\(\int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx\) [1329]

Optimal result

Integrand size = 15, antiderivative size = 33 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=-\frac {121}{250 (3+5 x)^2}+\frac {44}{125 (3+5 x)}+\frac {4}{125} \log (3+5 x) \]

[Out]

-121/250/(3+5*x)^2+44/125/(3+5*x)+4/125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, 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} \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {44}{125 (5 x+3)}-\frac {121}{250 (5 x+3)^2}+\frac {4}{125} \log (5 x+3) \]

[In]

Int[(1 - 2*x)^2/(3 + 5*x)^3,x]

[Out]

-121/(250*(3 + 5*x)^2) + 44/(125*(3 + 5*x)) + (4*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 {121}{25 (3+5 x)^3}-\frac {44}{25 (3+5 x)^2}+\frac {4}{25 (3+5 x)}\right ) \, dx \\ & = -\frac {121}{250 (3+5 x)^2}+\frac {44}{125 (3+5 x)}+\frac {4}{125} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {143+440 x+8 (3+5 x)^2 \log (6+10 x)}{250 (3+5 x)^2} \]

[In]

Integrate[(1 - 2*x)^2/(3 + 5*x)^3,x]

[Out]

(143 + 440*x + 8*(3 + 5*x)^2*Log[6 + 10*x])/(250*(3 + 5*x)^2)

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {\frac {44 x}{25}+\frac {143}{250}}{\left (3+5 x \right )^{2}}+\frac {4 \ln \left (3+5 x \right )}{125}\)	\(24\)
norman	\(\frac {-\frac {11}{75} x -\frac {143}{90} x^{2}}{\left (3+5 x \right )^{2}}+\frac {4 \ln \left (3+5 x \right )}{125}\)	\(27\)
default	\(-\frac {121}{250 \left (3+5 x \right )^{2}}+\frac {44}{125 \left (3+5 x \right )}+\frac {4 \ln \left (3+5 x \right )}{125}\)	\(28\)
parallelrisch	\(\frac {1800 \ln \left (x +\frac {3}{5}\right ) x^{2}+2160 \ln \left (x +\frac {3}{5}\right ) x -3575 x^{2}+648 \ln \left (x +\frac {3}{5}\right )-330 x}{2250 \left (3+5 x \right )^{2}}\)	\(41\)
meijerg	\(\frac {x \left (\frac {5 x}{3}+2\right )}{54 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {2 x^{2}}{27 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {2 x \left (15 x +6\right )}{225 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {4 \ln \left (1+\frac {5 x}{3}\right )}{125}\)	\(52\)

[In]

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

[Out]

25*(44/625*x+143/6250)/(3+5*x)^2+4/125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {8 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) + 440 \, x + 143}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

1/250*(8*(25*x^2 + 30*x + 9)*log(5*x + 3) + 440*x + 143)/(25*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {440 x + 143}{6250 x^{2} + 7500 x + 2250} + \frac {4 \log {\left (5 x + 3 \right )}}{125} \]

[In]

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

[Out]

(440*x + 143)/(6250*x**2 + 7500*x + 2250) + 4*log(5*x + 3)/125

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {11 \, {\left (40 \, x + 13\right )}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {4}{125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

11/250*(40*x + 13)/(25*x^2 + 30*x + 9) + 4/125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {11 \, {\left (40 \, x + 13\right )}}{250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {4}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

11/250*(40*x + 13)/(5*x + 3)^2 + 4/125*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^2}{(3+5 x)^3} \, dx=\frac {4\,\ln \left (x+\frac {3}{5}\right )}{125}+\frac {\frac {44\,x}{625}+\frac {143}{6250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]

[In]

int((2*x - 1)^2/(5*x + 3)^3,x)

[Out]

(4*log(x + 3/5))/125 + ((44*x)/625 + 143/6250)/((6*x)/5 + x^2 + 9/25)