\(\int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx\) [1494]

Optimal result

Integrand size = 22, antiderivative size = 31 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=-\frac {2}{77} \log (1-2 x)-\frac {3}{7} \log (2+3 x)+\frac {5}{11} \log (3+5 x) \]

[Out]

-2/77*ln(1-2*x)-3/7*ln(2+3*x)+5/11*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {84} \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=-\frac {2}{77} \log (1-2 x)-\frac {3}{7} \log (3 x+2)+\frac {5}{11} \log (5 x+3) \]

[In]

Int[1/((1 - 2*x)*(2 + 3*x)*(3 + 5*x)),x]

[Out]

(-2*Log[1 - 2*x])/77 - (3*Log[2 + 3*x])/7 + (5*Log[3 + 5*x])/11

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {4}{77 (-1+2 x)}-\frac {9}{7 (2+3 x)}+\frac {25}{11 (3+5 x)}\right ) \, dx \\ & = -\frac {2}{77} \log (1-2 x)-\frac {3}{7} \log (2+3 x)+\frac {5}{11} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=-\frac {2}{77} \log (1-2 x)-\frac {3}{7} \log (2+3 x)+\frac {5}{11} \log (3+5 x) \]

[In]

Integrate[1/((1 - 2*x)*(2 + 3*x)*(3 + 5*x)),x]

[Out]

(-2*Log[1 - 2*x])/77 - (3*Log[2 + 3*x])/7 + (5*Log[3 + 5*x])/11

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {3 \ln \left (\frac {2}{3}+x \right )}{7}+\frac {5 \ln \left (x +\frac {3}{5}\right )}{11}-\frac {2 \ln \left (x -\frac {1}{2}\right )}{77}\)	\(20\)
default	\(\frac {5 \ln \left (3+5 x \right )}{11}-\frac {2 \ln \left (-1+2 x \right )}{77}-\frac {3 \ln \left (2+3 x \right )}{7}\)	\(26\)
norman	\(\frac {5 \ln \left (3+5 x \right )}{11}-\frac {2 \ln \left (-1+2 x \right )}{77}-\frac {3 \ln \left (2+3 x \right )}{7}\)	\(26\)
risch	\(\frac {5 \ln \left (3+5 x \right )}{11}-\frac {2 \ln \left (-1+2 x \right )}{77}-\frac {3 \ln \left (2+3 x \right )}{7}\)	\(26\)

[In]

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

[Out]

-3/7*ln(2/3+x)+5/11*ln(x+3/5)-2/77*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=\frac {5}{11} \, \log \left (5 \, x + 3\right ) - \frac {3}{7} \, \log \left (3 \, x + 2\right ) - \frac {2}{77} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

5/11*log(5*x + 3) - 3/7*log(3*x + 2) - 2/77*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=- \frac {2 \log {\left (x - \frac {1}{2} \right )}}{77} + \frac {5 \log {\left (x + \frac {3}{5} \right )}}{11} - \frac {3 \log {\left (x + \frac {2}{3} \right )}}{7} \]

[In]

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

[Out]

-2*log(x - 1/2)/77 + 5*log(x + 3/5)/11 - 3*log(x + 2/3)/7

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=\frac {5}{11} \, \log \left (5 \, x + 3\right ) - \frac {3}{7} \, \log \left (3 \, x + 2\right ) - \frac {2}{77} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

5/11*log(5*x + 3) - 3/7*log(3*x + 2) - 2/77*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=\frac {5}{11} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {3}{7} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {2}{77} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

5/11*log(abs(5*x + 3)) - 3/7*log(abs(3*x + 2)) - 2/77*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)} \, dx=\frac {5\,\ln \left (x+\frac {3}{5}\right )}{11}-\frac {3\,\ln \left (x+\frac {2}{3}\right )}{7}-\frac {2\,\ln \left (x-\frac {1}{2}\right )}{77} \]

[In]

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

[Out]

(5*log(x + 3/5))/11 - (3*log(x + 2/3))/7 - (2*log(x - 1/2))/77