\(\int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx\) [1591]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {16807}{352 (1-2 x)}+\frac {152793 x}{2000}+\frac {567 x^2}{25}+\frac {81 x^3}{20}+\frac {156065 \log (1-2 x)}{1936}+\frac {\log (3+5 x)}{75625} \]

[Out]

16807/352/(1-2*x)+152793/2000*x+567/25*x^2+81/20*x^3+156065/1936*ln(1-2*x)+1/75625*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, 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} \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {81 x^3}{20}+\frac {567 x^2}{25}+\frac {152793 x}{2000}+\frac {16807}{352 (1-2 x)}+\frac {156065 \log (1-2 x)}{1936}+\frac {\log (5 x+3)}{75625} \]

[In]

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

[Out]

16807/(352*(1 - 2*x)) + (152793*x)/2000 + (567*x^2)/25 + (81*x^3)/20 + (156065*Log[1 - 2*x])/1936 + Log[3 + 5*
x]/75625

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 {152793}{2000}+\frac {1134 x}{25}+\frac {243 x^2}{20}+\frac {16807}{176 (-1+2 x)^2}+\frac {156065}{968 (-1+2 x)}+\frac {1}{15125 (3+5 x)}\right ) \, dx \\ & = \frac {16807}{352 (1-2 x)}+\frac {152793 x}{2000}+\frac {567 x^2}{25}+\frac {81 x^3}{20}+\frac {156065 \log (1-2 x)}{1936}+\frac {\log (3+5 x)}{75625} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {385479}{10000}+\frac {16807}{352-704 x}+\frac {152793 x}{2000}+\frac {567 x^2}{25}+\frac {81 x^3}{20}+\frac {156065 \log (5-10 x)}{1936}+\frac {\log (3+5 x)}{75625} \]

[In]

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

[Out]

385479/10000 + 16807/(352 - 704*x) + (152793*x)/2000 + (567*x^2)/25 + (81*x^3)/20 + (156065*Log[5 - 10*x])/193
6 + Log[3 + 5*x]/75625

Maple [A] (verified)

Time = 2.69 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {81 x^{3}}{20}+\frac {567 x^{2}}{25}+\frac {152793 x}{2000}-\frac {16807}{704 \left (x -\frac {1}{2}\right )}+\frac {156065 \ln \left (-1+2 x \right )}{1936}+\frac {\ln \left (3+5 x \right )}{75625}\)	\(38\)
default	\(\frac {81 x^{3}}{20}+\frac {567 x^{2}}{25}+\frac {152793 x}{2000}+\frac {\ln \left (3+5 x \right )}{75625}-\frac {16807}{352 \left (-1+2 x \right )}+\frac {156065 \ln \left (-1+2 x \right )}{1936}\)	\(40\)
norman	\(\frac {-\frac {1890799}{11000} x +\frac {130113}{1000} x^{2}+\frac {4131}{100} x^{3}+\frac {81}{10} x^{4}}{-1+2 x}+\frac {156065 \ln \left (-1+2 x \right )}{1936}+\frac {\ln \left (3+5 x \right )}{75625}\)	\(45\)
parallelrisch	\(\frac {9801000 x^{4}[phone] x^{3}+32 \ln \left (x +\frac {3}{5}\right ) x [phone] \ln \left (x -\frac {1}{2}\right ) x [phone] x^{2}-16 \ln \left (x +\frac {3}{5}\right )-97540625 \ln \left (x -\frac {1}{2}\right )-207987890 x}{-1210000+2420000 x}\)	\(55\)

[In]

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

[Out]

81/20*x^3+567/25*x^2+152793/2000*x-16807/704/(x-1/2)+156065/1936*ln(-1+2*x)+1/75625*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {19602000 \, x^{4} + 99970200 \, x^{3} + 314873460 \, x^{2} + 32 \, {\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) + 195081250 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 184879530 \, x - 115548125}{2420000 \, {\left (2 \, x - 1\right )}} \]

[In]

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

[Out]

1/2420000*(19602000*x^4 + 99970200*x^3 + 314873460*x^2 + 32*(2*x - 1)*log(5*x + 3) + 195081250*(2*x - 1)*log(2
*x - 1) - 184879530*x - 115548125)/(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {81 x^{3}}{20} + \frac {567 x^{2}}{25} + \frac {152793 x}{2000} + \frac {156065 \log {\left (x - \frac {1}{2} \right )}}{1936} + \frac {\log {\left (x + \frac {3}{5} \right )}}{75625} - \frac {16807}{704 x - 352} \]

[In]

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

[Out]

81*x**3/20 + 567*x**2/25 + 152793*x/2000 + 156065*log(x - 1/2)/1936 + log(x + 3/5)/75625 - 16807/(704*x - 352)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {81}{20} \, x^{3} + \frac {567}{25} \, x^{2} + \frac {152793}{2000} \, x - \frac {16807}{352 \, {\left (2 \, x - 1\right )}} + \frac {1}{75625} \, \log \left (5 \, x + 3\right ) + \frac {156065}{1936} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

81/20*x^3 + 567/25*x^2 + 152793/2000*x - 16807/352/(2*x - 1) + 1/75625*log(5*x + 3) + 156065/1936*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.41 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {27}{4000} \, {\left (2 \, x - 1\right )}^{3} {\left (\frac {1065}{2 \, x - 1} + \frac {7564}{{\left (2 \, x - 1\right )}^{2}} + 75\right )} - \frac {16807}{352 \, {\left (2 \, x - 1\right )}} - \frac {806121}{10000} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) + \frac {1}{75625} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \]

[In]

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

[Out]

27/4000*(2*x - 1)^3*(1065/(2*x - 1) + 7564/(2*x - 1)^2 + 75) - 16807/352/(2*x - 1) - 806121/10000*log(1/2*abs(
2*x - 1)/(2*x - 1)^2) + 1/75625*log(abs(-11/(2*x - 1) - 5))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^5}{(1-2 x)^2 (3+5 x)} \, dx=\frac {152793\,x}{2000}+\frac {156065\,\ln \left (x-\frac {1}{2}\right )}{1936}+\frac {\ln \left (x+\frac {3}{5}\right )}{75625}-\frac {16807}{704\,\left (x-\frac {1}{2}\right )}+\frac {567\,x^2}{25}+\frac {81\,x^3}{20} \]

[In]

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

[Out]

(152793*x)/2000 + (156065*log(x - 1/2))/1936 + log(x + 3/5)/75625 - 16807/(704*(x - 1/2)) + (567*x^2)/25 + (81
*x^3)/20