\(\int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx\) [161]

Optimal result

Integrand size = 13, antiderivative size = 20 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{a+\frac {b}{x^3}}}{3 b \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, 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 = {2240} \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{a+\frac {b}{x^3}}}{3 b \log (f)} \]

[In]

Int[f^(a + b/x^3)/x^4,x]

[Out]

-1/3*f^(a + b/x^3)/(b*Log[f])

Rule 2240

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

Rubi steps \begin{align*} \text {integral}& = -\frac {f^{a+\frac {b}{x^3}}}{3 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{a+\frac {b}{x^3}}}{3 b \log (f)} \]

[In]

Integrate[f^(a + b/x^3)/x^4,x]

[Out]

-1/3*f^(a + b/x^3)/(b*Log[f])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {f^{a +\frac {b}{x^{3}}}}{3 b \ln \left (f \right )}\)	\(19\)
default	\(-\frac {f^{a +\frac {b}{x^{3}}}}{3 b \ln \left (f \right )}\)	\(19\)
parallelrisch	\(-\frac {f^{a +\frac {b}{x^{3}}}}{3 b \ln \left (f \right )}\)	\(19\)
norman	\(-\frac {{\mathrm e}^{\left (a +\frac {b}{x^{3}}\right ) \ln \left (f \right )}}{3 b \ln \left (f \right )}\)	\(21\)
risch	\(-\frac {f^{\frac {a \,x^{3}+b}{x^{3}}}}{3 b \ln \left (f \right )}\)	\(23\)
meijerg	\(\frac {f^{a} \left (1-{\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}\right )}{3 b \ln \left (f \right )}\)	\(25\)

[In]

int(f^(a+b/x^3)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{\frac {a x^{3} + b}{x^{3}}}}{3 \, b \log \left (f\right )} \]

[In]

integrate(f^(a+b/x^3)/x^4,x, algorithm="fricas")

[Out]

-1/3*f^((a*x^3 + b)/x^3)/(b*log(f))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=\begin {cases} - \frac {f^{a + \frac {b}{x^{3}}}}{3 b \log {\left (f \right )}} & \text {for}\: b \log {\left (f \right )} \neq 0 \\- \frac {1}{3 x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(a+b/x**3)/x**4,x)

[Out]

Piecewise((-f**(a + b/x**3)/(3*b*log(f)), Ne(b*log(f), 0)), (-1/(3*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{a + \frac {b}{x^{3}}}}{3 \, b \log \left (f\right )} \]

[In]

integrate(f^(a+b/x^3)/x^4,x, algorithm="maxima")

[Out]

-1/3*f^(a + b/x^3)/(b*log(f))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{\frac {a x^{3} + b}{x^{3}}}}{3 \, b \log \left (f\right )} \]

[In]

integrate(f^(a+b/x^3)/x^4,x, algorithm="giac")

[Out]

-1/3*f^((a*x^3 + b)/x^3)/(b*log(f))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {f^{a+\frac {b}{x^3}}}{x^4} \, dx=-\frac {f^{a+\frac {b}{x^3}}}{3\,b\,\ln \left (f\right )} \]

[In]

int(f^(a + b/x^3)/x^4,x)

[Out]

-f^(a + b/x^3)/(3*b*log(f))