\(\int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx\) [1640]

Optimal result

Integrand size = 20, antiderivative size = 54 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {11}{98 (1-2 x)^2}+\frac {31}{343 (1-2 x)}+\frac {3}{343 (2+3 x)}-\frac {87 \log (1-2 x)}{2401}+\frac {87 \log (2+3 x)}{2401} \]

[Out]

11/98/(1-2*x)^2+31/343/(1-2*x)+3/343/(2+3*x)-87/2401*ln(1-2*x)+87/2401*ln(2+3*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} \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {31}{343 (1-2 x)}+\frac {3}{343 (3 x+2)}+\frac {11}{98 (1-2 x)^2}-\frac {87 \log (1-2 x)}{2401}+\frac {87 \log (3 x+2)}{2401} \]

[In]

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

[Out]

11/(98*(1 - 2*x)^2) + 31/(343*(1 - 2*x)) + 3/(343*(2 + 3*x)) - (87*Log[1 - 2*x])/2401 + (87*Log[2 + 3*x])/2401

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 {22}{49 (-1+2 x)^3}+\frac {62}{343 (-1+2 x)^2}-\frac {174}{2401 (-1+2 x)}-\frac {9}{343 (2+3 x)^2}+\frac {261}{2401 (2+3 x)}\right ) \, dx \\ & = \frac {11}{98 (1-2 x)^2}+\frac {31}{343 (1-2 x)}+\frac {3}{343 (2+3 x)}-\frac {87 \log (1-2 x)}{2401}+\frac {87 \log (2+3 x)}{2401} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {\frac {7 \left (284+145 x-348 x^2\right )}{(1-2 x)^2 (2+3 x)}-174 \log (1-2 x)+174 \log (4+6 x)}{4802} \]

[In]

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

[Out]

((7*(284 + 145*x - 348*x^2))/((1 - 2*x)^2*(2 + 3*x)) - 174*Log[1 - 2*x] + 174*Log[4 + 6*x])/4802

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {-\frac {174}{343} x^{2}+\frac {145}{686} x +\frac {142}{343}}{\left (-1+2 x \right )^{2} \left (2+3 x \right )}-\frac {87 \ln \left (-1+2 x \right )}{2401}+\frac {87 \ln \left (2+3 x \right )}{2401}\)	\(44\)
default	\(\frac {11}{98 \left (-1+2 x \right )^{2}}-\frac {31}{343 \left (-1+2 x \right )}-\frac {87 \ln \left (-1+2 x \right )}{2401}+\frac {3}{343 \left (2+3 x \right )}+\frac {87 \ln \left (2+3 x \right )}{2401}\)	\(45\)
norman	\(\frac {\frac {110}{343} x^{2}-\frac {852}{343} x^{3}+\frac {855}{686} x}{\left (-1+2 x \right )^{2} \left (2+3 x \right )}-\frac {87 \ln \left (-1+2 x \right )}{2401}+\frac {87 \ln \left (2+3 x \right )}{2401}\)	\(47\)
parallelrisch	\(\frac {2088 \ln \left (\frac {2}{3}+x \right ) x^{3}-2088 \ln \left (x -\frac {1}{2}\right ) x^{3}-696 \ln \left (\frac {2}{3}+x \right ) x^{2}+696 \ln \left (x -\frac {1}{2}\right ) x^{2}-11928 x^{3}-870 \ln \left (\frac {2}{3}+x \right ) x +870 \ln \left (x -\frac {1}{2}\right ) x +1540 x^{2}+348 \ln \left (\frac {2}{3}+x \right )-348 \ln \left (x -\frac {1}{2}\right )+5985 x}{4802 \left (-1+2 x \right )^{2} \left (2+3 x \right )}\)	\(93\)

[In]

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

[Out]

12*(-29/686*x^2+145/8232*x+71/2058)/(-1+2*x)^2/(2+3*x)-87/2401*ln(-1+2*x)+87/2401*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=-\frac {2436 \, x^{2} - 174 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 174 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (2 \, x - 1\right ) - 1015 \, x - 1988}{4802 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \]

[In]

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

[Out]

-1/4802*(2436*x^2 - 174*(12*x^3 - 4*x^2 - 5*x + 2)*log(3*x + 2) + 174*(12*x^3 - 4*x^2 - 5*x + 2)*log(2*x - 1)
- 1015*x - 1988)/(12*x^3 - 4*x^2 - 5*x + 2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=- \frac {348 x^{2} - 145 x - 284}{8232 x^{3} - 2744 x^{2} - 3430 x + 1372} - \frac {87 \log {\left (x - \frac {1}{2} \right )}}{2401} + \frac {87 \log {\left (x + \frac {2}{3} \right )}}{2401} \]

[In]

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

[Out]

-(348*x**2 - 145*x - 284)/(8232*x**3 - 2744*x**2 - 3430*x + 1372) - 87*log(x - 1/2)/2401 + 87*log(x + 2/3)/240
1

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=-\frac {348 \, x^{2} - 145 \, x - 284}{686 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} + \frac {87}{2401} \, \log \left (3 \, x + 2\right ) - \frac {87}{2401} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/686*(348*x^2 - 145*x - 284)/(12*x^3 - 4*x^2 - 5*x + 2) + 87/2401*log(3*x + 2) - 87/2401*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {3}{343 \, {\left (3 \, x + 2\right )}} + \frac {6 \, {\left (\frac {448}{3 \, x + 2} - 95\right )}}{2401 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}^{2}} - \frac {87}{2401} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]

[In]

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

[Out]

3/343/(3*x + 2) + 6/2401*(448/(3*x + 2) - 95)/(7/(3*x + 2) - 2)^2 - 87/2401*log(abs(-7/(3*x + 2) + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {3+5 x}{(1-2 x)^3 (2+3 x)^2} \, dx=\frac {174\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{2401}-\frac {-\frac {29\,x^2}{686}+\frac {145\,x}{8232}+\frac {71}{2058}}{-x^3+\frac {x^2}{3}+\frac {5\,x}{12}-\frac {1}{6}} \]

[In]

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

[Out]

(174*atanh((12*x)/7 + 1/7))/2401 - ((145*x)/8232 - (29*x^2)/686 + 71/2058)/((5*x)/12 + x^2/3 - x^3 - 1/6)