\(\int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx\) [1508]

Optimal result

Integrand size = 20, antiderivative size = 32 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 (3+5 x)}-\frac {7}{121} \log (1-2 x)+\frac {7}{121} \log (3+5 x) \]

[Out]

-1/55/(3+5*x)-7/121*ln(1-2*x)+7/121*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, 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} \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 (5 x+3)}-\frac {7}{121} \log (1-2 x)+\frac {7}{121} \log (5 x+3) \]

[In]

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

[Out]

-1/55*1/(3 + 5*x) - (7*Log[1 - 2*x])/121 + (7*Log[3 + 5*x])/121

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)}+\frac {1}{11 (3+5 x)^2}+\frac {35}{121 (3+5 x)}\right ) \, dx \\ & = -\frac {1}{55 (3+5 x)}-\frac {7}{121} \log (1-2 x)+\frac {7}{121} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=\frac {1}{605} \left (-\frac {11}{3+5 x}-35 \log (5-10 x)+35 \log (3+5 x)\right ) \]

[In]

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

[Out]

(-11/(3 + 5*x) - 35*Log[5 - 10*x] + 35*Log[3 + 5*x])/605

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {1}{275 \left (x +\frac {3}{5}\right )}-\frac {7 \ln \left (-1+2 x \right )}{121}+\frac {7 \ln \left (3+5 x \right )}{121}\)	\(25\)
default	\(-\frac {1}{55 \left (3+5 x \right )}+\frac {7 \ln \left (3+5 x \right )}{121}-\frac {7 \ln \left (-1+2 x \right )}{121}\)	\(27\)
norman	\(\frac {x}{99+165 x}-\frac {7 \ln \left (-1+2 x \right )}{121}+\frac {7 \ln \left (3+5 x \right )}{121}\)	\(28\)
parallelrisch	\(\frac {105 \ln \left (x +\frac {3}{5}\right ) x -105 \ln \left (x -\frac {1}{2}\right ) x +63 \ln \left (x +\frac {3}{5}\right )-63 \ln \left (x -\frac {1}{2}\right )+11 x}{1089+1815 x}\)	\(40\)

[In]

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

[Out]

-1/275/(x+3/5)-7/121*ln(-1+2*x)+7/121*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=\frac {35 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 35 \, {\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 11}{605 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

1/605*(35*(5*x + 3)*log(5*x + 3) - 35*(5*x + 3)*log(2*x - 1) - 11)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=- \frac {7 \log {\left (x - \frac {1}{2} \right )}}{121} + \frac {7 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {1}{275 x + 165} \]

[In]

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

[Out]

-7*log(x - 1/2)/121 + 7*log(x + 3/5)/121 - 1/(275*x + 165)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 \, {\left (5 \, x + 3\right )}} + \frac {7}{121} \, \log \left (5 \, x + 3\right ) - \frac {7}{121} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/55/(5*x + 3) + 7/121*log(5*x + 3) - 7/121*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=-\frac {1}{55 \, {\left (5 \, x + 3\right )}} - \frac {7}{121} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]

[In]

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

[Out]

-1/55/(5*x + 3) - 7/121*log(abs(-11/(5*x + 3) + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.56 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)^2} \, dx=\frac {14\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{121}-\frac {1}{275\,\left (x+\frac {3}{5}\right )} \]

[In]

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

[Out]

(14*atanh((20*x)/11 + 1/11))/121 - 1/(275*(x + 3/5))