3.78.67 \(\int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x^2+x \log (16)} \, dx\)

Optimal. Leaf size=22 \[ 4 \left (3 x+4 \left (x-\log \left (-\frac {9}{5} x (x+\log (16))\right )\right )\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 15, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1593, 1820} \begin {gather*} 28 x-16 \log (x)-16 \log (x+\log (16)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*x + 28*x^2 + (-16 + 28*x)*Log[16])/(x^2 + x*Log[16]),x]

[Out]

28*x - 16*Log[x] - 16*Log[x + Log[16]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-32 x+28 x^2+(-16+28 x) \log (16)}{x (x+\log (16))} \, dx\\ &=\int \left (28-\frac {16}{x}-\frac {16}{x+\log (16)}\right ) \, dx\\ &=28 x-16 \log (x)-16 \log (x+\log (16))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.77 \begin {gather*} 4 (7 x-4 \log (x)-4 \log (x+\log (16))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*x + 28*x^2 + (-16 + 28*x)*Log[16])/(x^2 + x*Log[16]),x]

[Out]

4*(7*x - 4*Log[x] - 4*Log[x + Log[16]])

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fricas [A]  time = 0.55, size = 16, normalized size = 0.73 \begin {gather*} 28 \, x - 16 \, \log \left (x^{2} + 4 \, x \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(28*x-16)*log(2)+28*x^2-32*x)/(4*x*log(2)+x^2),x, algorithm="fricas")

[Out]

28*x - 16*log(x^2 + 4*x*log(2))

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giac [A]  time = 0.12, size = 19, normalized size = 0.86 \begin {gather*} 28 \, x - 16 \, \log \left ({\left | x + 4 \, \log \relax (2) \right |}\right ) - 16 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(28*x-16)*log(2)+28*x^2-32*x)/(4*x*log(2)+x^2),x, algorithm="giac")

[Out]

28*x - 16*log(abs(x + 4*log(2))) - 16*log(abs(x))

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maple [A]  time = 0.23, size = 17, normalized size = 0.77

method	result	size
risch	\(28 x -16 \ln \left (4 x \ln \relax (2)+x^{2}\right )\)	\(17\)
default	\(28 x -16 \ln \relax (x )-16 \ln \left (x +4 \ln \relax (2)\right )\)	\(18\)
norman	\(28 x -16 \ln \relax (x )-16 \ln \left (x +4 \ln \relax (2)\right )\)	\(18\)
meijerg	\(-16 \ln \relax (x )+32 \ln \relax (2)+16 \ln \left (\ln \relax (2)\right )+16 \ln \left (1+\frac {x}{4 \ln \relax (2)}\right )+\left (112 \ln \relax (2)-32\right ) \ln \left (1+\frac {x}{4 \ln \relax (2)}\right )+112 \ln \relax (2) \left (\frac {x}{4 \ln \relax (2)}-\ln \left (1+\frac {x}{4 \ln \relax (2)}\right )\right )\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(28*x-16)*ln(2)+28*x^2-32*x)/(4*x*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

28*x-16*ln(4*x*ln(2)+x^2)

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maxima [A]  time = 0.36, size = 17, normalized size = 0.77 \begin {gather*} 28 \, x - 16 \, \log \left (x + 4 \, \log \relax (2)\right ) - 16 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(28*x-16)*log(2)+28*x^2-32*x)/(4*x*log(2)+x^2),x, algorithm="maxima")

[Out]

28*x - 16*log(x + 4*log(2)) - 16*log(x)

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mupad [B]  time = 0.12, size = 13, normalized size = 0.59 \begin {gather*} 28\,x-16\,\ln \left (x\,\left (x+\ln \left (16\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(2)*(28*x - 16) - 32*x + 28*x^2)/(4*x*log(2) + x^2),x)

[Out]

28*x - 16*log(x*(x + log(16)))

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sympy [A]  time = 0.15, size = 15, normalized size = 0.68 \begin {gather*} 28 x - 16 \log {\left (x^{2} + 4 x \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(28*x-16)*ln(2)+28*x**2-32*x)/(4*x*ln(2)+x**2),x)

[Out]

28*x - 16*log(x**2 + 4*x*log(2))

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