\(\int f^{a+\frac {b}{x^3}} x^m \, dx\) [154]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int f^{a+\frac {b}{x^3}} x^m \, dx=\frac {1}{3} f^a x^{1+m} \Gamma \left (\frac {1}{3} (-1-m),-\frac {b \log (f)}{x^3}\right ) \left (-\frac {b \log (f)}{x^3}\right )^{\frac {1+m}{3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, 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} \[ \int f^{a+\frac {b}{x^3}} x^m \, dx=\frac {1}{3} f^a x^{m+1} \left (-\frac {b \log (f)}{x^3}\right )^{\frac {m+1}{3}} \Gamma \left (\frac {1}{3} (-m-1),-\frac {b \log (f)}{x^3}\right ) \]

[In]

Int[f^(a + b/x^3)*x^m,x]

[Out]

(f^a*x^(1 + m)*Gamma[(-1 - m)/3, -((b*Log[f])/x^3)]*(-((b*Log[f])/x^3))^((1 + m)/3))/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 {1}{3} f^a x^{1+m} \Gamma \left (\frac {1}{3} (-1-m),-\frac {b \log (f)}{x^3}\right ) \left (-\frac {b \log (f)}{x^3}\right )^{\frac {1+m}{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^3}} x^m \, dx=\frac {1}{3} f^a x^{1+m} \Gamma \left (\frac {1}{3} (-1-m),-\frac {b \log (f)}{x^3}\right ) \left (-\frac {b \log (f)}{x^3}\right )^{\frac {1+m}{3}} \]

[In]

Integrate[f^(a + b/x^3)*x^m,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(168\) vs. \(2(38)=76\).

Time = 0.32 (sec) , antiderivative size = 169, normalized size of antiderivative = 3.67

method	result	size
meijerg	\(-\frac {f^{a} \left (-b \right )^{\frac {1}{3}+\frac {m}{3}} \ln \left (f \right )^{\frac {1}{3}+\frac {m}{3}} \left (\frac {3 x^{-2+m} \left (-b \right )^{-\frac {1}{3}-\frac {m}{3}} \ln \left (f \right )^{\frac {2}{3}-\frac {m}{3}} b \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )^{-\frac {2}{3}+\frac {m}{3}} \Gamma \left (\frac {2}{3}-\frac {m}{3}\right )}{1+m}-\frac {3 x^{1+m} \left (-b \right )^{-\frac {1}{3}-\frac {m}{3}} \ln \left (f \right )^{-\frac {1}{3}-\frac {m}{3}} {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}}{1+m}-\frac {3 x^{-2+m} \left (-b \right )^{-\frac {1}{3}-\frac {m}{3}} \ln \left (f \right )^{\frac {2}{3}-\frac {m}{3}} b \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )^{-\frac {2}{3}+\frac {m}{3}} \Gamma \left (\frac {2}{3}-\frac {m}{3}, -\frac {b \ln \left (f \right )}{x^{3}}\right )}{1+m}\right )}{3}\)	\(169\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(f^((a*x^3 + b)/x^3)*x^m, x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b/x**3)*x**m, x)

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int f^{a+\frac {b}{x^3}} x^m \, dx=\frac {1}{3} \, f^{a} x^{m + 1} \left (-\frac {b \log \left (f\right )}{x^{3}}\right )^{\frac {1}{3} \, m + \frac {1}{3}} \Gamma \left (-\frac {1}{3} \, m - \frac {1}{3}, -\frac {b \log \left (f\right )}{x^{3}}\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(a + b/x^3)*x^m, x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17 \[ \int f^{a+\frac {b}{x^3}} x^m \, dx=\frac {f^a\,x^{m+1}\,{\mathrm {e}}^{\frac {b\,\ln \left (f\right )}{2\,x^3}}\,{\mathrm {M}}_{\frac {m}{6}+\frac {2}{3},-\frac {m}{6}-\frac {1}{6}}\left (\frac {b\,\ln \left (f\right )}{x^3}\right )\,{\left (\frac {b\,\ln \left (f\right )}{x^3}\right )}^{\frac {m}{6}-\frac {1}{3}}}{m+1} \]

[In]

int(f^(a + b/x^3)*x^m,x)

[Out]

(f^a*x^(m + 1)*exp((b*log(f))/(2*x^3))*whittakerM(m/6 + 2/3, - m/6 - 1/6, (b*log(f))/x^3)*((b*log(f))/x^3)^(m/
6 - 1/3))/(m + 1)