3.102.62 \(\int \frac {4 e^{10}-4 e^x \log (5)}{\log (5)} \, dx\)

Optimal. Leaf size=17 \[ 4 \left (-e^x+\frac {e^{10} x}{\log (5)}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2194} \begin {gather*} \frac {4 e^{10} x}{\log (5)}-4 e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^10 - 4*E^x*Log[5])/Log[5],x]

[Out]

-4*E^x + (4*E^10*x)/Log[5]

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 {\int \left (4 e^{10}-4 e^x \log (5)\right ) \, dx}{\log (5)}\\ &=\frac {4 e^{10} x}{\log (5)}-4 \int e^x \, dx\\ &=-4 e^x+\frac {4 e^{10} x}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.94 \begin {gather*} -4 e^x+\frac {4 e^{10} x}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^10 - 4*E^x*Log[5])/Log[5],x]

[Out]

-4*E^x + (4*E^10*x)/Log[5]

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fricas [A]  time = 0.57, size = 17, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x e^{10} - e^{x} \log \relax (5)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*log(5)+4*exp(10))/log(5),x, algorithm="fricas")

[Out]

4*(x*e^10 - e^x*log(5))/log(5)

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giac [A]  time = 0.12, size = 17, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x e^{10} - e^{x} \log \relax (5)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*log(5)+4*exp(10))/log(5),x, algorithm="giac")

[Out]

4*(x*e^10 - e^x*log(5))/log(5)

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maple [A]  time = 0.08, size = 15, normalized size = 0.88

method	result	size
norman	\(\frac {4 \,{\mathrm e}^{10} x}{\ln \relax (5)}-4 \,{\mathrm e}^{x}\)	\(15\)
risch	\(\frac {4 \,{\mathrm e}^{10} x}{\ln \relax (5)}-4 \,{\mathrm e}^{x}\)	\(15\)
default	\(\frac {-4 \,{\mathrm e}^{x} \ln \relax (5)+4 x \,{\mathrm e}^{10}}{\ln \relax (5)}\)	\(18\)
derivativedivides	\(\frac {-4 \,{\mathrm e}^{x} \ln \relax (5)+4 \,{\mathrm e}^{10} \ln \left ({\mathrm e}^{x}\right )}{\ln \relax (5)}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(x)*ln(5)+4*exp(10))/ln(5),x,method=_RETURNVERBOSE)

[Out]

4*exp(10)/ln(5)*x-4*exp(x)

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maxima [A]  time = 0.34, size = 17, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x e^{10} - e^{x} \log \relax (5)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*log(5)+4*exp(10))/log(5),x, algorithm="maxima")

[Out]

4*(x*e^10 - e^x*log(5))/log(5)

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mupad [B]  time = 6.31, size = 14, normalized size = 0.82 \begin {gather*} \frac {4\,x\,{\mathrm {e}}^{10}}{\ln \relax (5)}-4\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(10) - 4*exp(x)*log(5))/log(5),x)

[Out]

(4*x*exp(10))/log(5) - 4*exp(x)

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sympy [A]  time = 0.09, size = 14, normalized size = 0.82 \begin {gather*} \frac {4 x e^{10}}{\log {\relax (5 )}} - 4 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*ln(5)+4*exp(10))/ln(5),x)

[Out]

4*x*exp(10)/log(5) - 4*exp(x)

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