\(\int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx\) [1399]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 59 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {343}{108 (2+3 x)^4}+\frac {1421}{81 (2+3 x)^3}+\frac {7189}{54 (2+3 x)^2}+\frac {1331}{2+3 x}-6655 \log (2+3 x)+6655 \log (3+5 x) \]

[Out]

343/108/(2+3*x)^4+1421/81/(2+3*x)^3+7189/54/(2+3*x)^2+1331/(2+3*x)-6655*ln(2+3*x)+6655*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, 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} \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {1331}{3 x+2}+\frac {7189}{54 (3 x+2)^2}+\frac {1421}{81 (3 x+2)^3}+\frac {343}{108 (3 x+2)^4}-6655 \log (3 x+2)+6655 \log (5 x+3) \]

[In]

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

[Out]

343/(108*(2 + 3*x)^4) + 1421/(81*(2 + 3*x)^3) + 7189/(54*(2 + 3*x)^2) + 1331/(2 + 3*x) - 6655*Log[2 + 3*x] + 6
655*Log[3 + 5*x]

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 {343}{9 (2+3 x)^5}-\frac {1421}{9 (2+3 x)^4}-\frac {7189}{9 (2+3 x)^3}-\frac {3993}{(2+3 x)^2}-\frac {19965}{2+3 x}+\frac {33275}{3+5 x}\right ) \, dx \\ & = \frac {343}{108 (2+3 x)^4}+\frac {1421}{81 (2+3 x)^3}+\frac {7189}{54 (2+3 x)^2}+\frac {1331}{2+3 x}-6655 \log (2+3 x)+6655 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {3634885+16059444 x+23675382 x^2+11643588 x^3}{324 (2+3 x)^4}-6655 \log (5 (2+3 x))+6655 \log (3+5 x) \]

[In]

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

[Out]

(3634885 + 16059444*x + 23675382*x^2 + 11643588*x^3)/(324*(2 + 3*x)^4) - 6655*Log[5*(2 + 3*x)] + 6655*Log[3 +
5*x]

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69

method result size
norman \(\frac {35937 x^{3}+\frac {438433}{6} x^{2}+\frac {1338287}{27} x +\frac {3634885}{324}}{\left (2+3 x \right )^{4}}-6655 \ln \left (2+3 x \right )+6655 \ln \left (3+5 x \right )\) \(41\)
risch \(\frac {35937 x^{3}+\frac {438433}{6} x^{2}+\frac {1338287}{27} x +\frac {3634885}{324}}{\left (2+3 x \right )^{4}}-6655 \ln \left (2+3 x \right )+6655 \ln \left (3+5 x \right )\) \(42\)
default \(\frac {343}{108 \left (2+3 x \right )^{4}}+\frac {1421}{81 \left (2+3 x \right )^{3}}+\frac {7189}{54 \left (2+3 x \right )^{2}}+\frac {1331}{2+3 x}-6655 \ln \left (2+3 x \right )+6655 \ln \left (3+5 x \right )\) \(54\)
parallelrisch \(-\frac {103498560 \ln \left (\frac {2}{3}+x \right ) x^{4}-103498560 \ln \left (x +\frac {3}{5}\right ) x^{4}+275996160 \ln \left (\frac {2}{3}+x \right ) x^{3}-275996160 \ln \left (x +\frac {3}{5}\right ) x^{3}+10904655 x^{4}+275996160 \ln \left (\frac {2}{3}+x \right ) x^{2}-275996160 \ln \left (x +\frac {3}{5}\right ) x^{2}+22179176 x^{3}+122664960 \ln \left (\frac {2}{3}+x \right ) x -122664960 \ln \left (x +\frac {3}{5}\right ) x +15049224 x^{2}+20444160 \ln \left (\frac {2}{3}+x \right )-20444160 \ln \left (x +\frac {3}{5}\right )+3407328 x}{192 \left (2+3 x \right )^{4}}\) \(109\)

[In]

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

[Out]

(35937*x^3+438433/6*x^2+1338287/27*x+3634885/324)/(2+3*x)^4-6655*ln(2+3*x)+6655*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.61 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {11643588 \, x^{3} + 23675382 \, x^{2} + 2156220 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (5 \, x + 3\right ) - 2156220 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 16059444 \, x + 3634885}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

1/324*(11643588*x^3 + 23675382*x^2 + 2156220*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(5*x + 3) - 2156220*(
81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 16059444*x + 3634885)/(81*x^4 + 216*x^3 + 216*x^2 + 96*
x + 16)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=- \frac {- 11643588 x^{3} - 23675382 x^{2} - 16059444 x - 3634885}{26244 x^{4} + 69984 x^{3} + 69984 x^{2} + 31104 x + 5184} + 6655 \log {\left (x + \frac {3}{5} \right )} - 6655 \log {\left (x + \frac {2}{3} \right )} \]

[In]

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

[Out]

-(-11643588*x**3 - 23675382*x**2 - 16059444*x - 3634885)/(26244*x**4 + 69984*x**3 + 69984*x**2 + 31104*x + 518
4) + 6655*log(x + 3/5) - 6655*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {11643588 \, x^{3} + 23675382 \, x^{2} + 16059444 \, x + 3634885}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} + 6655 \, \log \left (5 \, x + 3\right ) - 6655 \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

1/324*(11643588*x^3 + 23675382*x^2 + 16059444*x + 3634885)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) + 6655*log
(5*x + 3) - 6655*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {1331}{3 \, x + 2} + \frac {7189}{54 \, {\left (3 \, x + 2\right )}^{2}} + \frac {1421}{81 \, {\left (3 \, x + 2\right )}^{3}} + \frac {343}{108 \, {\left (3 \, x + 2\right )}^{4}} + 6655 \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) \]

[In]

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

[Out]

1331/(3*x + 2) + 7189/54/(3*x + 2)^2 + 1421/81/(3*x + 2)^3 + 343/108/(3*x + 2)^4 + 6655*log(abs(-1/(3*x + 2) +
 5))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^3}{(2+3 x)^5 (3+5 x)} \, dx=\frac {\frac {1331\,x^3}{3}+\frac {438433\,x^2}{486}+\frac {1338287\,x}{2187}+\frac {3634885}{26244}}{x^4+\frac {8\,x^3}{3}+\frac {8\,x^2}{3}+\frac {32\,x}{27}+\frac {16}{81}}-13310\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

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

[Out]

((1338287*x)/2187 + (438433*x^2)/486 + (1331*x^3)/3 + 3634885/26244)/((32*x)/27 + (8*x^2)/3 + (8*x^3)/3 + x^4
+ 16/81) - 13310*atanh(30*x + 19)