\(\int f^{a+\frac {b}{x}} x^4 \, dx\) [116]

Optimal result

Integrand size = 13, antiderivative size = 22 \[ \int f^{a+\frac {b}{x}} x^4 \, dx=-b^5 f^a \Gamma \left (-5,-\frac {b \log (f)}{x}\right ) \log ^5(f) \]

[Out]

f^a*x^5*Ei(6,-b*ln(f)/x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, 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} \[ \int f^{a+\frac {b}{x}} x^4 \, dx=-b^5 f^a \log ^5(f) \Gamma \left (-5,-\frac {b \log (f)}{x}\right ) \]

[In]

Int[f^(a + b/x)*x^4,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x}} x^4 \, dx=-b^5 f^a \Gamma \left (-5,-\frac {b \log (f)}{x}\right ) \log ^5(f) \]

[In]

Integrate[f^(a + b/x)*x^4,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(120\) vs. \(2(17)=34\).

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 5.50

method	result	size
risch	\(\frac {f^{a} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x}\right ) b^{5} \ln \left (f \right )^{5}}{120}+\frac {f^{a} f^{\frac {b}{x}} x \,b^{4} \ln \left (f \right )^{4}}{120}+\frac {f^{a} f^{\frac {b}{x}} x^{2} b^{3} \ln \left (f \right )^{3}}{120}+\frac {f^{a} f^{\frac {b}{x}} x^{3} b^{2} \ln \left (f \right )^{2}}{60}+\frac {f^{a} f^{\frac {b}{x}} x^{4} b \ln \left (f \right )}{20}+\frac {f^{a} f^{\frac {b}{x}} x^{5}}{5}\)	\(121\)
meijerg	\(f^{a} b^{5} \ln \left (f \right )^{5} \left (\frac {x^{5}}{5 b^{5} \ln \left (f \right )^{5}}+\frac {x^{4}}{4 b^{4} \ln \left (f \right )^{4}}+\frac {x^{3}}{6 b^{3} \ln \left (f \right )^{3}}+\frac {x^{2}}{12 b^{2} \ln \left (f \right )^{2}}+\frac {x}{24 b \ln \left (f \right )}+\frac {137}{7200}+\frac {\ln \left (x \right )}{120}-\frac {\ln \left (-b \right )}{120}-\frac {\ln \left (\ln \left (f \right )\right )}{120}-\frac {x^{5} \left (\frac {137 b^{5} \ln \left (f \right )^{5}}{x^{5}}+\frac {300 b^{4} \ln \left (f \right )^{4}}{x^{4}}+\frac {600 b^{3} \ln \left (f \right )^{3}}{x^{3}}+\frac {1200 b^{2} \ln \left (f \right )^{2}}{x^{2}}+\frac {1800 b \ln \left (f \right )}{x}+1440\right )}{7200 b^{5} \ln \left (f \right )^{5}}+\frac {x^{5} \left (\frac {6 b^{4} \ln \left (f \right )^{4}}{x^{4}}+\frac {6 b^{3} \ln \left (f \right )^{3}}{x^{3}}+\frac {12 b^{2} \ln \left (f \right )^{2}}{x^{2}}+\frac {36 b \ln \left (f \right )}{x}+144\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x}}}{720 b^{5} \ln \left (f \right )^{5}}+\frac {\ln \left (-\frac {b \ln \left (f \right )}{x}\right )}{120}+\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x}\right )}{120}\right )\)	\(246\)

[In]

int(f^(a+b/x)*x^4,x,method=_RETURNVERBOSE)

[Out]

1/120*f^a*Ei(1,-b*ln(f)/x)*b^5*ln(f)^5+1/120*f^a*f^(b/x)*x*b^4*ln(f)^4+1/120*f^a*f^(b/x)*x^2*b^3*ln(f)^3+1/60*
f^a*f^(b/x)*x^3*b^2*ln(f)^2+1/20*f^a*f^(b/x)*x^4*b*ln(f)+1/5*f^a*f^(b/x)*x^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (17) = 34\).

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.64 \[ \int f^{a+\frac {b}{x}} x^4 \, dx=-\frac {1}{120} \, b^{5} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x}\right ) \log \left (f\right )^{5} + \frac {1}{120} \, {\left (b^{4} x \log \left (f\right )^{4} + b^{3} x^{2} \log \left (f\right )^{3} + 2 \, b^{2} x^{3} \log \left (f\right )^{2} + 6 \, b x^{4} \log \left (f\right ) + 24 \, x^{5}\right )} f^{\frac {a x + b}{x}} \]

[In]

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

[Out]

-1/120*b^5*f^a*Ei(b*log(f)/x)*log(f)^5 + 1/120*(b^4*x*log(f)^4 + b^3*x^2*log(f)^3 + 2*b^2*x^3*log(f)^2 + 6*b*x
^4*log(f) + 24*x^5)*f^((a*x + b)/x)

Sympy [F]

\[ \int f^{a+\frac {b}{x}} x^4 \, dx=\int f^{a + \frac {b}{x}} x^{4}\, dx \]

[In]

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

[Out]

Integral(f**(a + b/x)*x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x}} x^4 \, dx=-b^{5} f^{a} \Gamma \left (-5, -\frac {b \log \left (f\right )}{x}\right ) \log \left (f\right )^{5} \]

[In]

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

[Out]

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

Giac [F]

\[ \int f^{a+\frac {b}{x}} x^4 \, dx=\int { f^{a + \frac {b}{x}} x^{4} \,d x } \]

[In]

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

[Out]

integrate(f^(a + b/x)*x^4, x)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.50 \[ \int f^{a+\frac {b}{x}} x^4 \, dx=\frac {b^5\,f^a\,{\ln \left (f\right )}^5\,\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x}\right )}{120}+b^5\,f^a\,f^{b/x}\,{\ln \left (f\right )}^5\,\left (\frac {x^2}{120\,b^2\,{\ln \left (f\right )}^2}+\frac {x^3}{60\,b^3\,{\ln \left (f\right )}^3}+\frac {x^4}{20\,b^4\,{\ln \left (f\right )}^4}+\frac {x^5}{5\,b^5\,{\ln \left (f\right )}^5}+\frac {x}{120\,b\,\ln \left (f\right )}\right ) \]

[In]

int(f^(a + b/x)*x^4,x)

[Out]

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