\(\int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx\) [1478]

Optimal result

Integrand size = 22, antiderivative size = 43 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=\frac {1}{378 (2+3 x)^2}-\frac {103}{1323 (2+3 x)}-\frac {1331}{686} \log (1-2 x)-\frac {3469 \log (2+3 x)}{9261} \]

[Out]

1/378/(2+3*x)^2-103/1323/(2+3*x)-1331/686*ln(1-2*x)-3469/9261*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} \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=-\frac {103}{1323 (3 x+2)}+\frac {1}{378 (3 x+2)^2}-\frac {1331}{686} \log (1-2 x)-\frac {3469 \log (3 x+2)}{9261} \]

[In]

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

[Out]

1/(378*(2 + 3*x)^2) - 103/(1323*(2 + 3*x)) - (1331*Log[1 - 2*x])/686 - (3469*Log[2 + 3*x])/9261

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 {1331}{343 (-1+2 x)}-\frac {1}{63 (2+3 x)^3}+\frac {103}{441 (2+3 x)^2}-\frac {3469}{3087 (2+3 x)}\right ) \, dx \\ & = \frac {1}{378 (2+3 x)^2}-\frac {103}{1323 (2+3 x)}-\frac {1331}{686} \log (1-2 x)-\frac {3469 \log (2+3 x)}{9261} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=\frac {-\frac {21 (135+206 x)}{(2+3 x)^2}-35937 \log (1-2 x)-6938 \log (4+6 x)}{18522} \]

[In]

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

[Out]

((-21*(135 + 206*x))/(2 + 3*x)^2 - 35937*Log[1 - 2*x] - 6938*Log[4 + 6*x])/18522

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {-\frac {103 x}{441}-\frac {15}{98}}{\left (2+3 x \right )^{2}}-\frac {1331 \ln \left (-1+2 x \right )}{686}-\frac {3469 \ln \left (2+3 x \right )}{9261}\)	\(32\)
norman	\(\frac {\frac {199}{882} x +\frac {135}{392} x^{2}}{\left (2+3 x \right )^{2}}-\frac {1331 \ln \left (-1+2 x \right )}{686}-\frac {3469 \ln \left (2+3 x \right )}{9261}\)	\(35\)
default	\(-\frac {1331 \ln \left (-1+2 x \right )}{686}+\frac {1}{378 \left (2+3 x \right )^{2}}-\frac {103}{1323 \left (2+3 x \right )}-\frac {3469 \ln \left (2+3 x \right )}{9261}\)	\(36\)
parallelrisch	\(-\frac {249768 \ln \left (\frac {2}{3}+x \right ) x^{2}+1293732 \ln \left (x -\frac {1}{2}\right ) x^{2}+333024 \ln \left (\frac {2}{3}+x \right ) x +1724976 \ln \left (x -\frac {1}{2}\right ) x -25515 x^{2}+111008 \ln \left (\frac {2}{3}+x \right )+574992 \ln \left (x -\frac {1}{2}\right )-16716 x}{74088 \left (2+3 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

9*(-103/3969*x-5/294)/(2+3*x)^2-1331/686*ln(-1+2*x)-3469/9261*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=-\frac {6938 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 35937 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (2 \, x - 1\right ) + 4326 \, x + 2835}{18522 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

-1/18522*(6938*(9*x^2 + 12*x + 4)*log(3*x + 2) + 35937*(9*x^2 + 12*x + 4)*log(2*x - 1) + 4326*x + 2835)/(9*x^2
+ 12*x + 4)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=- \frac {206 x + 135}{7938 x^{2} + 10584 x + 3528} - \frac {1331 \log {\left (x - \frac {1}{2} \right )}}{686} - \frac {3469 \log {\left (x + \frac {2}{3} \right )}}{9261} \]

[In]

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

[Out]

-(206*x + 135)/(7938*x**2 + 10584*x + 3528) - 1331*log(x - 1/2)/686 - 3469*log(x + 2/3)/9261

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=-\frac {206 \, x + 135}{882 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {3469}{9261} \, \log \left (3 \, x + 2\right ) - \frac {1331}{686} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/882*(206*x + 135)/(9*x^2 + 12*x + 4) - 3469/9261*log(3*x + 2) - 1331/686*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=-\frac {206 \, x + 135}{882 \, {\left (3 \, x + 2\right )}^{2}} - \frac {3469}{9261} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {1331}{686} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/882*(206*x + 135)/(3*x + 2)^2 - 3469/9261*log(abs(3*x + 2)) - 1331/686*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.70 \[ \int \frac {(3+5 x)^3}{(1-2 x) (2+3 x)^3} \, dx=-\frac {1331\,\ln \left (x-\frac {1}{2}\right )}{686}-\frac {3469\,\ln \left (x+\frac {2}{3}\right )}{9261}-\frac {\frac {103\,x}{3969}+\frac {5}{294}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]

[In]

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

[Out]

- (1331*log(x - 1/2))/686 - (3469*log(x + 2/3))/9261 - ((103*x)/3969 + 5/294)/((4*x)/3 + x^2 + 4/9)