\(\int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx\) [1476]

Optimal result

Integrand size = 22, antiderivative size = 33 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=-\frac {1225 x}{36}-\frac {125 x^2}{12}-\frac {1331}{56} \log (1-2 x)-\frac {1}{189} \log (2+3 x) \]

[Out]

-1225/36*x-125/12*x^2-1331/56*ln(1-2*x)-1/189*ln(2+3*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {84} \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=-\frac {125 x^2}{12}-\frac {1225 x}{36}-\frac {1331}{56} \log (1-2 x)-\frac {1}{189} \log (3 x+2) \]

[In]

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

[Out]

(-1225*x)/36 - (125*x^2)/12 - (1331*Log[1 - 2*x])/56 - Log[2 + 3*x]/189

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 {1225}{36}-\frac {125 x}{6}-\frac {1331}{28 (-1+2 x)}-\frac {1}{63 (2+3 x)}\right ) \, dx \\ & = -\frac {1225 x}{36}-\frac {125 x^2}{12}-\frac {1331}{56} \log (1-2 x)-\frac {1}{189} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=\frac {-1050 \left (24+49 x+15 x^2\right )-35937 \log (5-10 x)-8 \log (5 (2+3 x))}{1512} \]

[In]

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

[Out]

(-1050*(24 + 49*x + 15*x^2) - 35937*Log[5 - 10*x] - 8*Log[5*(2 + 3*x)])/1512

Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(-\frac {125 x^{2}}{12}-\frac {1225 x}{36}-\frac {\ln \left (\frac {2}{3}+x \right )}{189}-\frac {1331 \ln \left (x -\frac {1}{2}\right )}{56}\)	\(22\)
default	\(-\frac {125 x^{2}}{12}-\frac {1225 x}{36}-\frac {1331 \ln \left (-1+2 x \right )}{56}-\frac {\ln \left (2+3 x \right )}{189}\)	\(26\)
norman	\(-\frac {125 x^{2}}{12}-\frac {1225 x}{36}-\frac {1331 \ln \left (-1+2 x \right )}{56}-\frac {\ln \left (2+3 x \right )}{189}\)	\(26\)
risch	\(-\frac {125 x^{2}}{12}-\frac {1225 x}{36}-\frac {1331 \ln \left (-1+2 x \right )}{56}-\frac {\ln \left (2+3 x \right )}{189}\)	\(26\)

[In]

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

[Out]

-125/12*x^2-1225/36*x-1/189*ln(2/3+x)-1331/56*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=-\frac {125}{12} \, x^{2} - \frac {1225}{36} \, x - \frac {1}{189} \, \log \left (3 \, x + 2\right ) - \frac {1331}{56} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-125/12*x^2 - 1225/36*x - 1/189*log(3*x + 2) - 1331/56*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=- \frac {125 x^{2}}{12} - \frac {1225 x}{36} - \frac {1331 \log {\left (x - \frac {1}{2} \right )}}{56} - \frac {\log {\left (x + \frac {2}{3} \right )}}{189} \]

[In]

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

[Out]

-125*x**2/12 - 1225*x/36 - 1331*log(x - 1/2)/56 - log(x + 2/3)/189

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=-\frac {125}{12} \, x^{2} - \frac {1225}{36} \, x - \frac {1}{189} \, \log \left (3 \, x + 2\right ) - \frac {1331}{56} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-125/12*x^2 - 1225/36*x - 1/189*log(3*x + 2) - 1331/56*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=-\frac {125}{12} \, x^{2} - \frac {1225}{36} \, x - \frac {1}{189} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {1331}{56} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-125/12*x^2 - 1225/36*x - 1/189*log(abs(3*x + 2)) - 1331/56*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)} \, dx=-\frac {1225\,x}{36}-\frac {1331\,\ln \left (x-\frac {1}{2}\right )}{56}-\frac {\ln \left (x+\frac {2}{3}\right )}{189}-\frac {125\,x^2}{12} \]

[In]

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

[Out]

- (1225*x)/36 - (1331*log(x - 1/2))/56 - log(x + 2/3)/189 - (125*x^2)/12