\(\int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx\) [1231]

Optimal result

Integrand size = 20, antiderivative size = 46 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {21}{2+3 x}-\frac {11}{2 (3+5 x)^2}+\frac {68}{3+5 x}-309 \log (2+3 x)+309 \log (3+5 x) \]

[Out]

21/(2+3*x)-11/2/(3+5*x)^2+68/(3+5*x)-309*ln(2+3*x)+309*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, 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} \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {21}{3 x+2}+\frac {68}{5 x+3}-\frac {11}{2 (5 x+3)^2}-309 \log (3 x+2)+309 \log (5 x+3) \]

[In]

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

[Out]

21/(2 + 3*x) - 11/(2*(3 + 5*x)^2) + 68/(3 + 5*x) - 309*Log[2 + 3*x] + 309*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 {63}{(2+3 x)^2}-\frac {927}{2+3 x}+\frac {55}{(3+5 x)^3}-\frac {340}{(3+5 x)^2}+\frac {1545}{3+5 x}\right ) \, dx \\ & = \frac {21}{2+3 x}-\frac {11}{2 (3+5 x)^2}+\frac {68}{3+5 x}-309 \log (2+3 x)+309 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {21}{2+3 x}-\frac {11}{2 (3+5 x)^2}+\frac {68}{3+5 x}-309 \log (2+3 x)+309 \log (-3 (3+5 x)) \]

[In]

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

[Out]

21/(2 + 3*x) - 11/(2*(3 + 5*x)^2) + 68/(3 + 5*x) - 309*Log[2 + 3*x] + 309*Log[-3*(3 + 5*x)]

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {1545 x^{2}+\frac {3811}{2} x +586}{\left (2+3 x \right ) \left (3+5 x \right )^{2}}-309 \ln \left (2+3 x \right )+309 \ln \left (3+5 x \right )\)	\(44\)
default	\(\frac {21}{2+3 x}-\frac {11}{2 \left (3+5 x \right )^{2}}+\frac {68}{3+5 x}-309 \ln \left (2+3 x \right )+309 \ln \left (3+5 x \right )\)	\(45\)
norman	\(\frac {-\frac {7325}{3} x^{3}-\frac {5561}{6} x -\frac {27115}{9} x^{2}}{\left (2+3 x \right ) \left (3+5 x \right )^{2}}-309 \ln \left (2+3 x \right )+309 \ln \left (3+5 x \right )\)	\(47\)
parallelrisch	\(-\frac {417150 \ln \left (\frac {2}{3}+x \right ) x^{3}-417150 \ln \left (x +\frac {3}{5}\right ) x^{3}+778680 \ln \left (\frac {2}{3}+x \right ) x^{2}-778680 \ln \left (x +\frac {3}{5}\right ) x^{2}+43950 x^{3}+483894 \ln \left (\frac {2}{3}+x \right ) x -483894 \ln \left (x +\frac {3}{5}\right ) x +54230 x^{2}+100116 \ln \left (\frac {2}{3}+x \right )-100116 \ln \left (x +\frac {3}{5}\right )+16683 x}{18 \left (2+3 x \right ) \left (3+5 x \right )^{2}}\)	\(93\)

[In]

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

[Out]

75*(103/5*x^2+3811/150*x+586/75)/(2+3*x)/(3+5*x)^2-309*ln(2+3*x)+309*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.63 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {3090 \, x^{2} + 618 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (5 \, x + 3\right ) - 618 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (3 \, x + 2\right ) + 3811 \, x + 1172}{2 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \]

[In]

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

[Out]

1/2*(3090*x^2 + 618*(75*x^3 + 140*x^2 + 87*x + 18)*log(5*x + 3) - 618*(75*x^3 + 140*x^2 + 87*x + 18)*log(3*x +
2) + 3811*x + 1172)/(75*x^3 + 140*x^2 + 87*x + 18)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=- \frac {- 3090 x^{2} - 3811 x - 1172}{150 x^{3} + 280 x^{2} + 174 x + 36} + 309 \log {\left (x + \frac {3}{5} \right )} - 309 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(-3090*x**2 - 3811*x - 1172)/(150*x**3 + 280*x**2 + 174*x + 36) + 309*log(x + 3/5) - 309*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {3090 \, x^{2} + 3811 \, x + 1172}{2 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} + 309 \, \log \left (5 \, x + 3\right ) - 309 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/2*(3090*x^2 + 3811*x + 1172)/(75*x^3 + 140*x^2 + 87*x + 18) + 309*log(5*x + 3) - 309*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {21}{3 \, x + 2} - \frac {15 \, {\left (\frac {202}{3 \, x + 2} - 845\right )}}{2 \, {\left (\frac {1}{3 \, x + 2} - 5\right )}^{2}} + 309 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]

[In]

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

[Out]

21/(3*x + 2) - 15/2*(202/(3*x + 2) - 845)/(1/(3*x + 2) - 5)^2 + 309*log(abs(-1/(3*x + 2) + 5))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^3} \, dx=\frac {\frac {103\,x^2}{5}+\frac {3811\,x}{150}+\frac {586}{75}}{x^3+\frac {28\,x^2}{15}+\frac {29\,x}{25}+\frac {6}{25}}-618\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((3811*x)/150 + (103*x^2)/5 + 586/75)/((29*x)/25 + (28*x^2)/15 + x^3 + 6/25) - 618*atanh(30*x + 19)