\(\int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx\) [1460]

Optimal result

Integrand size = 22, antiderivative size = 26 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=-\frac {25 x}{6}-\frac {121}{28} \log (1-2 x)+\frac {1}{63} \log (2+3 x) \]

[Out]

-25/6*x-121/28*ln(1-2*x)+1/63*ln(2+3*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} \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=-\frac {25 x}{6}-\frac {121}{28} \log (1-2 x)+\frac {1}{63} \log (3 x+2) \]

[In]

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

[Out]

(-25*x)/6 - (121*Log[1 - 2*x])/28 + Log[2 + 3*x]/63

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 {25}{6}-\frac {121}{14 (-1+2 x)}+\frac {1}{21 (2+3 x)}\right ) \, dx \\ & = -\frac {25 x}{6}-\frac {121}{28} \log (1-2 x)+\frac {1}{63} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=-\frac {5}{6} (3+5 x)-\frac {121}{28} \log (5-10 x)+\frac {1}{63} \log (5 (2+3 x)) \]

[In]

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

[Out]

(-5*(3 + 5*x))/6 - (121*Log[5 - 10*x])/28 + Log[5*(2 + 3*x)]/63

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {25 x}{6}+\frac {\ln \left (\frac {2}{3}+x \right )}{63}-\frac {121 \ln \left (x -\frac {1}{2}\right )}{28}\)	\(17\)
default	\(-\frac {25 x}{6}-\frac {121 \ln \left (-1+2 x \right )}{28}+\frac {\ln \left (2+3 x \right )}{63}\)	\(21\)
norman	\(-\frac {25 x}{6}-\frac {121 \ln \left (-1+2 x \right )}{28}+\frac {\ln \left (2+3 x \right )}{63}\)	\(21\)
risch	\(-\frac {25 x}{6}-\frac {121 \ln \left (-1+2 x \right )}{28}+\frac {\ln \left (2+3 x \right )}{63}\)	\(21\)

[In]

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

[Out]

-25/6*x+1/63*ln(2/3+x)-121/28*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=-\frac {25}{6} \, x + \frac {1}{63} \, \log \left (3 \, x + 2\right ) - \frac {121}{28} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-25/6*x + 1/63*log(3*x + 2) - 121/28*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=- \frac {25 x}{6} - \frac {121 \log {\left (x - \frac {1}{2} \right )}}{28} + \frac {\log {\left (x + \frac {2}{3} \right )}}{63} \]

[In]

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

[Out]

-25*x/6 - 121*log(x - 1/2)/28 + log(x + 2/3)/63

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=-\frac {25}{6} \, x + \frac {1}{63} \, \log \left (3 \, x + 2\right ) - \frac {121}{28} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-25/6*x + 1/63*log(3*x + 2) - 121/28*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=-\frac {25}{6} \, x + \frac {1}{63} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {121}{28} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-25/6*x + 1/63*log(abs(3*x + 2)) - 121/28*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {(3+5 x)^2}{(1-2 x) (2+3 x)} \, dx=\frac {\ln \left (x+\frac {2}{3}\right )}{63}-\frac {121\,\ln \left (x-\frac {1}{2}\right )}{28}-\frac {25\,x}{6} \]

[In]

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

[Out]

log(x + 2/3)/63 - (121*log(x - 1/2))/28 - (25*x)/6