\(\int \frac {3+5 x}{1-2 x} \, dx\) [1442]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {3+5 x}{1-2 x} \, dx=-\frac {5 x}{2}-\frac {11}{4} \log (1-2 x) \]

[Out]

-5/2*x-11/4*ln(1-2*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} \[ \int \frac {3+5 x}{1-2 x} \, dx=-\frac {5 x}{2}-\frac {11}{4} \log (1-2 x) \]

[In]

Int[(3 + 5*x)/(1 - 2*x),x]

[Out]

(-5*x)/2 - (11*Log[1 - 2*x])/4

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 {5}{2}-\frac {11}{2 (-1+2 x)}\right ) \, dx \\ & = -\frac {5 x}{2}-\frac {11}{4} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {3+5 x}{1-2 x} \, dx=\frac {1}{4} (5-10 x-11 \log (1-2 x)) \]

[In]

Integrate[(3 + 5*x)/(1 - 2*x),x]

[Out]

(5 - 10*x - 11*Log[1 - 2*x])/4

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(-\frac {5 x}{2}-\frac {11 \ln \left (x -\frac {1}{2}\right )}{4}\)	\(11\)
default	\(-\frac {5 x}{2}-\frac {11 \ln \left (-1+2 x \right )}{4}\)	\(13\)
norman	\(-\frac {5 x}{2}-\frac {11 \ln \left (-1+2 x \right )}{4}\)	\(13\)
meijerg	\(-\frac {5 x}{2}-\frac {11 \ln \left (1-2 x \right )}{4}\)	\(13\)
risch	\(-\frac {5 x}{2}-\frac {11 \ln \left (-1+2 x \right )}{4}\)	\(13\)

[In]

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

[Out]

-5/2*x-11/4*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {3+5 x}{1-2 x} \, dx=-\frac {5}{2} \, x - \frac {11}{4} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-5/2*x - 11/4*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {3+5 x}{1-2 x} \, dx=- \frac {5 x}{2} - \frac {11 \log {\left (2 x - 1 \right )}}{4} \]

[In]

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

[Out]

-5*x/2 - 11*log(2*x - 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {3+5 x}{1-2 x} \, dx=-\frac {5}{2} \, x - \frac {11}{4} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-5/2*x - 11/4*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{1-2 x} \, dx=-\frac {5}{2} \, x - \frac {11}{4} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-5/2*x - 11/4*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {3+5 x}{1-2 x} \, dx=-\frac {5\,x}{2}-\frac {11\,\ln \left (x-\frac {1}{2}\right )}{4} \]

[In]

int(-(5*x + 3)/(2*x - 1),x)

[Out]

- (5*x)/2 - (11*log(x - 1/2))/4