\(\int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx\) [1204]

Optimal result

Integrand size = 20, antiderivative size = 59 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {7}{12 (2+3 x)^4}+\frac {11}{3 (2+3 x)^3}+\frac {55}{2 (2+3 x)^2}+\frac {275}{2+3 x}-1375 \log (2+3 x)+1375 \log (3+5 x) \]

[Out]

7/12/(2+3*x)^4+11/3/(2+3*x)^3+55/2/(2+3*x)^2+275/(2+3*x)-1375*ln(2+3*x)+1375*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, 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} \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {275}{3 x+2}+\frac {55}{2 (3 x+2)^2}+\frac {11}{3 (3 x+2)^3}+\frac {7}{12 (3 x+2)^4}-1375 \log (3 x+2)+1375 \log (5 x+3) \]

[In]

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

[Out]

7/(12*(2 + 3*x)^4) + 11/(3*(2 + 3*x)^3) + 55/(2*(2 + 3*x)^2) + 275/(2 + 3*x) - 1375*Log[2 + 3*x] + 1375*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)^5}-\frac {33}{(2+3 x)^4}-\frac {165}{(2+3 x)^3}-\frac {825}{(2+3 x)^2}-\frac {4125}{2+3 x}+\frac {6875}{3+5 x}\right ) \, dx \\ & = \frac {7}{12 (2+3 x)^4}+\frac {11}{3 (2+3 x)^3}+\frac {55}{2 (2+3 x)^2}+\frac {275}{2+3 x}-1375 \log (2+3 x)+1375 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {27815+122892 x+181170 x^2+89100 x^3}{12 (2+3 x)^4}-1375 \log (2+3 x)+1375 \log (-3 (3+5 x)) \]

[In]

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

[Out]

(27815 + 122892*x + 181170*x^2 + 89100*x^3)/(12*(2 + 3*x)^4) - 1375*Log[2 + 3*x] + 1375*Log[-3*(3 + 5*x)]

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69

method	result	size
norman	\(\frac {7425 x^{3}+10241 x +\frac {30195}{2} x^{2}+\frac {27815}{12}}{\left (2+3 x \right )^{4}}-1375 \ln \left (2+3 x \right )+1375 \ln \left (3+5 x \right )\)	\(41\)
risch	\(\frac {7425 x^{3}+10241 x +\frac {30195}{2} x^{2}+\frac {27815}{12}}{\left (2+3 x \right )^{4}}-1375 \ln \left (2+3 x \right )+1375 \ln \left (3+5 x \right )\)	\(42\)
default	\(\frac {7}{12 \left (2+3 x \right )^{4}}+\frac {11}{3 \left (2+3 x \right )^{3}}+\frac {55}{2 \left (2+3 x \right )^{2}}+\frac {275}{2+3 x}-1375 \ln \left (2+3 x \right )+1375 \ln \left (3+5 x \right )\)	\(54\)
parallelrisch	\(-\frac {7128000 \ln \left (\frac {2}{3}+x \right ) x^{4}-7128000 \ln \left (x +\frac {3}{5}\right ) x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{3}-19008000 \ln \left (x +\frac {3}{5}\right ) x^{3}+751005 x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{2}-19008000 \ln \left (x +\frac {3}{5}\right ) x^{2}+1527480 x^{3}+8448000 \ln \left (\frac {2}{3}+x \right ) x -8448000 \ln \left (x +\frac {3}{5}\right ) x +1036440 x^{2}+1408000 \ln \left (\frac {2}{3}+x \right )-1408000 \ln \left (x +\frac {3}{5}\right )+234656 x}{64 \left (2+3 x \right )^{4}}\)	\(109\)

[In]

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

[Out]

(7425*x^3+10241*x+30195/2*x^2+27815/12)/(2+3*x)^4-1375*ln(2+3*x)+1375*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.61 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {89100 \, x^{3} + 181170 \, x^{2} + 16500 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (5 \, x + 3\right ) - 16500 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 122892 \, x + 27815}{12 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

1/12*(89100*x^3 + 181170*x^2 + 16500*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(5*x + 3) - 16500*(81*x^4 + 2
16*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 122892*x + 27815)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=- \frac {- 89100 x^{3} - 181170 x^{2} - 122892 x - 27815}{972 x^{4} + 2592 x^{3} + 2592 x^{2} + 1152 x + 192} + 1375 \log {\left (x + \frac {3}{5} \right )} - 1375 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(-89100*x**3 - 181170*x**2 - 122892*x - 27815)/(972*x**4 + 2592*x**3 + 2592*x**2 + 1152*x + 192) + 1375*log(x
+ 3/5) - 1375*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {89100 \, x^{3} + 181170 \, x^{2} + 122892 \, x + 27815}{12 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} + 1375 \, \log \left (5 \, x + 3\right ) - 1375 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/12*(89100*x^3 + 181170*x^2 + 122892*x + 27815)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) + 1375*log(5*x + 3)
- 1375*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {275}{3 \, x + 2} + \frac {55}{2 \, {\left (3 \, x + 2\right )}^{2}} + \frac {11}{3 \, {\left (3 \, x + 2\right )}^{3}} + \frac {7}{12 \, {\left (3 \, x + 2\right )}^{4}} + 1375 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]

[In]

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

[Out]

275/(3*x + 2) + 55/2/(3*x + 2)^2 + 11/3/(3*x + 2)^3 + 7/12/(3*x + 2)^4 + 1375*log(abs(-1/(3*x + 2) + 5))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {1-2 x}{(2+3 x)^5 (3+5 x)} \, dx=\frac {\frac {275\,x^3}{3}+\frac {3355\,x^2}{18}+\frac {10241\,x}{81}+\frac {27815}{972}}{x^4+\frac {8\,x^3}{3}+\frac {8\,x^2}{3}+\frac {32\,x}{27}+\frac {16}{81}}-2750\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((10241*x)/81 + (3355*x^2)/18 + (275*x^3)/3 + 27815/972)/((32*x)/27 + (8*x^2)/3 + (8*x^3)/3 + x^4 + 16/81) - 2
750*atanh(30*x + 19)