\(\int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx\) [1458]

Optimal result

Integrand size = 20, antiderivative size = 30 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=-\frac {703 x}{8}-\frac {355 x^2}{8}-\frac {25 x^3}{2}-\frac {847}{16} \log (1-2 x) \]

[Out]

-703/8*x-355/8*x^2-25/2*x^3-847/16*ln(1-2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, 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} \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=-\frac {25 x^3}{2}-\frac {355 x^2}{8}-\frac {703 x}{8}-\frac {847}{16} \log (1-2 x) \]

[In]

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

[Out]

(-703*x)/8 - (355*x^2)/8 - (25*x^3)/2 - (847*Log[1 - 2*x])/16

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 {703}{8}-\frac {355 x}{4}-\frac {75 x^2}{2}-\frac {847}{8 (-1+2 x)}\right ) \, dx \\ & = -\frac {703 x}{8}-\frac {355 x^2}{8}-\frac {25 x^3}{2}-\frac {847}{16} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=\frac {1}{32} \left (1811-2812 x-1420 x^2-400 x^3-1694 \log (1-2 x)\right ) \]

[In]

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

[Out]

(1811 - 2812*x - 1420*x^2 - 400*x^3 - 1694*Log[1 - 2*x])/32

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {25 x^{3}}{2}-\frac {355 x^{2}}{8}-\frac {703 x}{8}-\frac {847 \ln \left (x -\frac {1}{2}\right )}{16}\)	\(21\)
default	\(-\frac {25 x^{3}}{2}-\frac {355 x^{2}}{8}-\frac {703 x}{8}-\frac {847 \ln \left (-1+2 x \right )}{16}\)	\(23\)
norman	\(-\frac {25 x^{3}}{2}-\frac {355 x^{2}}{8}-\frac {703 x}{8}-\frac {847 \ln \left (-1+2 x \right )}{16}\)	\(23\)
risch	\(-\frac {25 x^{3}}{2}-\frac {355 x^{2}}{8}-\frac {703 x}{8}-\frac {847 \ln \left (-1+2 x \right )}{16}\)	\(23\)
meijerg	\(-\frac {847 \ln \left (1-2 x \right )}{16}-\frac {87 x}{2}-\frac {35 x \left (6 x +6\right )}{6}-\frac {25 x \left (16 x^{2}+12 x +12\right )}{32}\)	\(34\)

[In]

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

[Out]

-25/2*x^3-355/8*x^2-703/8*x-847/16*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=-\frac {25}{2} \, x^{3} - \frac {355}{8} \, x^{2} - \frac {703}{8} \, x - \frac {847}{16} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-25/2*x^3 - 355/8*x^2 - 703/8*x - 847/16*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=- \frac {25 x^{3}}{2} - \frac {355 x^{2}}{8} - \frac {703 x}{8} - \frac {847 \log {\left (2 x - 1 \right )}}{16} \]

[In]

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

[Out]

-25*x**3/2 - 355*x**2/8 - 703*x/8 - 847*log(2*x - 1)/16

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=-\frac {25}{2} \, x^{3} - \frac {355}{8} \, x^{2} - \frac {703}{8} \, x - \frac {847}{16} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-25/2*x^3 - 355/8*x^2 - 703/8*x - 847/16*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=-\frac {25}{2} \, x^{3} - \frac {355}{8} \, x^{2} - \frac {703}{8} \, x - \frac {847}{16} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-25/2*x^3 - 355/8*x^2 - 703/8*x - 847/16*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x) (3+5 x)^2}{1-2 x} \, dx=-\frac {703\,x}{8}-\frac {847\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {355\,x^2}{8}-\frac {25\,x^3}{2} \]

[In]

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

[Out]

- (703*x)/8 - (847*log(x - 1/2))/16 - (355*x^2)/8 - (25*x^3)/2