\(\int (225+45 \log (2)) \, dx\) [7510]

Optimal result

Integrand size = 6, antiderivative size = 11 \[ \int (225+45 \log (2)) \, dx=5 (2+9 x) (5+\log (2)) \]

[Out]

5*(ln(2)+5)*(9*x+2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {8} \[ \int (225+45 \log (2)) \, dx=45 x (5+\log (2)) \]

[In]

Int[225 + 45*Log[2],x]

[Out]

45*x*(5 + Log[2])

Rule 8

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

Rubi steps \begin{align*} \text {integral}& = 45 x (5+\log (2)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int (225+45 \log (2)) \, dx=225 x+45 x \log (2) \]

[In]

Integrate[225 + 45*Log[2],x]

[Out]

225*x + 45*x*Log[2]

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73

method	result	size
default	\(45 x \left (\ln \left (2\right )+5\right )\)	\(8\)
norman	\(\left (45 \ln \left (2\right )+225\right ) x\)	\(9\)
parallelrisch	\(\left (45 \ln \left (2\right )+225\right ) x\)	\(9\)
risch	\(45 x \ln \left (2\right )+225 x\)	\(10\)
parts	\(45 x \ln \left (2\right )+225 x\)	\(10\)

[In]

int(45*ln(2)+225,x,method=_RETURNVERBOSE)

[Out]

45*x*(ln(2)+5)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int (225+45 \log (2)) \, dx=45 \, x \log \left (2\right ) + 225 \, x \]

[In]

integrate(45*log(2)+225,x, algorithm="fricas")

[Out]

45*x*log(2) + 225*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int (225+45 \log (2)) \, dx=x \left (45 \log {\left (2 \right )} + 225\right ) \]

[In]

integrate(45*ln(2)+225,x)

[Out]

x*(45*log(2) + 225)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int (225+45 \log (2)) \, dx=45 \, x {\left (\log \left (2\right ) + 5\right )} \]

[In]

integrate(45*log(2)+225,x, algorithm="maxima")

[Out]

45*x*(log(2) + 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int (225+45 \log (2)) \, dx=45 \, x {\left (\log \left (2\right ) + 5\right )} \]

[In]

integrate(45*log(2)+225,x, algorithm="giac")

[Out]

45*x*(log(2) + 5)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int (225+45 \log (2)) \, dx=x\,\left (45\,\ln \left (2\right )+225\right ) \]

[In]

int(45*log(2) + 225,x)

[Out]

x*(45*log(2) + 225)