3.74.56 \(\int (-10 e^8 x-5 \log (x)) \, dx\)

Optimal. Leaf size=18 \[ 5 \left (2+x-e^8 x^2-x \log (x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2295} \begin {gather*} -5 e^8 x^2+5 x-5 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-10*E^8*x - 5*Log[x],x]

[Out]

5*x - 5*E^8*x^2 - 5*x*Log[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} &=-5 e^8 x^2-5 \int \log (x) \, dx\\ &=5 x-5 e^8 x^2-5 x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.94 \begin {gather*} 5 x-5 e^8 x^2-5 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-10*E^8*x - 5*Log[x],x]

[Out]

5*x - 5*E^8*x^2 - 5*x*Log[x]

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fricas [A]  time = 0.72, size = 16, normalized size = 0.89 \begin {gather*} -5 \, x^{2} e^{8} - 5 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(x)-10*x*exp(4)^2,x, algorithm="fricas")

[Out]

-5*x^2*e^8 - 5*x*log(x) + 5*x

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giac [A]  time = 0.13, size = 16, normalized size = 0.89 \begin {gather*} -5 \, x^{2} e^{8} - 5 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(x)-10*x*exp(4)^2,x, algorithm="giac")

[Out]

-5*x^2*e^8 - 5*x*log(x) + 5*x

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maple [A]  time = 0.02, size = 17, normalized size = 0.94

method	result	size
risch	\(-5 x^{2} {\mathrm e}^{8}-5 x \ln \relax (x )+5 x\)	\(17\)
default	\(-5 x^{2} {\mathrm e}^{8}-5 x \ln \relax (x )+5 x\)	\(19\)
norman	\(-5 x^{2} {\mathrm e}^{8}-5 x \ln \relax (x )+5 x\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*ln(x)-10*x*exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

-5*x^2*exp(8)-5*x*ln(x)+5*x

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maxima [A]  time = 0.35, size = 16, normalized size = 0.89 \begin {gather*} -5 \, x^{2} e^{8} - 5 \, x \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*log(x)-10*x*exp(4)^2,x, algorithm="maxima")

[Out]

-5*x^2*e^8 - 5*x*log(x) + 5*x

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mupad [B]  time = 4.51, size = 11, normalized size = 0.61 \begin {gather*} -5\,x\,\left (\ln \relax (x)+x\,{\mathrm {e}}^8-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- 5*log(x) - 10*x*exp(8),x)

[Out]

-5*x*(log(x) + x*exp(8) - 1)

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sympy [A]  time = 0.09, size = 17, normalized size = 0.94 \begin {gather*} - 5 x^{2} e^{8} - 5 x \log {\relax (x )} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*ln(x)-10*x*exp(4)**2,x)

[Out]

-5*x**2*exp(8) - 5*x*log(x) + 5*x

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