(-{\mathrm e}^{-4} \left (\frac {\ln \left (\frac {5}{3 x}\right )}{x^{2}}-3 x \,{\mathrm e}^{4}-x^{2} {\mathrm e}^{4}-x^{3} {\mathrm e}^{4}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(5/3/x)+(3*x^5+2*x^4+3*x^3)*exp(4)+1)/x^3/exp(4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 32, normalized size = 1.33 \begin {gather*} {\left (x^{3} e^{4} + x^{2} e^{4} + 3 \, x e^{4} - \frac {\log \left (\frac {5}{3 \, x}\right )}{x^{2}}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(5/3/x)+(3*x^5+2*x^4+3*x^3)*exp(4)+1)/x^3/exp(4),x, algorithm="maxima")

[Out]

(x^3*e^4 + x^2*e^4 + 3*x*e^4 - log(5/3/x)/x^2)*e^(-4)

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mupad [B]  time = 0.54, size = 23, normalized size = 0.96 \begin {gather*} 3\,x+x^2+x^3-\frac {{\mathrm {e}}^{-4}\,\ln \left (\frac {5}{3\,x}\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(2*log(5/(3*x)) + exp(4)*(3*x^3 + 2*x^4 + 3*x^5) + 1))/x^3,x)

[Out]

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

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sympy [A]  time = 0.14, size = 22, normalized size = 0.92 \begin {gather*} x^{3} + x^{2} + 3 x - \frac {\log {\left (\frac {5}{3 x} \right )}}{x^{2} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(5/3/x)+(3*x**5+2*x**4+3*x**3)*exp(4)+1)/x**3/exp(4),x)

[Out]

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

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