\(\int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx\) [1202]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {7}{6 (2+3 x)^2}+\frac {11}{2+3 x}-55 \log (2+3 x)+55 \log (3+5 x) \]

[Out]

7/6/(2+3*x)^2+11/(2+3*x)-55*ln(2+3*x)+55*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, 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} \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {11}{3 x+2}+\frac {7}{6 (3 x+2)^2}-55 \log (3 x+2)+55 \log (5 x+3) \]

[In]

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

[Out]

7/(6*(2 + 3*x)^2) + 11/(2 + 3*x) - 55*Log[2 + 3*x] + 55*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)^3}-\frac {33}{(2+3 x)^2}-\frac {165}{2+3 x}+\frac {275}{3+5 x}\right ) \, dx \\ & = \frac {7}{6 (2+3 x)^2}+\frac {11}{2+3 x}-55 \log (2+3 x)+55 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {139+198 x}{6 (2+3 x)^2}-55 \log (2+3 x)+55 \log (-3 (3+5 x)) \]

[In]

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

[Out]

(139 + 198*x)/(6*(2 + 3*x)^2) - 55*Log[2 + 3*x] + 55*Log[-3*(3 + 5*x)]

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {33 x +\frac {139}{6}}{\left (2+3 x \right )^{2}}-55 \ln \left (2+3 x \right )+55 \ln \left (3+5 x \right )\)	\(32\)
norman	\(\frac {-\frac {73}{2} x -\frac {417}{8} x^{2}}{\left (2+3 x \right )^{2}}-55 \ln \left (2+3 x \right )+55 \ln \left (3+5 x \right )\)	\(35\)
default	\(\frac {7}{6 \left (2+3 x \right )^{2}}+\frac {11}{2+3 x}-55 \ln \left (2+3 x \right )+55 \ln \left (3+5 x \right )\)	\(36\)
parallelrisch	\(-\frac {3960 \ln \left (\frac {2}{3}+x \right ) x^{2}-3960 \ln \left (x +\frac {3}{5}\right ) x^{2}+5280 \ln \left (\frac {2}{3}+x \right ) x -5280 \ln \left (x +\frac {3}{5}\right ) x +417 x^{2}+1760 \ln \left (\frac {2}{3}+x \right )-1760 \ln \left (x +\frac {3}{5}\right )+292 x}{8 \left (2+3 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

9*(11/3*x+139/54)/(2+3*x)^2-55*ln(2+3*x)+55*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.49 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {330 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (5 \, x + 3\right ) - 330 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 198 \, x + 139}{6 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

1/6*(330*(9*x^2 + 12*x + 4)*log(5*x + 3) - 330*(9*x^2 + 12*x + 4)*log(3*x + 2) + 198*x + 139)/(9*x^2 + 12*x +
4)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=- \frac {- 198 x - 139}{54 x^{2} + 72 x + 24} + 55 \log {\left (x + \frac {3}{5} \right )} - 55 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(-198*x - 139)/(54*x**2 + 72*x + 24) + 55*log(x + 3/5) - 55*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {198 \, x + 139}{6 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + 55 \, \log \left (5 \, x + 3\right ) - 55 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/6*(198*x + 139)/(9*x^2 + 12*x + 4) + 55*log(5*x + 3) - 55*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {198 \, x + 139}{6 \, {\left (3 \, x + 2\right )}^{2}} + 55 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 55 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

1/6*(198*x + 139)/(3*x + 2)^2 + 55*log(abs(5*x + 3)) - 55*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {1-2 x}{(2+3 x)^3 (3+5 x)} \, dx=\frac {\frac {11\,x}{3}+\frac {139}{54}}{x^2+\frac {4\,x}{3}+\frac {4}{9}}-110\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((11*x)/3 + 139/54)/((4*x)/3 + x^2 + 4/9) - 110*atanh(30*x + 19)