3.51.13 \(\int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx\)

Optimal. Leaf size=12 \[ 5 x \left (-\frac {25}{6}-2 x+\log (4)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.25, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {9} \begin {gather*} -\frac {5}{288} (24 x+25-6 \log (4))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-125 - 120*x + 30*Log[4])/6,x]

[Out]

(-5*(25 + 24*x - 6*Log[4])^2)/288

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 {5}{288} (25+24 x-6 \log (4))^2\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 16, normalized size = 1.33 \begin {gather*} -\frac {125 x}{6}-10 x^2+5 x \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-125 - 120*x + 30*Log[4])/6,x]

[Out]

(-125*x)/6 - 10*x^2 + 5*x*Log[4]

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 14, normalized size = 1.17 \begin {gather*} -10 \, x^{2} + 10 \, x \log \relax (2) - \frac {125}{6} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*log(2)-20*x-125/6,x, algorithm="fricas")

[Out]

-10*x^2 + 10*x*log(2) - 125/6*x

________________________________________________________________________________________

giac [A]  time = 0.12, size = 14, normalized size = 1.17 \begin {gather*} -10 \, x^{2} + 10 \, x \log \relax (2) - \frac {125}{6} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*log(2)-20*x-125/6,x, algorithm="giac")

[Out]

-10*x^2 + 10*x*log(2) - 125/6*x

________________________________________________________________________________________

maple [A]  time = 0.03, size = 13, normalized size = 1.08

method	result	size
gosper	\(\frac {5 x \left (-12 x +12 \ln \relax (2)-25\right )}{6}\)	\(13\)
default	\(10 x \ln \relax (2)-10 x^{2}-\frac {125 x}{6}\)	\(15\)
norman	\(\left (10 \ln \relax (2)-\frac {125}{6}\right ) x -10 x^{2}\)	\(15\)
risch	\(10 x \ln \relax (2)-10 x^{2}-\frac {125 x}{6}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*ln(2)-20*x-125/6,x,method=_RETURNVERBOSE)

[Out]

5/6*x*(-12*x+12*ln(2)-25)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 14, normalized size = 1.17 \begin {gather*} -10 \, x^{2} + 10 \, x \log \relax (2) - \frac {125}{6} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*log(2)-20*x-125/6,x, algorithm="maxima")

[Out]

-10*x^2 + 10*x*log(2) - 125/6*x

________________________________________________________________________________________

mupad [B]  time = 3.17, size = 14, normalized size = 1.17 \begin {gather*} x\,\left (10\,\ln \relax (2)-\frac {125}{6}\right )-10\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*log(2) - 20*x - 125/6,x)

[Out]

x*(10*log(2) - 125/6) - 10*x^2

________________________________________________________________________________________

sympy [A]  time = 0.05, size = 14, normalized size = 1.17 \begin {gather*} - 10 x^{2} + x \left (- \frac {125}{6} + 10 \log {\relax (2 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*ln(2)-20*x-125/6,x)

[Out]

-10*x**2 + x*(-125/6 + 10*log(2))

________________________________________________________________________________________