\(\int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx\) [1312]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {4 x}{25}-\frac {121}{125 (3+5 x)}-\frac {44}{125} \log (3+5 x) \]

[Out]

4/25*x-121/125/(3+5*x)-44/125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {4 x}{25}-\frac {121}{125 (5 x+3)}-\frac {44}{125} \log (5 x+3) \]

[In]

Int[(1 - 2*x)^2/(3 + 5*x)^2,x]

[Out]

(4*x)/25 - 121/(125*(3 + 5*x)) - (44*Log[3 + 5*x])/125

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{25}+\frac {121}{25 (3+5 x)^2}-\frac {44}{25 (3+5 x)}\right ) \, dx \\ & = \frac {4 x}{25}-\frac {121}{125 (3+5 x)}-\frac {44}{125} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {-151+10 x+100 x^2-44 (3+5 x) \log (6+10 x)}{125 (3+5 x)} \]

[In]

Integrate[(1 - 2*x)^2/(3 + 5*x)^2,x]

[Out]

(-151 + 10*x + 100*x^2 - 44*(3 + 5*x)*Log[6 + 10*x])/(125*(3 + 5*x))

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {4 x}{25}-\frac {121}{625 \left (x +\frac {3}{5}\right )}-\frac {44 \ln \left (3+5 x \right )}{125}\)	\(20\)
default	\(\frac {4 x}{25}-\frac {121}{125 \left (3+5 x \right )}-\frac {44 \ln \left (3+5 x \right )}{125}\)	\(22\)
norman	\(\frac {\frac {157}{75} x +\frac {4}{5} x^{2}}{3+5 x}-\frac {44 \ln \left (3+5 x \right )}{125}\)	\(27\)
parallelrisch	\(-\frac {660 \ln \left (x +\frac {3}{5}\right ) x -300 x^{2}+396 \ln \left (x +\frac {3}{5}\right )-785 x}{375 \left (3+5 x \right )}\)	\(32\)
meijerg	\(\frac {17 x}{45 \left (1+\frac {5 x}{3}\right )}-\frac {44 \ln \left (1+\frac {5 x}{3}\right )}{125}+\frac {4 x \left (5 x +6\right )}{75 \left (1+\frac {5 x}{3}\right )}\)	\(35\)

[In]

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

[Out]

4/25*x-121/625/(x+3/5)-44/125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {100 \, x^{2} - 44 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 60 \, x - 121}{125 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

1/125*(100*x^2 - 44*(5*x + 3)*log(5*x + 3) + 60*x - 121)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {4 x}{25} - \frac {44 \log {\left (5 x + 3 \right )}}{125} - \frac {121}{625 x + 375} \]

[In]

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

[Out]

4*x/25 - 44*log(5*x + 3)/125 - 121/(625*x + 375)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {4}{25} \, x - \frac {121}{125 \, {\left (5 \, x + 3\right )}} - \frac {44}{125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

4/25*x - 121/125/(5*x + 3) - 44/125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {4}{25} \, x - \frac {121}{125 \, {\left (5 \, x + 3\right )}} + \frac {44}{125} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) + \frac {12}{125} \]

[In]

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

[Out]

4/25*x - 121/125/(5*x + 3) + 44/125*log(1/5*abs(5*x + 3)/(5*x + 3)^2) + 12/125

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^2}{(3+5 x)^2} \, dx=\frac {4\,x}{25}-\frac {44\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {121}{625\,\left (x+\frac {3}{5}\right )} \]

[In]

int((2*x - 1)^2/(5*x + 3)^2,x)

[Out]

(4*x)/25 - (44*log(x + 3/5))/125 - 121/(625*(x + 3/5))