\(\int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx\) [1510]

Optimal result

Integrand size = 22, antiderivative size = 42 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 (3+5 x)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (2+3 x)-\frac {155}{121} \log (3+5 x) \]

[Out]

-5/11/(3+5*x)-4/847*ln(1-2*x)+9/7*ln(2+3*x)-155/121*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} \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 (5 x+3)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (3 x+2)-\frac {155}{121} \log (5 x+3) \]

[In]

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

[Out]

-5/(11*(3 + 5*x)) - (4*Log[1 - 2*x])/847 + (9*Log[2 + 3*x])/7 - (155*Log[3 + 5*x])/121

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}{847 (-1+2 x)}+\frac {27}{7 (2+3 x)}+\frac {25}{11 (3+5 x)^2}-\frac {775}{121 (3+5 x)}\right ) \, dx \\ & = -\frac {5}{11 (3+5 x)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (2+3 x)-\frac {155}{121} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=\frac {1}{847} \left (-\frac {385}{3+5 x}-4 \log (1-2 x)+1089 \log (4+6 x)-1085 \log (6+10 x)\right ) \]

[In]

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

[Out]

(-385/(3 + 5*x) - 4*Log[1 - 2*x] + 1089*Log[4 + 6*x] - 1085*Log[6 + 10*x])/847

Maple [A] (verified)

Time = 2.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {1}{11 \left (x +\frac {3}{5}\right )}-\frac {4 \ln \left (-1+2 x \right )}{847}+\frac {9 \ln \left (2+3 x \right )}{7}-\frac {155 \ln \left (3+5 x \right )}{121}\)	\(33\)
default	\(-\frac {5}{11 \left (3+5 x \right )}-\frac {155 \ln \left (3+5 x \right )}{121}-\frac {4 \ln \left (-1+2 x \right )}{847}+\frac {9 \ln \left (2+3 x \right )}{7}\)	\(35\)
norman	\(\frac {25 x}{33 \left (3+5 x \right )}-\frac {4 \ln \left (-1+2 x \right )}{847}+\frac {9 \ln \left (2+3 x \right )}{7}-\frac {155 \ln \left (3+5 x \right )}{121}\)	\(36\)
parallelrisch	\(\frac {16335 \ln \left (\frac {2}{3}+x \right ) x -16275 \ln \left (x +\frac {3}{5}\right ) x -60 \ln \left (x -\frac {1}{2}\right ) x +9801 \ln \left (\frac {2}{3}+x \right )-9765 \ln \left (x +\frac {3}{5}\right )-36 \ln \left (x -\frac {1}{2}\right )+1925 x}{7623+12705 x}\)	\(53\)

[In]

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

[Out]

-1/11/(x+3/5)-4/847*ln(-1+2*x)+9/7*ln(2+3*x)-155/121*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {1085 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1089 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 4 \, {\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) + 385}{847 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

-1/847*(1085*(5*x + 3)*log(5*x + 3) - 1089*(5*x + 3)*log(3*x + 2) + 4*(5*x + 3)*log(2*x - 1) + 385)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=- \frac {4 \log {\left (x - \frac {1}{2} \right )}}{847} - \frac {155 \log {\left (x + \frac {3}{5} \right )}}{121} + \frac {9 \log {\left (x + \frac {2}{3} \right )}}{7} - \frac {5}{55 x + 33} \]

[In]

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

[Out]

-4*log(x - 1/2)/847 - 155*log(x + 3/5)/121 + 9*log(x + 2/3)/7 - 5/(55*x + 33)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 \, {\left (5 \, x + 3\right )}} - \frac {155}{121} \, \log \left (5 \, x + 3\right ) + \frac {9}{7} \, \log \left (3 \, x + 2\right ) - \frac {4}{847} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-5/11/(5*x + 3) - 155/121*log(5*x + 3) + 9/7*log(3*x + 2) - 4/847*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 \, {\left (5 \, x + 3\right )}} + \frac {9}{7} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) - \frac {4}{847} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]

[In]

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

[Out]

-5/11/(5*x + 3) + 9/7*log(abs(-1/(5*x + 3) - 3)) - 4/847*log(abs(-11/(5*x + 3) + 2))

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=\frac {9\,\ln \left (x+\frac {2}{3}\right )}{7}-\frac {4\,\ln \left (x-\frac {1}{2}\right )}{847}-\frac {155\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {1}{11\,\left (x+\frac {3}{5}\right )} \]

[In]

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

[Out]

(9*log(x + 2/3))/7 - (4*log(x - 1/2))/847 - (155*log(x + 3/5))/121 - 1/(11*(x + 3/5))