3.72.15 \(\int \frac {-4+4 x+20 x^2}{x+5 x^2} \, dx\)

Optimal. Leaf size=20 \[ -e^{25}+4 \left (-17+\log \left (e^x \left (5+\frac {1}{x}\right )\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1593, 893} \begin {gather*} 4 x-4 \log (x)+4 \log (5 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 4*x + 20*x^2)/(x + 5*x^2),x]

[Out]

4*x - 4*Log[x] + 4*Log[1 + 5*x]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+4 x+20 x^2}{x (1+5 x)} \, dx\\ &=\int \left (4-\frac {4}{x}+\frac {20}{1+5 x}\right ) \, dx\\ &=4 x-4 \log (x)+4 \log (1+5 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.70 \begin {gather*} 4 (x-\log (x)+\log (1+5 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 4*x + 20*x^2)/(x + 5*x^2),x]

[Out]

4*(x - Log[x] + Log[1 + 5*x])

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fricas [A]  time = 0.99, size = 16, normalized size = 0.80 \begin {gather*} 4 \, x + 4 \, \log \left (5 \, x + 1\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x^2+4*x-4)/(5*x^2+x),x, algorithm="fricas")

[Out]

4*x + 4*log(5*x + 1) - 4*log(x)

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giac [A]  time = 0.16, size = 18, normalized size = 0.90 \begin {gather*} 4 \, x + 4 \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - 4 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x^2+4*x-4)/(5*x^2+x),x, algorithm="giac")

[Out]

4*x + 4*log(abs(5*x + 1)) - 4*log(abs(x))

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maple [A]  time = 0.12, size = 17, normalized size = 0.85

method	result	size
default	\(4 x -4 \ln \relax (x )+4 \ln \left (1+5 x \right )\)	\(17\)
norman	\(4 x -4 \ln \relax (x )+4 \ln \left (1+5 x \right )\)	\(17\)
risch	\(4 x -4 \ln \relax (x )+4 \ln \left (1+5 x \right )\)	\(17\)
meijerg	\(4 x -4 \ln \relax (5)-4 \ln \relax (x )+4 \ln \left (1+5 x \right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x^2+4*x-4)/(5*x^2+x),x,method=_RETURNVERBOSE)

[Out]

4*x-4*ln(x)+4*ln(1+5*x)

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maxima [A]  time = 0.61, size = 16, normalized size = 0.80 \begin {gather*} 4 \, x + 4 \, \log \left (5 \, x + 1\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x^2+4*x-4)/(5*x^2+x),x, algorithm="maxima")

[Out]

4*x + 4*log(5*x + 1) - 4*log(x)

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mupad [B]  time = 0.05, size = 12, normalized size = 0.60 \begin {gather*} 4\,x+8\,\mathrm {atanh}\left (10\,x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + 20*x^2 - 4)/(x + 5*x^2),x)

[Out]

4*x + 8*atanh(10*x + 1)

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sympy [A]  time = 0.10, size = 15, normalized size = 0.75 \begin {gather*} 4 x - 4 \log {\relax (x )} + 4 \log {\left (x + \frac {1}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*x**2+4*x-4)/(5*x**2+x),x)

[Out]

4*x - 4*log(x) + 4*log(x + 1/5)

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