\(\int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx\) [1497]

Optimal result

Integrand size = 22, antiderivative size = 64 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {1}{7 (2+3 x)^3}+\frac {111}{98 (2+3 x)^2}+\frac {3897}{343 (2+3 x)}-\frac {16 \log (1-2 x)}{26411}-\frac {136419 \log (2+3 x)}{2401}+\frac {625}{11} \log (3+5 x) \]

[Out]

1/7/(2+3*x)^3+111/98/(2+3*x)^2+3897/343/(2+3*x)-16/26411*ln(1-2*x)-136419/2401*ln(2+3*x)+625/11*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, 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} \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {3897}{343 (3 x+2)}+\frac {111}{98 (3 x+2)^2}+\frac {1}{7 (3 x+2)^3}-\frac {16 \log (1-2 x)}{26411}-\frac {136419 \log (3 x+2)}{2401}+\frac {625}{11} \log (5 x+3) \]

[In]

Int[1/((1 - 2*x)*(2 + 3*x)^4*(3 + 5*x)),x]

[Out]

1/(7*(2 + 3*x)^3) + 111/(98*(2 + 3*x)^2) + 3897/(343*(2 + 3*x)) - (16*Log[1 - 2*x])/26411 - (136419*Log[2 + 3*
x])/2401 + (625*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 {32}{26411 (-1+2 x)}-\frac {9}{7 (2+3 x)^4}-\frac {333}{49 (2+3 x)^3}-\frac {11691}{343 (2+3 x)^2}-\frac {409257}{2401 (2+3 x)}+\frac {3125}{11 (3+5 x)}\right ) \, dx \\ & = \frac {1}{7 (2+3 x)^3}+\frac {111}{98 (2+3 x)^2}+\frac {3897}{343 (2+3 x)}-\frac {16 \log (1-2 x)}{26411}-\frac {136419 \log (2+3 x)}{2401}+\frac {625}{11} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {\frac {77 \left (32828+95859 x+70146 x^2\right )}{2 (2+3 x)^3}-16 \log (1-2 x)-1500609 \log (4+6 x)+1500625 \log (6+10 x)}{26411} \]

[In]

Integrate[1/((1 - 2*x)*(2 + 3*x)^4*(3 + 5*x)),x]

[Out]

((77*(32828 + 95859*x + 70146*x^2))/(2*(2 + 3*x)^3) - 16*Log[1 - 2*x] - 1500609*Log[4 + 6*x] + 1500625*Log[6 +
10*x])/26411

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.69

method	result	size
norman	\(\frac {\frac {35073}{343} x^{2}+\frac {95859}{686} x +\frac {16414}{343}}{\left (2+3 x \right )^{3}}-\frac {16 \ln \left (-1+2 x \right )}{26411}-\frac {136419 \ln \left (2+3 x \right )}{2401}+\frac {625 \ln \left (3+5 x \right )}{11}\)	\(44\)
risch	\(\frac {\frac {35073}{343} x^{2}+\frac {95859}{686} x +\frac {16414}{343}}{\left (2+3 x \right )^{3}}-\frac {16 \ln \left (-1+2 x \right )}{26411}-\frac {136419 \ln \left (2+3 x \right )}{2401}+\frac {625 \ln \left (3+5 x \right )}{11}\)	\(45\)
default	\(\frac {625 \ln \left (3+5 x \right )}{11}-\frac {16 \ln \left (-1+2 x \right )}{26411}+\frac {1}{7 \left (2+3 x \right )^{3}}+\frac {111}{98 \left (2+3 x \right )^{2}}+\frac {3897}{343 \left (2+3 x \right )}-\frac {136419 \ln \left (2+3 x \right )}{2401}\)	\(53\)
parallelrisch	\(-\frac {324131544 \ln \left (\frac {2}{3}+x \right ) x^{3}-324135000 \ln \left (x +\frac {3}{5}\right ) x^{3}+3456 \ln \left (x -\frac {1}{2}\right ) x^{3}[phone] \ln \left (\frac {2}{3}+x \right ) x^{2}-648270000 \ln \left (x +\frac {3}{5}\right ) x^{2}+6912 \ln \left (x -\frac {1}{2}\right ) x^{2}[phone] x^{3}[phone] \ln \left (\frac {2}{3}+x \right ) x -432180000 \ln \left (x +\frac {3}{5}\right ) x +4608 \ln \left (x -\frac {1}{2}\right ) x [phone] x^{2}[phone] \ln \left (\frac {2}{3}+x \right )-96040000 \ln \left (x +\frac {3}{5}\right )+1024 \ln \left (x -\frac {1}{2}\right )[phone] x}{211288 \left (2+3 x \right )^{3}}\)	\(117\)

