3.68.49 \(\int \frac {75+58 x+10 x^2}{20+5 x} \, dx\)

Optimal. Leaf size=17 \[ x^2+\frac {3}{5} (-3+6 x+\log (4+x)) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {698} \begin {gather*} x^2+\frac {18 x}{5}+\frac {3}{5} \log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75 + 58*x + 10*x^2)/(20 + 5*x),x]

[Out]

(18*x)/5 + x^2 + (3*Log[4 + x])/5

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 (\frac {18}{5}+2 x+\frac {3}{5 (4+x)}\right ) \, dx\\ &=\frac {18 x}{5}+x^2+\frac {3}{5} \log (4+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.29 \begin {gather*} \frac {1}{5} \left (-8+18 x+5 x^2+3 \log (5 (4+x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75 + 58*x + 10*x^2)/(20 + 5*x),x]

[Out]

(-8 + 18*x + 5*x^2 + 3*Log[5*(4 + x)])/5

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fricas [A]  time = 0.48, size = 13, normalized size = 0.76 \begin {gather*} x^{2} + \frac {18}{5} \, x + \frac {3}{5} \, \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2+58*x+75)/(20+5*x),x, algorithm="fricas")

[Out]

x^2 + 18/5*x + 3/5*log(x + 4)

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giac [A]  time = 0.14, size = 14, normalized size = 0.82 \begin {gather*} x^{2} + \frac {18}{5} \, x + \frac {3}{5} \, \log \left ({\left | x + 4 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2+58*x+75)/(20+5*x),x, algorithm="giac")

[Out]

x^2 + 18/5*x + 3/5*log(abs(x + 4))

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maple [A]  time = 0.08, size = 14, normalized size = 0.82

method	result	size
default	\(x^{2}+\frac {18 x}{5}+\frac {3 \ln \left (4+x \right )}{5}\)	\(14\)
risch	\(x^{2}+\frac {18 x}{5}+\frac {3 \ln \left (4+x \right )}{5}\)	\(14\)
norman	\(x^{2}+\frac {18 x}{5}+\frac {3 \ln \left (20+5 x \right )}{5}\)	\(16\)
meijerg	\(\frac {3 \ln \left (1+\frac {x}{4}\right )}{5}-\frac {4 x \left (-\frac {3 x}{4}+6\right )}{3}+\frac {58 x}{5}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x^2+58*x+75)/(20+5*x),x,method=_RETURNVERBOSE)

[Out]

x^2+18/5*x+3/5*ln(4+x)

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maxima [A]  time = 0.37, size = 13, normalized size = 0.76 \begin {gather*} x^{2} + \frac {18}{5} \, x + \frac {3}{5} \, \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2+58*x+75)/(20+5*x),x, algorithm="maxima")

[Out]

x^2 + 18/5*x + 3/5*log(x + 4)

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mupad [B]  time = 3.94, size = 13, normalized size = 0.76 \begin {gather*} \frac {18\,x}{5}+\frac {3\,\ln \left (x+4\right )}{5}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((58*x + 10*x^2 + 75)/(5*x + 20),x)

[Out]

(18*x)/5 + (3*log(x + 4))/5 + x^2

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sympy [A]  time = 0.07, size = 15, normalized size = 0.88 \begin {gather*} x^{2} + \frac {18 x}{5} + \frac {3 \log {\left (x + 4 \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x**2+58*x+75)/(20+5*x),x)

[Out]

x**2 + 18*x/5 + 3*log(x + 4)/5

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