\(\int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx\) [125]

Optimal result

Integrand size = 13, antiderivative size = 82 \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=\frac {6 f^{a+\frac {b}{x}}}{b^4 \log ^4(f)}-\frac {6 f^{a+\frac {b}{x}}}{b^3 x \log ^3(f)}+\frac {3 f^{a+\frac {b}{x}}}{b^2 x^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x}}}{b x^3 \log (f)} \]

[Out]

6*f^(a+b/x)/b^4/ln(f)^4-6*f^(a+b/x)/b^3/x/ln(f)^3+3*f^(a+b/x)/b^2/x^2/ln(f)^2-f^(a+b/x)/b/x^3/ln(f)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=\frac {6 f^{a+\frac {b}{x}}}{b^4 \log ^4(f)}-\frac {6 f^{a+\frac {b}{x}}}{b^3 x \log ^3(f)}+\frac {3 f^{a+\frac {b}{x}}}{b^2 x^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x}}}{b x^3 \log (f)} \]

[In]

Int[f^(a + b/x)/x^5,x]

[Out]

(6*f^(a + b/x))/(b^4*Log[f]^4) - (6*f^(a + b/x))/(b^3*x*Log[f]^3) + (3*f^(a + b/x))/(b^2*x^2*Log[f]^2) - f^(a
+ b/x)/(b*x^3*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]

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=\frac {f^{a+\frac {b}{x}} \left (6 x^3-6 b x^2 \log (f)+3 b^2 x \log ^2(f)-b^3 \log ^3(f)\right )}{b^4 x^3 \log ^4(f)} \]

[In]

Integrate[f^(a + b/x)/x^5,x]

[Out]

(f^(a + b/x)*(6*x^3 - 6*b*x^2*Log[f] + 3*b^2*x*Log[f]^2 - b^3*Log[f]^3))/(b^4*x^3*Log[f]^4)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {\left (\ln \left (f \right )^{3} b^{3}-3 \ln \left (f \right )^{2} b^{2} x +6 b \,x^{2} \ln \left (f \right )-6 x^{3}\right ) f^{\frac {a x +b}{x}}}{\ln \left (f \right )^{4} b^{4} x^{3}}\)	\(56\)
meijerg	\(-\frac {f^{a} \left (6-\frac {\left (-\frac {4 b^{3} \ln \left (f \right )^{3}}{x^{3}}+\frac {12 b^{2} \ln \left (f \right )^{2}}{x^{2}}-\frac {24 b \ln \left (f \right )}{x}+24\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x}}}{4}\right )}{b^{4} \ln \left (f \right )^{4}}\)	\(59\)
parallelrisch	\(\frac {-\ln \left (f \right )^{3} f^{a +\frac {b}{x}} b^{3}+3 b^{2} f^{a +\frac {b}{x}} x \ln \left (f \right )^{2}-6 b \,f^{a +\frac {b}{x}} x^{2} \ln \left (f \right )+6 f^{a +\frac {b}{x}} x^{3}}{x^{3} \ln \left (f \right )^{4} b^{4}}\)	\(81\)
norman	\(\frac {-\frac {x \,{\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \left (f \right )}}{b \ln \left (f \right )}+\frac {3 x^{2} {\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \left (f \right )}}{b^{2} \ln \left (f \right )^{2}}-\frac {6 x^{3} {\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \left (f \right )}}{b^{3} \ln \left (f \right )^{3}}+\frac {6 x^{4} {\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \left (f \right )}}{b^{4} \ln \left (f \right )^{4}}}{x^{4}}\)	\(96\)

[In]

int(f^(a+b/x)/x^5,x,method=_RETURNVERBOSE)

[Out]

-(ln(f)^3*b^3-3*ln(f)^2*b^2*x+6*b*x^2*ln(f)-6*x^3)/ln(f)^4/b^4/x^3*f^((a*x+b)/x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.67 \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=-\frac {{\left (b^{3} \log \left (f\right )^{3} - 3 \, b^{2} x \log \left (f\right )^{2} + 6 \, b x^{2} \log \left (f\right ) - 6 \, x^{3}\right )} f^{\frac {a x + b}{x}}}{b^{4} x^{3} \log \left (f\right )^{4}} \]

[In]

integrate(f^(a+b/x)/x^5,x, algorithm="fricas")

[Out]

-(b^3*log(f)^3 - 3*b^2*x*log(f)^2 + 6*b*x^2*log(f) - 6*x^3)*f^((a*x + b)/x)/(b^4*x^3*log(f)^4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=\frac {f^{a + \frac {b}{x}} \left (- b^{3} \log {\left (f \right )}^{3} + 3 b^{2} x \log {\left (f \right )}^{2} - 6 b x^{2} \log {\left (f \right )} + 6 x^{3}\right )}{b^{4} x^{3} \log {\left (f \right )}^{4}} \]

[In]

integrate(f**(a+b/x)/x**5,x)

[Out]

f**(a + b/x)*(-b**3*log(f)**3 + 3*b**2*x*log(f)**2 - 6*b*x**2*log(f) + 6*x**3)/(b**4*x**3*log(f)**4)

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.26 \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=\frac {f^{a} \Gamma \left (4, -\frac {b \log \left (f\right )}{x}\right )}{b^{4} \log \left (f\right )^{4}} \]

[In]

integrate(f^(a+b/x)/x^5,x, algorithm="maxima")

[Out]

f^a*gamma(4, -b*log(f)/x)/(b^4*log(f)^4)

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=\int { \frac {f^{a + \frac {b}{x}}}{x^{5}} \,d x } \]

[In]

integrate(f^(a+b/x)/x^5,x, algorithm="giac")

[Out]

integrate(f^(a + b/x)/x^5, x)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \frac {f^{a+\frac {b}{x}}}{x^5} \, dx=-\frac {f^{a+\frac {b}{x}}\,\left (\frac {1}{b\,\ln \left (f\right )}+\frac {6\,x^2}{b^3\,{\ln \left (f\right )}^3}-\frac {6\,x^3}{b^4\,{\ln \left (f\right )}^4}-\frac {3\,x}{b^2\,{\ln \left (f\right )}^2}\right )}{x^3} \]

[In]

int(f^(a + b/x)/x^5,x)

[Out]

-(f^(a + b/x)*(1/(b*log(f)) + (6*x^2)/(b^3*log(f)^3) - (6*x^3)/(b^4*log(f)^4) - (3*x)/(b^2*log(f)^2)))/x^3