3.61.2 \(\int \frac {1}{25} (-26-25 e^x+25 \log (3)) \, dx\)

Optimal. Leaf size=25 \[ 2-e^x+\frac {1}{25} (-1-x)-x+x \log (3)+\log (4) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 2194} \begin {gather*} -e^x-\frac {1}{25} x (26-25 \log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-26 - 25*E^x + 25*Log[3])/25,x]

[Out]

-E^x - (x*(26 - 25*Log[3]))/25

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \left (-26-25 e^x+25 \log (3)\right ) \, dx\\ &=-\frac {1}{25} x (26-25 \log (3))-\int e^x \, dx\\ &=-e^x-\frac {1}{25} x (26-25 \log (3))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.60 \begin {gather*} -e^x-\frac {26 x}{25}+x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-26 - 25*E^x + 25*Log[3])/25,x]

[Out]

-E^x - (26*x)/25 + x*Log[3]

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fricas [A]  time = 0.63, size = 12, normalized size = 0.48 \begin {gather*} x \log \relax (3) - \frac {26}{25} \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)+log(3)-26/25,x, algorithm="fricas")

[Out]

x*log(3) - 26/25*x - e^x

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giac [A]  time = 0.13, size = 12, normalized size = 0.48 \begin {gather*} x \log \relax (3) - \frac {26}{25} \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)+log(3)-26/25,x, algorithm="giac")

[Out]

x*log(3) - 26/25*x - e^x

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maple [A]  time = 0.04, size = 12, normalized size = 0.48

method	result	size
norman	\(\left (\ln \relax (3)-\frac {26}{25}\right ) x -{\mathrm e}^{x}\)	\(12\)
default	\(-\frac {26 x}{25}-{\mathrm e}^{x}+x \ln \relax (3)\)	\(13\)
risch	\(-\frac {26 x}{25}-{\mathrm e}^{x}+x \ln \relax (3)\)	\(13\)
derivativedivides	\(-{\mathrm e}^{x}+\frac {\left (25 \ln \relax (3)-26\right ) \ln \left ({\mathrm e}^{x}\right )}{25}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x)+ln(3)-26/25,x,method=_RETURNVERBOSE)

[Out]

(ln(3)-26/25)*x-exp(x)

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maxima [A]  time = 0.34, size = 12, normalized size = 0.48 \begin {gather*} x \log \relax (3) - \frac {26}{25} \, x - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)+log(3)-26/25,x, algorithm="maxima")

[Out]

x*log(3) - 26/25*x - e^x

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mupad [B]  time = 0.05, size = 11, normalized size = 0.44 \begin {gather*} x\,\left (\ln \relax (3)-\frac {26}{25}\right )-{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(3) - exp(x) - 26/25,x)

[Out]

x*(log(3) - 26/25) - exp(x)

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sympy [A]  time = 0.11, size = 10, normalized size = 0.40 \begin {gather*} x \left (- \frac {26}{25} + \log {\relax (3 )}\right ) - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(x)+ln(3)-26/25,x)

[Out]

x*(-26/25 + log(3)) - exp(x)

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