\(\int f^{a+\frac {b}{x^3}} x^{14} \, dx\) [155]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=-\frac {1}{3} b^5 f^a \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \log ^5(f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2250} \[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=-\frac {1}{3} b^5 f^a \log ^5(f) \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \]

[In]

Int[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)

Rule 2250

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3} b^5 f^a \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \log ^5(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=-\frac {1}{3} b^5 f^a \Gamma \left (-5,-\frac {b \log (f)}{x^3}\right ) \log ^5(f) \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(18)=36\).

Time = 0.95 (sec) , antiderivative size = 249, normalized size of antiderivative = 10.38

method	result	size
meijerg	\(\frac {f^{a} b^{5} \ln \left (f \right )^{5} \left (\frac {x^{15}}{5 b^{5} \ln \left (f \right )^{5}}+\frac {x^{12}}{4 b^{4} \ln \left (f \right )^{4}}+\frac {x^{9}}{6 b^{3} \ln \left (f \right )^{3}}+\frac {x^{6}}{12 b^{2} \ln \left (f \right )^{2}}+\frac {x^{3}}{24 b \ln \left (f \right )}+\frac {137}{7200}+\frac {\ln \left (x \right )}{40}-\frac {\ln \left (-b \right )}{120}-\frac {\ln \left (\ln \left (f \right )\right )}{120}-\frac {x^{15} \left (\frac {137 b^{5} \ln \left (f \right )^{5}}{x^{15}}+\frac {300 b^{4} \ln \left (f \right )^{4}}{x^{12}}+\frac {600 b^{3} \ln \left (f \right )^{3}}{x^{9}}+\frac {1200 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {1800 b \ln \left (f \right )}{x^{3}}+1440\right )}{7200 b^{5} \ln \left (f \right )^{5}}+\frac {x^{15} \left (\frac {6 b^{4} \ln \left (f \right )^{4}}{x^{12}}+\frac {6 b^{3} \ln \left (f \right )^{3}}{x^{9}}+\frac {12 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {36 b \ln \left (f \right )}{x^{3}}+144\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}}{720 b^{5} \ln \left (f \right )^{5}}+\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{120}+\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{120}\right )}{3}\)	\(249\)

[In]

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

[Out]

1/3*f^a*b^5*ln(f)^5*(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)+137/7200+1/40*ln(x)-1/120*ln(-b)-1/120*ln(ln(f))-1/7200/b^5/ln(f)^5*x^15*(137*b^5*ln(f)^5/x^15+300*
b^4*ln(f)^4/x^12+600*b^3*ln(f)^3/x^9+1200*b^2*ln(f)^2/x^6+1800*b*ln(f)/x^3+1440)+1/720/b^5/ln(f)^5*x^15*(6*b^4
*ln(f)^4/x^12+6*b^3*ln(f)^3/x^9+12*b^2*ln(f)^2/x^6+36*b*ln(f)/x^3+144)*exp(b*ln(f)/x^3)+1/120*ln(-b*ln(f)/x^3)
+1/120*Ei(1,-b*ln(f)/x^3))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.50 \[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=-\frac {1}{360} \, b^{5} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{5} + \frac {1}{360} \, {\left (24 \, x^{15} + 6 \, b x^{12} \log \left (f\right ) + 2 \, b^{2} x^{9} \log \left (f\right )^{2} + b^{3} x^{6} \log \left (f\right )^{3} + b^{4} x^{3} \log \left (f\right )^{4}\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

[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)

Sympy [F]

\[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=\int f^{a + \frac {b}{x^{3}}} x^{14}\, dx \]

[In]

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

[Out]

Integral(f**(a + b/x**3)*x**14, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=-\frac {1}{3} \, b^{5} f^{a} \Gamma \left (-5, -\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{5} \]

[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

Giac [F]

\[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=\int { f^{a + \frac {b}{x^{3}}} x^{14} \,d x } \]

[In]

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

[Out]

integrate(f^(a + b/x^3)*x^14, x)

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 102, normalized size of antiderivative = 4.25 \[ \int f^{a+\frac {b}{x^3}} x^{14} \, dx=\frac {b^5\,f^a\,{\ln \left (f\right )}^5\,\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^3}\right )}{360}+\frac {b^5\,f^a\,f^{\frac {b}{x^3}}\,{\ln \left (f\right )}^5\,\left (\frac {x^3}{120\,b\,\ln \left (f\right )}+\frac {x^6}{120\,b^2\,{\ln \left (f\right )}^2}+\frac {x^9}{60\,b^3\,{\ln \left (f\right )}^3}+\frac {x^{12}}{20\,b^4\,{\ln \left (f\right )}^4}+\frac {x^{15}}{5\,b^5\,{\ln \left (f\right )}^5}\right )}{3} \]

[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