\(\int \frac {1}{(1-2 x) (3+5 x)^3} \, dx\) [1524]

Optimal result

Integrand size = 15, antiderivative size = 43 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=-\frac {1}{22 (3+5 x)^2}-\frac {2}{121 (3+5 x)}-\frac {4 \log (1-2 x)}{1331}+\frac {4 \log (3+5 x)}{1331} \]

[Out]

-1/22/(3+5*x)^2-2/121/(3+5*x)-4/1331*ln(1-2*x)+4/1331*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {46} \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=-\frac {2}{121 (5 x+3)}-\frac {1}{22 (5 x+3)^2}-\frac {4 \log (1-2 x)}{1331}+\frac {4 \log (5 x+3)}{1331} \]

[In]

Int[1/((1 - 2*x)*(3 + 5*x)^3),x]

[Out]

-1/22*1/(3 + 5*x)^2 - 2/(121*(3 + 5*x)) - (4*Log[1 - 2*x])/1331 + (4*Log[3 + 5*x])/1331

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{1331 (-1+2 x)}+\frac {5}{11 (3+5 x)^3}+\frac {10}{121 (3+5 x)^2}+\frac {20}{1331 (3+5 x)}\right ) \, dx \\ & = -\frac {1}{22 (3+5 x)^2}-\frac {2}{121 (3+5 x)}-\frac {4 \log (1-2 x)}{1331}+\frac {4 \log (3+5 x)}{1331} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=\frac {-\frac {11 (23+20 x)}{(3+5 x)^2}-8 \log (5-10 x)+8 \log (3+5 x)}{2662} \]

[In]

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

[Out]

((-11*(23 + 20*x))/(3 + 5*x)^2 - 8*Log[5 - 10*x] + 8*Log[3 + 5*x])/2662

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {-\frac {10 x}{121}-\frac {23}{242}}{\left (3+5 x \right )^{2}}-\frac {4 \ln \left (-1+2 x \right )}{1331}+\frac {4 \ln \left (3+5 x \right )}{1331}\)	\(32\)
norman	\(\frac {\frac {85}{363} x +\frac {575}{2178} x^{2}}{\left (3+5 x \right )^{2}}-\frac {4 \ln \left (-1+2 x \right )}{1331}+\frac {4 \ln \left (3+5 x \right )}{1331}\)	\(35\)
default	\(-\frac {1}{22 \left (3+5 x \right )^{2}}-\frac {2}{121 \left (3+5 x \right )}+\frac {4 \ln \left (3+5 x \right )}{1331}-\frac {4 \ln \left (-1+2 x \right )}{1331}\)	\(36\)
parallelrisch	\(\frac {1800 \ln \left (x +\frac {3}{5}\right ) x^{2}-1800 \ln \left (x -\frac {1}{2}\right ) x^{2}+2160 \ln \left (x +\frac {3}{5}\right ) x -2160 \ln \left (x -\frac {1}{2}\right ) x +6325 x^{2}+648 \ln \left (x +\frac {3}{5}\right )-648 \ln \left (x -\frac {1}{2}\right )+5610 x}{23958 \left (3+5 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

25*(-2/605*x-23/6050)/(3+5*x)^2-4/1331*ln(-1+2*x)+4/1331*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=\frac {8 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 8 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (2 \, x - 1\right ) - 220 \, x - 253}{2662 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/2662*(8*(25*x^2 + 30*x + 9)*log(5*x + 3) - 8*(25*x^2 + 30*x + 9)*log(2*x - 1) - 220*x - 253)/(25*x^2 + 30*x
+ 9)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=- \frac {20 x + 23}{6050 x^{2} + 7260 x + 2178} - \frac {4 \log {\left (x - \frac {1}{2} \right )}}{1331} + \frac {4 \log {\left (x + \frac {3}{5} \right )}}{1331} \]

[In]

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

[Out]

-(20*x + 23)/(6050*x**2 + 7260*x + 2178) - 4*log(x - 1/2)/1331 + 4*log(x + 3/5)/1331

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=-\frac {20 \, x + 23}{242 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {4}{1331} \, \log \left (5 \, x + 3\right ) - \frac {4}{1331} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/242*(20*x + 23)/(25*x^2 + 30*x + 9) + 4/1331*log(5*x + 3) - 4/1331*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=-\frac {20 \, x + 23}{242 \, {\left (5 \, x + 3\right )}^{2}} + \frac {4}{1331} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {4}{1331} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/242*(20*x + 23)/(5*x + 3)^2 + 4/1331*log(abs(5*x + 3)) - 4/1331*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.60 \[ \int \frac {1}{(1-2 x) (3+5 x)^3} \, dx=\frac {8\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{1331}-\frac {\frac {2\,x}{605}+\frac {23}{6050}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]

[In]

int(-1/((2*x - 1)*(5*x + 3)^3),x)

[Out]

(8*atanh((20*x)/11 + 1/11))/1331 - ((2*x)/605 + 23/6050)/((6*x)/5 + x^2 + 9/25)