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

Optimal. Leaf size=34 \[ \frac {1}{3} x^5 f^a \left (-\frac {b \log (f)}{x^3}\right )^{5/3} \Gamma \left (-\frac {5}{3},-\frac {b \log (f)}{x^3}\right ) \]

[Out]

1/3*f^a*x^5*GAMMA(-5/3,-b*ln(f)/x^3)*(-b*ln(f)/x^3)^(5/3)

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Rubi [A]  time = 0.03, antiderivative size = 34, 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} \[ \frac {1}{3} x^5 f^a \left (-\frac {b \log (f)}{x^3}\right )^{5/3} \text {Gamma}\left (-\frac {5}{3},-\frac {b \log (f)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^3)*x^4,x]

[Out]

(f^a*x^5*Gamma[-5/3, -((b*Log[f])/x^3)]*(-((b*Log[f])/x^3))^(5/3))/3

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 f^{a+\frac {b}{x^3}} x^4 \, dx &=\frac {1}{3} f^a x^5 \Gamma \left (-\frac {5}{3},-\frac {b \log (f)}{x^3}\right ) \left (-\frac {b \log (f)}{x^3}\right )^{5/3}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 34, normalized size = 1.00 \[ \frac {1}{3} x^5 f^a \left (-\frac {b \log (f)}{x^3}\right )^{5/3} \Gamma \left (-\frac {5}{3},-\frac {b \log (f)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^3)*x^4,x]

[Out]

(f^a*x^5*Gamma[-5/3, -((b*Log[f])/x^3)]*(-((b*Log[f])/x^3))^(5/3))/3

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fricas [A]  time = 0.43, size = 55, normalized size = 1.62 \[ -\frac {3}{10} \, \left (-b \log \relax (f)\right )^{\frac {2}{3}} b f^{a} \Gamma \left (\frac {1}{3}, -\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f) + \frac {1}{10} \, {\left (2 \, x^{5} + 3 \, b x^{2} \log \relax (f)\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/10*(-b*log(f))^(2/3)*b*f^a*gamma(1/3, -b*log(f)/x^3)*log(f) + 1/10*(2*x^5 + 3*b*x^2*log(f))*f^((a*x^3 + b)/
x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^3)*x^4, x)

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maple [B]  time = 0.06, size = 120, normalized size = 3.53 \[ -\frac {\left (-b \right )^{\frac {5}{3}} \left (-\frac {3 \left (\frac {3 b \ln \relax (f )}{2 x^{3}}+1\right ) x^{5} {\mathrm e}^{\frac {b \ln \relax (f )}{x^{3}}}}{5 \left (-b \right )^{\frac {5}{3}} \ln \relax (f )^{\frac {5}{3}}}-\frac {9 b^{2} \Gamma \left (\frac {1}{3}, -\frac {b \ln \relax (f )}{x^{3}}\right ) \ln \relax (f )^{\frac {1}{3}}}{10 \left (-b \right )^{\frac {5}{3}} \left (-\frac {b \ln \relax (f )}{x^{3}}\right )^{\frac {1}{3}} x}+\frac {3 \pi \sqrt {3}\, b^{2} \ln \relax (f )^{\frac {1}{3}}}{5 \left (-b \right )^{\frac {5}{3}} \Gamma \left (\frac {2}{3}\right ) \left (-\frac {b \ln \relax (f )}{x^{3}}\right )^{\frac {1}{3}} x}\right ) f^{a} \ln \relax (f )^{\frac {5}{3}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*f^a*(-b)^(5/3)*ln(f)^(5/3)*(3/5/x/(-b)^(5/3)*ln(f)^(1/3)*b^2*Pi*3^(1/2)/GAMMA(2/3)/(-b/x^3*ln(f))^(1/3)-3
/5*x^5/(-b)^(5/3)/ln(f)^(5/3)*(3/2*b/x^3*ln(f)+1)*exp(b/x^3*ln(f))-9/10/x/(-b)^(5/3)*ln(f)^(1/3)*b^2/(-b/x^3*l
n(f))^(1/3)*GAMMA(1/3,-b/x^3*ln(f)))

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maxima [A]  time = 1.33, size = 28, normalized size = 0.82 \[ \frac {1}{3} \, f^{a} x^{5} \left (-\frac {b \log \relax (f)}{x^{3}}\right )^{\frac {5}{3}} \Gamma \left (-\frac {5}{3}, -\frac {b \log \relax (f)}{x^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*f^a*x^5*(-b*log(f)/x^3)^(5/3)*gamma(-5/3, -b*log(f)/x^3)

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mupad [B]  time = 3.57, size = 88, normalized size = 2.59 \[ \frac {f^a\,f^{\frac {b}{x^3}}\,x^5}{5}+\frac {3\,f^a\,x^5\,\Gamma \left (\frac {1}{3},-\frac {b\,\ln \relax (f)}{x^3}\right )\,{\left (-\frac {b\,\ln \relax (f)}{x^3}\right )}^{5/3}}{10}+\frac {3\,b\,f^a\,f^{\frac {b}{x^3}}\,x^2\,\ln \relax (f)}{10}-\frac {\pi \,\sqrt {3}\,f^a\,x^5\,{\left (-\frac {b\,\ln \relax (f)}{x^3}\right )}^{5/3}}{5\,\Gamma \left (\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^3)*x^4,x)

[Out]

(f^a*f^(b/x^3)*x^5)/5 + (3*f^a*x^5*igamma(1/3, -(b*log(f))/x^3)*(-(b*log(f))/x^3)^(5/3))/10 + (3*b*f^a*f^(b/x^
3)*x^2*log(f))/10 - (3^(1/2)*f^a*x^5*pi*(-(b*log(f))/x^3)^(5/3))/(5*gamma(2/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**3)*x**4, x)

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