3.17.25 $\int \frac{12+x^2+x^2\log (3)}{4x^2}dx$

Optimal. Leaf size=19 $$-5+\frac{-3+x}{x}+\frac{1}{4}(x+x\log (3))$$

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, integrand size = 18, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.167, Rules used = {6, 12, 14} $$\begin{array}{c}\frac{1}{4}x(1+\log (3))-\frac{3}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(12 + x^2 + x^2*Log[3])/(4*x^2),x]

[Out]

-3/x + (x*(1 + Log[3]))/4

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{12+x^2(1+\log (3))}{4x^2}dx\\ & =\frac{1}{4}\int \frac{12+x^2(1+\log (3))}{x^2}dx\\ & =\frac{1}{4}\int (1+\frac{12}{x^2}+\log (3))dx\\ & =-\frac{3}{x}+\frac{1}{4}x(1+\log (3))\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.79 $$\begin{array}{c}\frac{1}{4}(-\frac{12}{x}+x+x\log (3))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(12 + x^2 + x^2*Log[3])/(4*x^2),x]

[Out]

(-12/x + x + x*Log[3])/4

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fricas [A]  time = 0.88, size = 16, normalized size = 0.84 $$\begin{array}{c}\frac{x^2\log (3)+x^2-12}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x^2*log(3)+x^2+12)/x^2,x, algorithm="fricas")

[Out]

1/4*(x^2*log(3) + x^2 - 12)/x

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giac [A]  time = 0.29, size = 14, normalized size = 0.74 $$\begin{array}{c}\frac{1}{4}x\log (3)+\frac{1}{4}x-\frac{3}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x^2*log(3)+x^2+12)/x^2,x, algorithm="giac")

[Out]

1/4*x*log(3) + 1/4*x - 3/x

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maple [A]  time = 0.03, size = 15, normalized size = 0.79

method	result	size
default	$\frac{x\ln (3)}{4}+\frac{x}{4}-\frac{3}{x}$	$15$
risch	$\frac{x\ln (3)}{4}+\frac{x}{4}-\frac{3}{x}$	$15$
gosper	$\frac{x^2\ln (3)+x^2-12}{4x}$	$17$
norman	$\frac{-3+(\frac{\ln (3)}{4}+\frac{1}{4})x^2}{x}$	$17$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(x^2*ln(3)+x^2+12)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.34, size = 13, normalized size = 0.68 $$\begin{array}{c}\frac{1}{4}x(\log (3)+1)-\frac{3}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x^2*log(3)+x^2+12)/x^2,x, algorithm="maxima")

[Out]

1/4*x*(log(3) + 1) - 3/x

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mupad [B]  time = 0.04, size = 14, normalized size = 0.74 $$\begin{array}{c}x(\frac{\ln (3)}{4}+\frac{1}{4})-\frac{3}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*log(3))/4 + x^2/4 + 3)/x^2,x)

[Out]

x*(log(3)/4 + 1/4) - 3/x

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sympy [A]  time = 0.08, size = 10, normalized size = 0.53 $$\begin{array}{c}\frac{x(1+\log (3))}{4}-\frac{3}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(x**2*ln(3)+x**2+12)/x**2,x)

[Out]

x*(1 + log(3))/4 - 3/x

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