\(\int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx\) [1462]

Optimal result

Integrand size = 22, antiderivative size = 43 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=-\frac {1}{126 (2+3 x)^2}+\frac {68}{441 (2+3 x)}-\frac {121}{343} \log (1-2 x)+\frac {121}{343} \log (2+3 x) \]

[Out]

-1/126/(2+3*x)^2+68/441/(2+3*x)-121/343*ln(1-2*x)+121/343*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, 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} \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=\frac {68}{441 (3 x+2)}-\frac {1}{126 (3 x+2)^2}-\frac {121}{343} \log (1-2 x)+\frac {121}{343} \log (3 x+2) \]

[In]

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

[Out]

-1/126*1/(2 + 3*x)^2 + 68/(441*(2 + 3*x)) - (121*Log[1 - 2*x])/343 + (121*Log[2 + 3*x])/343

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 {242}{343 (-1+2 x)}+\frac {1}{21 (2+3 x)^3}-\frac {68}{147 (2+3 x)^2}+\frac {363}{343 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{126 (2+3 x)^2}+\frac {68}{441 (2+3 x)}-\frac {121}{343} \log (1-2 x)+\frac {121}{343} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=\frac {\frac {7 (265+408 x)}{(2+3 x)^2}-2178 \log (1-2 x)+2178 \log (4+6 x)}{6174} \]

[In]

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

[Out]

((7*(265 + 408*x))/(2 + 3*x)^2 - 2178*Log[1 - 2*x] + 2178*Log[4 + 6*x])/6174

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\frac {68 x}{147}+\frac {265}{882}}{\left (2+3 x \right )^{2}}-\frac {121 \ln \left (-1+2 x \right )}{343}+\frac {121 \ln \left (2+3 x \right )}{343}\)	\(32\)
norman	\(\frac {-\frac {265}{392} x^{2}-\frac {43}{98} x}{\left (2+3 x \right )^{2}}-\frac {121 \ln \left (-1+2 x \right )}{343}+\frac {121 \ln \left (2+3 x \right )}{343}\)	\(35\)
default	\(-\frac {121 \ln \left (-1+2 x \right )}{343}-\frac {1}{126 \left (2+3 x \right )^{2}}+\frac {68}{441 \left (2+3 x \right )}+\frac {121 \ln \left (2+3 x \right )}{343}\)	\(36\)
parallelrisch	\(\frac {8712 \ln \left (\frac {2}{3}+x \right ) x^{2}-8712 \ln \left (x -\frac {1}{2}\right ) x^{2}+11616 \ln \left (\frac {2}{3}+x \right ) x -11616 \ln \left (x -\frac {1}{2}\right ) x -1855 x^{2}+3872 \ln \left (\frac {2}{3}+x \right )-3872 \ln \left (x -\frac {1}{2}\right )-1204 x}{2744 \left (2+3 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

9*(68/1323*x+265/7938)/(2+3*x)^2-121/343*ln(-1+2*x)+121/343*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=\frac {2178 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 2178 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (2 \, x - 1\right ) + 2856 \, x + 1855}{6174 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

1/6174*(2178*(9*x^2 + 12*x + 4)*log(3*x + 2) - 2178*(9*x^2 + 12*x + 4)*log(2*x - 1) + 2856*x + 1855)/(9*x^2 +
12*x + 4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=- \frac {- 408 x - 265}{7938 x^{2} + 10584 x + 3528} - \frac {121 \log {\left (x - \frac {1}{2} \right )}}{343} + \frac {121 \log {\left (x + \frac {2}{3} \right )}}{343} \]

[In]

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

[Out]

-(-408*x - 265)/(7938*x**2 + 10584*x + 3528) - 121*log(x - 1/2)/343 + 121*log(x + 2/3)/343

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=\frac {408 \, x + 265}{882 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {121}{343} \, \log \left (3 \, x + 2\right ) - \frac {121}{343} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

1/882*(408*x + 265)/(9*x^2 + 12*x + 4) + 121/343*log(3*x + 2) - 121/343*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=\frac {408 \, x + 265}{882 \, {\left (3 \, x + 2\right )}^{2}} + \frac {121}{343} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {121}{343} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

1/882*(408*x + 265)/(3*x + 2)^2 + 121/343*log(abs(3*x + 2)) - 121/343*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.58 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)^3} \, dx=\frac {242\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{343}+\frac {\frac {68\,x}{1323}+\frac {265}{7938}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]

[In]

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

[Out]

(242*atanh((12*x)/7 + 1/7))/343 + ((68*x)/1323 + 265/7938)/((4*x)/3 + x^2 + 4/9)