\(\int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx\) [1492]

Optimal result

Integrand size = 20, antiderivative size = 21 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=-\frac {7}{22} \log (1-2 x)+\frac {1}{55} \log (3+5 x) \]

[Out]

-7/22*ln(1-2*x)+1/55*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, 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} \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=\frac {1}{55} \log (5 x+3)-\frac {7}{22} \log (1-2 x) \]

[In]

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

[Out]

(-7*Log[1 - 2*x])/22 + Log[3 + 5*x]/55

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 {7}{11 (-1+2 x)}+\frac {1}{11 (3+5 x)}\right ) \, dx \\ & = -\frac {7}{22} \log (1-2 x)+\frac {1}{55} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=-\frac {7}{22} \log (1-2 x)+\frac {1}{55} \log (3+5 x) \]

[In]

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

[Out]

(-7*Log[1 - 2*x])/22 + Log[3 + 5*x]/55

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {\ln \left (x +\frac {3}{5}\right )}{55}-\frac {7 \ln \left (x -\frac {1}{2}\right )}{22}\)	\(14\)
default	\(\frac {\ln \left (3+5 x \right )}{55}-\frac {7 \ln \left (-1+2 x \right )}{22}\)	\(18\)
norman	\(\frac {\ln \left (3+5 x \right )}{55}-\frac {7 \ln \left (-1+2 x \right )}{22}\)	\(18\)
risch	\(\frac {\ln \left (3+5 x \right )}{55}-\frac {7 \ln \left (-1+2 x \right )}{22}\)	\(18\)

[In]

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

[Out]

1/55*ln(x+3/5)-7/22*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=\frac {1}{55} \, \log \left (5 \, x + 3\right ) - \frac {7}{22} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

1/55*log(5*x + 3) - 7/22*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=- \frac {7 \log {\left (x - \frac {1}{2} \right )}}{22} + \frac {\log {\left (x + \frac {3}{5} \right )}}{55} \]

[In]

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

[Out]

-7*log(x - 1/2)/22 + log(x + 3/5)/55

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=\frac {1}{55} \, \log \left (5 \, x + 3\right ) - \frac {7}{22} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

1/55*log(5*x + 3) - 7/22*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=\frac {1}{55} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {7}{22} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

1/55*log(abs(5*x + 3)) - 7/22*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.45 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx=\frac {\ln \left (x+\frac {3}{5}\right )}{55}-\frac {7\,\ln \left (x-\frac {1}{2}\right )}{22} \]

[In]

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

[Out]

log(x + 3/5)/55 - (7*log(x - 1/2))/22