\(\int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx\) [1447]

Optimal result

Integrand size = 20, antiderivative size = 65 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=\frac {1}{84 (2+3 x)^4}-\frac {11}{147 (2+3 x)^3}-\frac {11}{343 (2+3 x)^2}-\frac {44}{2401 (2+3 x)}-\frac {88 \log (1-2 x)}{16807}+\frac {88 \log (2+3 x)}{16807} \]

[Out]

1/84/(2+3*x)^4-11/147/(2+3*x)^3-11/343/(2+3*x)^2-44/2401/(2+3*x)-88/16807*ln(1-2*x)+88/16807*ln(2+3*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, 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} \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=-\frac {44}{2401 (3 x+2)}-\frac {11}{343 (3 x+2)^2}-\frac {11}{147 (3 x+2)^3}+\frac {1}{84 (3 x+2)^4}-\frac {88 \log (1-2 x)}{16807}+\frac {88 \log (3 x+2)}{16807} \]

[In]

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

[Out]

1/(84*(2 + 3*x)^4) - 11/(147*(2 + 3*x)^3) - 11/(343*(2 + 3*x)^2) - 44/(2401*(2 + 3*x)) - (88*Log[1 - 2*x])/168
07 + (88*Log[2 + 3*x])/16807

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 {176}{16807 (-1+2 x)}-\frac {1}{7 (2+3 x)^5}+\frac {33}{49 (2+3 x)^4}+\frac {66}{343 (2+3 x)^3}+\frac {132}{2401 (2+3 x)^2}+\frac {264}{16807 (2+3 x)}\right ) \, dx \\ & = \frac {1}{84 (2+3 x)^4}-\frac {11}{147 (2+3 x)^3}-\frac {11}{343 (2+3 x)^2}-\frac {44}{2401 (2+3 x)}-\frac {88 \log (1-2 x)}{16807}+\frac {88 \log (2+3 x)}{16807} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=\frac {-\frac {7 \left (3963+12188 x+12276 x^2+4752 x^3\right )}{(2+3 x)^4}-352 \log (3-6 x)+352 \log (2+3 x)}{67228} \]

[In]

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

[Out]

((-7*(3963 + 12188*x + 12276*x^2 + 4752*x^3))/(2 + 3*x)^4 - 352*Log[3 - 6*x] + 352*Log[2 + 3*x])/67228

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.63

method	result	size
norman	\(\frac {-\frac {3069}{2401} x^{2}-\frac {3047}{2401} x -\frac {1188}{2401} x^{3}-\frac {3963}{9604}}{\left (2+3 x \right )^{4}}-\frac {88 \ln \left (-1+2 x \right )}{16807}+\frac {88 \ln \left (2+3 x \right )}{16807}\)	\(41\)
risch	\(\frac {-\frac {3069}{2401} x^{2}-\frac {3047}{2401} x -\frac {1188}{2401} x^{3}-\frac {3963}{9604}}{\left (2+3 x \right )^{4}}-\frac {88 \ln \left (-1+2 x \right )}{16807}+\frac {88 \ln \left (2+3 x \right )}{16807}\)	\(42\)
default	\(-\frac {88 \ln \left (-1+2 x \right )}{16807}+\frac {1}{84 \left (2+3 x \right )^{4}}-\frac {11}{147 \left (2+3 x \right )^{3}}-\frac {11}{343 \left (2+3 x \right )^{2}}-\frac {44}{2401 \left (2+3 x \right )}+\frac {88 \ln \left (2+3 x \right )}{16807}\)	\(54\)
parallelrisch	\(\frac {456192 \ln \left (\frac {2}{3}+x \right ) x^{4}-456192 \ln \left (x -\frac {1}{2}\right ) x^{4}+1216512 \ln \left (\frac {2}{3}+x \right ) x^{3}-1216512 \ln \left (x -\frac {1}{2}\right ) x^{3}+2247021 x^{4}+1216512 \ln \left (\frac {2}{3}+x \right ) x^{2}-1216512 \ln \left (x -\frac {1}{2}\right ) x^{2}+5459832 x^{3}+540672 \ln \left (\frac {2}{3}+x \right ) x -540672 \ln \left (x -\frac {1}{2}\right ) x +4617144 x^{2}+90112 \ln \left (\frac {2}{3}+x \right )-90112 \ln \left (x -\frac {1}{2}\right )+1298080 x}{1075648 \left (2+3 x \right )^{4}}\)	\(109\)

[In]

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

[Out]

(-3069/2401*x^2-3047/2401*x-1188/2401*x^3-3963/9604)/(2+3*x)^4-88/16807*ln(-1+2*x)+88/16807*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.46 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=-\frac {33264 \, x^{3} + 85932 \, x^{2} - 352 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 352 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (2 \, x - 1\right ) + 85316 \, x + 27741}{67228 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

-1/67228*(33264*x^3 + 85932*x^2 - 352*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 352*(81*x^4 + 21
6*x^3 + 216*x^2 + 96*x + 16)*log(2*x - 1) + 85316*x + 27741)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=- \frac {4752 x^{3} + 12276 x^{2} + 12188 x + 3963}{777924 x^{4} + 2074464 x^{3} + 2074464 x^{2} + 921984 x + 153664} - \frac {88 \log {\left (x - \frac {1}{2} \right )}}{16807} + \frac {88 \log {\left (x + \frac {2}{3} \right )}}{16807} \]

[In]

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

[Out]

-(4752*x**3 + 12276*x**2 + 12188*x + 3963)/(777924*x**4 + 2074464*x**3 + 2074464*x**2 + 921984*x + 153664) - 8
8*log(x - 1/2)/16807 + 88*log(x + 2/3)/16807

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=-\frac {4752 \, x^{3} + 12276 \, x^{2} + 12188 \, x + 3963}{9604 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} + \frac {88}{16807} \, \log \left (3 \, x + 2\right ) - \frac {88}{16807} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-1/9604*(4752*x^3 + 12276*x^2 + 12188*x + 3963)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) + 88/16807*log(3*x +
2) - 88/16807*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=-\frac {44}{2401 \, {\left (3 \, x + 2\right )}} - \frac {11}{343 \, {\left (3 \, x + 2\right )}^{2}} - \frac {11}{147 \, {\left (3 \, x + 2\right )}^{3}} + \frac {1}{84 \, {\left (3 \, x + 2\right )}^{4}} - \frac {88}{16807} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]

[In]

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

[Out]

-44/2401/(3*x + 2) - 11/343/(3*x + 2)^2 - 11/147/(3*x + 2)^3 + 1/84/(3*x + 2)^4 - 88/16807*log(abs(-7/(3*x + 2
) + 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \frac {3+5 x}{(1-2 x) (2+3 x)^5} \, dx=\frac {176\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{16807}-\frac {\frac {44\,x^3}{7203}+\frac {341\,x^2}{21609}+\frac {3047\,x}{194481}+\frac {1321}{259308}}{x^4+\frac {8\,x^3}{3}+\frac {8\,x^2}{3}+\frac {32\,x}{27}+\frac {16}{81}} \]

[In]

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

[Out]

(176*atanh((12*x)/7 + 1/7))/16807 - ((3047*x)/194481 + (341*x^2)/21609 + (44*x^3)/7203 + 1321/259308)/((32*x)/
27 + (8*x^2)/3 + (8*x^3)/3 + x^4 + 16/81)