3.180 \(\int \frac {f^{a+b x^n}}{x^2} \, dx\)

Optimal. Leaf size=37 \[ -\frac {f^a \left (-b \log (f) x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right )}{n x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ -\frac {f^a \left (-b \log (f) x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},-b \log (f) x^n\right )}{n x} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^n)/x^2,x]

[Out]

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

Rule 2218

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

Rubi steps

\begin {align*} \int \frac {f^{a+b x^n}}{x^2} \, dx &=-\frac {f^a \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{\frac {1}{n}}}{n x}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 37, normalized size = 1.00 \[ -\frac {f^a \left (-b \log (f) x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n \log (f)\right )}{n x} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^n)/x^2,x]

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {f^{b x^{n} + a}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(f^(b*x^n + a)/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{n} + a}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^n + a)/x^2, x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {f^{b \,x^{n}+a}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^n+a)/x^2,x)

[Out]

int(f^(b*x^n+a)/x^2,x)

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maxima [A]  time = 1.23, size = 37, normalized size = 1.00 \[ -\frac {\left (-b x^{n} \log \relax (f)\right )^{\left (\frac {1}{n}\right )} f^{a} \Gamma \left (-\frac {1}{n}, -b x^{n} \log \relax (f)\right )}{n x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.53, size = 52, normalized size = 1.41 \[ -\frac {f^a\,{\mathrm {e}}^{\frac {b\,x^n\,\ln \relax (f)}{2}}\,{\mathrm {M}}_{\frac {1}{2\,n}+\frac {1}{2},-\frac {1}{2\,n}}\left (b\,x^n\,\ln \relax (f)\right )\,{\left (b\,x^n\,\ln \relax (f)\right )}^{\frac {1}{2\,n}-\frac {1}{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^n)/x^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {b f^{a} f^{b x^{n}} n x^{n} \log {\relax (f )}}{n x - x} - \frac {f^{a} f^{b x^{n}} n}{n x - x} + \frac {f^{a} f^{b x^{n}}}{n x - x} & \text {for}\: n \neq 1 \\\int \frac {f^{a + b x}}{x^{2}}\, dx & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((b*f**a*f**(b*x**n)*n*x**n*log(f)/(n*x - x) - f**a*f**(b*x**n)*n/(n*x - x) + f**a*f**(b*x**n)/(n*x -
 x), Ne(n, 1)), (Integral(f**(a + b*x)/x**2, x), True))

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