3.40.83 \(\int \frac {-2+2 x-3 e^x x-3 e^{x^3} x^3}{x} \, dx\)

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

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Rubi [A]  time = 0.04, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {14, 2209, 2194, 43} \begin {gather*} -e^{x^3}+2 x-3 e^x-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3*E^x - E^x^3 + 2*x - 2*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]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.83 \begin {gather*} -3 e^x-e^{x^3}+2 x-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3*E^x - E^x^3 + 2*x - 2*Log[x]

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fricas [A]  time = 0.74, size = 18, normalized size = 0.75 \begin {gather*} 2 \, x - e^{\left (x^{3}\right )} - 3 \, e^{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^(x^3) - 3*e^x - 2*log(x)

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giac [A]  time = 0.13, size = 18, normalized size = 0.75 \begin {gather*} 2 \, x - e^{\left (x^{3}\right )} - 3 \, e^{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^(x^3) - 3*e^x - 2*log(x)

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maple [A]  time = 0.03, size = 19, normalized size = 0.79

method	result	size
default	\(2 x -2 \ln \relax (x )-{\mathrm e}^{x^{3}}-3 \,{\mathrm e}^{x}\)	\(19\)
norman	\(2 x -2 \ln \relax (x )-{\mathrm e}^{x^{3}}-3 \,{\mathrm e}^{x}\)	\(19\)
risch	\(2 x -2 \ln \relax (x )-{\mathrm e}^{x^{3}}-3 \,{\mathrm e}^{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x-2*ln(x)-exp(x^3)-3*exp(x)

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maxima [A]  time = 0.52, size = 18, normalized size = 0.75 \begin {gather*} 2 \, x - e^{\left (x^{3}\right )} - 3 \, e^{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^(x^3) - 3*e^x - 2*log(x)

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mupad [B]  time = 0.06, size = 18, normalized size = 0.75 \begin {gather*} 2\,x-{\mathrm {e}}^{x^3}-3\,{\mathrm {e}}^x-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - exp(x^3) - 3*exp(x) - 2*log(x)

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sympy [A]  time = 0.16, size = 17, normalized size = 0.71 \begin {gather*} 2 x - 3 e^{x} - e^{x^{3}} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - 3*exp(x) - exp(x**3) - 2*log(x)

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