\(\int \frac {1}{3} (15-4 \log (176)) \, dx\) [1618]

Optimal result

Integrand size = 10, antiderivative size = 12 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=-1+5 x-\frac {4}{3} x \log (176) \]

[Out]

-4/3*x*ln(176)+5*x-1

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {8} \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=\frac {1}{3} x (15-4 \log (176)) \]

[In]

Int[(15 - 4*Log[176])/3,x]

[Out]

(x*(15 - 4*Log[176]))/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x (15-4 \log (176)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=5 x-\frac {4}{3} x \log (176) \]

[In]

Integrate[(15 - 4*Log[176])/3,x]

[Out]

5*x - (4*x*Log[176])/3

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75

method	result	size
norman	\(\left (-\frac {4 \ln \left (176\right )}{3}+5\right ) x\)	\(9\)
parallelrisch	\(\left (-\frac {4 \ln \left (176\right )}{3}+5\right ) x\)	\(9\)
default	\(\frac {x \left (-4 \ln \left (176\right )+15\right )}{3}\)	\(10\)
parts	\(5 x -\frac {4 x \ln \left (176\right )}{3}\)	\(10\)
risch	\(-\frac {16 x \ln \left (2\right )}{3}-\frac {4 x \ln \left (11\right )}{3}+5 x\)	\(15\)

[In]

int(-4/3*ln(176)+5,x,method=_RETURNVERBOSE)

[Out]

(-4/3*ln(176)+5)*x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=-\frac {4}{3} \, x \log \left (176\right ) + 5 \, x \]

[In]

integrate(-4/3*log(176)+5,x, algorithm="fricas")

[Out]

-4/3*x*log(176) + 5*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=x \left (5 - \frac {4 \log {\left (176 \right )}}{3}\right ) \]

[In]

integrate(-4/3*ln(176)+5,x)

[Out]

x*(5 - 4*log(176)/3)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=-\frac {1}{3} \, x {\left (4 \, \log \left (176\right ) - 15\right )} \]

[In]

integrate(-4/3*log(176)+5,x, algorithm="maxima")

[Out]

-1/3*x*(4*log(176) - 15)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=-\frac {1}{3} \, x {\left (4 \, \log \left (176\right ) - 15\right )} \]

[In]

integrate(-4/3*log(176)+5,x, algorithm="giac")

[Out]

-1/3*x*(4*log(176) - 15)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{3} (15-4 \log (176)) \, dx=-x\,\left (\frac {4\,\ln \left (176\right )}{3}-5\right ) \]

[In]

int(5 - (4*log(176))/3,x)

[Out]

-x*((4*log(176))/3 - 5)