3.47.61 \(\int (-1+16 e^4+2 x) \log (5+\log (16)) \, dx\)

Optimal. Leaf size=20 \[ \left (2-x+x \left (16 e^4+x\right )\right ) \log (5+\log (16)) \]

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Rubi [A]  time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {9} \begin {gather*} \frac {1}{4} \left (-2 x-16 e^4+1\right )^2 \log (5+\log (16)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 16*E^4 + 2*x)*Log[5 + Log[16]],x]

[Out]

((1 - 16*E^4 - 2*x)^2*Log[5 + Log[16]])/4

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \left (1-16 e^4-2 x\right )^2 \log (5+\log (16))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 0.95 \begin {gather*} \left (-x+16 e^4 x+x^2\right ) \log (5+\log (16)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 16*E^4 + 2*x)*Log[5 + Log[16]],x]

[Out]

(-x + 16*E^4*x + x^2)*Log[5 + Log[16]]

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fricas [A]  time = 1.01, size = 20, normalized size = 1.00 \begin {gather*} {\left (x^{2} + 16 \, x e^{4} - x\right )} \log \left (4 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(4)+2*x-1)*log(4*log(2)+5),x, algorithm="fricas")

[Out]

(x^2 + 16*x*e^4 - x)*log(4*log(2) + 5)

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giac [A]  time = 0.21, size = 20, normalized size = 1.00 \begin {gather*} {\left (x^{2} + 16 \, x e^{4} - x\right )} \log \left (4 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(4)+2*x-1)*log(4*log(2)+5),x, algorithm="giac")

[Out]

(x^2 + 16*x*e^4 - x)*log(4*log(2) + 5)

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maple [A]  time = 0.02, size = 17, normalized size = 0.85

method	result	size
gosper	\(x \left (x +16 \,{\mathrm e}^{4}-1\right ) \ln \left (4 \ln \relax (2)+5\right )\)	\(17\)
default	\(\ln \left (4 \ln \relax (2)+5\right ) \left (16 x \,{\mathrm e}^{4}+x^{2}-x \right )\)	\(21\)
risch	\(16 \ln \left (4 \ln \relax (2)+5\right ) x \,{\mathrm e}^{4}+\ln \left (4 \ln \relax (2)+5\right ) x^{2}-\ln \left (4 \ln \relax (2)+5\right ) x\)	\(35\)
norman	\(\left (16 \ln \left (4 \ln \relax (2)+5\right ) {\mathrm e}^{4}-\ln \left (4 \ln \relax (2)+5\right )\right ) x +\ln \left (4 \ln \relax (2)+5\right ) x^{2}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*exp(4)+2*x-1)*ln(4*ln(2)+5),x,method=_RETURNVERBOSE)

[Out]

x*(x+16*exp(4)-1)*ln(4*ln(2)+5)

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maxima [A]  time = 0.37, size = 20, normalized size = 1.00 \begin {gather*} {\left (x^{2} + 16 \, x e^{4} - x\right )} \log \left (4 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(4)+2*x-1)*log(4*log(2)+5),x, algorithm="maxima")

[Out]

(x^2 + 16*x*e^4 - x)*log(4*log(2) + 5)

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mupad [B]  time = 0.21, size = 18, normalized size = 0.90 \begin {gather*} \frac {\ln \left (\ln \left (16\right )+5\right )\,{\left (2\,x+16\,{\mathrm {e}}^4-1\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(4*log(2) + 5)*(2*x + 16*exp(4) - 1),x)

[Out]

(log(log(16) + 5)*(2*x + 16*exp(4) - 1)^2)/4

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sympy [A]  time = 0.06, size = 34, normalized size = 1.70 \begin {gather*} x^{2} \log {\left (4 \log {\relax (2 )} + 5 \right )} + x \left (- \log {\left (4 \log {\relax (2 )} + 5 \right )} + 16 e^{4} \log {\left (4 \log {\relax (2 )} + 5 \right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(4)+2*x-1)*ln(4*ln(2)+5),x)

[Out]

x**2*log(4*log(2) + 5) + x*(-log(4*log(2) + 5) + 16*exp(4)*log(4*log(2) + 5))

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