\(\int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx\) [1330]

Optimal result

Integrand size = 22, antiderivative size = 39 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=-\frac {121}{50 (3+5 x)^2}+\frac {407}{25 (3+5 x)}-49 \log (2+3 x)+49 \log (3+5 x) \]

[Out]

-121/50/(3+5*x)^2+407/25/(3+5*x)-49*ln(2+3*x)+49*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} \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {407}{25 (5 x+3)}-\frac {121}{50 (5 x+3)^2}-49 \log (3 x+2)+49 \log (5 x+3) \]

[In]

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

[Out]

-121/(50*(3 + 5*x)^2) + 407/(25*(3 + 5*x)) - 49*Log[2 + 3*x] + 49*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 {147}{2+3 x}+\frac {121}{5 (3+5 x)^3}-\frac {407}{5 (3+5 x)^2}+\frac {245}{3+5 x}\right ) \, dx \\ & = -\frac {121}{50 (3+5 x)^2}+\frac {407}{25 (3+5 x)}-49 \log (2+3 x)+49 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {2321+4070 x-2450 (3+5 x)^2 \log (2+3 x)+2450 (3+5 x)^2 \log (-3 (3+5 x))}{50 (3+5 x)^2} \]

[In]

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

[Out]

(2321 + 4070*x - 2450*(3 + 5*x)^2*Log[2 + 3*x] + 2450*(3 + 5*x)^2*Log[-3*(3 + 5*x)])/(50*(3 + 5*x)^2)

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {\frac {407 x}{5}+\frac {2321}{50}}{\left (3+5 x \right )^{2}}-49 \ln \left (2+3 x \right )+49 \ln \left (3+5 x \right )\)	\(32\)
norman	\(\frac {-\frac {2321}{18} x^{2}-\frac {220}{3} x}{\left (3+5 x \right )^{2}}-49 \ln \left (2+3 x \right )+49 \ln \left (3+5 x \right )\)	\(35\)
default	\(-\frac {121}{50 \left (3+5 x \right )^{2}}+\frac {407}{25 \left (3+5 x \right )}-49 \ln \left (2+3 x \right )+49 \ln \left (3+5 x \right )\)	\(36\)
parallelrisch	\(-\frac {22050 \ln \left (\frac {2}{3}+x \right ) x^{2}-22050 \ln \left (x +\frac {3}{5}\right ) x^{2}+26460 \ln \left (\frac {2}{3}+x \right ) x -26460 \ln \left (x +\frac {3}{5}\right ) x +2321 x^{2}+7938 \ln \left (\frac {2}{3}+x \right )-7938 \ln \left (x +\frac {3}{5}\right )+1320 x}{18 \left (3+5 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

25*(407/125*x+2321/1250)/(3+5*x)^2-49*ln(2+3*x)+49*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {2450 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 2450 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (3 \, x + 2\right ) + 4070 \, x + 2321}{50 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/50*(2450*(25*x^2 + 30*x + 9)*log(5*x + 3) - 2450*(25*x^2 + 30*x + 9)*log(3*x + 2) + 4070*x + 2321)/(25*x^2 +
30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {4070 x + 2321}{1250 x^{2} + 1500 x + 450} + 49 \log {\left (x + \frac {3}{5} \right )} - 49 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

(4070*x + 2321)/(1250*x**2 + 1500*x + 450) + 49*log(x + 3/5) - 49*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {11 \, {\left (370 \, x + 211\right )}}{50 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + 49 \, \log \left (5 \, x + 3\right ) - 49 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

11/50*(370*x + 211)/(25*x^2 + 30*x + 9) + 49*log(5*x + 3) - 49*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {11 \, {\left (370 \, x + 211\right )}}{50 \, {\left (5 \, x + 3\right )}^{2}} + 49 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 49 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

11/50*(370*x + 211)/(5*x + 3)^2 + 49*log(abs(5*x + 3)) - 49*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^3} \, dx=\frac {\frac {407\,x}{125}+\frac {2321}{1250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}}-98\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((407*x)/125 + 2321/1250)/((6*x)/5 + x^2 + 9/25) - 98*atanh(30*x + 19)