\(\int f^{a+\frac {b}{x^3}} x^{11} \, dx\) [156]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\frac {1}{3} b^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \log ^4(f) \]

[Out]

1/3*f^a*x^12*Ei(5,-b*ln(f)/x^3)

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} \[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\frac {1}{3} b^4 f^a \log ^4(f) \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \]

[In]

Int[f^(a + b/x^3)*x^11,x]

[Out]

(b^4*f^a*Gamma[-4, -((b*Log[f])/x^3)]*Log[f]^4)/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^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \log ^4(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\frac {1}{3} b^4 f^a \Gamma \left (-4,-\frac {b \log (f)}{x^3}\right ) \log ^4(f) \]

[In]

Integrate[f^(a + b/x^3)*x^11,x]

[Out]

(b^4*f^a*Gamma[-4, -((b*Log[f])/x^3)]*Log[f]^4)/3

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(212\) vs. \(2(18)=36\).

Time = 0.52 (sec) , antiderivative size = 213, normalized size of antiderivative = 8.88

method	result	size
meijerg	\(-\frac {f^{a} b^{4} \ln \left (f \right )^{4} \left (-\frac {x^{12}}{4 b^{4} \ln \left (f \right )^{4}}-\frac {x^{9}}{3 b^{3} \ln \left (f \right )^{3}}-\frac {x^{6}}{4 b^{2} \ln \left (f \right )^{2}}-\frac {x^{3}}{6 b \ln \left (f \right )}-\frac {25}{288}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (-b \right )}{24}+\frac {\ln \left (\ln \left (f \right )\right )}{24}+\frac {x^{12} \left (\frac {125 b^{4} \ln \left (f \right )^{4}}{x^{12}}+\frac {240 b^{3} \ln \left (f \right )^{3}}{x^{9}}+\frac {360 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {480 b \ln \left (f \right )}{x^{3}}+360\right )}{1440 b^{4} \ln \left (f \right )^{4}}-\frac {x^{12} \left (\frac {5 b^{3} \ln \left (f \right )^{3}}{x^{9}}+\frac {5 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {10 b \ln \left (f \right )}{x^{3}}+30\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}}{120 b^{4} \ln \left (f \right )^{4}}-\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{24}-\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{24}\right )}{3}\)	\(213\)

[In]

int(f^(a+b/x^3)*x^11,x,method=_RETURNVERBOSE)

[Out]

-1/3*f^a*b^4*ln(f)^4*(-1/4*x^12/b^4/ln(f)^4-1/3*x^9/b^3/ln(f)^3-1/4*x^6/b^2/ln(f)^2-1/6*x^3/b/ln(f)-25/288-1/8
*ln(x)+1/24*ln(-b)+1/24*ln(ln(f))+1/1440/b^4/ln(f)^4*x^12*(125*b^4*ln(f)^4/x^12+240*b^3*ln(f)^3/x^9+360*b^2*ln
(f)^2/x^6+480*b*ln(f)/x^3+360)-1/120/b^4/ln(f)^4*x^12*(5*b^3*ln(f)^3/x^9+5*b^2*ln(f)^2/x^6+10*b*ln(f)/x^3+30)*
exp(b*ln(f)/x^3)-1/24*ln(-b*ln(f)/x^3)-1/24*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. 72 vs. \(2 (18) = 36\).

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=-\frac {1}{72} \, b^{4} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{4} + \frac {1}{72} \, {\left (6 \, x^{12} + 2 \, b x^{9} \log \left (f\right ) + b^{2} x^{6} \log \left (f\right )^{2} + b^{3} x^{3} \log \left (f\right )^{3}\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

[In]

integrate(f^(a+b/x^3)*x^11,x, algorithm="fricas")

[Out]

-1/72*b^4*f^a*Ei(b*log(f)/x^3)*log(f)^4 + 1/72*(6*x^12 + 2*b*x^9*log(f) + b^2*x^6*log(f)^2 + b^3*x^3*log(f)^3)
*f^((a*x^3 + b)/x^3)

Sympy [F]

\[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\int f^{a + \frac {b}{x^{3}}} x^{11}\, dx \]

[In]

integrate(f**(a+b/x**3)*x**11,x)

[Out]

Integral(f**(a + b/x**3)*x**11, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\frac {1}{3} \, b^{4} f^{a} \Gamma \left (-4, -\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{4} \]

[In]

integrate(f^(a+b/x^3)*x^11,x, algorithm="maxima")

[Out]

1/3*b^4*f^a*gamma(-4, -b*log(f)/x^3)*log(f)^4

Giac [F]

\[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\int { f^{a + \frac {b}{x^{3}}} x^{11} \,d x } \]

[In]

integrate(f^(a+b/x^3)*x^11,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)*x^11, x)

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.75 \[ \int f^{a+\frac {b}{x^3}} x^{11} \, dx=\frac {b^4\,f^a\,{\ln \left (f\right )}^4\,\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^3}\right )}{72}+\frac {b^4\,f^a\,f^{\frac {b}{x^3}}\,{\ln \left (f\right )}^4\,\left (\frac {x^3}{24\,b\,\ln \left (f\right )}+\frac {x^6}{24\,b^2\,{\ln \left (f\right )}^2}+\frac {x^9}{12\,b^3\,{\ln \left (f\right )}^3}+\frac {x^{12}}{4\,b^4\,{\ln \left (f\right )}^4}\right )}{3} \]

[In]

int(f^(a + b/x^3)*x^11,x)

[Out]

(b^4*f^a*log(f)^4*expint(-(b*log(f))/x^3))/72 + (b^4*f^a*f^(b/x^3)*log(f)^4*(x^3/(24*b*log(f)) + x^6/(24*b^2*l
og(f)^2) + x^9/(12*b^3*log(f)^3) + x^12/(4*b^4*log(f)^4)))/3