3.1548 \(\int \frac {3+5 x}{(1-2 x)^2 (2+3 x)^2} \, dx\)

Optimal. Leaf size=43 \[ \frac {11}{49 (1-2 x)}+\frac {1}{49 (3 x+2)}-\frac {31}{343} \log (1-2 x)+\frac {31}{343} \log (3 x+2) \]

[Out]

11/49/(1-2*x)+1/49/(2+3*x)-31/343*ln(1-2*x)+31/343*ln(2+3*x)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ \frac {11}{49 (1-2 x)}+\frac {1}{49 (3 x+2)}-\frac {31}{343} \log (1-2 x)+\frac {31}{343} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

11/(49*(1 - 2*x)) + 1/(49*(2 + 3*x)) - (31*Log[1 - 2*x])/343 + (31*Log[2 + 3*x])/343

Rule 77

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*} \int \frac {3+5 x}{(1-2 x)^2 (2+3 x)^2} \, dx &=\int \left (\frac {22}{49 (-1+2 x)^2}-\frac {62}{343 (-1+2 x)}-\frac {3}{49 (2+3 x)^2}+\frac {93}{343 (2+3 x)}\right ) \, dx\\ &=\frac {11}{49 (1-2 x)}+\frac {1}{49 (2+3 x)}-\frac {31}{343} \log (1-2 x)+\frac {31}{343} \log (2+3 x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 40, normalized size = 0.93 \[ \frac {-31 x-23}{49 \left (6 x^2+x-2\right )}-\frac {31}{343} \log (1-2 x)+\frac {31}{343} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-23 - 31*x)/(49*(-2 + x + 6*x^2)) - (31*Log[1 - 2*x])/343 + (31*Log[2 + 3*x])/343

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 49, normalized size = 1.14 \[ \frac {31 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (3 \, x + 2\right ) - 31 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (2 \, x - 1\right ) - 217 \, x - 161}{343 \, {\left (6 \, x^{2} + x - 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/343*(31*(6*x^2 + x - 2)*log(3*x + 2) - 31*(6*x^2 + x - 2)*log(2*x - 1) - 217*x - 161)/(6*x^2 + x - 2)

________________________________________________________________________________________

giac [A]  time = 0.93, size = 40, normalized size = 0.93 \[ \frac {1}{49 \, {\left (3 \, x + 2\right )}} + \frac {66}{343 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}} - \frac {31}{343} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/49/(3*x + 2) + 66/343/(7/(3*x + 2) - 2) - 31/343*log(abs(-7/(3*x + 2) + 2))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 36, normalized size = 0.84 \[ -\frac {31 \ln \left (2 x -1\right )}{343}+\frac {31 \ln \left (3 x +2\right )}{343}+\frac {1}{147 x +98}-\frac {11}{49 \left (2 x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/49/(3*x+2)+31/343*ln(3*x+2)-11/49/(2*x-1)-31/343*ln(2*x-1)

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 34, normalized size = 0.79 \[ -\frac {31 \, x + 23}{49 \, {\left (6 \, x^{2} + x - 2\right )}} + \frac {31}{343} \, \log \left (3 \, x + 2\right ) - \frac {31}{343} \, \log \left (2 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/49*(31*x + 23)/(6*x^2 + x - 2) + 31/343*log(3*x + 2) - 31/343*log(2*x - 1)

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 26, normalized size = 0.60 \[ \frac {62\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{343}-\frac {\frac {31\,x}{294}+\frac {23}{294}}{x^2+\frac {x}{6}-\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(62*atanh((12*x)/7 + 1/7))/343 - ((31*x)/294 + 23/294)/(x/6 + x^2 - 1/3)

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 36, normalized size = 0.84 \[ \frac {- 31 x - 23}{294 x^{2} + 49 x - 98} - \frac {31 \log {\left (x - \frac {1}{2} \right )}}{343} + \frac {31 \log {\left (x + \frac {2}{3} \right )}}{343} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-31*x - 23)/(294*x**2 + 49*x - 98) - 31*log(x - 1/2)/343 + 31*log(x + 2/3)/343

________________________________________________________________________________________