3.17.70 $\int \frac{1}{25}(40-2x-10\log (2))dx$

Optimal. Leaf size=17 $$\frac{11}{3}-{(-4+\frac{x}{5}+\log (2))}^2$$

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 0.94, 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} $$\begin{array}{c}-\frac{1}{25}(x-5(4-\log (2)){)}^2\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(40 - 2*x - 10*Log[2])/25,x]

[Out]

-1/25*(x - 5*(4 - Log[2]))^2

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-\frac{1}{25}(x-5(4-\log (2)){)}^2\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.12 $$\begin{array}{c}-\frac{2}{25}(-20x+\frac{x^2}{2}+x\log (32))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(40 - 2*x - 10*Log[2])/25,x]

[Out]

(-2*(-20*x + x^2/2 + x*Log[32]))/25

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fricas [A]  time = 0.77, size = 14, normalized size = 0.82 $$\begin{array}{c}-\frac{1}{25}{x}^2-\frac{2}{5}x\log (2)+\frac{8}{5}x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*log(2)-2/25*x+8/5,x, algorithm="fricas")

[Out]

-1/25*x^2 - 2/5*x*log(2) + 8/5*x

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giac [A]  time = 0.29, size = 14, normalized size = 0.82 $$\begin{array}{c}-\frac{1}{25}{x}^2-\frac{2}{5}x\log (2)+\frac{8}{5}x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*log(2)-2/25*x+8/5,x, algorithm="giac")

[Out]

-1/25*x^2 - 2/5*x*log(2) + 8/5*x

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maple [A]  time = 0.02, size = 11, normalized size = 0.65

method	result	size
gosper	$-\frac{x(x+10\ln (2)-40)}{25}$	$11$
default	$-\frac{2x\ln (2)}{5}-\frac{x^2}{25}+\frac{8x}{5}$	$15$
norman	$(-\frac{2\ln (2)}{5}+\frac{8}{5})x-\frac{x^2}{25}$	$15$
risch	$-\frac{2x\ln (2)}{5}-\frac{x^2}{25}+\frac{8x}{5}$	$15$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2/5*ln(2)-2/25*x+8/5,x,method=_RETURNVERBOSE)

[Out]

-1/25*x*(x+10*ln(2)-40)

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maxima [A]  time = 0.37, size = 14, normalized size = 0.82 $$\begin{array}{c}-\frac{1}{25}{x}^2-\frac{2}{5}x\log (2)+\frac{8}{5}x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*log(2)-2/25*x+8/5,x, algorithm="maxima")

[Out]

-1/25*x^2 - 2/5*x*log(2) + 8/5*x

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mupad [B]  time = 0.04, size = 10, normalized size = 0.59 $$\begin{array}{c}-\frac{x(x+10\ln (2)-40)}{25}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8/5 - (2*log(2))/5 - (2*x)/25,x)

[Out]

-(x*(x + 10*log(2) - 40))/25

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sympy [A]  time = 0.06, size = 15, normalized size = 0.88 $$\begin{array}{c}-\frac{x^2}{25}+x(\frac{8}{5}-\frac{2\log (2)}{5})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*ln(2)-2/25*x+8/5,x)

[Out]

-x**2/25 + x*(8/5 - 2*log(2)/5)

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