3.77.92 \(\int (-902+e^{5 x} (-1-5 x)+2 x-\log (x)) \, dx\)

Optimal. Leaf size=16 \[ x \left (-901-e^{5 x}+x-\log (x)\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 2.19, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2176, 2194, 2295} \begin {gather*} x^2-901 x+\frac {e^{5 x}}{5}-\frac {1}{5} e^{5 x} (5 x+1)-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-902 + E^(5*x)*(-1 - 5*x) + 2*x - Log[x],x]

[Out]

E^(5*x)/5 - 901*x + x^2 - (E^(5*x)*(1 + 5*x))/5 - x*Log[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} &=-902 x+x^2+\int e^{5 x} (-1-5 x) \, dx-\int \log (x) \, dx\\ &=-901 x+x^2-\frac {1}{5} e^{5 x} (1+5 x)-x \log (x)+\int e^{5 x} \, dx\\ &=\frac {e^{5 x}}{5}-901 x+x^2-\frac {1}{5} e^{5 x} (1+5 x)-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.25 \begin {gather*} -901 x-e^{5 x} x+x^2-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-902 + E^(5*x)*(-1 - 5*x) + 2*x - Log[x],x]

[Out]

-901*x - E^(5*x)*x + x^2 - x*Log[x]

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fricas [A]  time = 0.71, size = 19, normalized size = 1.19 \begin {gather*} x^{2} - x e^{\left (5 \, x\right )} - x \log \relax (x) - 901 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)+(-5*x-1)*exp(5*x)+2*x-902,x, algorithm="fricas")

[Out]

x^2 - x*e^(5*x) - x*log(x) - 901*x

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giac [A]  time = 0.23, size = 19, normalized size = 1.19 \begin {gather*} x^{2} - x e^{\left (5 \, x\right )} - x \log \relax (x) - 901 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)+(-5*x-1)*exp(5*x)+2*x-902,x, algorithm="giac")

[Out]

x^2 - x*e^(5*x) - x*log(x) - 901*x

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maple [A]  time = 0.02, size = 20, normalized size = 1.25

method	result	size
default	\(-901 x -x \,{\mathrm e}^{5 x}+x^{2}-x \ln \relax (x )\)	\(20\)
norman	\(-901 x -x \,{\mathrm e}^{5 x}+x^{2}-x \ln \relax (x )\)	\(20\)
risch	\(-901 x -x \,{\mathrm e}^{5 x}+x^{2}-x \ln \relax (x )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(x)+(-5*x-1)*exp(5*x)+2*x-902,x,method=_RETURNVERBOSE)

[Out]

-901*x-x*exp(5*x)+x^2-x*ln(x)

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maxima [A]  time = 0.36, size = 19, normalized size = 1.19 \begin {gather*} x^{2} - x e^{\left (5 \, x\right )} - x \log \relax (x) - 901 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)+(-5*x-1)*exp(5*x)+2*x-902,x, algorithm="maxima")

[Out]

x^2 - x*e^(5*x) - x*log(x) - 901*x

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mupad [B]  time = 4.89, size = 14, normalized size = 0.88 \begin {gather*} -x\,\left ({\mathrm {e}}^{5\,x}-x+\ln \relax (x)+901\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - log(x) - exp(5*x)*(5*x + 1) - 902,x)

[Out]

-x*(exp(5*x) - x + log(x) + 901)

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sympy [A]  time = 0.25, size = 17, normalized size = 1.06 \begin {gather*} x^{2} - x e^{5 x} - x \log {\relax (x )} - 901 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(x)+(-5*x-1)*exp(5*x)+2*x-902,x)

[Out]

x**2 - x*exp(5*x) - x*log(x) - 901*x

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