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

Optimal. Leaf size=81 \[ -\frac{1}{18} b^3 f^a \log ^3(f) \text{Ei}\left (\frac{b \log (f)}{x^3}\right )+\frac{1}{18} b^2 x^3 \log ^2(f) f^{a+\frac{b}{x^3}}+\frac{1}{9} x^9 f^{a+\frac{b}{x^3}}+\frac{1}{18} b x^6 \log (f) f^{a+\frac{b}{x^3}} \]

[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

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Rubi [A]  time = 0.114718, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2214, 2210} \[ -\frac{1}{18} b^3 f^a \log ^3(f) \text{Ei}\left (\frac{b \log (f)}{x^3}\right )+\frac{1}{18} b^2 x^3 \log ^2(f) f^{a+\frac{b}{x^3}}+\frac{1}{9} x^9 f^{a+\frac{b}{x^3}}+\frac{1}{18} b x^6 \log (f) f^{a+\frac{b}{x^3}} \]

Antiderivative was successfully verified.

[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 2214

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]))

Rule 2210

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]

Rubi steps

\begin{align*} \int f^{a+\frac{b}{x^3}} x^8 \, dx &=\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]  time = 0.019582, size = 57, normalized size = 0.7 \[ \frac{1}{18} f^a \left (x^3 f^{\frac{b}{x^3}} \left (b^2 \log ^2(f)+b x^3 \log (f)+2 x^6\right )-b^3 \log ^3(f) \text{Ei}\left (\frac{b \log (f)}{x^3}\right )\right ) \]

Antiderivative was successfully verified.

[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

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Maple [B]  time = 0.043, size = 177, normalized size = 2.2 \begin{align*}{\frac{{f}^{a}{b}^{3} \left ( \ln \left ( f \right ) \right ) ^{3}}{3} \left ({\frac{{x}^{9}}{3\, \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}}}+{\frac{{x}^{6}}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{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}}{72\, \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}} \left ( 22\,{\frac{ \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}}{{x}^{9}}}+36\,{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}}{{x}^{6}}}+36\,{\frac{b\ln \left ( f \right ) }{{x}^{3}}}+24 \right ) }+{\frac{{x}^{9}}{24\, \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}} \left ( 4\,{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}}{{x}^{6}}}+4\,{\frac{b\ln \left ( f \right ) }{{x}^{3}}}+8 \right ){{\rm e}^{{\frac{b\ln \left ( f \right ) }{{x}^{3}}}}}}+{\frac{1}{6}\ln \left ( -{\frac{b\ln \left ( f \right ) }{{x}^{3}}} \right ) }+{\frac{1}{6}{\it Ei} \left ( 1,-{\frac{b\ln \left ( f \right ) }{{x}^{3}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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Maxima [A]  time = 1.29929, size = 30, normalized size = 0.37 \begin{align*} -\frac{1}{3} \, b^{3} f^{a} \Gamma \left (-3, -\frac{b \log \left (f\right )}{x^{3}}\right ) \log \left (f\right )^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.80731, size = 149, normalized size = 1.84 \begin{align*} -\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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{3}}} x^{8}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**3)*x**8, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{3}}} x^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^3)*x^8, x)