3.9.62 \(\int \frac {1}{16} (1-16 e^{4+x}+\log (-\frac {x}{20})) \, dx\)

Optimal. Leaf size=19 \[ -e^{4+x}+\frac {1}{16} x \log \left (-\frac {x}{20}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2194, 2295} \begin {gather*} \frac {1}{16} x \log \left (-\frac {x}{20}\right )-e^{x+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 16*E^(4 + x) + Log[-1/20*x])/16,x]

[Out]

-E^(4 + x) + (x*Log[-1/20*x])/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \left (1-16 e^{4+x}+\log \left (-\frac {x}{20}\right )\right ) \, dx\\ &=\frac {x}{16}+\frac {1}{16} \int \log \left (-\frac {x}{20}\right ) \, dx-\int e^{4+x} \, dx\\ &=-e^{4+x}+\frac {1}{16} x \log \left (-\frac {x}{20}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.05 \begin {gather*} \frac {1}{16} \left (-16 e^{4+x}+x \log \left (-\frac {x}{20}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 16*E^(4 + x) + Log[-1/20*x])/16,x]

[Out]

(-16*E^(4 + x) + x*Log[-1/20*x])/16

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fricas [A]  time = 0.58, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{16} \, x \log \left (-\frac {1}{20} \, x\right ) - e^{\left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*log(-1/20*x)-exp(4+x)+1/16,x, algorithm="fricas")

[Out]

1/16*x*log(-1/20*x) - e^(x + 4)

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giac [A]  time = 0.40, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{16} \, x \log \left (-\frac {1}{20} \, x\right ) - e^{\left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*log(-1/20*x)-exp(4+x)+1/16,x, algorithm="giac")

[Out]

1/16*x*log(-1/20*x) - e^(x + 4)

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maple [A]  time = 0.01, size = 15, normalized size = 0.79

method	result	size
default	\(\frac {x \ln \left (-\frac {x}{20}\right )}{16}-{\mathrm e}^{4+x}\)	\(15\)
norman	\(\frac {x \ln \left (-\frac {x}{20}\right )}{16}-{\mathrm e}^{4+x}\)	\(15\)
risch	\(\frac {x \ln \left (-\frac {x}{20}\right )}{16}-{\mathrm e}^{4+x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*ln(-1/20*x)-exp(4+x)+1/16,x,method=_RETURNVERBOSE)

[Out]

1/16*x*ln(-1/20*x)-exp(4+x)

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maxima [A]  time = 0.66, size = 14, normalized size = 0.74 \begin {gather*} \frac {1}{16} \, x \log \left (-\frac {1}{20} \, x\right ) - e^{\left (x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*log(-1/20*x)-exp(4+x)+1/16,x, algorithm="maxima")

[Out]

1/16*x*log(-1/20*x) - e^(x + 4)

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mupad [B]  time = 0.63, size = 14, normalized size = 0.74 \begin {gather*} \frac {x\,\ln \left (-\frac {x}{20}\right )}{16}-{\mathrm {e}}^{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(-x/20)/16 - exp(x + 4) + 1/16,x)

[Out]

(x*log(-x/20))/16 - exp(x + 4)

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sympy [A]  time = 0.24, size = 14, normalized size = 0.74 \begin {gather*} \frac {x \log {\left (- \frac {x}{20} \right )}}{16} - e^{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*ln(-1/20*x)-exp(4+x)+1/16,x)

[Out]

x*log(-x/20)/16 - exp(x + 4)

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