3.95.1 \(\int (e^{5 x} (1+5 x)+e^x (-1-x) \log (4)) \, dx\)

Optimal. Leaf size=20 \[ x \left (e^{5 x}+\frac {3}{x}-e^x \log (4)\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2176, 2194} \begin {gather*} -\frac {e^{5 x}}{5}+\frac {1}{5} e^{5 x} (5 x+1)-e^x (x+1) \log (4)+e^x \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*E^(5*x) + (E^(5*x)*(1 + 5*x))/5 + E^x*Log[4] - E^x*(1 + x)*Log[4]

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.80 \begin {gather*} e^{5 x} x-e^x x \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(5*x)*x - E^x*x*Log[4]

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fricas [A]  time = 0.61, size = 14, normalized size = 0.70 \begin {gather*} -2 \, x e^{x} \log \relax (2) + x e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^x*log(2) + x*e^(5*x)

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giac [A]  time = 0.19, size = 14, normalized size = 0.70 \begin {gather*} -2 \, x e^{x} \log \relax (2) + x e^{\left (5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^x*log(2) + x*e^(5*x)

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maple [A]  time = 0.04, size = 15, normalized size = 0.75

method	result	size
default	\(x \,{\mathrm e}^{5 x}-2 x \ln \relax (2) {\mathrm e}^{x}\)	\(15\)
norman	\(x \,{\mathrm e}^{5 x}-2 x \ln \relax (2) {\mathrm e}^{x}\)	\(15\)
risch	\(x \,{\mathrm e}^{5 x}-2 x \ln \relax (2) {\mathrm e}^{x}\)	\(15\)
meijerg	\(\frac {{\mathrm e}^{5 x}}{5}-\frac {\left (2-10 x \right ) {\mathrm e}^{5 x}}{10}+2 \left (1-{\mathrm e}^{x}\right ) \ln \relax (2)-2 \ln \relax (2) \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(5*x)-2*x*ln(2)*exp(x)

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maxima [A]  time = 0.35, size = 20, normalized size = 1.00 \begin {gather*} x e^{\left (5 \, x\right )} - 2 \, {\left ({\left (x - 1\right )} e^{x} + e^{x}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.13, size = 13, normalized size = 0.65 \begin {gather*} x\,{\mathrm {e}}^x\,\left ({\mathrm {e}}^{4\,x}-2\,\ln \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(x)*(exp(4*x) - 2*log(2))

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sympy [A]  time = 0.14, size = 15, normalized size = 0.75 \begin {gather*} x e^{5 x} - 2 x e^{x} \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(5*x) - 2*x*exp(x)*log(2)

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