\(\int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx\) [171]

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

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=\frac {f^a \Gamma \left (\frac {1}{3},-\frac {b \log (f)}{x^3}\right )}{3 x \sqrt [3]{-\frac {b \log (f)}{x^3}}} \]

[Out]

1/3*f^a*GAMMA(1/3,-b*ln(f)/x^3)/x/(-b*ln(f)/x^3)^(1/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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 = {2250} \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=\frac {f^a \Gamma \left (\frac {1}{3},-\frac {b \log (f)}{x^3}\right )}{3 x \sqrt [3]{-\frac {b \log (f)}{x^3}}} \]

[In]

Int[f^(a + b/x^3)/x^2,x]

[Out]

(f^a*Gamma[1/3, -((b*Log[f])/x^3)])/(3*x*(-((b*Log[f])/x^3))^(1/3))

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {f^a \Gamma \left (\frac {1}{3},-\frac {b \log (f)}{x^3}\right )}{3 x \sqrt [3]{-\frac {b \log (f)}{x^3}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=\frac {f^a \Gamma \left (\frac {1}{3},-\frac {b \log (f)}{x^3}\right )}{3 x \sqrt [3]{-\frac {b \log (f)}{x^3}}} \]

[In]

Integrate[f^(a + b/x^3)/x^2,x]

[Out]

(f^a*Gamma[1/3, -((b*Log[f])/x^3)])/(3*x*(-((b*Log[f])/x^3))^(1/3))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(81\) vs. \(2(28)=56\).

Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.41

method result size
meijerg \(-\frac {f^{a} \left (\frac {2 \left (-b \right )^{\frac {1}{3}} \ln \left (f \right )^{\frac {1}{3}} \pi \sqrt {3}}{3 x \Gamma \left (\frac {2}{3}\right ) \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )^{\frac {1}{3}}}-\frac {\left (-b \right )^{\frac {1}{3}} \ln \left (f \right )^{\frac {1}{3}} \Gamma \left (\frac {1}{3}, -\frac {b \ln \left (f \right )}{x^{3}}\right )}{x \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )^{\frac {1}{3}}}\right )}{3 \left (-b \right )^{\frac {1}{3}} \ln \left (f \right )^{\frac {1}{3}}}\) \(82\)

[In]

int(f^(a+b/x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*f^a/(-b)^(1/3)/ln(f)^(1/3)*(2/3/x*(-b)^(1/3)*ln(f)^(1/3)*Pi*3^(1/2)/GAMMA(2/3)/(-b*ln(f)/x^3)^(1/3)-1/x*(
-b)^(1/3)*ln(f)^(1/3)/(-b*ln(f)/x^3)^(1/3)*GAMMA(1/3,-b*ln(f)/x^3))

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=-\frac {\left (-b \log \left (f\right )\right )^{\frac {2}{3}} f^{a} \Gamma \left (\frac {1}{3}, -\frac {b \log \left (f\right )}{x^{3}}\right )}{3 \, b \log \left (f\right )} \]

[In]

integrate(f^(a+b/x^3)/x^2,x, algorithm="fricas")

[Out]

-1/3*(-b*log(f))^(2/3)*f^a*gamma(1/3, -b*log(f)/x^3)/(b*log(f))

Sympy [F]

\[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=\int \frac {f^{a + \frac {b}{x^{3}}}}{x^{2}}\, dx \]

[In]

integrate(f**(a+b/x**3)/x**2,x)

[Out]

Integral(f**(a + b/x**3)/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=\frac {f^{a} \Gamma \left (\frac {1}{3}, -\frac {b \log \left (f\right )}{x^{3}}\right )}{3 \, x \left (-\frac {b \log \left (f\right )}{x^{3}}\right )^{\frac {1}{3}}} \]

[In]

integrate(f^(a+b/x^3)/x^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=\int { \frac {f^{a + \frac {b}{x^{3}}}}{x^{2}} \,d x } \]

[In]

integrate(f^(a+b/x^3)/x^2,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)/x^2, x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^2} \, dx=-\frac {2\,\pi \,\sqrt {3}\,f^a-3\,f^a\,\Gamma \left (\frac {2}{3}\right )\,\Gamma \left (\frac {1}{3},-\frac {b\,\ln \left (f\right )}{x^3}\right )}{9\,x\,\Gamma \left (\frac {2}{3}\right )\,{\left (-\frac {b\,\ln \left (f\right )}{x^3}\right )}^{1/3}} \]

[In]

int(f^(a + b/x^3)/x^2,x)

[Out]

-(2*3^(1/2)*f^a*pi - 3*f^a*gamma(2/3)*igamma(1/3, -(b*log(f))/x^3))/(9*x*gamma(2/3)*(-(b*log(f))/x^3)^(1/3))