\(\int \frac {1}{(2+x)^3 (3+x)^4} \, dx\) [187]

Optimal result

Integrand size = 11, antiderivative size = 54 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=-\frac {1}{2 (2+x)^2}+\frac {4}{2+x}+\frac {1}{3 (3+x)^3}+\frac {3}{2 (3+x)^2}+\frac {6}{3+x}+10 \log (2+x)-10 \log (3+x) \]

[Out]

-1/2/(2+x)^2+4/(2+x)+1/3/(3+x)^3+3/2/(3+x)^2+6/(3+x)+10*ln(2+x)-10*ln(3+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=\frac {4}{x+2}+\frac {6}{x+3}-\frac {1}{2 (x+2)^2}+\frac {3}{2 (x+3)^2}+\frac {1}{3 (x+3)^3}+10 \log (x+2)-10 \log (x+3) \]

[In]

Int[1/((2 + x)^3*(3 + x)^4),x]

[Out]

-1/2*1/(2 + x)^2 + 4/(2 + x) + 1/(3*(3 + x)^3) + 3/(2*(3 + x)^2) + 6/(3 + x) + 10*Log[2 + x] - 10*Log[3 + x]

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{(2+x)^3}-\frac {4}{(2+x)^2}+\frac {10}{2+x}-\frac {1}{(3+x)^4}-\frac {3}{(3+x)^3}-\frac {6}{(3+x)^2}-\frac {10}{3+x}\right ) \, dx \\ & = -\frac {1}{2 (2+x)^2}+\frac {4}{2+x}+\frac {1}{3 (3+x)^3}+\frac {3}{2 (3+x)^2}+\frac {6}{3+x}+10 \log (2+x)-10 \log (3+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=-\frac {1}{2 (2+x)^2}+\frac {4}{2+x}+\frac {1}{3 (3+x)^3}+\frac {3}{2 (3+x)^2}+\frac {6}{3+x}+10 \log (2+x)-10 \log (3+x) \]

[In]

Integrate[1/((2 + x)^3*(3 + x)^4),x]

[Out]

-1/2*1/(2 + x)^2 + 4/(2 + x) + 1/(3*(3 + x)^3) + 3/(2*(3 + x)^2) + 6/(3 + x) + 10*Log[2 + x] - 10*Log[3 + x]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
norman	\(\frac {10 x^{4}+105 x^{3}+\frac {1225}{3} x^{2}+\frac {4175}{6} x +\frac {2627}{6}}{\left (2+x \right )^{2} \left (3+x \right )^{3}}+10 \ln \left (2+x \right )-10 \ln \left (3+x \right )\)	\(45\)
risch	\(\frac {10 x^{4}+105 x^{3}+\frac {1225}{3} x^{2}+\frac {4175}{6} x +\frac {2627}{6}}{\left (2+x \right )^{2} \left (3+x \right )^{3}}+10 \ln \left (2+x \right )-10 \ln \left (3+x \right )\)	\(45\)
default	\(-\frac {1}{2 \left (2+x \right )^{2}}+\frac {4}{2+x}+\frac {1}{3 \left (3+x \right )^{3}}+\frac {3}{2 \left (3+x \right )^{2}}+\frac {6}{3+x}+10 \ln \left (2+x \right )-10 \ln \left (3+x \right )\)	\(49\)
parallelrisch	\(\frac {60 \ln \left (2+x \right ) x^{5}-60 \ln \left (3+x \right ) x^{5}+2627+780 \ln \left (2+x \right ) x^{4}-780 \ln \left (3+x \right ) x^{4}+4020 \ln \left (2+x \right ) x^{3}-4020 \ln \left (3+x \right ) x^{3}+60 x^{4}+10260 \ln \left (2+x \right ) x^{2}-10260 \ln \left (3+x \right ) x^{2}+630 x^{3}+12960 \ln \left (2+x \right ) x -12960 \ln \left (3+x \right ) x +2450 x^{2}+6480 \ln \left (2+x \right )-6480 \ln \left (3+x \right )+4175 x}{6 \left (2+x \right )^{2} \left (3+x \right )^{3}}\)	\(131\)

[In]

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

[Out]

(10*x^4+105*x^3+1225/3*x^2+4175/6*x+2627/6)/(2+x)^2/(3+x)^3+10*ln(2+x)-10*ln(3+x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (48) = 96\).

Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.94 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=\frac {60 \, x^{4} + 630 \, x^{3} + 2450 \, x^{2} - 60 \, {\left (x^{5} + 13 \, x^{4} + 67 \, x^{3} + 171 \, x^{2} + 216 \, x + 108\right )} \log \left (x + 3\right ) + 60 \, {\left (x^{5} + 13 \, x^{4} + 67 \, x^{3} + 171 \, x^{2} + 216 \, x + 108\right )} \log \left (x + 2\right ) + 4175 \, x + 2627}{6 \, {\left (x^{5} + 13 \, x^{4} + 67 \, x^{3} + 171 \, x^{2} + 216 \, x + 108\right )}} \]

[In]

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

[Out]

1/6*(60*x^4 + 630*x^3 + 2450*x^2 - 60*(x^5 + 13*x^4 + 67*x^3 + 171*x^2 + 216*x + 108)*log(x + 3) + 60*(x^5 + 1
3*x^4 + 67*x^3 + 171*x^2 + 216*x + 108)*log(x + 2) + 4175*x + 2627)/(x^5 + 13*x^4 + 67*x^3 + 171*x^2 + 216*x +
108)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=\frac {60 x^{4} + 630 x^{3} + 2450 x^{2} + 4175 x + 2627}{6 x^{5} + 78 x^{4} + 402 x^{3} + 1026 x^{2} + 1296 x + 648} + 10 \log {\left (x + 2 \right )} - 10 \log {\left (x + 3 \right )} \]

[In]

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

[Out]

(60*x**4 + 630*x**3 + 2450*x**2 + 4175*x + 2627)/(6*x**5 + 78*x**4 + 402*x**3 + 1026*x**2 + 1296*x + 648) + 10
*log(x + 2) - 10*log(x + 3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=\frac {60 \, x^{4} + 630 \, x^{3} + 2450 \, x^{2} + 4175 \, x + 2627}{6 \, {\left (x^{5} + 13 \, x^{4} + 67 \, x^{3} + 171 \, x^{2} + 216 \, x + 108\right )}} - 10 \, \log \left (x + 3\right ) + 10 \, \log \left (x + 2\right ) \]

[In]

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

[Out]

1/6*(60*x^4 + 630*x^3 + 2450*x^2 + 4175*x + 2627)/(x^5 + 13*x^4 + 67*x^3 + 171*x^2 + 216*x + 108) - 10*log(x +
3) + 10*log(x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=\frac {60 \, x^{4} + 630 \, x^{3} + 2450 \, x^{2} + 4175 \, x + 2627}{6 \, {\left (x + 3\right )}^{3} {\left (x + 2\right )}^{2}} - 10 \, \log \left ({\left | x + 3 \right |}\right ) + 10 \, \log \left ({\left | x + 2 \right |}\right ) \]

[In]

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

[Out]

1/6*(60*x^4 + 630*x^3 + 2450*x^2 + 4175*x + 2627)/((x + 3)^3*(x + 2)^2) - 10*log(abs(x + 3)) + 10*log(abs(x +
2))

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(2+x)^3 (3+x)^4} \, dx=\frac {10\,x^4+105\,x^3+\frac {1225\,x^2}{3}+\frac {4175\,x}{6}+\frac {2627}{6}}{x^5+13\,x^4+67\,x^3+171\,x^2+216\,x+108}-20\,\mathrm {atanh}\left (2\,x+5\right ) \]

[In]

int(1/((x + 2)^3*(x + 3)^4),x)

[Out]

((4175*x)/6 + (1225*x^2)/3 + 105*x^3 + 10*x^4 + 2627/6)/(216*x + 171*x^2 + 67*x^3 + 13*x^4 + x^5 + 108) - 20*a
tanh(2*x + 5)