\(\int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx\) [1495]

Optimal result

Integrand size = 22, antiderivative size = 42 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {3}{7 (2+3 x)}-\frac {4}{539} \log (1-2 x)-\frac {111}{49} \log (2+3 x)+\frac {25}{11} \log (3+5 x) \]

[Out]

3/7/(2+3*x)-4/539*ln(1-2*x)-111/49*ln(2+3*x)+25/11*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, 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} \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {3}{7 (3 x+2)}-\frac {4}{539} \log (1-2 x)-\frac {111}{49} \log (3 x+2)+\frac {25}{11} \log (5 x+3) \]

[In]

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

[Out]

3/(7*(2 + 3*x)) - (4*Log[1 - 2*x])/539 - (111*Log[2 + 3*x])/49 + (25*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 {8}{539 (-1+2 x)}-\frac {9}{7 (2+3 x)^2}-\frac {333}{49 (2+3 x)}+\frac {125}{11 (3+5 x)}\right ) \, dx \\ & = \frac {3}{7 (2+3 x)}-\frac {4}{539} \log (1-2 x)-\frac {111}{49} \log (2+3 x)+\frac {25}{11} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {1}{539} \left (\frac {231}{2+3 x}-4 \log (1-2 x)-1221 \log (4+6 x)+1225 \log (6+10 x)\right ) \]

[In]

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

[Out]

(231/(2 + 3*x) - 4*Log[1 - 2*x] - 1221*Log[4 + 6*x] + 1225*Log[6 + 10*x])/539

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {1}{\frac {14}{3}+7 x}-\frac {4 \ln \left (-1+2 x \right )}{539}-\frac {111 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{11}\)	\(33\)
default	\(\frac {25 \ln \left (3+5 x \right )}{11}-\frac {4 \ln \left (-1+2 x \right )}{539}+\frac {3}{7 \left (2+3 x \right )}-\frac {111 \ln \left (2+3 x \right )}{49}\)	\(35\)
norman	\(-\frac {9 x}{14 \left (2+3 x \right )}-\frac {4 \ln \left (-1+2 x \right )}{539}-\frac {111 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{11}\)	\(36\)
parallelrisch	\(-\frac {7326 \ln \left (\frac {2}{3}+x \right ) x -7350 \ln \left (x +\frac {3}{5}\right ) x +24 \ln \left (x -\frac {1}{2}\right ) x +4884 \ln \left (\frac {2}{3}+x \right )-4900 \ln \left (x +\frac {3}{5}\right )+16 \ln \left (x -\frac {1}{2}\right )+693 x}{1078 \left (2+3 x \right )}\)	\(53\)

[In]

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

[Out]

1/7/(2/3+x)-4/539*ln(-1+2*x)-111/49*ln(2+3*x)+25/11*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {1225 \, {\left (3 \, x + 2\right )} \log \left (5 \, x + 3\right ) - 1221 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 4 \, {\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 231}{539 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

1/539*(1225*(3*x + 2)*log(5*x + 3) - 1221*(3*x + 2)*log(3*x + 2) - 4*(3*x + 2)*log(2*x - 1) + 231)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=- \frac {4 \log {\left (x - \frac {1}{2} \right )}}{539} + \frac {25 \log {\left (x + \frac {3}{5} \right )}}{11} - \frac {111 \log {\left (x + \frac {2}{3} \right )}}{49} + \frac {3}{21 x + 14} \]

[In]

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

[Out]

-4*log(x - 1/2)/539 + 25*log(x + 3/5)/11 - 111*log(x + 2/3)/49 + 3/(21*x + 14)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {3}{7 \, {\left (3 \, x + 2\right )}} + \frac {25}{11} \, \log \left (5 \, x + 3\right ) - \frac {111}{49} \, \log \left (3 \, x + 2\right ) - \frac {4}{539} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

3/7/(3*x + 2) + 25/11*log(5*x + 3) - 111/49*log(3*x + 2) - 4/539*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {3}{7 \, {\left (3 \, x + 2\right )}} + \frac {25}{11} \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) - \frac {4}{539} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]

[In]

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

[Out]

3/7/(3*x + 2) + 25/11*log(abs(-1/(3*x + 2) + 5)) - 4/539*log(abs(-7/(3*x + 2) + 2))

Mupad [B] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(1-2 x) (2+3 x)^2 (3+5 x)} \, dx=\frac {25\,\ln \left (x+\frac {3}{5}\right )}{11}-\frac {111\,\ln \left (x+\frac {2}{3}\right )}{49}-\frac {4\,\ln \left (x-\frac {1}{2}\right )}{539}+\frac {1}{7\,\left (x+\frac {2}{3}\right )} \]

[In]

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

[Out]

(25*log(x + 3/5))/11 - (111*log(x + 2/3))/49 - (4*log(x - 1/2))/539 + 1/(7*(x + 2/3))