\(\int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx\) [1622]

Optimal result

Integrand size = 22, antiderivative size = 54 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {343}{2662 (1-2 x)}-\frac {1}{6050 (3+5 x)^2}-\frac {103}{33275 (3+5 x)}-\frac {147 \log (1-2 x)}{14641}+\frac {147 \log (3+5 x)}{14641} \]

[Out]

343/2662/(1-2*x)-1/6050/(3+5*x)^2-103/33275/(3+5*x)-147/14641*ln(1-2*x)+147/14641*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, 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} \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {343}{2662 (1-2 x)}-\frac {103}{33275 (5 x+3)}-\frac {1}{6050 (5 x+3)^2}-\frac {147 \log (1-2 x)}{14641}+\frac {147 \log (5 x+3)}{14641} \]

[In]

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

[Out]

343/(2662*(1 - 2*x)) - 1/(6050*(3 + 5*x)^2) - 103/(33275*(3 + 5*x)) - (147*Log[1 - 2*x])/14641 + (147*Log[3 +
5*x])/14641

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 {343}{1331 (-1+2 x)^2}-\frac {294}{14641 (-1+2 x)}+\frac {1}{605 (3+5 x)^3}+\frac {103}{6655 (3+5 x)^2}+\frac {735}{14641 (3+5 x)}\right ) \, dx \\ & = \frac {343}{2662 (1-2 x)}-\frac {1}{6050 (3+5 x)^2}-\frac {103}{33275 (3+5 x)}-\frac {147 \log (1-2 x)}{14641}+\frac {147 \log (3+5 x)}{14641} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {-\frac {11 \left (76546+257478 x+216435 x^2\right )}{(-1+2 x) (3+5 x)^2}-7350 \log (1-2 x)+7350 \log (6+10 x)}{732050} \]

[In]

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

[Out]

((-11*(76546 + 257478*x + 216435*x^2))/((-1 + 2*x)*(3 + 5*x)^2) - 7350*Log[1 - 2*x] + 7350*Log[6 + 10*x])/7320
50

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {-\frac {43287}{13310} x^{2}-\frac {128739}{33275} x -\frac {38273}{33275}}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {147 \ln \left (-1+2 x \right )}{14641}+\frac {147 \ln \left (3+5 x \right )}{14641}\)	\(44\)
default	\(-\frac {1}{6050 \left (3+5 x \right )^{2}}-\frac {103}{33275 \left (3+5 x \right )}+\frac {147 \ln \left (3+5 x \right )}{14641}-\frac {343}{2662 \left (-1+2 x \right )}-\frac {147 \ln \left (-1+2 x \right )}{14641}\)	\(45\)
norman	\(\frac {-\frac {76546}{11979} x^{3}-\frac {185081}{23958} x^{2}-\frac {9325}{3993} x}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {147 \ln \left (-1+2 x \right )}{14641}+\frac {147 \ln \left (3+5 x \right )}{14641}\)	\(47\)
parallelrisch	\(\frac {132300 \ln \left (x +\frac {3}{5}\right ) x^{3}-132300 \ln \left (x -\frac {1}{2}\right ) x^{3}+92610 \ln \left (x +\frac {3}{5}\right ) x^{2}-92610 \ln \left (x -\frac {1}{2}\right ) x^{2}-1684012 x^{3}-31752 \ln \left (x +\frac {3}{5}\right ) x +31752 \ln \left (x -\frac {1}{2}\right ) x -2035891 x^{2}-23814 \ln \left (x +\frac {3}{5}\right )+23814 \ln \left (x -\frac {1}{2}\right )-615450 x}{263538 \left (-1+2 x \right ) \left (3+5 x \right )^{2}}\)	\(93\)

[In]

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

[Out]

50*(-43287/665500*x^2-128739/1663750*x-38273/1663750)/(-1+2*x)/(3+5*x)^2-147/14641*ln(-1+2*x)+147/14641*ln(3+5
*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {2380785 \, x^{2} - 7350 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) + 7350 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 2832258 \, x + 842006}{732050 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

[In]

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

[Out]

-1/732050*(2380785*x^2 - 7350*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) + 7350*(50*x^3 + 35*x^2 - 12*x - 9)*lo
g(2*x - 1) + 2832258*x + 842006)/(50*x^3 + 35*x^2 - 12*x - 9)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {- 216435 x^{2} - 257478 x - 76546}{3327500 x^{3} + 2329250 x^{2} - 798600 x - 598950} - \frac {147 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {147 \log {\left (x + \frac {3}{5} \right )}}{14641} \]

[In]

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

[Out]

(-216435*x**2 - 257478*x - 76546)/(3327500*x**3 + 2329250*x**2 - 798600*x - 598950) - 147*log(x - 1/2)/14641 +
147*log(x + 3/5)/14641

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {216435 \, x^{2} + 257478 \, x + 76546}{66550 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {147}{14641} \, \log \left (5 \, x + 3\right ) - \frac {147}{14641} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/66550*(216435*x^2 + 257478*x + 76546)/(50*x^3 + 35*x^2 - 12*x - 9) + 147/14641*log(5*x + 3) - 147/14641*log
(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {343}{2662 \, {\left (2 \, x - 1\right )}} + \frac {2 \, {\left (\frac {231}{2 \, x - 1} + 104\right )}}{14641 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} + \frac {147}{14641} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \]

[In]

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

[Out]

-343/2662/(2*x - 1) + 2/14641*(231/(2*x - 1) + 104)/(11/(2*x - 1) + 5)^2 + 147/14641*log(abs(-11/(2*x - 1) - 5
))

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {294\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}+\frac {\frac {43287\,x^2}{665500}+\frac {128739\,x}{1663750}+\frac {38273}{1663750}}{-x^3-\frac {7\,x^2}{10}+\frac {6\,x}{25}+\frac {9}{50}} \]

[In]

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

[Out]

(294*atanh((20*x)/11 + 1/11))/14641 + ((128739*x)/1663750 + (43287*x^2)/665500 + 38273/1663750)/((6*x)/25 - (7
*x^2)/10 - x^3 + 9/50)