3.24.25 \(\int (-4+e^{2 x} (-25-50 x)-4 x-5 \log (5 x)) \, dx\)

Optimal. Leaf size=23 \[ -4+x-x \left (25 e^{2 x}+2 x+5 \log (5 x)\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2176, 2194, 2295} \begin {gather*} -2 x^2+x+\frac {25 e^{2 x}}{2}-\frac {25}{2} e^{2 x} (2 x+1)-5 x \log (5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-4 + E^(2*x)*(-25 - 50*x) - 4*x - 5*Log[5*x],x]

[Out]

(25*E^(2*x))/2 + x - 2*x^2 - (25*E^(2*x)*(1 + 2*x))/2 - 5*x*Log[5*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]

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} &=-4 x-2 x^2-5 \int \log (5 x) \, dx+\int e^{2 x} (-25-50 x) \, dx\\ &=x-2 x^2-\frac {25}{2} e^{2 x} (1+2 x)-5 x \log (5 x)+25 \int e^{2 x} \, dx\\ &=\frac {25 e^{2 x}}{2}+x-2 x^2-\frac {25}{2} e^{2 x} (1+2 x)-5 x \log (5 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.96 \begin {gather*} x-25 e^{2 x} x-2 x^2-5 x \log (5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-4 + E^(2*x)*(-25 - 50*x) - 4*x - 5*Log[5*x],x]

[Out]

x - 25*E^(2*x)*x - 2*x^2 - 5*x*Log[5*x]

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fricas [A]  time = 0.56, size = 21, normalized size = 0.91 \begin {gather*} -2 \, x^{2} - 25 \, x e^{\left (2 \, x\right )} - 5 \, x \log \left (5 \, x\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(5*x)+(-50*x-25)*exp(x)^2-4*x-4,x, algorithm="fricas")

[Out]

-2*x^2 - 25*x*e^(2*x) - 5*x*log(5*x) + x

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giac [A]  time = 0.58, size = 21, normalized size = 0.91 \begin {gather*} -2 \, x^{2} - 25 \, x e^{\left (2 \, x\right )} - 5 \, x \log \left (5 \, x\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(5*x)+(-50*x-25)*exp(x)^2-4*x-4,x, algorithm="giac")

[Out]

-2*x^2 - 25*x*e^(2*x) - 5*x*log(5*x) + x

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maple [A]  time = 0.03, size = 22, normalized size = 0.96

method	result	size
default	\(x -25 x \,{\mathrm e}^{2 x}-2 x^{2}-5 x \ln \left (5 x \right )\)	\(22\)
norman	\(x -25 x \,{\mathrm e}^{2 x}-2 x^{2}-5 x \ln \left (5 x \right )\)	\(22\)
risch	\(x -25 x \,{\mathrm e}^{2 x}-2 x^{2}-5 x \ln \left (5 x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*ln(5*x)+(-50*x-25)*exp(x)^2-4*x-4,x,method=_RETURNVERBOSE)

[Out]

x-25*x*exp(x)^2-2*x^2-5*x*ln(5*x)

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maxima [A]  time = 0.56, size = 21, normalized size = 0.91 \begin {gather*} -2 \, x^{2} - 25 \, x e^{\left (2 \, x\right )} - 5 \, x \log \left (5 \, x\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(5*x)+(-50*x-25)*exp(x)^2-4*x-4,x, algorithm="maxima")

[Out]

-2*x^2 - 25*x*e^(2*x) - 5*x*log(5*x) + x

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mupad [B]  time = 1.39, size = 22, normalized size = 0.96 \begin {gather*} -x\,\left (2\,x+25\,{\mathrm {e}}^{2\,x}+5\,\ln \relax (5)+5\,\ln \relax (x)-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- 4*x - 5*log(5*x) - exp(2*x)*(50*x + 25) - 4,x)

[Out]

-x*(2*x + 25*exp(2*x) + 5*log(5) + 5*log(x) - 1)

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sympy [A]  time = 0.26, size = 22, normalized size = 0.96 \begin {gather*} - 2 x^{2} - 25 x e^{2 x} - 5 x \log {\left (5 x \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*ln(5*x)+(-50*x-25)*exp(x)**2-4*x-4,x)

[Out]

-2*x**2 - 25*x*exp(2*x) - 5*x*log(5*x) + x

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