3.54.95 \(\int \frac {-10-50 x+5 e^{-5+x} x-400 x^2}{x} \, dx\)

Optimal. Leaf size=24 \[ 5 \left (e^{-5+x}+10 (3-4 x) (1+x)-\log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14, 2194} \begin {gather*} -200 x^2-50 x+5 e^{x-5}-10 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 - 50*x + 5*E^(-5 + x)*x - 400*x^2)/x,x]

[Out]

5*E^(-5 + x) - 50*x - 200*x^2 - 10*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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} &=\int \left (5 e^{-5+x}-\frac {10 \left (1+5 x+40 x^2\right )}{x}\right ) \, dx\\ &=5 \int e^{-5+x} \, dx-10 \int \frac {1+5 x+40 x^2}{x} \, dx\\ &=5 e^{-5+x}-10 \int \left (5+\frac {1}{x}+40 x\right ) \, dx\\ &=5 e^{-5+x}-50 x-200 x^2-10 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.83 \begin {gather*} 5 \left (e^{-5+x}-10 x-40 x^2-2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 50*x + 5*E^(-5 + x)*x - 400*x^2)/x,x]

[Out]

5*(E^(-5 + x) - 10*x - 40*x^2 - 2*Log[x])

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fricas [A]  time = 0.47, size = 19, normalized size = 0.79 \begin {gather*} -200 \, x^{2} - 50 \, x + 5 \, e^{\left (x - 5\right )} - 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x-5)-400*x^2-50*x-10)/x,x, algorithm="fricas")

[Out]

-200*x^2 - 50*x + 5*e^(x - 5) - 10*log(x)

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giac [A]  time = 0.21, size = 27, normalized size = 1.12 \begin {gather*} -5 \, {\left (40 \, x^{2} e^{5} + 10 \, x e^{5} + 2 \, e^{5} \log \relax (x) - e^{x}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x-5)-400*x^2-50*x-10)/x,x, algorithm="giac")

[Out]

-5*(40*x^2*e^5 + 10*x*e^5 + 2*e^5*log(x) - e^x)*e^(-5)

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maple [A]  time = 0.07, size = 20, normalized size = 0.83

method	result	size
norman	\(-50 x -200 x^{2}+5 \,{\mathrm e}^{x -5}-10 \ln \relax (x )\)	\(20\)
risch	\(-50 x -200 x^{2}+5 \,{\mathrm e}^{x -5}-10 \ln \relax (x )\)	\(20\)
derivativedivides	\(5 \,{\mathrm e}^{x -5}-10 \ln \relax (x )-2050 x [phone] \left (x -5\right )^{2}\)	\(23\)
default	\(5 \,{\mathrm e}^{x -5}-10 \ln \relax (x )-2050 x [phone] \left (x -5\right )^{2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*exp(x-5)-400*x^2-50*x-10)/x,x,method=_RETURNVERBOSE)

[Out]

-50*x-200*x^2+5*exp(x-5)-10*ln(x)

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maxima [A]  time = 0.37, size = 19, normalized size = 0.79 \begin {gather*} -200 \, x^{2} - 50 \, x + 5 \, e^{\left (x - 5\right )} - 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x-5)-400*x^2-50*x-10)/x,x, algorithm="maxima")

[Out]

-200*x^2 - 50*x + 5*e^(x - 5) - 10*log(x)

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mupad [B]  time = 3.49, size = 19, normalized size = 0.79 \begin {gather*} 5\,{\mathrm {e}}^{x-5}-50\,x-10\,\ln \relax (x)-200\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*x - 5*x*exp(x - 5) + 400*x^2 + 10)/x,x)

[Out]

5*exp(x - 5) - 50*x - 10*log(x) - 200*x^2

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sympy [A]  time = 0.10, size = 19, normalized size = 0.79 \begin {gather*} - 200 x^{2} - 50 x + 5 e^{x - 5} - 10 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x-5)-400*x**2-50*x-10)/x,x)

[Out]

-200*x**2 - 50*x + 5*exp(x - 5) - 10*log(x)

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