\(\int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx\) [1201]

Optimal result

Integrand size = 20, antiderivative size = 28 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {7}{3 (2+3 x)}-11 \log (2+3 x)+11 \log (3+5 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, 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} \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {7}{3 (3 x+2)}-11 \log (3 x+2)+11 \log (5 x+3) \]

[In]

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

[Out]

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

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)^2}-\frac {33}{2+3 x}+\frac {55}{3+5 x}\right ) \, dx \\ & = \frac {7}{3 (2+3 x)}-11 \log (2+3 x)+11 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {7-33 (2+3 x) \log (2+3 x)+33 (2+3 x) \log (-3 (3+5 x))}{6+9 x} \]

[In]

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

[Out]

(7 - 33*(2 + 3*x)*Log[2 + 3*x] + 33*(2 + 3*x)*Log[-3*(3 + 5*x)])/(6 + 9*x)

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {7}{9 \left (\frac {2}{3}+x \right )}-11 \ln \left (2+3 x \right )+11 \ln \left (3+5 x \right )\)	\(25\)
default	\(\frac {7}{3 \left (2+3 x \right )}-11 \ln \left (2+3 x \right )+11 \ln \left (3+5 x \right )\)	\(27\)
norman	\(-\frac {7 x}{2 \left (2+3 x \right )}-11 \ln \left (2+3 x \right )+11 \ln \left (3+5 x \right )\)	\(28\)
parallelrisch	\(-\frac {66 \ln \left (\frac {2}{3}+x \right ) x -66 \ln \left (x +\frac {3}{5}\right ) x +44 \ln \left (\frac {2}{3}+x \right )-44 \ln \left (x +\frac {3}{5}\right )+7 x}{2 \left (2+3 x \right )}\)	\(40\)

[In]

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

[Out]

7/9/(2/3+x)-11*ln(2+3*x)+11*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {33 \, {\left (3 \, x + 2\right )} \log \left (5 \, x + 3\right ) - 33 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 7}{3 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

1/3*(33*(3*x + 2)*log(5*x + 3) - 33*(3*x + 2)*log(3*x + 2) + 7)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=11 \log {\left (x + \frac {3}{5} \right )} - 11 \log {\left (x + \frac {2}{3} \right )} + \frac {7}{9 x + 6} \]

[In]

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

[Out]

11*log(x + 3/5) - 11*log(x + 2/3) + 7/(9*x + 6)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {7}{3 \, {\left (3 \, x + 2\right )}} + 11 \, \log \left (5 \, x + 3\right ) - 11 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {7}{3 \, {\left (3 \, x + 2\right )}} + 11 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)} \, dx=\frac {7}{9\,\left (x+\frac {2}{3}\right )}-22\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

7/(9*(x + 2/3)) - 22*atanh(30*x + 19)