\(\int \frac {75+58 x+10 x^2}{20+5 x} \, dx\) [6749]

Optimal result

Integrand size = 18, antiderivative size = 17 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=x^2+\frac {3}{5} (-3+6 x+\log (4+x)) \]

[Out]

3/5*ln(4+x)-9/5+18/5*x+x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {712} \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=x^2+\frac {18 x}{5}+\frac {3}{5} \log (x+4) \]

[In]

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

[Out]

(18*x)/5 + x^2 + (3*Log[4 + x])/5

Rule 712

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=\frac {1}{5} \left (-8+18 x+5 x^2+3 \log (5 (4+x))\right ) \]

[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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=x^{2} + \frac {18}{5} \, x + \frac {3}{5} \, \log \left (x + 4\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=x^{2} + \frac {18 x}{5} + \frac {3 \log {\left (x + 4 \right )}}{5} \]

[In]

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

[Out]

x**2 + 18*x/5 + 3*log(x + 4)/5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=x^{2} + \frac {18}{5} \, x + \frac {3}{5} \, \log \left (x + 4\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=x^{2} + \frac {18}{5} \, x + \frac {3}{5} \, \log \left ({\left | x + 4 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 11.93 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {75+58 x+10 x^2}{20+5 x} \, dx=\frac {18\,x}{5}+\frac {3\,\ln \left (x+4\right )}{5}+x^2 \]

[In]

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

[Out]

(18*x)/5 + (3*log(x + 4))/5 + x^2