\(\int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx\) [1215]

Optimal result

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

[Out]

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

[In]

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

[Out]

-11/(5*(3 + 5*x)) + 7*Log[2 + 3*x] - 7*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 {21}{2+3 x}+\frac {11}{(3+5 x)^2}-\frac {35}{3+5 x}\right ) \, dx \\ & = -\frac {11}{5 (3+5 x)}+7 \log (2+3 x)-7 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx=-\frac {11}{5 (3+5 x)}+7 \log (5 (2+3 x))-7 \log (3+5 x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {11}{25 \left (x +\frac {3}{5}\right )}+7 \ln \left (2+3 x \right )-7 \ln \left (3+5 x \right )\)	\(25\)
default	\(-\frac {11}{5 \left (3+5 x \right )}+7 \ln \left (2+3 x \right )-7 \ln \left (3+5 x \right )\)	\(27\)
norman	\(\frac {11 x}{3 \left (3+5 x \right )}+7 \ln \left (2+3 x \right )-7 \ln \left (3+5 x \right )\)	\(28\)
parallelrisch	\(\frac {105 \ln \left (\frac {2}{3}+x \right ) x -105 \ln \left (x +\frac {3}{5}\right ) x +63 \ln \left (\frac {2}{3}+x \right )-63 \ln \left (x +\frac {3}{5}\right )+11 x}{9+15 x}\)	\(40\)

[In]

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

[Out]

-11/25/(x+3/5)+7*ln(2+3*x)-7*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx=-\frac {35 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 35 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 11}{5 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx=- 7 \log {\left (x + \frac {3}{5} \right )} + 7 \log {\left (x + \frac {2}{3} \right )} - \frac {11}{25 x + 15} \]

[In]

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

[Out]

-7*log(x + 3/5) + 7*log(x + 2/3) - 11/(25*x + 15)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx=-\frac {11}{5 \, {\left (5 \, x + 3\right )}} - 7 \, \log \left (5 \, x + 3\right ) + 7 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx=-\frac {11}{5 \, {\left (5 \, x + 3\right )}} + 7 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^2} \, dx=14\,\mathrm {atanh}\left (30\,x+19\right )-\frac {11}{25\,\left (x+\frac {3}{5}\right )} \]

[In]

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

[Out]

14*atanh(30*x + 19) - 11/(25*(x + 3/5))