\(\int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx\) [1491]

Optimal result

Integrand size = 22, antiderivative size = 26 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=-\frac {9 x}{10}-\frac {49}{44} \log (1-2 x)+\frac {1}{275} \log (3+5 x) \]

[Out]

-9/10*x-49/44*ln(1-2*x)+1/275*ln(3+5*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} \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=-\frac {9 x}{10}-\frac {49}{44} \log (1-2 x)+\frac {1}{275} \log (5 x+3) \]

[In]

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

[Out]

(-9*x)/10 - (49*Log[1 - 2*x])/44 + Log[3 + 5*x]/275

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 {9}{10}-\frac {49}{22 (-1+2 x)}+\frac {1}{55 (3+5 x)}\right ) \, dx \\ & = -\frac {9 x}{10}-\frac {49}{44} \log (1-2 x)+\frac {1}{275} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=-\frac {3}{5}-\frac {9 x}{10}-\frac {49}{44} \log (3-6 x)+\frac {1}{275} \log (-3 (3+5 x)) \]

[In]

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

[Out]

-3/5 - (9*x)/10 - (49*Log[3 - 6*x])/44 + Log[-3*(3 + 5*x)]/275

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {9 x}{10}+\frac {\ln \left (x +\frac {3}{5}\right )}{275}-\frac {49 \ln \left (x -\frac {1}{2}\right )}{44}\)	\(17\)
default	\(-\frac {9 x}{10}+\frac {\ln \left (3+5 x \right )}{275}-\frac {49 \ln \left (-1+2 x \right )}{44}\)	\(21\)
norman	\(-\frac {9 x}{10}+\frac {\ln \left (3+5 x \right )}{275}-\frac {49 \ln \left (-1+2 x \right )}{44}\)	\(21\)
risch	\(-\frac {9 x}{10}+\frac {\ln \left (3+5 x \right )}{275}-\frac {49 \ln \left (-1+2 x \right )}{44}\)	\(21\)

[In]

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

[Out]

-9/10*x+1/275*ln(x+3/5)-49/44*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=-\frac {9}{10} \, x + \frac {1}{275} \, \log \left (5 \, x + 3\right ) - \frac {49}{44} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-9/10*x + 1/275*log(5*x + 3) - 49/44*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=- \frac {9 x}{10} - \frac {49 \log {\left (x - \frac {1}{2} \right )}}{44} + \frac {\log {\left (x + \frac {3}{5} \right )}}{275} \]

[In]

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

[Out]

-9*x/10 - 49*log(x - 1/2)/44 + log(x + 3/5)/275

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=-\frac {9}{10} \, x + \frac {1}{275} \, \log \left (5 \, x + 3\right ) - \frac {49}{44} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-9/10*x + 1/275*log(5*x + 3) - 49/44*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=-\frac {9}{10} \, x + \frac {1}{275} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {49}{44} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-9/10*x + 1/275*log(abs(5*x + 3)) - 49/44*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)} \, dx=\frac {\ln \left (x+\frac {3}{5}\right )}{275}-\frac {49\,\ln \left (x-\frac {1}{2}\right )}{44}-\frac {9\,x}{10} \]

[In]

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

[Out]

log(x + 3/5)/275 - (49*log(x - 1/2))/44 - (9*x)/10