\(\int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx\) [1445]

Optimal result

Integrand size = 20, antiderivative size = 43 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=\frac {1}{42 (2+3 x)^2}-\frac {11}{49 (2+3 x)}-\frac {22}{343} \log (1-2 x)+\frac {22}{343} \log (2+3 x) \]

[Out]

1/42/(2+3*x)^2-11/49/(2+3*x)-22/343*ln(1-2*x)+22/343*ln(2+3*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=-\frac {11}{49 (3 x+2)}+\frac {1}{42 (3 x+2)^2}-\frac {22}{343} \log (1-2 x)+\frac {22}{343} \log (3 x+2) \]

[In]

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

[Out]

1/(42*(2 + 3*x)^2) - 11/(49*(2 + 3*x)) - (22*Log[1 - 2*x])/343 + (22*Log[2 + 3*x])/343

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 {44}{343 (-1+2 x)}-\frac {1}{7 (2+3 x)^3}+\frac {33}{49 (2+3 x)^2}+\frac {66}{343 (2+3 x)}\right ) \, dx \\ & = \frac {1}{42 (2+3 x)^2}-\frac {11}{49 (2+3 x)}-\frac {22}{343} \log (1-2 x)+\frac {22}{343} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=\frac {-\frac {7 (125+198 x)}{(2+3 x)^2}-132 \log (3-6 x)+132 \log (2+3 x)}{2058} \]

[In]

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

[Out]

((-7*(125 + 198*x))/(2 + 3*x)^2 - 132*Log[3 - 6*x] + 132*Log[2 + 3*x])/2058

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {-\frac {33 x}{49}-\frac {125}{294}}{\left (2+3 x \right )^{2}}-\frac {22 \ln \left (-1+2 x \right )}{343}+\frac {22 \ln \left (2+3 x \right )}{343}\)	\(32\)
norman	\(\frac {\frac {59}{98} x +\frac {375}{392} x^{2}}{\left (2+3 x \right )^{2}}-\frac {22 \ln \left (-1+2 x \right )}{343}+\frac {22 \ln \left (2+3 x \right )}{343}\)	\(35\)
default	\(-\frac {22 \ln \left (-1+2 x \right )}{343}+\frac {1}{42 \left (2+3 x \right )^{2}}-\frac {11}{49 \left (2+3 x \right )}+\frac {22 \ln \left (2+3 x \right )}{343}\)	\(36\)
parallelrisch	\(\frac {1584 \ln \left (\frac {2}{3}+x \right ) x^{2}-1584 \ln \left (x -\frac {1}{2}\right ) x^{2}+2112 \ln \left (\frac {2}{3}+x \right ) x -2112 \ln \left (x -\frac {1}{2}\right ) x +2625 x^{2}+704 \ln \left (\frac {2}{3}+x \right )-704 \ln \left (x -\frac {1}{2}\right )+1652 x}{2744 \left (2+3 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

9*(-11/147*x-125/2646)/(2+3*x)^2-22/343*ln(-1+2*x)+22/343*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=\frac {132 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 132 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (2 \, x - 1\right ) - 1386 \, x - 875}{2058 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

1/2058*(132*(9*x^2 + 12*x + 4)*log(3*x + 2) - 132*(9*x^2 + 12*x + 4)*log(2*x - 1) - 1386*x - 875)/(9*x^2 + 12*
x + 4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=- \frac {198 x + 125}{2646 x^{2} + 3528 x + 1176} - \frac {22 \log {\left (x - \frac {1}{2} \right )}}{343} + \frac {22 \log {\left (x + \frac {2}{3} \right )}}{343} \]

[In]

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

[Out]

-(198*x + 125)/(2646*x**2 + 3528*x + 1176) - 22*log(x - 1/2)/343 + 22*log(x + 2/3)/343

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=-\frac {198 \, x + 125}{294 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {22}{343} \, \log \left (3 \, x + 2\right ) - \frac {22}{343} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/294*(198*x + 125)/(9*x^2 + 12*x + 4) + 22/343*log(3*x + 2) - 22/343*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=-\frac {198 \, x + 125}{294 \, {\left (3 \, x + 2\right )}^{2}} + \frac {22}{343} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {22}{343} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/294*(198*x + 125)/(3*x + 2)^2 + 22/343*log(abs(3*x + 2)) - 22/343*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.60 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^3} \, dx=\frac {44\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{343}-\frac {\frac {11\,x}{147}+\frac {125}{2646}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]

[In]

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

[Out]

(44*atanh((12*x)/7 + 1/7))/343 - ((11*x)/147 + 125/2646)/((4*x)/3 + x^2 + 4/9)