\(\int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx\) [1691]

Optimal result

Integrand size = 20, antiderivative size = 54 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {7}{242 (1-2 x)^2}+\frac {37}{1331 (1-2 x)}-\frac {5}{1331 (3+5 x)}-\frac {195 \log (1-2 x)}{14641}+\frac {195 \log (3+5 x)}{14641} \]

[Out]

7/242/(1-2*x)^2+37/1331/(1-2*x)-5/1331/(3+5*x)-195/14641*ln(1-2*x)+195/14641*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {37}{1331 (1-2 x)}-\frac {5}{1331 (5 x+3)}+\frac {7}{242 (1-2 x)^2}-\frac {195 \log (1-2 x)}{14641}+\frac {195 \log (5 x+3)}{14641} \]

[In]

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

[Out]

7/(242*(1 - 2*x)^2) + 37/(1331*(1 - 2*x)) - 5/(1331*(3 + 5*x)) - (195*Log[1 - 2*x])/14641 + (195*Log[3 + 5*x])
/14641

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {14}{121 (-1+2 x)^3}+\frac {74}{1331 (-1+2 x)^2}-\frac {390}{14641 (-1+2 x)}+\frac {25}{1331 (3+5 x)^2}+\frac {975}{14641 (3+5 x)}\right ) \, dx \\ & = \frac {7}{242 (1-2 x)^2}+\frac {37}{1331 (1-2 x)}-\frac {5}{1331 (3+5 x)}-\frac {195 \log (1-2 x)}{14641}+\frac {195 \log (3+5 x)}{14641} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.15 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {10}{1331 (-11+5 (1-2 x))}+\frac {7}{242 (1-2 x)^2}+\frac {37}{1331 (1-2 x)}+\frac {195 \log (11-5 (1-2 x))}{14641}-\frac {195 \log (1-2 x)}{14641} \]

[In]

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

[Out]

10/(1331*(-11 + 5*(1 - 2*x))) + 7/(242*(1 - 2*x)^2) + 37/(1331*(1 - 2*x)) + (195*Log[11 - 5*(1 - 2*x)])/14641
- (195*Log[1 - 2*x])/14641

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {-\frac {390}{1331} x^{2}+\frac {351}{2662} x +\frac {443}{2662}}{\left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {195 \ln \left (-1+2 x \right )}{14641}+\frac {195 \ln \left (3+5 x \right )}{14641}\)	\(44\)
default	\(-\frac {5}{1331 \left (3+5 x \right )}+\frac {195 \ln \left (3+5 x \right )}{14641}+\frac {7}{242 \left (-1+2 x \right )^{2}}-\frac {37}{1331 \left (-1+2 x \right )}-\frac {195 \ln \left (-1+2 x \right )}{14641}\)	\(45\)
norman	\(\frac {\frac {602}{3993} x^{2}+\frac {2077}{3993} x -\frac {4430}{3993} x^{3}}{\left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {195 \ln \left (-1+2 x \right )}{14641}+\frac {195 \ln \left (3+5 x \right )}{14641}\)	\(47\)
parallelrisch	\(\frac {11700 \ln \left (x +\frac {3}{5}\right ) x^{3}-11700 \ln \left (x -\frac {1}{2}\right ) x^{3}-4680 \ln \left (x +\frac {3}{5}\right ) x^{2}+4680 \ln \left (x -\frac {1}{2}\right ) x^{2}-48730 x^{3}-4095 \ln \left (x +\frac {3}{5}\right ) x +4095 \ln \left (x -\frac {1}{2}\right ) x +6622 x^{2}+1755 \ln \left (x +\frac {3}{5}\right )-1755 \ln \left (x -\frac {1}{2}\right )+22847 x}{43923 \left (-1+2 x \right )^{2} \left (3+5 x \right )}\)	\(93\)

[In]

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

[Out]

20*(-39/2662*x^2+351/53240*x+443/53240)/(-1+2*x)^2/(3+5*x)-195/14641*ln(-1+2*x)+195/14641*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {8580 \, x^{2} - 390 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 390 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 3861 \, x - 4873}{29282 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]

[In]

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

[Out]

-1/29282*(8580*x^2 - 390*(20*x^3 - 8*x^2 - 7*x + 3)*log(5*x + 3) + 390*(20*x^3 - 8*x^2 - 7*x + 3)*log(2*x - 1)
- 3861*x - 4873)/(20*x^3 - 8*x^2 - 7*x + 3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=- \frac {780 x^{2} - 351 x - 443}{53240 x^{3} - 21296 x^{2} - 18634 x + 7986} - \frac {195 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {195 \log {\left (x + \frac {3}{5} \right )}}{14641} \]

[In]

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

[Out]

-(780*x**2 - 351*x - 443)/(53240*x**3 - 21296*x**2 - 18634*x + 7986) - 195*log(x - 1/2)/14641 + 195*log(x + 3/
5)/14641

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {780 \, x^{2} - 351 \, x - 443}{2662 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} + \frac {195}{14641} \, \log \left (5 \, x + 3\right ) - \frac {195}{14641} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/2662*(780*x^2 - 351*x - 443)/(20*x^3 - 8*x^2 - 7*x + 3) + 195/14641*log(5*x + 3) - 195/14641*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {5}{1331 \, {\left (5 \, x + 3\right )}} + \frac {10 \, {\left (\frac {792}{5 \, x + 3} - 109\right )}}{14641 \, {\left (\frac {11}{5 \, x + 3} - 2\right )}^{2}} - \frac {195}{14641} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]

[In]

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

[Out]

-5/1331/(5*x + 3) + 10/14641*(792/(5*x + 3) - 109)/(11/(5*x + 3) - 2)^2 - 195/14641*log(abs(-11/(5*x + 3) + 2)
)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {2+3 x}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {390\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}-\frac {-\frac {39\,x^2}{2662}+\frac {351\,x}{53240}+\frac {443}{53240}}{-x^3+\frac {2\,x^2}{5}+\frac {7\,x}{20}-\frac {3}{20}} \]

[In]

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

[Out]

(390*atanh((20*x)/11 + 1/11))/14641 - ((351*x)/53240 - (39*x^2)/2662 + 443/53240)/((7*x)/20 + (2*x^2)/5 - x^3
- 3/20)