3.123 \(\int \frac {f^{a+\frac {b}{x}}}{x^3} \, dx\)

Optimal. Leaf size=39 \[ \frac {f^{a+\frac {b}{x}}}{b^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x}}}{b x \log (f)} \]

[Out]

f^(a+b/x)/b^2/ln(f)^2-f^(a+b/x)/b/x/ln(f)

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Rubi [A]  time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2212, 2209} \[ \frac {f^{a+\frac {b}{x}}}{b^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x}}}{b x \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x)/x^3,x]

[Out]

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

Rule 2209

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 2212

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*} \int \frac {f^{a+\frac {b}{x}}}{x^3} \, dx &=-\frac {f^{a+\frac {b}{x}}}{b x \log (f)}-\frac {\int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx}{b \log (f)}\\ &=\frac {f^{a+\frac {b}{x}}}{b^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x}}}{b x \log (f)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.69 \[ \frac {f^{a+\frac {b}{x}} (x-b \log (f))}{b^2 x \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x)/x^3,x]

[Out]

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

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fricas [A]  time = 0.40, size = 31, normalized size = 0.79 \[ -\frac {{\left (b \log \relax (f) - x\right )} f^{\frac {a x + b}{x}}}{b^{2} x \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + \frac {b}{x}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x)/x^3, x)

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maple [A]  time = 0.02, size = 49, normalized size = 1.26 \[ \frac {-\frac {x \,{\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \relax (f )}}{b \ln \relax (f )}+\frac {x^{2} {\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \relax (f )}}{b^{2} \ln \relax (f )^{2}}}{x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x)/x^3,x)

[Out]

(1/b^2/ln(f)^2*x^2*exp((a+b/x)*ln(f))-1/b/ln(f)*x*exp((a+b/x)*ln(f)))/x^2

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maxima [C]  time = 1.29, size = 21, normalized size = 0.54 \[ \frac {f^{a} \Gamma \left (2, -\frac {b \log \relax (f)}{x}\right )}{b^{2} \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.55, size = 27, normalized size = 0.69 \[ \frac {f^{a+\frac {b}{x}}\,\left (x-b\,\ln \relax (f)\right )}{b^2\,x\,{\ln \relax (f)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x)/x^3,x)

[Out]

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

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sympy [A]  time = 0.12, size = 22, normalized size = 0.56 \[ \frac {f^{a + \frac {b}{x}} \left (- b \log {\relax (f )} + x\right )}{b^{2} x \log {\relax (f )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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