3.27.83 \(\int \frac {1+128 x+32 x^2}{4+x} \, dx\)

Optimal. Leaf size=10 \[ 16 x^2+\log (4+x) \]

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {698} \begin {gather*} 16 x^2+\log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 128*x + 32*x^2)/(4 + x),x]

[Out]

16*x^2 + Log[4 + x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (32 x+\frac {1}{4+x}\right ) \, dx\\ &=16 x^2+\log (4+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.20 \begin {gather*} 16 \left (-16+x^2\right )+\log (4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 128*x + 32*x^2)/(4 + x),x]

[Out]

16*(-16 + x^2) + Log[4 + x]

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fricas [A]  time = 0.48, size = 10, normalized size = 1.00 \begin {gather*} 16 \, x^{2} + \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2+128*x+1)/(4+x),x, algorithm="fricas")

[Out]

16*x^2 + log(x + 4)

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giac [A]  time = 0.23, size = 11, normalized size = 1.10 \begin {gather*} 16 \, x^{2} + \log \left ({\left | x + 4 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2+128*x+1)/(4+x),x, algorithm="giac")

[Out]

16*x^2 + log(abs(x + 4))

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maple [A]  time = 0.41, size = 11, normalized size = 1.10

method	result	size
default	\(16 x^{2}+\ln \left (4+x \right )\)	\(11\)
norman	\(16 x^{2}+\ln \left (4+x \right )\)	\(11\)
risch	\(16 x^{2}+\ln \left (4+x \right )\)	\(11\)
meijerg	\(\ln \left (1+\frac {x}{4}\right )-\frac {64 x \left (-\frac {3 x}{4}+6\right )}{3}+128 x\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x^2+128*x+1)/(4+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 10, normalized size = 1.00 \begin {gather*} 16 \, x^{2} + \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2+128*x+1)/(4+x),x, algorithm="maxima")

[Out]

16*x^2 + log(x + 4)

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mupad [B]  time = 0.03, size = 10, normalized size = 1.00 \begin {gather*} \ln \left (x+4\right )+16\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((128*x + 32*x^2 + 1)/(x + 4),x)

[Out]

log(x + 4) + 16*x^2

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sympy [A]  time = 0.07, size = 8, normalized size = 0.80 \begin {gather*} 16 x^{2} + \log {\left (x + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x**2+128*x+1)/(4+x),x)

[Out]

16*x**2 + log(x + 4)

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