\(\int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx\) [1461]

Optimal result

Integrand size = 22, antiderivative size = 32 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=-\frac {1}{63 (2+3 x)}-\frac {121}{98} \log (1-2 x)-\frac {68}{441} \log (2+3 x) \]

[Out]

-1/63/(2+3*x)-121/98*ln(1-2*x)-68/441*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, 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} \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=-\frac {1}{63 (3 x+2)}-\frac {121}{98} \log (1-2 x)-\frac {68}{441} \log (3 x+2) \]

[In]

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

[Out]

-1/63*1/(2 + 3*x) - (121*Log[1 - 2*x])/98 - (68*Log[2 + 3*x])/441

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 {121}{49 (-1+2 x)}+\frac {1}{21 (2+3 x)^2}-\frac {68}{147 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{63 (2+3 x)}-\frac {121}{98} \log (1-2 x)-\frac {68}{441} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=\frac {1}{882} \left (-\frac {14}{2+3 x}-1089 \log (1-2 x)-136 \log (4+6 x)\right ) \]

[In]

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

[Out]

(-14/(2 + 3*x) - 1089*Log[1 - 2*x] - 136*Log[4 + 6*x])/882

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {1}{189 \left (\frac {2}{3}+x \right )}-\frac {121 \ln \left (-1+2 x \right )}{98}-\frac {68 \ln \left (2+3 x \right )}{441}\)	\(25\)
default	\(-\frac {121 \ln \left (-1+2 x \right )}{98}-\frac {1}{63 \left (2+3 x \right )}-\frac {68 \ln \left (2+3 x \right )}{441}\)	\(27\)
norman	\(\frac {x}{84+126 x}-\frac {121 \ln \left (-1+2 x \right )}{98}-\frac {68 \ln \left (2+3 x \right )}{441}\)	\(28\)
parallelrisch	\(-\frac {408 \ln \left (\frac {2}{3}+x \right ) x +3267 \ln \left (x -\frac {1}{2}\right ) x +272 \ln \left (\frac {2}{3}+x \right )+2178 \ln \left (x -\frac {1}{2}\right )-21 x}{882 \left (2+3 x \right )}\)	\(40\)

[In]

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

[Out]

-1/189/(2/3+x)-121/98*ln(-1+2*x)-68/441*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=-\frac {136 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 1089 \, {\left (3 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 14}{882 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

-1/882*(136*(3*x + 2)*log(3*x + 2) + 1089*(3*x + 2)*log(2*x - 1) + 14)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=- \frac {121 \log {\left (x - \frac {1}{2} \right )}}{98} - \frac {68 \log {\left (x + \frac {2}{3} \right )}}{441} - \frac {1}{189 x + 126} \]

[In]

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

[Out]

-121*log(x - 1/2)/98 - 68*log(x + 2/3)/441 - 1/(189*x + 126)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=-\frac {1}{63 \, {\left (3 \, x + 2\right )}} - \frac {68}{441} \, \log \left (3 \, x + 2\right ) - \frac {121}{98} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/63/(3*x + 2) - 68/441*log(3*x + 2) - 121/98*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=-\frac {1}{63 \, {\left (3 \, x + 2\right )}} + \frac {25}{18} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) - \frac {121}{98} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]

[In]

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

[Out]

-1/63/(3*x + 2) + 25/18*log(1/3*abs(3*x + 2)/(3*x + 2)^2) - 121/98*log(abs(-7/(3*x + 2) + 2))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^2} \, dx=-\frac {121\,\ln \left (x-\frac {1}{2}\right )}{98}-\frac {68\,\ln \left (x+\frac {2}{3}\right )}{441}-\frac {1}{189\,\left (x+\frac {2}{3}\right )} \]

[In]

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

[Out]

- (121*log(x - 1/2))/98 - (68*log(x + 2/3))/441 - 1/(189*(x + 2/3))