\(\int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx\) [1438]

Optimal result

Integrand size = 20, antiderivative size = 44 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=-\frac {24875 x}{32}-\frac {18987 x^2}{32}-\frac {3321 x^3}{8}-\frac {3051 x^4}{16}-\frac {81 x^5}{2}-\frac {26411}{64} \log (1-2 x) \]

[Out]

-24875/32*x-18987/32*x^2-3321/8*x^3-3051/16*x^4-81/2*x^5-26411/64*ln(1-2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=-\frac {81 x^5}{2}-\frac {3051 x^4}{16}-\frac {3321 x^3}{8}-\frac {18987 x^2}{32}-\frac {24875 x}{32}-\frac {26411}{64} \log (1-2 x) \]

[In]

Int[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x),x]

[Out]

(-24875*x)/32 - (18987*x^2)/32 - (3321*x^3)/8 - (3051*x^4)/16 - (81*x^5)/2 - (26411*Log[1 - 2*x])/64

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {24875}{32}-\frac {18987 x}{16}-\frac {9963 x^2}{8}-\frac {3051 x^3}{4}-\frac {405 x^4}{2}-\frac {26411}{32 (-1+2 x)}\right ) \, dx \\ & = -\frac {24875 x}{32}-\frac {18987 x^2}{32}-\frac {3321 x^3}{8}-\frac {3051 x^4}{16}-\frac {81 x^5}{2}-\frac {26411}{64} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=\frac {1}{256} \left (154133-199000 x-151896 x^2-106272 x^3-48816 x^4-10368 x^5-105644 \log (1-2 x)\right ) \]

[In]

Integrate[((2 + 3*x)^4*(3 + 5*x))/(1 - 2*x),x]

[Out]

(154133 - 199000*x - 151896*x^2 - 106272*x^3 - 48816*x^4 - 10368*x^5 - 105644*Log[1 - 2*x])/256

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {81 x^{5}}{2}-\frac {3051 x^{4}}{16}-\frac {3321 x^{3}}{8}-\frac {18987 x^{2}}{32}-\frac {24875 x}{32}-\frac {26411 \ln \left (x -\frac {1}{2}\right )}{64}\)	\(31\)
default	\(-\frac {81 x^{5}}{2}-\frac {3051 x^{4}}{16}-\frac {3321 x^{3}}{8}-\frac {18987 x^{2}}{32}-\frac {24875 x}{32}-\frac {26411 \ln \left (-1+2 x \right )}{64}\)	\(33\)
norman	\(-\frac {81 x^{5}}{2}-\frac {3051 x^{4}}{16}-\frac {3321 x^{3}}{8}-\frac {18987 x^{2}}{32}-\frac {24875 x}{32}-\frac {26411 \ln \left (-1+2 x \right )}{64}\)	\(33\)
risch	\(-\frac {81 x^{5}}{2}-\frac {3051 x^{4}}{16}-\frac {3321 x^{3}}{8}-\frac {18987 x^{2}}{32}-\frac {24875 x}{32}-\frac {26411 \ln \left (-1+2 x \right )}{64}\)	\(33\)
meijerg	\(-\frac {26411 \ln \left (1-2 x \right )}{64}-184 x -47 x \left (6 x +6\right )-18 x \left (16 x^{2}+12 x +12\right )-\frac {441 x \left (120 x^{3}+80 x^{2}+60 x +60\right )}{320}-\frac {27 x \left (192 x^{4}+120 x^{3}+80 x^{2}+60 x +60\right )}{128}\)	\(75\)

[In]

int((2+3*x)^4*(3+5*x)/(1-2*x),x,method=_RETURNVERBOSE)

[Out]

-81/2*x^5-3051/16*x^4-3321/8*x^3-18987/32*x^2-24875/32*x-26411/64*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=-\frac {81}{2} \, x^{5} - \frac {3051}{16} \, x^{4} - \frac {3321}{8} \, x^{3} - \frac {18987}{32} \, x^{2} - \frac {24875}{32} \, x - \frac {26411}{64} \, \log \left (2 \, x - 1\right ) \]

[In]

integrate((2+3*x)^4*(3+5*x)/(1-2*x),x, algorithm="fricas")

[Out]

-81/2*x^5 - 3051/16*x^4 - 3321/8*x^3 - 18987/32*x^2 - 24875/32*x - 26411/64*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=- \frac {81 x^{5}}{2} - \frac {3051 x^{4}}{16} - \frac {3321 x^{3}}{8} - \frac {18987 x^{2}}{32} - \frac {24875 x}{32} - \frac {26411 \log {\left (2 x - 1 \right )}}{64} \]

[In]

integrate((2+3*x)**4*(3+5*x)/(1-2*x),x)

[Out]

-81*x**5/2 - 3051*x**4/16 - 3321*x**3/8 - 18987*x**2/32 - 24875*x/32 - 26411*log(2*x - 1)/64

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=-\frac {81}{2} \, x^{5} - \frac {3051}{16} \, x^{4} - \frac {3321}{8} \, x^{3} - \frac {18987}{32} \, x^{2} - \frac {24875}{32} \, x - \frac {26411}{64} \, \log \left (2 \, x - 1\right ) \]

[In]

integrate((2+3*x)^4*(3+5*x)/(1-2*x),x, algorithm="maxima")

[Out]

-81/2*x^5 - 3051/16*x^4 - 3321/8*x^3 - 18987/32*x^2 - 24875/32*x - 26411/64*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=-\frac {81}{2} \, x^{5} - \frac {3051}{16} \, x^{4} - \frac {3321}{8} \, x^{3} - \frac {18987}{32} \, x^{2} - \frac {24875}{32} \, x - \frac {26411}{64} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

integrate((2+3*x)^4*(3+5*x)/(1-2*x),x, algorithm="giac")

[Out]

-81/2*x^5 - 3051/16*x^4 - 3321/8*x^3 - 18987/32*x^2 - 24875/32*x - 26411/64*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^4 (3+5 x)}{1-2 x} \, dx=-\frac {24875\,x}{32}-\frac {26411\,\ln \left (x-\frac {1}{2}\right )}{64}-\frac {18987\,x^2}{32}-\frac {3321\,x^3}{8}-\frac {3051\,x^4}{16}-\frac {81\,x^5}{2} \]

[In]

int(-((3*x + 2)^4*(5*x + 3))/(2*x - 1),x)

[Out]

- (24875*x)/32 - (26411*log(x - 1/2))/64 - (18987*x^2)/32 - (3321*x^3)/8 - (3051*x^4)/16 - (81*x^5)/2