\(\int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx\) [1216]

Optimal result

Integrand size = 20, antiderivative size = 35 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {7}{2+3 x}-\frac {11}{3+5 x}+68 \log (2+3 x)-68 \log (3+5 x) \]

[Out]

-7/(2+3*x)-11/(3+5*x)+68*ln(2+3*x)-68*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, 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} \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {7}{3 x+2}-\frac {11}{5 x+3}+68 \log (3 x+2)-68 \log (5 x+3) \]

[In]

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

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*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)^2}+\frac {204}{2+3 x}+\frac {55}{(3+5 x)^2}-\frac {340}{3+5 x}\right ) \, dx \\ & = -\frac {7}{2+3 x}-\frac {11}{3+5 x}+68 \log (2+3 x)-68 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {7}{2+3 x}-\frac {11}{3+5 x}+68 \log (2+3 x)-68 \log (-3 (3+5 x)) \]

[In]

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

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*Log[-3*(3 + 5*x)]

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
default	\(-\frac {7}{2+3 x}-\frac {11}{3+5 x}+68 \ln \left (2+3 x \right )-68 \ln \left (3+5 x \right )\)	\(36\)
risch	\(\frac {-68 x -43}{\left (2+3 x \right ) \left (3+5 x \right )}+68 \ln \left (2+3 x \right )-68 \ln \left (3+5 x \right )\)	\(39\)
norman	\(\frac {\frac {215}{2} x^{2}+\frac {409}{6} x}{\left (2+3 x \right ) \left (3+5 x \right )}+68 \ln \left (2+3 x \right )-68 \ln \left (3+5 x \right )\)	\(42\)
parallelrisch	\(\frac {6120 \ln \left (\frac {2}{3}+x \right ) x^{2}-6120 \ln \left (x +\frac {3}{5}\right ) x^{2}+7752 \ln \left (\frac {2}{3}+x \right ) x -7752 \ln \left (x +\frac {3}{5}\right ) x +645 x^{2}+2448 \ln \left (\frac {2}{3}+x \right )-2448 \ln \left (x +\frac {3}{5}\right )+409 x}{6 \left (2+3 x \right ) \left (3+5 x \right )}\)	\(70\)

[In]

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

[Out]

-7/(2+3*x)-11/(3+5*x)+68*ln(2+3*x)-68*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {68 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (5 \, x + 3\right ) - 68 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) + 68 \, x + 43}{15 \, x^{2} + 19 \, x + 6} \]

[In]

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

[Out]

-(68*(15*x^2 + 19*x + 6)*log(5*x + 3) - 68*(15*x^2 + 19*x + 6)*log(3*x + 2) + 68*x + 43)/(15*x^2 + 19*x + 6)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=- \frac {68 x + 43}{15 x^{2} + 19 x + 6} - 68 \log {\left (x + \frac {3}{5} \right )} + 68 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(68*x + 43)/(15*x**2 + 19*x + 6) - 68*log(x + 3/5) + 68*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {68 \, x + 43}{15 \, x^{2} + 19 \, x + 6} - 68 \, \log \left (5 \, x + 3\right ) + 68 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-(68*x + 43)/(15*x^2 + 19*x + 6) - 68*log(5*x + 3) + 68*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {11}{5 \, x + 3} + \frac {105}{\frac {1}{5 \, x + 3} + 3} + 68 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

[In]

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

[Out]

-11/(5*x + 3) + 105/(1/(5*x + 3) + 3) + 68*log(abs(-1/(5*x + 3) - 3))

Mupad [B] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx=136\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {68\,x}{15}+\frac {43}{15}}{x^2+\frac {19\,x}{15}+\frac {2}{5}} \]

[In]

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

[Out]

136*atanh(30*x + 19) - ((68*x)/15 + 43/15)/((19*x)/15 + x^2 + 2/5)