\(\int \frac {1}{(1-2 x) (3+5 x)} \, dx\) [1493]

Optimal result

Integrand size = 15, antiderivative size = 21 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=-\frac {1}{11} \log (1-2 x)+\frac {1}{11} \log (3+5 x) \]

[Out]

-1/11*ln(1-2*x)+1/11*ln(3+5*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {36, 31} \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=\frac {1}{11} \log (5 x+3)-\frac {1}{11} \log (1-2 x) \]

[In]

Int[1/((1 - 2*x)*(3 + 5*x)),x]

[Out]

-1/11*Log[1 - 2*x] + Log[3 + 5*x]/11

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{11} \int \frac {1}{1-2 x} \, dx+\frac {5}{11} \int \frac {1}{3+5 x} \, dx \\ & = -\frac {1}{11} \log (1-2 x)+\frac {1}{11} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=-\frac {1}{11} \log (1-2 x)+\frac {1}{11} \log (3+5 x) \]

[In]

Integrate[1/((1 - 2*x)*(3 + 5*x)),x]

[Out]

-1/11*Log[1 - 2*x] + Log[3 + 5*x]/11

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {\ln \left (x +\frac {3}{5}\right )}{11}-\frac {\ln \left (x -\frac {1}{2}\right )}{11}\)	\(14\)
default	\(\frac {\ln \left (3+5 x \right )}{11}-\frac {\ln \left (-1+2 x \right )}{11}\)	\(18\)
norman	\(\frac {\ln \left (3+5 x \right )}{11}-\frac {\ln \left (-1+2 x \right )}{11}\)	\(18\)
risch	\(\frac {\ln \left (3+5 x \right )}{11}-\frac {\ln \left (-1+2 x \right )}{11}\)	\(18\)

[In]

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

[Out]

1/11*ln(x+3/5)-1/11*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=\frac {1}{11} \, \log \left (5 \, x + 3\right ) - \frac {1}{11} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

1/11*log(5*x + 3) - 1/11*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=- \frac {\log {\left (x - \frac {1}{2} \right )}}{11} + \frac {\log {\left (x + \frac {3}{5} \right )}}{11} \]

[In]

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

[Out]

-log(x - 1/2)/11 + log(x + 3/5)/11

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=\frac {1}{11} \, \log \left (5 \, x + 3\right ) - \frac {1}{11} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

1/11*log(5*x + 3) - 1/11*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=\frac {1}{11} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {1}{11} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

1/11*log(abs(5*x + 3)) - 1/11*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(1-2 x) (3+5 x)} \, dx=\frac {\ln \left (\frac {5\,x+3}{2\,x-1}\right )}{11} \]

[In]

int(-1/((2*x - 1)*(5*x + 3)),x)

[Out]

log((5*x + 3)/(2*x - 1))/11