\(\int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx\) [1302]

Optimal result

Integrand size = 22, antiderivative size = 70 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {49}{45 (2+3 x)^5}+\frac {217}{36 (2+3 x)^4}+\frac {121}{3 (2+3 x)^3}+\frac {605}{2 (2+3 x)^2}+\frac {3025}{2+3 x}-15125 \log (2+3 x)+15125 \log (3+5 x) \]

[Out]

49/45/(2+3*x)^5+217/36/(2+3*x)^4+121/3/(2+3*x)^3+605/2/(2+3*x)^2+3025/(2+3*x)-15125*ln(2+3*x)+15125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {3025}{3 x+2}+\frac {605}{2 (3 x+2)^2}+\frac {121}{3 (3 x+2)^3}+\frac {217}{36 (3 x+2)^4}+\frac {49}{45 (3 x+2)^5}-15125 \log (3 x+2)+15125 \log (5 x+3) \]

[In]

Int[(1 - 2*x)^2/((2 + 3*x)^6*(3 + 5*x)),x]

[Out]

49/(45*(2 + 3*x)^5) + 217/(36*(2 + 3*x)^4) + 121/(3*(2 + 3*x)^3) + 605/(2*(2 + 3*x)^2) + 3025/(2 + 3*x) - 1512
5*Log[2 + 3*x] + 15125*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 {49}{3 (2+3 x)^6}-\frac {217}{3 (2+3 x)^5}-\frac {363}{(2+3 x)^4}-\frac {1815}{(2+3 x)^3}-\frac {9075}{(2+3 x)^2}-\frac {45375}{2+3 x}+\frac {75625}{3+5 x}\right ) \, dx \\ & = \frac {49}{45 (2+3 x)^5}+\frac {217}{36 (2+3 x)^4}+\frac {121}{3 (2+3 x)^3}+\frac {605}{2 (2+3 x)^2}+\frac {3025}{2+3 x}-15125 \log (2+3 x)+15125 \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=-15125 \log (5 (2+3 x))+\frac {9179006+54322575 x+120617640 x^2+119082150 x^3+44104500 x^4+2722500 (2+3 x)^5 \log (3+5 x)}{180 (2+3 x)^5} \]

[In]

Integrate[(1 - 2*x)^2/((2 + 3*x)^6*(3 + 5*x)),x]

[Out]

-15125*Log[5*(2 + 3*x)] + (9179006 + 54322575*x + 120617640*x^2 + 119082150*x^3 + 44104500*x^4 + 2722500*(2 +
3*x)^5*Log[3 + 5*x])/(180*(2 + 3*x)^5)

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.66

method	result	size
norman	\(\frac {245025 x^{4}+670098 x^{2}+\frac {1323135}{2} x^{3}+\frac {3621505}{12} x +\frac {4589503}{90}}{\left (2+3 x \right )^{5}}-15125 \ln \left (2+3 x \right )+15125 \ln \left (3+5 x \right )\)	\(46\)
risch	\(\frac {245025 x^{4}+670098 x^{2}+\frac {1323135}{2} x^{3}+\frac {3621505}{12} x +\frac {4589503}{90}}{\left (2+3 x \right )^{5}}-15125 \ln \left (2+3 x \right )+15125 \ln \left (3+5 x \right )\)	\(47\)
default	\(\frac {49}{45 \left (2+3 x \right )^{5}}+\frac {217}{36 \left (2+3 x \right )^{4}}+\frac {121}{3 \left (2+3 x \right )^{3}}+\frac {605}{2 \left (2+3 x \right )^{2}}+\frac {3025}{2+3 x}-15125 \ln \left (2+3 x \right )+15125 \ln \left (3+5 x \right )\)	\(63\)
parallelrisch	\(-\frac {1176120000 \ln \left (\frac {2}{3}+x \right ) x^{5}-1176120000 \ln \left (x +\frac {3}{5}\right ) x^{5}[phone] \ln \left (\frac {2}{3}+x \right ) x^{4}-3920400000 \ln \left (x +\frac {3}{5}\right ) x^{4}[phone] x^{5}[phone] \ln \left (\frac {2}{3}+x \right ) x^{3}-5227200000 \ln \left (x +\frac {3}{5}\right ) x^{3}[phone] x^{4}[phone] \ln \left (\frac {2}{3}+x \right ) x^{2}-3484800000 \ln \left (x +\frac {3}{5}\right ) x^{2}[phone] x^{3}[phone] \ln \left (\frac {2}{3}+x \right ) x -1161600000 \ln \left (x +\frac {3}{5}\right ) x [phone] x^{2}[phone] \ln \left (\frac {2}{3}+x \right )-154880000 \ln \left (x +\frac {3}{5}\right )[phone] x}{320 \left (2+3 x \right )^{5}}\)	\(132\)

