3.25.64 \(\int \frac {4 e^{5/2}+6 x^2+4 x^3}{x^3} \, dx\)

Optimal. Leaf size=23 \[ 2 \left (-3-\frac {e^{5/2}}{x^2}+2 x+\log \left (25 x^3\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {14} \begin {gather*} -\frac {2 e^{5/2}}{x^2}+4 x+6 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^(5/2) + 6*x^2 + 4*x^3)/x^3,x]

[Out]

(-2*E^(5/2))/x^2 + 4*x + 6*Log[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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (4+\frac {4 e^{5/2}}{x^3}+\frac {6}{x}\right ) \, dx\\ &=-\frac {2 e^{5/2}}{x^2}+4 x+6 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.78 \begin {gather*} -\frac {2 e^{5/2}}{x^2}+4 x+6 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(5/2) + 6*x^2 + 4*x^3)/x^3,x]

[Out]

(-2*E^(5/2))/x^2 + 4*x + 6*Log[x]

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fricas [A]  time = 1.08, size = 22, normalized size = 0.96 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} + 3 \, x^{2} \log \relax (x) - e^{\frac {5}{2}}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5/4)^2+4*x^3+6*x^2)/x^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 16, normalized size = 0.70 \begin {gather*} 4 \, x - \frac {2 \, e^{\frac {5}{2}}}{x^{2}} + 6 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5/4)^2+4*x^3+6*x^2)/x^3,x, algorithm="giac")

[Out]

4*x - 2*e^(5/2)/x^2 + 6*log(abs(x))

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maple [A]  time = 0.03, size = 16, normalized size = 0.70

method	result	size
default	\(4 x +6 \ln \relax (x )-\frac {2 \,{\mathrm e}^{\frac {5}{2}}}{x^{2}}\)	\(16\)
risch	\(4 x +6 \ln \relax (x )-\frac {2 \,{\mathrm e}^{\frac {5}{2}}}{x^{2}}\)	\(16\)
norman	\(\frac {4 x^{3}-2 \,{\mathrm e}^{\frac {5}{2}}}{x^{2}}+6 \ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(5/4)^2+4*x^3+6*x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

4*x+6*ln(x)-2*exp(5/2)/x^2

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maxima [A]  time = 0.54, size = 15, normalized size = 0.65 \begin {gather*} 4 \, x - \frac {2 \, e^{\frac {5}{2}}}{x^{2}} + 6 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5/4)^2+4*x^3+6*x^2)/x^3,x, algorithm="maxima")

[Out]

4*x - 2*e^(5/2)/x^2 + 6*log(x)

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mupad [B]  time = 0.03, size = 15, normalized size = 0.65 \begin {gather*} 4\,x+6\,\ln \relax (x)-\frac {2\,{\mathrm {e}}^{5/2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(5/2) + 6*x^2 + 4*x^3)/x^3,x)

[Out]

4*x + 6*log(x) - (2*exp(5/2))/x^2

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sympy [A]  time = 0.10, size = 17, normalized size = 0.74 \begin {gather*} 4 x + 6 \log {\relax (x )} - \frac {2 e^{\frac {5}{2}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5/4)**2+4*x**3+6*x**2)/x**3,x)

[Out]

4*x + 6*log(x) - 2*exp(5/2)/x**2

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