3.1216 \(\int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx\)

Optimal. Leaf size=35 \[ -\frac {7}{3 x+2}-\frac {11}{5 x+3}+68 \log (3 x+2)-68 \log (5 x+3) \]

[Out]

-7/(2+3*x)-11/(3+5*x)+68*ln(2+3*x)-68*ln(3+5*x)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ -\frac {7}{3 x+2}-\frac {11}{5 x+3}+68 \log (3 x+2)-68 \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*Log[3 + 5*x]

Rule 77

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*} \int \frac {1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx &=\int \left (\frac {21}{(2+3 x)^2}+\frac {204}{2+3 x}+\frac {55}{(3+5 x)^2}-\frac {340}{3+5 x}\right ) \, dx\\ &=-\frac {7}{2+3 x}-\frac {11}{3+5 x}+68 \log (2+3 x)-68 \log (3+5 x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 37, normalized size = 1.06 \[ -\frac {7}{3 x+2}-\frac {11}{5 x+3}+68 \log (3 x+2)-68 \log (-3 (5 x+3)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*Log[-3*(3 + 5*x)]

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 55, normalized size = 1.57 \[ -\frac {68 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (5 \, x + 3\right ) - 68 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) + 68 \, x + 43}{15 \, x^{2} + 19 \, x + 6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(68*(15*x^2 + 19*x + 6)*log(5*x + 3) - 68*(15*x^2 + 19*x + 6)*log(3*x + 2) + 68*x + 43)/(15*x^2 + 19*x + 6)

________________________________________________________________________________________

giac [A]  time = 1.26, size = 38, normalized size = 1.09 \[ -\frac {11}{5 \, x + 3} + \frac {105}{\frac {1}{5 \, x + 3} + 3} + 68 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/(5*x + 3) + 105/(1/(5*x + 3) + 3) + 68*log(abs(-1/(5*x + 3) - 3))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 36, normalized size = 1.03 \[ 68 \ln \left (3 x +2\right )-68 \ln \left (5 x +3\right )-\frac {7}{3 x +2}-\frac {11}{5 x +3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/(3*x+2)-11/(5*x+3)+68*ln(3*x+2)-68*ln(5*x+3)

________________________________________________________________________________________

maxima [A]  time = 0.70, size = 36, normalized size = 1.03 \[ -\frac {68 \, x + 43}{15 \, x^{2} + 19 \, x + 6} - 68 \, \log \left (5 \, x + 3\right ) + 68 \, \log \left (3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(68*x + 43)/(15*x^2 + 19*x + 6) - 68*log(5*x + 3) + 68*log(3*x + 2)

________________________________________________________________________________________

mupad [B]  time = 1.11, size = 26, normalized size = 0.74 \[ 136\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {68\,x}{15}+\frac {43}{15}}{x^2+\frac {19\,x}{15}+\frac {2}{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

136*atanh(30*x + 19) - ((68*x)/15 + 43/15)/((19*x)/15 + x^2 + 2/5)

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 31, normalized size = 0.89 \[ - \frac {68 x + 43}{15 x^{2} + 19 x + 6} - 68 \log {\left (x + \frac {3}{5} \right )} + 68 \log {\left (x + \frac {2}{3} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(68*x + 43)/(15*x**2 + 19*x + 6) - 68*log(x + 3/5) + 68*log(x + 2/3)

________________________________________________________________________________________