\(\int f^{a+\frac {b}{x^3}} x^8 \, dx\) [157]

Optimal result

Integrand size = 13, antiderivative size = 81 \[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=\frac {1}{9} f^{a+\frac {b}{x^3}} x^9+\frac {1}{18} b f^{a+\frac {b}{x^3}} x^6 \log (f)+\frac {1}{18} b^2 f^{a+\frac {b}{x^3}} x^3 \log ^2(f)-\frac {1}{18} b^3 f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^3}\right ) \log ^3(f) \]

[Out]

1/9*f^(a+b/x^3)*x^9+1/18*b*f^(a+b/x^3)*x^6*ln(f)+1/18*b^2*f^(a+b/x^3)*x^3*ln(f)^2-1/18*b^3*f^a*Ei(b*ln(f)/x^3)
*ln(f)^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2241} \[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=-\frac {1}{18} b^3 f^a \log ^3(f) \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^3}\right )+\frac {1}{18} b^2 x^3 \log ^2(f) f^{a+\frac {b}{x^3}}+\frac {1}{9} x^9 f^{a+\frac {b}{x^3}}+\frac {1}{18} b x^6 \log (f) f^{a+\frac {b}{x^3}} \]

[In]

Int[f^(a + b/x^3)*x^8,x]

[Out]

(f^(a + b/x^3)*x^9)/9 + (b*f^(a + b/x^3)*x^6*Log[f])/18 + (b^2*f^(a + b/x^3)*x^3*Log[f]^2)/18 - (b^3*f^a*ExpIn
tegralEi[(b*Log[f])/x^3]*Log[f]^3)/18

Rule 2241

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

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} f^{a+\frac {b}{x^3}} x^9+\frac {1}{3} (b \log (f)) \int f^{a+\frac {b}{x^3}} x^5 \, dx \\ & = \frac {1}{9} f^{a+\frac {b}{x^3}} x^9+\frac {1}{18} b f^{a+\frac {b}{x^3}} x^6 \log (f)+\frac {1}{6} \left (b^2 \log ^2(f)\right ) \int f^{a+\frac {b}{x^3}} x^2 \, dx \\ & = \frac {1}{9} f^{a+\frac {b}{x^3}} x^9+\frac {1}{18} b f^{a+\frac {b}{x^3}} x^6 \log (f)+\frac {1}{18} b^2 f^{a+\frac {b}{x^3}} x^3 \log ^2(f)+\frac {1}{6} \left (b^3 \log ^3(f)\right ) \int \frac {f^{a+\frac {b}{x^3}}}{x} \, dx \\ & = \frac {1}{9} f^{a+\frac {b}{x^3}} x^9+\frac {1}{18} b f^{a+\frac {b}{x^3}} x^6 \log (f)+\frac {1}{18} b^2 f^{a+\frac {b}{x^3}} x^3 \log ^2(f)-\frac {1}{18} b^3 f^a \text {Ei}\left (\frac {b \log (f)}{x^3}\right ) \log ^3(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=\frac {1}{18} f^a \left (-b^3 \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^3}\right ) \log ^3(f)+f^{\frac {b}{x^3}} x^3 \left (2 x^6+b x^3 \log (f)+b^2 \log ^2(f)\right )\right ) \]

[In]

Integrate[f^(a + b/x^3)*x^8,x]

[Out]

(f^a*(-(b^3*ExpIntegralEi[(b*Log[f])/x^3]*Log[f]^3) + f^(b/x^3)*x^3*(2*x^6 + b*x^3*Log[f] + b^2*Log[f]^2)))/18

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(176\) vs. \(2(73)=146\).

Time = 0.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.19

method	result	size
meijerg	\(\frac {f^{a} b^{3} \ln \left (f \right )^{3} \left (\frac {x^{9}}{3 b^{3} \ln \left (f \right )^{3}}+\frac {x^{6}}{2 b^{2} \ln \left (f \right )^{2}}+\frac {x^{3}}{2 b \ln \left (f \right )}+\frac {11}{36}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (-b \right )}{6}-\frac {\ln \left (\ln \left (f \right )\right )}{6}-\frac {x^{9} \left (\frac {22 b^{3} \ln \left (f \right )^{3}}{x^{9}}+\frac {36 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {36 b \ln \left (f \right )}{x^{3}}+24\right )}{72 b^{3} \ln \left (f \right )^{3}}+\frac {x^{9} \left (\frac {4 b^{2} \ln \left (f \right )^{2}}{x^{6}}+\frac {4 b \ln \left (f \right )}{x^{3}}+8\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{3}}}}{24 b^{3} \ln \left (f \right )^{3}}+\frac {\ln \left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{6}+\frac {\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{3}}\right )}{6}\right )}{3}\)	\(177\)

[In]

int(f^(a+b/x^3)*x^8,x,method=_RETURNVERBOSE)

[Out]

1/3*f^a*b^3*ln(f)^3*(1/3*x^9/b^3/ln(f)^3+1/2*x^6/b^2/ln(f)^2+1/2*x^3/b/ln(f)+11/36+1/2*ln(x)-1/6*ln(-b)-1/6*ln
(ln(f))-1/72/b^3/ln(f)^3*x^9*(22*b^3*ln(f)^3/x^9+36*b^2*ln(f)^2/x^6+36*b*ln(f)/x^3+24)+1/24/b^3/ln(f)^3*x^9*(4
*b^2*ln(f)^2/x^6+4*b*ln(f)/x^3+8)*exp(b*ln(f)/x^3)+1/6*ln(-b*ln(f)/x^3)+1/6*Ei(1,-b*ln(f)/x^3))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74 \[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=-\frac {1}{18} \, b^{3} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{3} + \frac {1}{18} \, {\left (2 \, x^{9} + b x^{6} \log \left (f\right ) + b^{2} x^{3} \log \left (f\right )^{2}\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

[In]

integrate(f^(a+b/x^3)*x^8,x, algorithm="fricas")

[Out]

-1/18*b^3*f^a*Ei(b*log(f)/x^3)*log(f)^3 + 1/18*(2*x^9 + b*x^6*log(f) + b^2*x^3*log(f)^2)*f^((a*x^3 + b)/x^3)

Sympy [F]

\[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=\int f^{a + \frac {b}{x^{3}}} x^{8}\, dx \]

[In]

integrate(f**(a+b/x**3)*x**8,x)

[Out]

Integral(f**(a + b/x**3)*x**8, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.27 \[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=-\frac {1}{3} \, b^{3} f^{a} \Gamma \left (-3, -\frac {b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{3} \]

[In]

integrate(f^(a+b/x^3)*x^8,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=\int { f^{a + \frac {b}{x^{3}}} x^{8} \,d x } \]

[In]

integrate(f^(a+b/x^3)*x^8,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^3)*x^8, x)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int f^{a+\frac {b}{x^3}} x^8 \, dx=\frac {b^3\,f^a\,{\ln \left (f\right )}^3\,\left (f^{\frac {b}{x^3}}\,\left (\frac {x^3}{6\,b\,\ln \left (f\right )}+\frac {x^6}{6\,b^2\,{\ln \left (f\right )}^2}+\frac {x^9}{3\,b^3\,{\ln \left (f\right )}^3}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^3}\right )}{6}\right )}{3} \]

[In]

int(f^(a + b/x^3)*x^8,x)

[Out]

(b^3*f^a*log(f)^3*(f^(b/x^3)*(x^3/(6*b*log(f)) + x^6/(6*b^2*log(f)^2) + x^9/(3*b^3*log(f)^3)) + expint(-(b*log
(f))/x^3)/6))/3