\(\int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx\) [1203]

Optimal result

Integrand size = 20, antiderivative size = 48 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {7}{9 (2+3 x)^3}+\frac {11}{2 (2+3 x)^2}+\frac {55}{2+3 x}-275 \log (2+3 x)+275 \log (3+5 x) \]

[Out]

7/9/(2+3*x)^3+11/2/(2+3*x)^2+55/(2+3*x)-275*ln(2+3*x)+275*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, 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} \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {55}{3 x+2}+\frac {11}{2 (3 x+2)^2}+\frac {7}{9 (3 x+2)^3}-275 \log (3 x+2)+275 \log (5 x+3) \]

[In]

Int[(1 - 2*x)/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

7/(9*(2 + 3*x)^3) + 11/(2*(2 + 3*x)^2) + 55/(2 + 3*x) - 275*Log[2 + 3*x] + 275*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)^4}-\frac {33}{(2+3 x)^3}-\frac {165}{(2+3 x)^2}-\frac {825}{2+3 x}+\frac {1375}{3+5 x}\right ) \, dx \\ & = \frac {7}{9 (2+3 x)^3}+\frac {11}{2 (2+3 x)^2}+\frac {55}{2+3 x}-275 \log (2+3 x)+275 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.83 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {4172+12177 x+8910 x^2}{18 (2+3 x)^3}-275 \log (2+3 x)+275 \log (-3 (3+5 x)) \]

[In]

Integrate[(1 - 2*x)/((2 + 3*x)^4*(3 + 5*x)),x]

[Out]

(4172 + 12177*x + 8910*x^2)/(18*(2 + 3*x)^3) - 275*Log[2 + 3*x] + 275*Log[-3*(3 + 5*x)]

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75

method	result	size
norman	\(\frac {495 x^{2}+\frac {1353}{2} x +\frac {2086}{9}}{\left (2+3 x \right )^{3}}-275 \ln \left (2+3 x \right )+275 \ln \left (3+5 x \right )\)	\(36\)
risch	\(\frac {495 x^{2}+\frac {1353}{2} x +\frac {2086}{9}}{\left (2+3 x \right )^{3}}-275 \ln \left (2+3 x \right )+275 \ln \left (3+5 x \right )\)	\(37\)
default	\(\frac {7}{9 \left (2+3 x \right )^{3}}+\frac {11}{2 \left (2+3 x \right )^{2}}+\frac {55}{2+3 x}-275 \ln \left (2+3 x \right )+275 \ln \left (3+5 x \right )\)	\(45\)
parallelrisch	\(-\frac {59400 \ln \left (\frac {2}{3}+x \right ) x^{3}-59400 \ln \left (x +\frac {3}{5}\right ) x^{3}+118800 \ln \left (\frac {2}{3}+x \right ) x^{2}-118800 \ln \left (x +\frac {3}{5}\right ) x^{2}+6258 x^{3}+79200 \ln \left (\frac {2}{3}+x \right ) x -79200 \ln \left (x +\frac {3}{5}\right ) x +8556 x^{2}+17600 \ln \left (\frac {2}{3}+x \right )-17600 \ln \left (x +\frac {3}{5}\right )+2932 x}{8 \left (2+3 x \right )^{3}}\)	\(86\)

[In]

int((1-2*x)/(2+3*x)^4/(3+5*x),x,method=_RETURNVERBOSE)

[Out]

(495*x^2+1353/2*x+2086/9)/(2+3*x)^3-275*ln(2+3*x)+275*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {8910 \, x^{2} + 4950 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (5 \, x + 3\right ) - 4950 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) + 12177 \, x + 4172}{18 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

integrate((1-2*x)/(2+3*x)^4/(3+5*x),x, algorithm="fricas")

[Out]

1/18*(8910*x^2 + 4950*(27*x^3 + 54*x^2 + 36*x + 8)*log(5*x + 3) - 4950*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x +
2) + 12177*x + 4172)/(27*x^3 + 54*x^2 + 36*x + 8)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=- \frac {- 8910 x^{2} - 12177 x - 4172}{486 x^{3} + 972 x^{2} + 648 x + 144} + 275 \log {\left (x + \frac {3}{5} \right )} - 275 \log {\left (x + \frac {2}{3} \right )} \]

[In]

integrate((1-2*x)/(2+3*x)**4/(3+5*x),x)

[Out]

-(-8910*x**2 - 12177*x - 4172)/(486*x**3 + 972*x**2 + 648*x + 144) + 275*log(x + 3/5) - 275*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {8910 \, x^{2} + 12177 \, x + 4172}{18 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + 275 \, \log \left (5 \, x + 3\right ) - 275 \, \log \left (3 \, x + 2\right ) \]

[In]

integrate((1-2*x)/(2+3*x)^4/(3+5*x),x, algorithm="maxima")

[Out]

1/18*(8910*x^2 + 12177*x + 4172)/(27*x^3 + 54*x^2 + 36*x + 8) + 275*log(5*x + 3) - 275*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {8910 \, x^{2} + 12177 \, x + 4172}{18 \, {\left (3 \, x + 2\right )}^{3}} + 275 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 275 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

integrate((1-2*x)/(2+3*x)^4/(3+5*x),x, algorithm="giac")

[Out]

1/18*(8910*x^2 + 12177*x + 4172)/(3*x + 2)^3 + 275*log(abs(5*x + 3)) - 275*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)} \, dx=\frac {\frac {55\,x^2}{3}+\frac {451\,x}{18}+\frac {2086}{243}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}}-550\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

int(-(2*x - 1)/((3*x + 2)^4*(5*x + 3)),x)

[Out]

((451*x)/18 + (55*x^2)/3 + 2086/243)/((4*x)/3 + 2*x^2 + x^3 + 8/27) - 550*atanh(30*x + 19)