\(\int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx\) [166]

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

Optimal result

Integrand size = 13, antiderivative size = 82 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\frac {f^{a+\frac {b}{x^3}} \left (120 x^{15}-120 b x^{12} \log (f)+60 b^2 x^9 \log ^2(f)-20 b^3 x^6 \log ^3(f)+5 b^4 x^3 \log ^4(f)-b^5 \log ^5(f)\right )}{3 b^6 x^{15} \log ^6(f)} \]

[Out]

1/3*f^(a+b/x^3)*(120*x^15-120*b*x^12*ln(f)+60*b^2*x^9*ln(f)^2-20*b^3*x^6*ln(f)^3+5*b^4*x^3*ln(f)^4-b^5*ln(f)^5
)/b^6/x^15/ln(f)^6

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2249} \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\frac {f^{a+\frac {b}{x^3}} \left (-b^5 \log ^5(f)+5 b^4 x^3 \log ^4(f)-20 b^3 x^6 \log ^3(f)+60 b^2 x^9 \log ^2(f)-120 b x^{12} \log (f)+120 x^{15}\right )}{3 b^6 x^{15} \log ^6(f)} \]

[In]

Int[f^(a + b/x^3)/x^19,x]

[Out]

(f^(a + b/x^3)*(120*x^15 - 120*b*x^12*Log[f] + 60*b^2*x^9*Log[f]^2 - 20*b^3*x^6*Log[f]^3 + 5*b^4*x^3*Log[f]^4
- b^5*Log[f]^5))/(3*b^6*x^15*Log[f]^6)

Rule 2249

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{p = Simplify
[(m + 1)/n]}, Simp[(-F^a)*((f/d)^m/(d*n*((-b)*Log[F])^p))*Simplify[FunctionExpand[Gamma[p, (-b)*(c + d*x)^n*Lo
g[F]]]], x] /; IGtQ[p, 0]] /; FreeQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+\frac {b}{x^3}} \left (120 x^{15}-120 b x^{12} \log (f)+60 b^2 x^9 \log ^2(f)-20 b^3 x^6 \log ^3(f)+5 b^4 x^3 \log ^4(f)-b^5 \log ^5(f)\right )}{3 b^6 x^{15} \log ^6(f)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.29 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\frac {f^a \Gamma \left (6,-\frac {b \log (f)}{x^3}\right )}{3 b^6 \log ^6(f)} \]

[In]

Integrate[f^(a + b/x^3)/x^19,x]

[Out]

(f^a*Gamma[6, -((b*Log[f])/x^3)])/(3*b^6*Log[f]^6)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01

method result size
meijerg \(-\frac {f^{a} \left (120-\frac {\left (-\frac {6 b^{5} \ln \left (f \right )^{5}}{x^{15}}+\frac {30 b^{4} \ln \left (f \right )^{4}}{x^{12}}-\frac {120 b^{3} \ln \left (f \right )^{3}}{x^{9}}+\frac {360 b^{2} \ln \left (f \right )^{2}}{x^{6}}-\frac {720 b \ln \left (f \right )}{x^{3}}+720\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}}{6}\right )}{3 b^{6} \ln \left (f \right )^{6}}\) \(83\)
risch \(-\frac {\left (-120 x^{15}+120 b \,x^{12} \ln \left (f \right )-60 b^{2} x^{9} \ln \left (f \right )^{2}+20 b^{3} x^{6} \ln \left (f \right )^{3}-5 b^{4} x^{3} \ln \left (f \right )^{4}+b^{5} \ln \left (f \right )^{5}\right ) f^{\frac {a \,x^{3}+b}{x^{3}}}}{3 \ln \left (f \right )^{6} b^{6} x^{15}}\) \(84\)
parallelrisch \(\frac {120 f^{a +\frac {b}{x^{3}}} x^{15}-120 f^{a +\frac {b}{x^{3}}} x^{12} b \ln \left (f \right )+60 f^{a +\frac {b}{x^{3}}} x^{9} b^{2} \ln \left (f \right )^{2}-20 f^{a +\frac {b}{x^{3}}} x^{6} b^{3} \ln \left (f \right )^{3}+5 f^{a +\frac {b}{x^{3}}} x^{3} b^{4} \ln \left (f \right )^{4}-f^{a +\frac {b}{x^{3}}} b^{5} \ln \left (f \right )^{5}}{3 x^{15} b^{6} \ln \left (f \right )^{6}}\) \(126\)

