\(\int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx\) [1230]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=-\frac {11}{10 (3+5 x)^2}+\frac {7}{3+5 x}-21 \log (2+3 x)+21 \log (3+5 x) \]

[Out]

-11/10/(3+5*x)^2+7/(3+5*x)-21*ln(2+3*x)+21*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} \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=\frac {7}{5 x+3}-\frac {11}{10 (5 x+3)^2}-21 \log (3 x+2)+21 \log (5 x+3) \]

[In]

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

[Out]

-11/(10*(3 + 5*x)^2) + 7/(3 + 5*x) - 21*Log[2 + 3*x] + 21*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}+\frac {11}{(3+5 x)^3}-\frac {35}{(3+5 x)^2}+\frac {105}{3+5 x}\right ) \, dx \\ & = -\frac {11}{10 (3+5 x)^2}+\frac {7}{3+5 x}-21 \log (2+3 x)+21 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=\frac {199+350 x-210 (3+5 x)^2 \log (5 (2+3 x))+210 (3+5 x)^2 \log (3+5 x)}{10 (3+5 x)^2} \]

[In]

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

[Out]

(199 + 350*x - 210*(3 + 5*x)^2*Log[5*(2 + 3*x)] + 210*(3 + 5*x)^2*Log[3 + 5*x])/(10*(3 + 5*x)^2)

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {35 x +\frac {199}{10}}{\left (3+5 x \right )^{2}}-21 \ln \left (2+3 x \right )+21 \ln \left (3+5 x \right )\)	\(32\)
norman	\(\frac {-\frac {94}{3} x -\frac {995}{18} x^{2}}{\left (3+5 x \right )^{2}}-21 \ln \left (2+3 x \right )+21 \ln \left (3+5 x \right )\)	\(35\)
default	\(-\frac {11}{10 \left (3+5 x \right )^{2}}+\frac {7}{3+5 x}-21 \ln \left (2+3 x \right )+21 \ln \left (3+5 x \right )\)	\(36\)
parallelrisch	\(-\frac {9450 \ln \left (\frac {2}{3}+x \right ) x^{2}-9450 \ln \left (x +\frac {3}{5}\right ) x^{2}+11340 \ln \left (\frac {2}{3}+x \right ) x -11340 \ln \left (x +\frac {3}{5}\right ) x +995 x^{2}+3402 \ln \left (\frac {2}{3}+x \right )-3402 \ln \left (x +\frac {3}{5}\right )+564 x}{18 \left (3+5 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

25*(7/5*x+199/250)/(3+5*x)^2-21*ln(2+3*x)+21*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.49 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=\frac {210 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 210 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (3 \, x + 2\right ) + 350 \, x + 199}{10 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/10*(210*(25*x^2 + 30*x + 9)*log(5*x + 3) - 210*(25*x^2 + 30*x + 9)*log(3*x + 2) + 350*x + 199)/(25*x^2 + 30*
x + 9)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=- \frac {- 350 x - 199}{250 x^{2} + 300 x + 90} + 21 \log {\left (x + \frac {3}{5} \right )} - 21 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(-350*x - 199)/(250*x**2 + 300*x + 90) + 21*log(x + 3/5) - 21*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=\frac {350 \, x + 199}{10 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + 21 \, \log \left (5 \, x + 3\right ) - 21 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/10*(350*x + 199)/(25*x^2 + 30*x + 9) + 21*log(5*x + 3) - 21*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=\frac {350 \, x + 199}{10 \, {\left (5 \, x + 3\right )}^{2}} + 21 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 21 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

1/10*(350*x + 199)/(5*x + 3)^2 + 21*log(abs(5*x + 3)) - 21*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {1-2 x}{(2+3 x) (3+5 x)^3} \, dx=\frac {\frac {7\,x}{5}+\frac {199}{250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}}-42\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((7*x)/5 + 199/250)/((6*x)/5 + x^2 + 9/25) - 42*atanh(30*x + 19)