3.38.91 \(\int \frac {64-16 e^9+32 x}{\log (\log (4))} \, dx\)

Optimal. Leaf size=16 \[ \frac {16 x \left (4-e^9+x\right )}{\log (\log (4))} \]

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Rubi [A]  time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.19, 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 {4 \left (2 x-e^9+4\right )^2}{\log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64 - 16*E^9 + 32*x)/Log[Log[4]],x]

[Out]

(4*(4 - E^9 + 2*x)^2)/Log[Log[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 {4 \left (4-e^9+2 x\right )^2}{\log (\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.31 \begin {gather*} -\frac {16 \left (-4 x+e^9 x-x^2\right )}{\log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64 - 16*E^9 + 32*x)/Log[Log[4]],x]

[Out]

(-16*(-4*x + E^9*x - x^2))/Log[Log[4]]

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fricas [A]  time = 0.72, size = 21, normalized size = 1.31 \begin {gather*} \frac {16 \, {\left (x^{2} - x e^{9} + 4 \, x\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*exp(9)+32*x+64)/log(2*log(2)),x, algorithm="fricas")

[Out]

16*(x^2 - x*e^9 + 4*x)/log(2*log(2))

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giac [A]  time = 0.23, size = 21, normalized size = 1.31 \begin {gather*} \frac {16 \, {\left (x^{2} - x e^{9} + 4 \, x\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*exp(9)+32*x+64)/log(2*log(2)),x, algorithm="giac")

[Out]

16*(x^2 - x*e^9 + 4*x)/log(2*log(2))

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maple [A]  time = 0.03, size = 18, normalized size = 1.12

method	result	size
gosper	\(-\frac {16 x \left (-x +{\mathrm e}^{9}-4\right )}{\ln \left (2 \ln \relax (2)\right )}\)	\(18\)
default	\(\frac {-16 x \,{\mathrm e}^{9}+16 x^{2}+64 x}{\ln \left (2 \ln \relax (2)\right )}\)	\(23\)
norman	\(\frac {16 x^{2}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}-\frac {16 \left ({\mathrm e}^{9}-4\right ) x}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\)	\(30\)
risch	\(-\frac {16 x \,{\mathrm e}^{9}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}+\frac {16 x^{2}}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}+\frac {64 x}{\ln \relax (2)+\ln \left (\ln \relax (2)\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*exp(9)+32*x+64)/ln(2*ln(2)),x,method=_RETURNVERBOSE)

[Out]

-16*x*(-x+exp(9)-4)/ln(2*ln(2))

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maxima [A]  time = 0.35, size = 21, normalized size = 1.31 \begin {gather*} \frac {16 \, {\left (x^{2} - x e^{9} + 4 \, x\right )}}{\log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*exp(9)+32*x+64)/log(2*log(2)),x, algorithm="maxima")

[Out]

16*(x^2 - x*e^9 + 4*x)/log(2*log(2))

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mupad [B]  time = 2.18, size = 18, normalized size = 1.12 \begin {gather*} \frac {{\left (32\,x-16\,{\mathrm {e}}^9+64\right )}^2}{64\,\ln \left (\ln \relax (4)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x - 16*exp(9) + 64)/log(2*log(2)),x)

[Out]

(32*x - 16*exp(9) + 64)^2/(64*log(log(4)))

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sympy [A]  time = 0.06, size = 29, normalized size = 1.81 \begin {gather*} \frac {16 x^{2}}{\log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}} + \frac {x \left (64 - 16 e^{9}\right )}{\log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*exp(9)+32*x+64)/ln(2*ln(2)),x)

[Out]

16*x**2/(log(log(2)) + log(2)) + x*(64 - 16*exp(9))/(log(log(2)) + log(2))

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