\(\int \frac {f^{a+b x^3}}{x^{16}} \, dx\) [107]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\frac {1}{3} b^5 f^a \Gamma \left (-5,-b x^3 \log (f)\right ) \log ^5(f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (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} \[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\frac {1}{3} b^5 f^a \log ^5(f) \Gamma \left (-5,-b x^3 \log (f)\right ) \]

[In]

Int[f^(a + b*x^3)/x^16,x]

[Out]

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

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,-b x^3 \log (f)\right ) \log ^5(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\frac {1}{3} b^5 f^a \Gamma \left (-5,-b x^3 \log (f)\right ) \log ^5(f) \]

[In]

Integrate[f^(a + b*x^3)/x^16,x]

[Out]

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

Maple [B] (verified)

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

Time = 0.58 (sec) , antiderivative size = 249, normalized size of antiderivative = 10.38

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

[In]

int(f^(b*x^3+a)/x^16,x,method=_RETURNVERBOSE)

[Out]

-1/3*f^a*b^5*ln(f)^5*(1/5/b^5/x^15/ln(f)^5+1/4/b^4/x^12/ln(f)^4+1/6/b^3/x^9/ln(f)^3+1/12/b^2/x^6/ln(f)^2+1/24/
b/x^3/ln(f)+137/7200-1/40*ln(x)-1/120*ln(-b)-1/120*ln(ln(f))-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+1200*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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.46 \[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\frac {b^{5} f^{a} x^{15} {\rm Ei}\left (b x^{3} \log \left (f\right )\right ) \log \left (f\right )^{5} - {\left (b^{4} x^{12} \log \left (f\right )^{4} + b^{3} x^{9} \log \left (f\right )^{3} + 2 \, b^{2} x^{6} \log \left (f\right )^{2} + 6 \, b x^{3} \log \left (f\right ) + 24\right )} f^{b x^{3} + a}}{360 \, x^{15}} \]

[In]

integrate(f^(b*x^3+a)/x^16,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\int \frac {f^{a + b x^{3}}}{x^{16}}\, dx \]

[In]

integrate(f**(b*x**3+a)/x**16,x)

[Out]

Integral(f**(a + b*x**3)/x**16, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\frac {1}{3} \, b^{5} f^{a} \Gamma \left (-5, -b x^{3} \log \left (f\right )\right ) \log \left (f\right )^{5} \]

[In]

integrate(f^(b*x^3+a)/x^16,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=\int { \frac {f^{b x^{3} + a}}{x^{16}} \,d x } \]

[In]

integrate(f^(b*x^3+a)/x^16,x, algorithm="giac")

[Out]

integrate(f^(b*x^3 + a)/x^16, x)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 4.25 \[ \int \frac {f^{a+b x^3}}{x^{16}} \, dx=-\frac {b^5\,f^a\,{\ln \left (f\right )}^5\,\mathrm {expint}\left (-b\,x^3\,\ln \left (f\right )\right )}{360}-\frac {b^5\,f^a\,f^{b\,x^3}\,{\ln \left (f\right )}^5\,\left (\frac {1}{120\,b\,x^3\,\ln \left (f\right )}+\frac {1}{120\,b^2\,x^6\,{\ln \left (f\right )}^2}+\frac {1}{60\,b^3\,x^9\,{\ln \left (f\right )}^3}+\frac {1}{20\,b^4\,x^{12}\,{\ln \left (f\right )}^4}+\frac {1}{5\,b^5\,x^{15}\,{\ln \left (f\right )}^5}\right )}{3} \]

[In]

int(f^(a + b*x^3)/x^16,x)

[Out]

- (b^5*f^a*log(f)^5*expint(-b*x^3*log(f)))/360 - (b^5*f^a*f^(b*x^3)*log(f)^5*(1/(120*b*x^3*log(f)) + 1/(120*b^
2*x^6*log(f)^2) + 1/(60*b^3*x^9*log(f)^3) + 1/(20*b^4*x^12*log(f)^4) + 1/(5*b^5*x^15*log(f)^5)))/3