\(\int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx\) [1314]

Optimal result

Integrand size = 22, antiderivative size = 39 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {49}{3 (2+3 x)}-\frac {121}{5 (3+5 x)}+154 \log (2+3 x)-154 \log (3+5 x) \]

[Out]

-49/3/(2+3*x)-121/5/(3+5*x)+154*ln(2+3*x)-154*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {49}{3 (3 x+2)}-\frac {121}{5 (5 x+3)}+154 \log (3 x+2)-154 \log (5 x+3) \]

[In]

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

[Out]

-49/(3*(2 + 3*x)) - 121/(5*(3 + 5*x)) + 154*Log[2 + 3*x] - 154*Log[3 + 5*x]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {49}{(2+3 x)^2}+\frac {462}{2+3 x}+\frac {121}{(3+5 x)^2}-\frac {770}{3+5 x}\right ) \, dx \\ & = -\frac {49}{3 (2+3 x)}-\frac {121}{5 (3+5 x)}+154 \log (2+3 x)-154 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.56 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {1461+2314 x-2310 \left (6+19 x+15 x^2\right ) \log (5 (2+3 x))+2310 \left (6+19 x+15 x^2\right ) \log (3+5 x)}{15 (2+3 x) (3+5 x)} \]

[In]

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

[Out]

-1/15*(1461 + 2314*x - 2310*(6 + 19*x + 15*x^2)*Log[5*(2 + 3*x)] + 2310*(6 + 19*x + 15*x^2)*Log[3 + 5*x])/((2
+ 3*x)*(3 + 5*x))

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {49}{3 \left (2+3 x \right )}-\frac {121}{5 \left (3+5 x \right )}+154 \ln \left (2+3 x \right )-154 \ln \left (3+5 x \right )\)	\(36\)
risch	\(\frac {-\frac {2314 x}{15}-\frac {487}{5}}{\left (2+3 x \right ) \left (3+5 x \right )}+154 \ln \left (2+3 x \right )-154 \ln \left (3+5 x \right )\)	\(39\)
norman	\(\frac {\frac {487}{2} x^{2}+\frac {925}{6} x}{\left (2+3 x \right ) \left (3+5 x \right )}+154 \ln \left (2+3 x \right )-154 \ln \left (3+5 x \right )\)	\(42\)
parallelrisch	\(\frac {13860 \ln \left (\frac {2}{3}+x \right ) x^{2}-13860 \ln \left (x +\frac {3}{5}\right ) x^{2}+17556 \ln \left (\frac {2}{3}+x \right ) x -17556 \ln \left (x +\frac {3}{5}\right ) x +1461 x^{2}+5544 \ln \left (\frac {2}{3}+x \right )-5544 \ln \left (x +\frac {3}{5}\right )+925 x}{6 \left (2+3 x \right ) \left (3+5 x \right )}\)	\(70\)

[In]

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

[Out]

-49/3/(2+3*x)-121/5/(3+5*x)+154*ln(2+3*x)-154*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {2310 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (5 \, x + 3\right ) - 2310 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) + 2314 \, x + 1461}{15 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \]

[In]

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

[Out]

-1/15*(2310*(15*x^2 + 19*x + 6)*log(5*x + 3) - 2310*(15*x^2 + 19*x + 6)*log(3*x + 2) + 2314*x + 1461)/(15*x^2
+ 19*x + 6)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {- 2314 x - 1461}{225 x^{2} + 285 x + 90} - 154 \log {\left (x + \frac {3}{5} \right )} + 154 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

(-2314*x - 1461)/(225*x**2 + 285*x + 90) - 154*log(x + 3/5) + 154*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {2314 \, x + 1461}{15 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} - 154 \, \log \left (5 \, x + 3\right ) + 154 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-1/15*(2314*x + 1461)/(15*x^2 + 19*x + 6) - 154*log(5*x + 3) + 154*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {121}{5 \, {\left (5 \, x + 3\right )}} + \frac {245}{\frac {1}{5 \, x + 3} + 3} + 154 \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]

[In]

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

[Out]

-121/5/(5*x + 3) + 245/(1/(5*x + 3) + 3) + 154*log(abs(-1/(5*x + 3) - 3))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^2}{(2+3 x)^2 (3+5 x)^2} \, dx=308\,\mathrm {atanh}\left (30\,x+19\right )-\frac {\frac {2314\,x}{225}+\frac {487}{75}}{x^2+\frac {19\,x}{15}+\frac {2}{5}} \]

[In]

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

[Out]

308*atanh(30*x + 19) - ((2314*x)/225 + 487/75)/((19*x)/15 + x^2 + 2/5)