[In]

int(1/(1-2*x)/(2+3*x)^4/(3+5*x),x,method=_RETURNVERBOSE)

[Out]

(35073/343*x^2+95859/686*x+16414/343)/(2+3*x)^3-16/26411*ln(-1+2*x)-136419/2401*ln(2+3*x)+625/11*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {5401242 \, x^{2} + 3001250 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (5 \, x + 3\right ) - 3001218 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) - 32 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (2 \, x - 1\right ) + 7381143 \, x + 2527756}{52822 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

integrate(1/(1-2*x)/(2+3*x)^4/(3+5*x),x, algorithm="fricas")

[Out]

1/52822*(5401242*x^2 + 3001250*(27*x^3 + 54*x^2 + 36*x + 8)*log(5*x + 3) - 3001218*(27*x^3 + 54*x^2 + 36*x + 8
)*log(3*x + 2) - 32*(27*x^3 + 54*x^2 + 36*x + 8)*log(2*x - 1) + 7381143*x + 2527756)/(27*x^3 + 54*x^2 + 36*x +
8)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=- \frac {- 70146 x^{2} - 95859 x - 32828}{18522 x^{3} + 37044 x^{2} + 24696 x + 5488} - \frac {16 \log {\left (x - \frac {1}{2} \right )}}{26411} + \frac {625 \log {\left (x + \frac {3}{5} \right )}}{11} - \frac {136419 \log {\left (x + \frac {2}{3} \right )}}{2401} \]

[In]

integrate(1/(1-2*x)/(2+3*x)**4/(3+5*x),x)

[Out]

-(-70146*x**2 - 95859*x - 32828)/(18522*x**3 + 37044*x**2 + 24696*x + 5488) - 16*log(x - 1/2)/26411 + 625*log(
x + 3/5)/11 - 136419*log(x + 2/3)/2401

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {70146 \, x^{2} + 95859 \, x + 32828}{686 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + \frac {625}{11} \, \log \left (5 \, x + 3\right ) - \frac {136419}{2401} \, \log \left (3 \, x + 2\right ) - \frac {16}{26411} \, \log \left (2 \, x - 1\right ) \]

[In]

integrate(1/(1-2*x)/(2+3*x)^4/(3+5*x),x, algorithm="maxima")

[Out]

1/686*(70146*x^2 + 95859*x + 32828)/(27*x^3 + 54*x^2 + 36*x + 8) + 625/11*log(5*x + 3) - 136419/2401*log(3*x +
2) - 16/26411*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {70146 \, x^{2} + 95859 \, x + 32828}{686 \, {\left (3 \, x + 2\right )}^{3}} + \frac {625}{11} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {136419}{2401} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {16}{26411} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

integrate(1/(1-2*x)/(2+3*x)^4/(3+5*x),x, algorithm="giac")

[Out]

1/686*(70146*x^2 + 95859*x + 32828)/(3*x + 2)^3 + 625/11*log(abs(5*x + 3)) - 136419/2401*log(abs(3*x + 2)) - 1
6/26411*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(1-2 x) (2+3 x)^4 (3+5 x)} \, dx=\frac {625\,\ln \left (x+\frac {3}{5}\right )}{11}-\frac {136419\,\ln \left (x+\frac {2}{3}\right )}{2401}-\frac {16\,\ln \left (x-\frac {1}{2}\right )}{26411}+\frac {\frac {1299\,x^2}{343}+\frac {10651\,x}{2058}+\frac {16414}{9261}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}} \]

[In]

int(-1/((2*x - 1)*(3*x + 2)^4*(5*x + 3)),x)

[Out]

(625*log(x + 3/5))/11 - (136419*log(x + 2/3))/2401 - (16*log(x - 1/2))/26411 + ((10651*x)/2058 + (1299*x^2)/34
3 + 16414/9261)/((4*x)/3 + 2*x^2 + x^3 + 8/27)