\(\int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx\) [1443]

Optimal result

Integrand size = 20, antiderivative size = 21 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {11}{14} \log (1-2 x)-\frac {1}{21} \log (2+3 x) \]

[Out]

-11/14*ln(1-2*x)-1/21*ln(2+3*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} \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {11}{14} \log (1-2 x)-\frac {1}{21} \log (3 x+2) \]

[In]

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

[Out]

(-11*Log[1 - 2*x])/14 - Log[2 + 3*x]/21

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 {11}{7 (-1+2 x)}-\frac {1}{7 (2+3 x)}\right ) \, dx \\ & = -\frac {11}{14} \log (1-2 x)-\frac {1}{21} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {11}{14} \log (1-2 x)-\frac {1}{21} \log (2+3 x) \]

[In]

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

[Out]

(-11*Log[1 - 2*x])/14 - Log[2 + 3*x]/21

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {\ln \left (\frac {2}{3}+x \right )}{21}-\frac {11 \ln \left (x -\frac {1}{2}\right )}{14}\)	\(14\)
default	\(-\frac {11 \ln \left (-1+2 x \right )}{14}-\frac {\ln \left (2+3 x \right )}{21}\)	\(18\)
norman	\(-\frac {11 \ln \left (-1+2 x \right )}{14}-\frac {\ln \left (2+3 x \right )}{21}\)	\(18\)
risch	\(-\frac {11 \ln \left (-1+2 x \right )}{14}-\frac {\ln \left (2+3 x \right )}{21}\)	\(18\)

[In]

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

[Out]

-1/21*ln(2/3+x)-11/14*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {1}{21} \, \log \left (3 \, x + 2\right ) - \frac {11}{14} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/21*log(3*x + 2) - 11/14*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=- \frac {11 \log {\left (x - \frac {1}{2} \right )}}{14} - \frac {\log {\left (x + \frac {2}{3} \right )}}{21} \]

[In]

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

[Out]

-11*log(x - 1/2)/14 - log(x + 2/3)/21

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {1}{21} \, \log \left (3 \, x + 2\right ) - \frac {11}{14} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/21*log(3*x + 2) - 11/14*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {1}{21} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {11}{14} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/21*log(abs(3*x + 2)) - 11/14*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)} \, dx=-\frac {11\,\ln \left (x-\frac {1}{2}\right )}{14}-\frac {\ln \left (x+\frac {2}{3}\right )}{21} \]

[In]

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

[Out]

- (11*log(x - 1/2))/14 - log(x + 2/3)/21