\(\int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx\) [1525]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=-\frac {5}{22 (3+5 x)^2}+\frac {155}{121 (3+5 x)}-\frac {8 \log (1-2 x)}{9317}-\frac {27}{7} \log (2+3 x)+\frac {5135 \log (3+5 x)}{1331} \]

[Out]

-5/22/(3+5*x)^2+155/121/(3+5*x)-8/9317*ln(1-2*x)-27/7*ln(2+3*x)+5135/1331*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, 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} \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=\frac {155}{121 (5 x+3)}-\frac {5}{22 (5 x+3)^2}-\frac {8 \log (1-2 x)}{9317}-\frac {27}{7} \log (3 x+2)+\frac {5135 \log (5 x+3)}{1331} \]

[In]

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

[Out]

-5/(22*(3 + 5*x)^2) + 155/(121*(3 + 5*x)) - (8*Log[1 - 2*x])/9317 - (27*Log[2 + 3*x])/7 + (5135*Log[3 + 5*x])/
1331

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 {16}{9317 (-1+2 x)}-\frac {81}{7 (2+3 x)}+\frac {25}{11 (3+5 x)^3}-\frac {775}{121 (3+5 x)^2}+\frac {25675}{1331 (3+5 x)}\right ) \, dx \\ & = -\frac {5}{22 (3+5 x)^2}+\frac {155}{121 (3+5 x)}-\frac {8 \log (1-2 x)}{9317}-\frac {27}{7} \log (2+3 x)+\frac {5135 \log (3+5 x)}{1331} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=\frac {\frac {1925 (35+62 x)}{(3+5 x)^2}-16 \log (1-2 x)-71874 \log (4+6 x)+71890 \log (6+10 x)}{18634} \]

[In]

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

[Out]

((1925*(35 + 62*x))/(3 + 5*x)^2 - 16*Log[1 - 2*x] - 71874*Log[4 + 6*x] + 71890*Log[6 + 10*x])/18634

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {\frac {775 x}{121}+\frac {875}{242}}{\left (3+5 x \right )^{2}}-\frac {8 \ln \left (-1+2 x \right )}{9317}-\frac {27 \ln \left (2+3 x \right )}{7}+\frac {5135 \ln \left (3+5 x \right )}{1331}\)	\(40\)
norman	\(\frac {-\frac {2050}{363} x -\frac {21875}{2178} x^{2}}{\left (3+5 x \right )^{2}}-\frac {8 \ln \left (-1+2 x \right )}{9317}-\frac {27 \ln \left (2+3 x \right )}{7}+\frac {5135 \ln \left (3+5 x \right )}{1331}\)	\(43\)
default	\(-\frac {5}{22 \left (3+5 x \right )^{2}}+\frac {155}{121 \left (3+5 x \right )}+\frac {5135 \ln \left (3+5 x \right )}{1331}-\frac {8 \ln \left (-1+2 x \right )}{9317}-\frac {27 \ln \left (2+3 x \right )}{7}\)	\(44\)
parallelrisch	\(-\frac {16171650 \ln \left (\frac {2}{3}+x \right ) x^{2}-16175250 \ln \left (x +\frac {3}{5}\right ) x^{2}+3600 \ln \left (x -\frac {1}{2}\right ) x^{2}[phone] \ln \left (\frac {2}{3}+x \right ) x -19410300 \ln \left (x +\frac {3}{5}\right ) x +4320 \ln \left (x -\frac {1}{2}\right ) x +1684375 x^{2}+5821794 \ln \left (\frac {2}{3}+x \right )-5823090 \ln \left (x +\frac {3}{5}\right )+1296 \ln \left (x -\frac {1}{2}\right )+947100 x}{167706 \left (3+5 x \right )^{2}}\)	\(85\)

[In]

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

[Out]

25*(31/121*x+35/242)/(3+5*x)^2-8/9317*ln(-1+2*x)-27/7*ln(2+3*x)+5135/1331*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=\frac {71890 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 71874 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (3 \, x + 2\right ) - 16 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (2 \, x - 1\right ) + 119350 \, x + 67375}{18634 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/18634*(71890*(25*x^2 + 30*x + 9)*log(5*x + 3) - 71874*(25*x^2 + 30*x + 9)*log(3*x + 2) - 16*(25*x^2 + 30*x +
9)*log(2*x - 1) + 119350*x + 67375)/(25*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=- \frac {- 1550 x - 875}{6050 x^{2} + 7260 x + 2178} - \frac {8 \log {\left (x - \frac {1}{2} \right )}}{9317} + \frac {5135 \log {\left (x + \frac {3}{5} \right )}}{1331} - \frac {27 \log {\left (x + \frac {2}{3} \right )}}{7} \]

[In]

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

[Out]

-(-1550*x - 875)/(6050*x**2 + 7260*x + 2178) - 8*log(x - 1/2)/9317 + 5135*log(x + 3/5)/1331 - 27*log(x + 2/3)/
7

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=\frac {25 \, {\left (62 \, x + 35\right )}}{242 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {5135}{1331} \, \log \left (5 \, x + 3\right ) - \frac {27}{7} \, \log \left (3 \, x + 2\right ) - \frac {8}{9317} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

25/242*(62*x + 35)/(25*x^2 + 30*x + 9) + 5135/1331*log(5*x + 3) - 27/7*log(3*x + 2) - 8/9317*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=\frac {25 \, {\left (62 \, x + 35\right )}}{242 \, {\left (5 \, x + 3\right )}^{2}} + \frac {5135}{1331} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {27}{7} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {8}{9317} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

25/242*(62*x + 35)/(5*x + 3)^2 + 5135/1331*log(abs(5*x + 3)) - 27/7*log(abs(3*x + 2)) - 8/9317*log(abs(2*x - 1
))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^3} \, dx=\frac {5135\,\ln \left (x+\frac {3}{5}\right )}{1331}-\frac {27\,\ln \left (x+\frac {2}{3}\right )}{7}-\frac {8\,\ln \left (x-\frac {1}{2}\right )}{9317}+\frac {\frac {31\,x}{121}+\frac {35}{242}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]

[In]

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

[Out]

(5135*log(x + 3/5))/1331 - (27*log(x + 2/3))/7 - (8*log(x - 1/2))/9317 + ((31*x)/121 + 35/242)/((6*x)/5 + x^2
+ 9/25)