\(\int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx\) [1200]

Optimal result

Integrand size = 20, antiderivative size = 21 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=-\frac {7}{3} \log (2+3 x)+\frac {11}{5} \log (3+5 x) \]

[Out]

-7/3*ln(2+3*x)+11/5*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} \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=\frac {11}{5} \log (5 x+3)-\frac {7}{3} \log (3 x+2) \]

[In]

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

[Out]

(-7*Log[2 + 3*x])/3 + (11*Log[3 + 5*x])/5

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=-\frac {7}{3} \log (2+3 x)+\frac {11}{5} \log (3+5 x) \]

[In]

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

[Out]

(-7*Log[2 + 3*x])/3 + (11*Log[3 + 5*x])/5

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {7 \ln \left (\frac {2}{3}+x \right )}{3}+\frac {11 \ln \left (x +\frac {3}{5}\right )}{5}\)	\(14\)
default	\(-\frac {7 \ln \left (2+3 x \right )}{3}+\frac {11 \ln \left (3+5 x \right )}{5}\)	\(18\)
norman	\(-\frac {7 \ln \left (2+3 x \right )}{3}+\frac {11 \ln \left (3+5 x \right )}{5}\)	\(18\)
risch	\(-\frac {7 \ln \left (2+3 x \right )}{3}+\frac {11 \ln \left (3+5 x \right )}{5}\)	\(18\)

[In]

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

[Out]

-7/3*ln(2/3+x)+11/5*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=\frac {11}{5} \, \log \left (5 \, x + 3\right ) - \frac {7}{3} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

11/5*log(5*x + 3) - 7/3*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=\frac {11 \log {\left (x + \frac {3}{5} \right )}}{5} - \frac {7 \log {\left (x + \frac {2}{3} \right )}}{3} \]

[In]

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

[Out]

11*log(x + 3/5)/5 - 7*log(x + 2/3)/3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=\frac {11}{5} \, \log \left (5 \, x + 3\right ) - \frac {7}{3} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

11/5*log(5*x + 3) - 7/3*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=\frac {11}{5} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {7}{3} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)} \, dx=\frac {11\,\ln \left (x+\frac {3}{5}\right )}{5}-\frac {7\,\ln \left (x+\frac {2}{3}\right )}{3} \]

[In]

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

[Out]

(11*log(x + 3/5))/5 - (7*log(x + 2/3))/3