\(\int \frac {1}{2} (-6+9 x+2 \log ^2(16)) \, dx\) [4748]

Optimal result

Integrand size = 15, antiderivative size = 13 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=x \left (-3+\frac {9 x}{4}+\log ^2(16)\right ) \]

[Out]

(16*ln(2)^2-3+9/4*x)*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.54, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {9} \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=\frac {1}{36} \left (9 x-2 \left (3-\log ^2(16)\right )\right )^2 \]

[In]

Int[(-6 + 9*x + 2*Log[16]^2)/2,x]

[Out]

(9*x - 2*(3 - Log[16]^2))^2/36

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{align*} \text {integral}& = \frac {1}{36} \left (9 x-2 \left (3-\log ^2(16)\right )\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=-3 x+\frac {9 x^2}{4}+x \log ^2(16) \]

[In]

Integrate[(-6 + 9*x + 2*Log[16]^2)/2,x]

[Out]

-3*x + (9*x^2)/4 + x*Log[16]^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15

method	result	size
gosper	\(\frac {x \left (64 \ln \left (2\right )^{2}+9 x -12\right )}{4}\)	\(15\)
default	\(16 x \ln \left (2\right )^{2}+\frac {9 x^{2}}{4}-3 x\)	\(17\)
norman	\(\left (-3+16 \ln \left (2\right )^{2}\right ) x +\frac {9 x^{2}}{4}\)	\(17\)
risch	\(16 x \ln \left (2\right )^{2}+\frac {9 x^{2}}{4}-3 x\)	\(17\)
parallelrisch	\(\left (-3+16 \ln \left (2\right )^{2}\right ) x +\frac {9 x^{2}}{4}\)	\(17\)
parts	\(16 x \ln \left (2\right )^{2}+\frac {9 x^{2}}{4}-3 x\)	\(17\)

[In]

int(16*ln(2)^2+9/2*x-3,x,method=_RETURNVERBOSE)

[Out]

1/4*x*(64*ln(2)^2+9*x-12)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=16 \, x \log \left (2\right )^{2} + \frac {9}{4} \, x^{2} - 3 \, x \]

[In]

integrate(16*log(2)^2+9/2*x-3,x, algorithm="fricas")

[Out]

16*x*log(2)^2 + 9/4*x^2 - 3*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=\frac {9 x^{2}}{4} + x \left (-3 + 16 \log {\left (2 \right )}^{2}\right ) \]

[In]

integrate(16*ln(2)**2+9/2*x-3,x)

[Out]

9*x**2/4 + x*(-3 + 16*log(2)**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=16 \, x \log \left (2\right )^{2} + \frac {9}{4} \, x^{2} - 3 \, x \]

[In]

integrate(16*log(2)^2+9/2*x-3,x, algorithm="maxima")

[Out]

16*x*log(2)^2 + 9/4*x^2 - 3*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=16 \, x \log \left (2\right )^{2} + \frac {9}{4} \, x^{2} - 3 \, x \]

[In]

integrate(16*log(2)^2+9/2*x-3,x, algorithm="giac")

[Out]

16*x*log(2)^2 + 9/4*x^2 - 3*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {1}{2} \left (-6+9 x+2 \log ^2(16)\right ) \, dx=\frac {9\,x^2}{4}+\left (16\,{\ln \left (2\right )}^2-3\right )\,x \]

[In]

int((9*x)/2 + 16*log(2)^2 - 3,x)

[Out]

x*(16*log(2)^2 - 3) + (9*x^2)/4