3.85.45 \(\int \frac {-2-15 x-e^x x-4 x^2-2 \log (x)}{x} \, dx\)

Optimal. Leaf size=24 \[ 3-e^x+(1-2 x) (8+x)-(1+\log (x))^2 \]

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 2194, 2301} \begin {gather*} -2 x^2-15 x-e^x-\log ^2(x)-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 15*x - E^x*x - 4*x^2 - 2*Log[x])/x,x]

[Out]

-E^x - 15*x - 2*x^2 - 2*Log[x] - Log[x]^2

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]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^x+\frac {-2-15 x-4 x^2-2 \log (x)}{x}\right ) \, dx\\ &=-\int e^x \, dx+\int \frac {-2-15 x-4 x^2-2 \log (x)}{x} \, dx\\ &=-e^x+\int \left (\frac {-2-15 x-4 x^2}{x}-\frac {2 \log (x)}{x}\right ) \, dx\\ &=-e^x-2 \int \frac {\log (x)}{x} \, dx+\int \frac {-2-15 x-4 x^2}{x} \, dx\\ &=-e^x-\log ^2(x)+\int \left (-15-\frac {2}{x}-4 x\right ) \, dx\\ &=-e^x-15 x-2 x^2-2 \log (x)-\log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.92 \begin {gather*} -e^x-15 x-2 x^2-(1+\log (x))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 15*x - E^x*x - 4*x^2 - 2*Log[x])/x,x]

[Out]

-E^x - 15*x - 2*x^2 - (1 + Log[x])^2

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fricas [A]  time = 0.88, size = 23, normalized size = 0.96 \begin {gather*} -2 \, x^{2} - \log \relax (x)^{2} - 15 \, x - e^{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)-exp(x)*x-4*x^2-15*x-2)/x,x, algorithm="fricas")

[Out]

-2*x^2 - log(x)^2 - 15*x - e^x - 2*log(x)

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giac [A]  time = 0.14, size = 23, normalized size = 0.96 \begin {gather*} -2 \, x^{2} - \log \relax (x)^{2} - 15 \, x - e^{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)-exp(x)*x-4*x^2-15*x-2)/x,x, algorithm="giac")

[Out]

-2*x^2 - log(x)^2 - 15*x - e^x - 2*log(x)

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maple [A]  time = 0.03, size = 24, normalized size = 1.00

method	result	size
default	\(-2 x^{2}-15 x -2 \ln \relax (x )-\ln \relax (x )^{2}-{\mathrm e}^{x}\)	\(24\)
norman	\(-2 x^{2}-15 x -2 \ln \relax (x )-\ln \relax (x )^{2}-{\mathrm e}^{x}\)	\(24\)
risch	\(-2 x^{2}-15 x -2 \ln \relax (x )-\ln \relax (x )^{2}-{\mathrm e}^{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(x)-exp(x)*x-4*x^2-15*x-2)/x,x,method=_RETURNVERBOSE)

[Out]

-2*x^2-15*x-2*ln(x)-ln(x)^2-exp(x)

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maxima [A]  time = 0.36, size = 23, normalized size = 0.96 \begin {gather*} -2 \, x^{2} - \log \relax (x)^{2} - 15 \, x - e^{x} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)-exp(x)*x-4*x^2-15*x-2)/x,x, algorithm="maxima")

[Out]

-2*x^2 - log(x)^2 - 15*x - e^x - 2*log(x)

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mupad [B]  time = 5.16, size = 23, normalized size = 0.96 \begin {gather*} -15\,x-{\mathrm {e}}^x-2\,\ln \relax (x)-{\ln \relax (x)}^2-2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x + 2*log(x) + x*exp(x) + 4*x^2 + 2)/x,x)

[Out]

- 15*x - exp(x) - 2*log(x) - log(x)^2 - 2*x^2

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sympy [A]  time = 0.27, size = 22, normalized size = 0.92 \begin {gather*} - 2 x^{2} - 15 x - e^{x} - \log {\relax (x )}^{2} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(x)-exp(x)*x-4*x**2-15*x-2)/x,x)

[Out]

-2*x**2 - 15*x - exp(x) - log(x)**2 - 2*log(x)

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