\(\int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx\) [1297]

Optimal result

Integrand size = 22, antiderivative size = 26 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4 x}{15}-\frac {49}{9} \log (2+3 x)+\frac {121}{25} \log (3+5 x) \]

[Out]

4/15*x-49/9*ln(2+3*x)+121/25*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, 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 = {84} \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4 x}{15}-\frac {49}{9} \log (3 x+2)+\frac {121}{25} \log (5 x+3) \]

[In]

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

[Out]

(4*x)/15 - (49*Log[2 + 3*x])/9 + (121*Log[3 + 5*x])/25

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{15}-\frac {49}{3 (2+3 x)}+\frac {121}{5 (3+5 x)}\right ) \, dx \\ & = \frac {4 x}{15}-\frac {49}{9} \log (2+3 x)+\frac {121}{25} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {1}{225} (40+60 x-1225 \log (2+3 x)+1089 \log (-3 (3+5 x))) \]

[In]

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

[Out]

(40 + 60*x - 1225*Log[2 + 3*x] + 1089*Log[-3*(3 + 5*x)])/225

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {4 x}{15}-\frac {49 \ln \left (\frac {2}{3}+x \right )}{9}+\frac {121 \ln \left (x +\frac {3}{5}\right )}{25}\)	\(17\)
default	\(\frac {4 x}{15}-\frac {49 \ln \left (2+3 x \right )}{9}+\frac {121 \ln \left (3+5 x \right )}{25}\)	\(21\)
norman	\(\frac {4 x}{15}-\frac {49 \ln \left (2+3 x \right )}{9}+\frac {121 \ln \left (3+5 x \right )}{25}\)	\(21\)
risch	\(\frac {4 x}{15}-\frac {49 \ln \left (2+3 x \right )}{9}+\frac {121 \ln \left (3+5 x \right )}{25}\)	\(21\)

[In]

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

[Out]

4/15*x-49/9*ln(2/3+x)+121/25*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4}{15} \, x + \frac {121}{25} \, \log \left (5 \, x + 3\right ) - \frac {49}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

4/15*x + 121/25*log(5*x + 3) - 49/9*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4 x}{15} + \frac {121 \log {\left (x + \frac {3}{5} \right )}}{25} - \frac {49 \log {\left (x + \frac {2}{3} \right )}}{9} \]

[In]

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

[Out]

4*x/15 + 121*log(x + 3/5)/25 - 49*log(x + 2/3)/9

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4}{15} \, x + \frac {121}{25} \, \log \left (5 \, x + 3\right ) - \frac {49}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

4/15*x + 121/25*log(5*x + 3) - 49/9*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4}{15} \, x + \frac {121}{25} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {49}{9} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

4/15*x + 121/25*log(abs(5*x + 3)) - 49/9*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)} \, dx=\frac {4\,x}{15}-\frac {49\,\ln \left (x+\frac {2}{3}\right )}{9}+\frac {121\,\ln \left (x+\frac {3}{5}\right )}{25} \]

[In]

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

[Out]

(4*x)/15 - (49*log(x + 2/3))/9 + (121*log(x + 3/5))/25