3.71.29 \(\int \frac {-5+4 e^{-2 x} (-1+2 x)}{\log (15)} \, dx\)

Optimal. Leaf size=15 \[ \frac {\left (-5-4 e^{-2 x}\right ) x}{\log (15)} \]

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Rubi [B]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 2.33, 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, 2176, 2194} \begin {gather*} \frac {2 e^{-2 x} (1-2 x)}{\log (15)}-\frac {2 e^{-2 x}}{\log (15)}-\frac {5 x}{\log (15)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + (4*(-1 + 2*x))/E^(2*x))/Log[15],x]

[Out]

-2/(E^(2*x)*Log[15]) + (2*(1 - 2*x))/(E^(2*x)*Log[15]) - (5*x)/Log[15]

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 (-5+4 e^{-2 x} (-1+2 x)\right ) \, dx}{\log (15)}\\ &=-\frac {5 x}{\log (15)}+\frac {4 \int e^{-2 x} (-1+2 x) \, dx}{\log (15)}\\ &=\frac {2 e^{-2 x} (1-2 x)}{\log (15)}-\frac {5 x}{\log (15)}+\frac {4 \int e^{-2 x} \, dx}{\log (15)}\\ &=-\frac {2 e^{-2 x}}{\log (15)}+\frac {2 e^{-2 x} (1-2 x)}{\log (15)}-\frac {5 x}{\log (15)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.13 \begin {gather*} \frac {-5 x-4 e^{-2 x} x}{\log (15)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + (4*(-1 + 2*x))/E^(2*x))/Log[15],x]

[Out]

(-5*x - (4*x)/E^(2*x))/Log[15]

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fricas [A]  time = 0.54, size = 21, normalized size = 1.40 \begin {gather*} -\frac {x e^{\left (-2 \, x + 2 \, \log \relax (2)\right )} + 5 \, x}{\log \left (15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(2*log(2)-2*x)-5)/log(15),x, algorithm="fricas")

[Out]

-(x*e^(-2*x + 2*log(2)) + 5*x)/log(15)

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giac [A]  time = 0.23, size = 21, normalized size = 1.40 \begin {gather*} -\frac {x e^{\left (-2 \, x + 2 \, \log \relax (2)\right )} + 5 \, x}{\log \left (15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(2*log(2)-2*x)-5)/log(15),x, algorithm="giac")

[Out]

-(x*e^(-2*x + 2*log(2)) + 5*x)/log(15)

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maple [A]  time = 0.04, size = 25, normalized size = 1.67

method	result	size
norman	\(-\frac {5 x}{\ln \left (15\right )}-\frac {x \,{\mathrm e}^{2 \ln \relax (2)-2 x}}{\ln \left (15\right )}\)	\(25\)
risch	\(-\frac {4 \,{\mathrm e}^{-2 x} x}{\ln \relax (3)+\ln \relax (5)}-\frac {5 x}{\ln \relax (3)+\ln \relax (5)}\)	\(26\)
default	\(\frac {-5 x +\frac {{\mathrm e}^{2 \ln \relax (2)-2 x} \left (2 \ln \relax (2)-2 x \right )}{2}-{\mathrm e}^{2 \ln \relax (2)-2 x} \ln \relax (2)}{\ln \left (15\right )}\)	\(42\)
derivativedivides	\(-\frac {-10 \ln \relax (2)+10 x -{\mathrm e}^{2 \ln \relax (2)-2 x} \left (2 \ln \relax (2)-2 x \right )+2 \,{\mathrm e}^{2 \ln \relax (2)-2 x} \ln \relax (2)}{2 \ln \left (15\right )}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-1)*exp(2*ln(2)-2*x)-5)/ln(15),x,method=_RETURNVERBOSE)

[Out]

-5/ln(15)*x-1/ln(15)*x*exp(2*ln(2)-2*x)

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maxima [A]  time = 0.37, size = 27, normalized size = 1.80 \begin {gather*} -\frac {2 \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + 5 \, x - 2 \, e^{\left (-2 \, x\right )}}{\log \left (15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(2*log(2)-2*x)-5)/log(15),x, algorithm="maxima")

[Out]

-(2*(2*x + 1)*e^(-2*x) + 5*x - 2*e^(-2*x))/log(15)

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mupad [B]  time = 0.06, size = 15, normalized size = 1.00 \begin {gather*} -\frac {x\,\left (4\,{\mathrm {e}}^{-2\,x}+5\right )}{\ln \left (15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*log(2) - 2*x)*(2*x - 1) - 5)/log(15),x)

[Out]

-(x*(4*exp(-2*x) + 5))/log(15)

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sympy [A]  time = 0.11, size = 19, normalized size = 1.27 \begin {gather*} - \frac {5 x}{\log {\left (15 \right )}} - \frac {4 x e^{- 2 x}}{\log {\left (15 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-1)*exp(2*ln(2)-2*x)-5)/ln(15),x)

[Out]

-5*x/log(15) - 4*x*exp(-2*x)/log(15)

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