[In]

int(f^(a+b/x^3)/x^19,x,method=_RETURNVERBOSE)

[Out]

-1/3*f^a/b^6/ln(f)^6*(120-1/6*(-6*b^5*ln(f)^5/x^15+30*b^4*ln(f)^4/x^12-120*b^3*ln(f)^3/x^9+360*b^2*ln(f)^2/x^6
-720*b*ln(f)/x^3+720)*exp(b*ln(f)/x^3))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\frac {{\left (120 \, x^{15} - 120 \, b x^{12} \log \left (f\right ) + 60 \, b^{2} x^{9} \log \left (f\right )^{2} - 20 \, b^{3} x^{6} \log \left (f\right )^{3} + 5 \, b^{4} x^{3} \log \left (f\right )^{4} - b^{5} \log \left (f\right )^{5}\right )} f^{\frac {a x^{3} + b}{x^{3}}}}{3 \, b^{6} x^{15} \log \left (f\right )^{6}} \]

[In]

integrate(f^(a+b/x^3)/x^19,x, algorithm="fricas")

[Out]

1/3*(120*x^15 - 120*b*x^12*log(f) + 60*b^2*x^9*log(f)^2 - 20*b^3*x^6*log(f)^3 + 5*b^4*x^3*log(f)^4 - b^5*log(f
)^5)*f^((a*x^3 + b)/x^3)/(b^6*x^15*log(f)^6)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\frac {f^{a + \frac {b}{x^{3}}} \left (- b^{5} \log {\left (f \right )}^{5} + 5 b^{4} x^{3} \log {\left (f \right )}^{4} - 20 b^{3} x^{6} \log {\left (f \right )}^{3} + 60 b^{2} x^{9} \log {\left (f \right )}^{2} - 120 b x^{12} \log {\left (f \right )} + 120 x^{15}\right )}{3 b^{6} x^{15} \log {\left (f \right )}^{6}} \]

[In]

integrate(f**(a+b/x**3)/x**19,x)

[Out]

f**(a + b/x**3)*(-b**5*log(f)**5 + 5*b**4*x**3*log(f)**4 - 20*b**3*x**6*log(f)**3 + 60*b**2*x**9*log(f)**2 - 1
20*b*x**12*log(f) + 120*x**15)/(3*b**6*x**15*log(f)**6)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.27 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\frac {f^{a} \Gamma \left (6, -\frac {b \log \left (f\right )}{x^{3}}\right )}{3 \, b^{6} \log \left (f\right )^{6}} \]

[In]

integrate(f^(a+b/x^3)/x^19,x, algorithm="maxima")

[Out]

1/3*f^a*gamma(6, -b*log(f)/x^3)/(b^6*log(f)^6)

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=\int { \frac {f^{a + \frac {b}{x^{3}}}}{x^{19}} \,d x } \]

[In]

integrate(f^(a+b/x^3)/x^19,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)/x^19, x)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^{19}} \, dx=-\frac {f^{a+\frac {b}{x^3}}\,\left (\frac {1}{3\,b\,\ln \left (f\right )}-\frac {5\,x^3}{3\,b^2\,{\ln \left (f\right )}^2}+\frac {20\,x^6}{3\,b^3\,{\ln \left (f\right )}^3}-\frac {20\,x^9}{b^4\,{\ln \left (f\right )}^4}+\frac {40\,x^{12}}{b^5\,{\ln \left (f\right )}^5}-\frac {40\,x^{15}}{b^6\,{\ln \left (f\right )}^6}\right )}{x^{15}} \]

[In]

int(f^(a + b/x^3)/x^19,x)

[Out]

-(f^(a + b/x^3)*(1/(3*b*log(f)) - (5*x^3)/(3*b^2*log(f)^2) + (20*x^6)/(3*b^3*log(f)^3) - (20*x^9)/(b^4*log(f)^
4) + (40*x^12)/(b^5*log(f)^5) - (40*x^15)/(b^6*log(f)^6)))/x^15