\(\int \frac {1-2 x}{3+5 x} \, dx\) [1199]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {1-2 x}{3+5 x} \, dx=-\frac {2 x}{5}+\frac {11}{25} \log (3+5 x) \]

[Out]

-2/5*x+11/25*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int \frac {1-2 x}{3+5 x} \, dx=\frac {11}{25} \log (5 x+3)-\frac {2 x}{5} \]

[In]

Int[(1 - 2*x)/(3 + 5*x),x]

[Out]

(-2*x)/5 + (11*Log[3 + 5*x])/25

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2}{5}+\frac {11}{5 (3+5 x)}\right ) \, dx \\ & = -\frac {2 x}{5}+\frac {11}{25} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {1-2 x}{3+5 x} \, dx=\frac {1}{25} (-6-10 x+11 \log (3+5 x)) \]

[In]

Integrate[(1 - 2*x)/(3 + 5*x),x]

[Out]

(-6 - 10*x + 11*Log[3 + 5*x])/25

Maple [A] (verified)

Time = 1.81 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(-\frac {2 x}{5}+\frac {11 \ln \left (x +\frac {3}{5}\right )}{25}\)	\(11\)
default	\(-\frac {2 x}{5}+\frac {11 \ln \left (3+5 x \right )}{25}\)	\(13\)
norman	\(-\frac {2 x}{5}+\frac {11 \ln \left (3+5 x \right )}{25}\)	\(13\)
meijerg	\(\frac {11 \ln \left (1+\frac {5 x}{3}\right )}{25}-\frac {2 x}{5}\)	\(13\)
risch	\(-\frac {2 x}{5}+\frac {11 \ln \left (3+5 x \right )}{25}\)	\(13\)

[In]

int((1-2*x)/(3+5*x),x,method=_RETURNVERBOSE)

[Out]

-2/5*x+11/25*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1-2 x}{3+5 x} \, dx=-\frac {2}{5} \, x + \frac {11}{25} \, \log \left (5 \, x + 3\right ) \]

[In]

integrate((1-2*x)/(3+5*x),x, algorithm="fricas")

[Out]

-2/5*x + 11/25*log(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1-2 x}{3+5 x} \, dx=- \frac {2 x}{5} + \frac {11 \log {\left (5 x + 3 \right )}}{25} \]

[In]

integrate((1-2*x)/(3+5*x),x)

[Out]

-2*x/5 + 11*log(5*x + 3)/25

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1-2 x}{3+5 x} \, dx=-\frac {2}{5} \, x + \frac {11}{25} \, \log \left (5 \, x + 3\right ) \]

[In]

integrate((1-2*x)/(3+5*x),x, algorithm="maxima")

[Out]

-2/5*x + 11/25*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {1-2 x}{3+5 x} \, dx=-\frac {2}{5} \, x + \frac {11}{25} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

integrate((1-2*x)/(3+5*x),x, algorithm="giac")

[Out]

-2/5*x + 11/25*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {1-2 x}{3+5 x} \, dx=\frac {11\,\ln \left (x+\frac {3}{5}\right )}{25}-\frac {2\,x}{5} \]

[In]

int(-(2*x - 1)/(5*x + 3),x)

[Out]

(11*log(x + 3/5))/25 - (2*x)/5