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

Optimal. Leaf size=13 \[ x \left (-3+\frac {9 x}{4}+\log ^2(16)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.54, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {9} \begin {gather*} \frac {1}{36} \left (9 x-2 \left (3-\log ^2(16)\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

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

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 17, normalized size = 1.31 \begin {gather*} -3 x+\frac {9 x^2}{4}+x \log ^2(16) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3*x + (9*x^2)/4 + x*Log[16]^2

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 16, normalized size = 1.23 \begin {gather*} 16 \, x \log \relax (2)^{2} + \frac {9}{4} \, x^{2} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [A]  time = 0.13, size = 16, normalized size = 1.23 \begin {gather*} 16 \, x \log \relax (2)^{2} + \frac {9}{4} \, x^{2} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 15, normalized size = 1.15

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 16, normalized size = 1.23 \begin {gather*} 16 \, x \log \relax (2)^{2} + \frac {9}{4} \, x^{2} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 16, normalized size = 1.23 \begin {gather*} \frac {9\,x^2}{4}+\left (16\,{\ln \relax (2)}^2-3\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x)/2 + 16*log(2)^2 - 3,x)

[Out]

x*(16*log(2)^2 - 3) + (9*x^2)/4

________________________________________________________________________________________

sympy [A]  time = 0.05, size = 15, normalized size = 1.15 \begin {gather*} \frac {9 x^{2}}{4} + x \left (-3 + 16 \log {\relax (2 )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*x**2/4 + x*(-3 + 16*log(2)**2)

________________________________________________________________________________________