[In]

int((1-2*x)^2/(2+3*x)^6/(3+5*x),x,method=_RETURNVERBOSE)

[Out]

(245025*x^4+670098*x^2+1323135/2*x^3+3621505/12*x+4589503/90)/(2+3*x)^5-15125*ln(2+3*x)+15125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.64 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {44104500 \, x^{4} + 119082150 \, x^{3} + 120617640 \, x^{2} + 2722500 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )} \log \left (5 \, x + 3\right ) - 2722500 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )} \log \left (3 \, x + 2\right ) + 54322575 \, x + 9179006}{180 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

integrate((1-2*x)^2/(2+3*x)^6/(3+5*x),x, algorithm="fricas")

[Out]

1/180*(44104500*x^4 + 119082150*x^3 + 120617640*x^2 + 2722500*(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x
+ 32)*log(5*x + 3) - 2722500*(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x + 32)*log(3*x + 2) + 54322575*x +
9179006)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x + 32)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {44104500 x^{4} + 119082150 x^{3} + 120617640 x^{2} + 54322575 x + 9179006}{43740 x^{5} + 145800 x^{4} + 194400 x^{3} + 129600 x^{2} + 43200 x + 5760} + 15125 \log {\left (x + \frac {3}{5} \right )} - 15125 \log {\left (x + \frac {2}{3} \right )} \]

[In]

integrate((1-2*x)**2/(2+3*x)**6/(3+5*x),x)

[Out]

(44104500*x**4 + 119082150*x**3 + 120617640*x**2 + 54322575*x + 9179006)/(43740*x**5 + 145800*x**4 + 194400*x*
*3 + 129600*x**2 + 43200*x + 5760) + 15125*log(x + 3/5) - 15125*log(x + 2/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {44104500 \, x^{4} + 119082150 \, x^{3} + 120617640 \, x^{2} + 54322575 \, x + 9179006}{180 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} + 15125 \, \log \left (5 \, x + 3\right ) - 15125 \, \log \left (3 \, x + 2\right ) \]

[In]

integrate((1-2*x)^2/(2+3*x)^6/(3+5*x),x, algorithm="maxima")

[Out]

1/180*(44104500*x^4 + 119082150*x^3 + 120617640*x^2 + 54322575*x + 9179006)/(243*x^5 + 810*x^4 + 1080*x^3 + 72
0*x^2 + 240*x + 32) + 15125*log(5*x + 3) - 15125*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {44104500 \, x^{4} + 119082150 \, x^{3} + 120617640 \, x^{2} + 54322575 \, x + 9179006}{180 \, {\left (3 \, x + 2\right )}^{5}} + 15125 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 15125 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

integrate((1-2*x)^2/(2+3*x)^6/(3+5*x),x, algorithm="giac")

[Out]

1/180*(44104500*x^4 + 119082150*x^3 + 120617640*x^2 + 54322575*x + 9179006)/(3*x + 2)^5 + 15125*log(abs(5*x +
3)) - 15125*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2}{(2+3 x)^6 (3+5 x)} \, dx=\frac {\frac {3025\,x^4}{3}+\frac {5445\,x^3}{2}+\frac {223366\,x^2}{81}+\frac {3621505\,x}{2916}+\frac {4589503}{21870}}{x^5+\frac {10\,x^4}{3}+\frac {40\,x^3}{9}+\frac {80\,x^2}{27}+\frac {80\,x}{81}+\frac {32}{243}}-30250\,\mathrm {atanh}\left (30\,x+19\right ) \]

[In]

int((2*x - 1)^2/((3*x + 2)^6*(5*x + 3)),x)

[Out]

((3621505*x)/2916 + (223366*x^2)/81 + (5445*x^3)/2 + (3025*x^4)/3 + 4589503/21870)/((80*x)/81 + (80*x^2)/27 +
(40*x^3)/9 + (10*x^4)/3 + x^5 + 32/243) - 30250*atanh(30*x + 19)