\(\int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx\) [1560]

Optimal result

Integrand size = 22, antiderivative size = 41 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {5929}{32 (1-2 x)}+\frac {5119 x}{16}+\frac {795 x^2}{8}+\frac {75 x^3}{4}+\frac {1309}{4} \log (1-2 x) \]

[Out]

5929/32/(1-2*x)+5119/16*x+795/8*x^2+75/4*x^3+1309/4*ln(1-2*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, 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} \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {75 x^3}{4}+\frac {795 x^2}{8}+\frac {5119 x}{16}+\frac {5929}{32 (1-2 x)}+\frac {1309}{4} \log (1-2 x) \]

[In]

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

[Out]

5929/(32*(1 - 2*x)) + (5119*x)/16 + (795*x^2)/8 + (75*x^3)/4 + (1309*Log[1 - 2*x])/4

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 {5119}{16}+\frac {795 x}{4}+\frac {225 x^2}{4}+\frac {5929}{16 (-1+2 x)^2}+\frac {1309}{2 (-1+2 x)}\right ) \, dx \\ & = \frac {5929}{32 (1-2 x)}+\frac {5119 x}{16}+\frac {795 x^2}{8}+\frac {75 x^3}{4}+\frac {1309}{4} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {15-5554 x+4324 x^2+1440 x^3+300 x^4+2618 (-1+2 x) \log (1-2 x)}{-8+16 x} \]

[In]

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

[Out]

(15 - 5554*x + 4324*x^2 + 1440*x^3 + 300*x^4 + 2618*(-1 + 2*x)*Log[1 - 2*x])/(-8 + 16*x)

Maple [A] (verified)

Time = 2.64 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {75 x^{3}}{4}+\frac {795 x^{2}}{8}+\frac {5119 x}{16}-\frac {5929}{64 \left (x -\frac {1}{2}\right )}+\frac {1309 \ln \left (-1+2 x \right )}{4}\)	\(30\)
default	\(\frac {75 x^{3}}{4}+\frac {795 x^{2}}{8}+\frac {5119 x}{16}+\frac {1309 \ln \left (-1+2 x \right )}{4}-\frac {5929}{32 \left (-1+2 x \right )}\)	\(32\)
norman	\(\frac {-\frac {1381}{2} x +\frac {1081}{2} x^{2}+180 x^{3}+\frac {75}{2} x^{4}}{-1+2 x}+\frac {1309 \ln \left (-1+2 x \right )}{4}\)	\(37\)
parallelrisch	\(\frac {150 x^{4}+720 x^{3}+2618 \ln \left (x -\frac {1}{2}\right ) x +2162 x^{2}-1309 \ln \left (x -\frac {1}{2}\right )-2762 x}{-4+8 x}\)	\(42\)
meijerg	\(\frac {150 x}{1-2 x}+\frac {1309 \ln \left (1-2 x \right )}{4}+\frac {541 x \left (-6 x +6\right )}{12 \left (1-2 x \right )}+\frac {285 x \left (-8 x^{2}-12 x +12\right )}{16 \left (1-2 x \right )}+\frac {15 x \left (-40 x^{3}-40 x^{2}-60 x +60\right )}{16 \left (1-2 x \right )}\)	\(80\)

[In]

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

[Out]

75/4*x^3+795/8*x^2+5119/16*x-5929/64/(x-1/2)+1309/4*ln(-1+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {1200 \, x^{4} + 5760 \, x^{3} + 17296 \, x^{2} + 10472 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 10238 \, x - 5929}{32 \, {\left (2 \, x - 1\right )}} \]

[In]

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

[Out]

1/32*(1200*x^4 + 5760*x^3 + 17296*x^2 + 10472*(2*x - 1)*log(2*x - 1) - 10238*x - 5929)/(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {75 x^{3}}{4} + \frac {795 x^{2}}{8} + \frac {5119 x}{16} + \frac {1309 \log {\left (2 x - 1 \right )}}{4} - \frac {5929}{64 x - 32} \]

[In]

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

[Out]

75*x**3/4 + 795*x**2/8 + 5119*x/16 + 1309*log(2*x - 1)/4 - 5929/(64*x - 32)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {75}{4} \, x^{3} + \frac {795}{8} \, x^{2} + \frac {5119}{16} \, x - \frac {5929}{32 \, {\left (2 \, x - 1\right )}} + \frac {1309}{4} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

75/4*x^3 + 795/8*x^2 + 5119/16*x - 5929/32/(2*x - 1) + 1309/4*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {1}{32} \, {\left (2 \, x - 1\right )}^{3} {\left (\frac {1020}{2 \, x - 1} + \frac {6934}{{\left (2 \, x - 1\right )}^{2}} + 75\right )} - \frac {5929}{32 \, {\left (2 \, x - 1\right )}} - \frac {1309}{4} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) \]

[In]

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

[Out]

1/32*(2*x - 1)^3*(1020/(2*x - 1) + 6934/(2*x - 1)^2 + 75) - 5929/32/(2*x - 1) - 1309/4*log(1/2*abs(2*x - 1)/(2
*x - 1)^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^2 (3+5 x)^2}{(1-2 x)^2} \, dx=\frac {5119\,x}{16}+\frac {1309\,\ln \left (x-\frac {1}{2}\right )}{4}-\frac {5929}{64\,\left (x-\frac {1}{2}\right )}+\frac {795\,x^2}{8}+\frac {75\,x^3}{4} \]

[In]

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

[Out]

(5119*x)/16 + (1309*log(x - 1/2))/4 - 5929/(64*(x - 1/2)) + (795*x^2)/8 + (75*x^3)/4