\(\int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx\) [1218]

Optimal result

Integrand size = 20, antiderivative size = 55 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=-\frac {7}{3 (2+3 x)^3}-\frac {34}{(2+3 x)^2}-\frac {505}{2+3 x}-\frac {275}{3+5 x}+3350 \log (2+3 x)-3350 \log (3+5 x) \]

[Out]

-7/3/(2+3*x)^3-34/(2+3*x)^2-505/(2+3*x)-275/(3+5*x)+3350*ln(2+3*x)-3350*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, 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} \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=-\frac {505}{3 x+2}-\frac {275}{5 x+3}-\frac {34}{(3 x+2)^2}-\frac {7}{3 (3 x+2)^3}+3350 \log (3 x+2)-3350 \log (5 x+3) \]

[In]

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

[Out]

-7/(3*(2 + 3*x)^3) - 34/(2 + 3*x)^2 - 505/(2 + 3*x) - 275/(3 + 5*x) + 3350*Log[2 + 3*x] - 3350*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 {21}{(2+3 x)^4}+\frac {204}{(2+3 x)^3}+\frac {1515}{(2+3 x)^2}+\frac {10050}{2+3 x}+\frac {1375}{(3+5 x)^2}-\frac {16750}{3+5 x}\right ) \, dx \\ & = -\frac {7}{3 (2+3 x)^3}-\frac {34}{(2+3 x)^2}-\frac {505}{2+3 x}-\frac {275}{3+5 x}+3350 \log (2+3 x)-3350 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=-\frac {7}{3 (2+3 x)^3}-\frac {34}{(2+3 x)^2}-\frac {505}{2+3 x}-\frac {275}{3+5 x}+3350 \log (2+3 x)-3350 \log (-3 (3+5 x)) \]

[In]

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

[Out]

-7/(3*(2 + 3*x)^3) - 34/(2 + 3*x)^2 - 505/(2 + 3*x) - 275/(3 + 5*x) + 3350*Log[2 + 3*x] - 3350*Log[-3*(3 + 5*x
)]

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87

method	result	size
norman	\(\frac {-59295 x^{2}-30150 x^{3}-\frac {116513}{3} x -8471}{\left (2+3 x \right )^{3} \left (3+5 x \right )}+3350 \ln \left (2+3 x \right )-3350 \ln \left (3+5 x \right )\)	\(48\)
risch	\(\frac {-59295 x^{2}-30150 x^{3}-\frac {116513}{3} x -8471}{\left (2+3 x \right )^{3} \left (3+5 x \right )}+3350 \ln \left (2+3 x \right )-3350 \ln \left (3+5 x \right )\)	\(49\)
default	\(-\frac {7}{3 \left (2+3 x \right )^{3}}-\frac {34}{\left (2+3 x \right )^{2}}-\frac {505}{2+3 x}-\frac {275}{3+5 x}+3350 \ln \left (2+3 x \right )-3350 \ln \left (3+5 x \right )\)	\(54\)
parallelrisch	\(\frac {10854000 \ln \left (\frac {2}{3}+x \right ) x^{4}-10854000 \ln \left (x +\frac {3}{5}\right ) x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{3}-28220400 \ln \left (x +\frac {3}{5}\right ) x^{3}+1143585 x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{2}-27496800 \ln \left (x +\frac {3}{5}\right ) x^{2}+2249721 x^{3}[phone] \ln \left (\frac {2}{3}+x \right ) x -11899200 \ln \left (x +\frac {3}{5}\right ) x +1474002 x^{2}+1929600 \ln \left (\frac {2}{3}+x \right )-1929600 \ln \left (x +\frac {3}{5}\right )+321604 x}{24 \left (2+3 x \right )^{3} \left (3+5 x \right )}\)	\(116\)

[In]

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

[Out]

(-59295*x^2-30150*x^3-116513/3*x-8471)/(2+3*x)^3/(3+5*x)+3350*ln(2+3*x)-3350*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=-\frac {90450 \, x^{3} + 177885 \, x^{2} + 10050 \, {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )} \log \left (5 \, x + 3\right ) - 10050 \, {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )} \log \left (3 \, x + 2\right ) + 116513 \, x + 25413}{3 \, {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )}} \]

[In]

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

[Out]

-1/3*(90450*x^3 + 177885*x^2 + 10050*(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)*log(5*x + 3) - 10050*(135*x^4
+ 351*x^3 + 342*x^2 + 148*x + 24)*log(3*x + 2) + 116513*x + 25413)/(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=- \frac {90450 x^{3} + 177885 x^{2} + 116513 x + 25413}{405 x^{4} + 1053 x^{3} + 1026 x^{2} + 444 x + 72} - 3350 \log {\left (x + \frac {3}{5} \right )} + 3350 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(90450*x**3 + 177885*x**2 + 116513*x + 25413)/(405*x**4 + 1053*x**3 + 1026*x**2 + 444*x + 72) - 3350*log(x +
3/5) + 3350*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=-\frac {90450 \, x^{3} + 177885 \, x^{2} + 116513 \, x + 25413}{3 \, {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )}} - 3350 \, \log \left (5 \, x + 3\right ) + 3350 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-1/3*(90450*x^3 + 177885*x^2 + 116513*x + 25413)/(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24) - 3350*log(5*x + 3
) + 3350*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=-\frac {275}{5 \, x + 3} + \frac {225 \, {\left (\frac {339}{5 \, x + 3} + \frac {68}{{\left (5 \, x + 3\right )}^{2}} + 440\right )}}{{\left (\frac {1}{5 \, x + 3} + 3\right )}^{3}} + 3350 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

[In]

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

[Out]

-275/(5*x + 3) + 225*(339/(5*x + 3) + 68/(5*x + 3)^2 + 440)/(1/(5*x + 3) + 3)^3 + 3350*log(abs(-1/(5*x + 3) -
3))

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {1-2 x}{(2+3 x)^4 (3+5 x)^2} \, dx=6700\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {670\,x^3}{3}+\frac {3953\,x^2}{9}+\frac {116513\,x}{405}+\frac {8471}{135}}{x^4+\frac {13\,x^3}{5}+\frac {38\,x^2}{15}+\frac {148\,x}{135}+\frac {8}{45}} \]

[In]

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

[Out]

6700*atanh(30*x + 19) - ((116513*x)/405 + (3953*x^2)/9 + (670*x^3)/3 + 8471/135)/((148*x)/135 + (38*x^2)/15 +
(13*x^3)/5 + x^4 + 8/45)