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

Optimal. Leaf size=30 \[ x^3+\frac {\log (4)+\frac {-e^x+x (4+x)+x \log (x)}{x}}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 25, normalized size of antiderivative = 0.83, number of steps used = 9, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 14, 2197, 2304} \begin {gather*} x^3-\frac {e^x}{x^2}+\frac {4}{x}+\frac {\log (4 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^x/x^2) + 4/x + x^3 + Log[4*x]/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 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

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

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Mathematica [A]  time = 0.04, size = 22, normalized size = 0.73 \begin {gather*} \frac {-e^x+4 x+x^5+x \log (4 x)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-E^x + 4*x + x^5 + x*Log[4*x])/x^2

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fricas [A]  time = 0.50, size = 24, normalized size = 0.80 \begin {gather*} \frac {x^{5} + 2 \, x \log \relax (2) + x \log \relax (x) + 4 \, x - e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 24, normalized size = 0.80 \begin {gather*} \frac {x^{5} + 2 \, x \log \relax (2) + x \log \relax (x) + 4 \, x - e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 28, normalized size = 0.93

method	result	size
risch	\(\frac {\ln \relax (x )}{x}+\frac {x^{5}+2 x \ln \relax (2)+4 x -{\mathrm e}^{x}}{x^{2}}\)	\(28\)
default	\(x^{3}+\frac {2 \ln \relax (2)}{x}+\frac {4}{x}+\frac {\ln \relax (x )}{x}-\frac {{\mathrm e}^{x}}{x^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.47, size = 36, normalized size = 1.20 \begin {gather*} x^{3} + \frac {2 \, \log \relax (2)}{x} + \frac {\log \relax (x)}{x} + \frac {4}{x} - \Gamma \left (-1, -x\right ) - 2 \, \Gamma \left (-2, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3 + 2*log(2)/x + log(x)/x + 4/x - gamma(-1, -x) - 2*gamma(-2, -x)

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mupad [B]  time = 2.00, size = 21, normalized size = 0.70 \begin {gather*} x^3-\frac {{\mathrm {e}}^x-x\,\left (\ln \left (4\,x\right )+4\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3 - (exp(x) - x*(log(4*x) + 4))/x^2

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sympy [A]  time = 0.31, size = 22, normalized size = 0.73 \begin {gather*} x^{3} + \frac {\log {\relax (x )}}{x} + \frac {2 \log {\relax (2 )} + 4}{x} - \frac {e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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