3.155 \(\int f^{a+\frac {b}{x^3}} x^{14} \, dx\)

Optimal. Leaf size=24 \[ -\frac {1}{3} b^5 f^a \log ^5(f) \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \]

[Out]

1/3*f^a*x^15*Ei(6,-b*ln(f)/x^3)

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Rubi [A]  time = 0.03, antiderivative size = 24, 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} \[ -\frac {1}{3} b^5 f^a \log ^5(f) \text {Gamma}\left (-5,-\frac {b \log (f)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^3)*x^14,x]

[Out]

-(b^5*f^a*Gamma[-5, -((b*Log[f])/x^3)]*Log[f]^5)/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^{14} \, dx &=-\frac {1}{3} b^5 f^a \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \log ^5(f)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ -\frac {1}{3} b^5 f^a \log ^5(f) \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^3)*x^14,x]

[Out]

-1/3*(b^5*f^a*Gamma[-5, -((b*Log[f])/x^3)]*Log[f]^5)

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fricas [B]  time = 0.44, size = 84, normalized size = 3.50 \[ -\frac {1}{360} \, b^{5} f^{a} {\rm Ei}\left (\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{5} + \frac {1}{360} \, {\left (24 \, x^{15} + 6 \, b x^{12} \log \relax (f) + 2 \, b^{2} x^{9} \log \relax (f)^{2} + b^{3} x^{6} \log \relax (f)^{3} + b^{4} x^{3} \log \relax (f)^{4}\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^14,x, algorithm="fricas")

[Out]

-1/360*b^5*f^a*Ei(b*log(f)/x^3)*log(f)^5 + 1/360*(24*x^15 + 6*b*x^12*log(f) + 2*b^2*x^9*log(f)^2 + b^3*x^6*log
(f)^3 + b^4*x^3*log(f)^4)*f^((a*x^3 + b)/x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{14}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^14,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)*x^14, x)

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maple [B]  time = 0.09, size = 249, normalized size = 10.38 \[ \frac {\left (\frac {\left (\frac {36 b \ln \relax (f )}{x^{3}}+\frac {12 b^{2} \ln \relax (f )^{2}}{x^{6}}+\frac {6 b^{3} \ln \relax (f )^{3}}{x^{9}}+\frac {6 b^{4} \ln \relax (f )^{4}}{x^{12}}+144\right ) x^{15} {\mathrm e}^{\frac {b \ln \relax (f )}{x^{3}}}}{720 b^{5} \ln \relax (f )^{5}}-\frac {\left (\frac {1800 b \ln \relax (f )}{x^{3}}+\frac {1200 b^{2} \ln \relax (f )^{2}}{x^{6}}+\frac {600 b^{3} \ln \relax (f )^{3}}{x^{9}}+\frac {300 b^{4} \ln \relax (f )^{4}}{x^{12}}+\frac {137 b^{5} \ln \relax (f )^{5}}{x^{15}}+1440\right ) x^{15}}{7200 b^{5} \ln \relax (f )^{5}}+\frac {x^{15}}{5 b^{5} \ln \relax (f )^{5}}+\frac {x^{12}}{4 b^{4} \ln \relax (f )^{4}}+\frac {x^{9}}{6 b^{3} \ln \relax (f )^{3}}+\frac {x^{6}}{12 b^{2} \ln \relax (f )^{2}}+\frac {x^{3}}{24 b \ln \relax (f )}+\frac {\Ei \left (1, -\frac {b \ln \relax (f )}{x^{3}}\right )}{120}+\frac {\ln \relax (x )}{40}-\frac {\ln \left (-b \right )}{120}+\frac {\ln \left (-\frac {b \ln \relax (f )}{x^{3}}\right )}{120}-\frac {\ln \left (\ln \relax (f )\right )}{120}+\frac {137}{7200}\right ) b^{5} f^{a} \ln \relax (f )^{5}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^3)*x^14,x)

[Out]

1/3*f^a*b^5*ln(f)^5*(-1/7200/b^5*x^15/ln(f)^5*(137*b^5/x^15*ln(f)^5+300*b^4/x^12*ln(f)^4+600*b^3/x^9*ln(f)^3+1
200*b^2/x^6*ln(f)^2+1800*b/x^3*ln(f)+1440)+1/720/b^5*x^15/ln(f)^5*(6*b^4/x^12*ln(f)^4+6*b^3/x^9*ln(f)^3+12*b^2
/x^6*ln(f)^2+36*b/x^3*ln(f)+144)*exp(b/x^3*ln(f))+1/120*ln(-b/x^3*ln(f))+1/120*Ei(1,-b/x^3*ln(f))+137/7200+1/4
0*ln(x)-1/120*ln(-b)-1/120*ln(ln(f))+1/5*x^15/b^5/ln(f)^5+1/4*x^12/b^4/ln(f)^4+1/6*x^9/b^3/ln(f)^3+1/12*x^6/b^
2/ln(f)^2+1/24*x^3/b/ln(f))

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maxima [B]  time = 1.63, size = 22, normalized size = 0.92 \[ -\frac {1}{3} \, b^{5} f^{a} \Gamma \left (-5, -\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^3)*x^14,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.83, size = 102, normalized size = 4.25 \[ \frac {b^5\,f^a\,{\ln \relax (f)}^5\,\mathrm {expint}\left (-\frac {b\,\ln \relax (f)}{x^3}\right )}{360}+\frac {b^5\,f^a\,f^{\frac {b}{x^3}}\,{\ln \relax (f)}^5\,\left (\frac {x^3}{120\,b\,\ln \relax (f)}+\frac {x^6}{120\,b^2\,{\ln \relax (f)}^2}+\frac {x^9}{60\,b^3\,{\ln \relax (f)}^3}+\frac {x^{12}}{20\,b^4\,{\ln \relax (f)}^4}+\frac {x^{15}}{5\,b^5\,{\ln \relax (f)}^5}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^3)*x^14,x)

[Out]

(b^5*f^a*log(f)^5*expint(-(b*log(f))/x^3))/360 + (b^5*f^a*f^(b/x^3)*log(f)^5*(x^3/(120*b*log(f)) + x^6/(120*b^
2*log(f)^2) + x^9/(60*b^3*log(f)^3) + x^12/(20*b^4*log(f)^4) + x^15/(5*b^5*log(f)^5)))/3

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**3)*x**14,x)

[Out]

Timed out

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