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]

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

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Rubi [A]  time = 0.11, antiderivative size = 81, normalized size of antiderivative = 1.00, 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 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]

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]))

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*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.70 \[ \frac {1}{18} f^a \left (x^3 f^{\frac {b}{x^3}} \left (b^2 \log ^2(f)+b x^3 \log (f)+2 x^6\right )-b^3 \log ^3(f) \text {Ei}\left (\frac {b \log (f)}{x^3}\right )\right ) \]

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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fricas [A]  time = 0.45, size = 60, normalized size = 0.74 \[ -\frac {1}{18} \, b^{3} f^{a} {\rm Ei}\left (\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{3} + \frac {1}{18} \, {\left (2 \, x^{9} + b x^{6} \log \relax (f) + b^{2} x^{3} \log \relax (f)^{2}\right )} f^{\frac {a x^{3} + b}{x^{3}}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{8}\,{d x} \]

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)

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maple [B]  time = 0.07, size = 177, normalized size = 2.19 \[ \frac {\left (\frac {\left (\frac {4 b \ln \relax (f )}{x^{3}}+\frac {4 b^{2} \ln \relax (f )^{2}}{x^{6}}+8\right ) x^{9} {\mathrm e}^{\frac {b \ln \relax (f )}{x^{3}}}}{24 b^{3} \ln \relax (f )^{3}}-\frac {\left (\frac {36 b \ln \relax (f )}{x^{3}}+\frac {36 b^{2} \ln \relax (f )^{2}}{x^{6}}+\frac {22 b^{3} \ln \relax (f )^{3}}{x^{9}}+24\right ) x^{9}}{72 b^{3} \ln \relax (f )^{3}}+\frac {x^{9}}{3 b^{3} \ln \relax (f )^{3}}+\frac {x^{6}}{2 b^{2} \ln \relax (f )^{2}}+\frac {x^{3}}{2 b \ln \relax (f )}+\frac {\Ei \left (1, -\frac {b \ln \relax (f )}{x^{3}}\right )}{6}+\frac {\ln \relax (x )}{2}-\frac {\ln \left (-b \right )}{6}+\frac {\ln \left (-\frac {b \ln \relax (f )}{x^{3}}\right )}{6}-\frac {\ln \left (\ln \relax (f )\right )}{6}+\frac {11}{36}\right ) b^{3} f^{a} \ln \relax (f )^{3}}{3} \]

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

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maxima [A]  time = 1.27, size = 22, normalized size = 0.27 \[ -\frac {1}{3} \, b^{3} f^{a} \Gamma \left (-3, -\frac {b \log \relax (f)}{x^{3}}\right ) \log \relax (f)^{3} \]

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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mupad [B]  time = 3.70, size = 69, normalized size = 0.85 \[ \frac {b^3\,f^a\,{\ln \relax (f)}^3\,\left (f^{\frac {b}{x^3}}\,\left (\frac {x^3}{6\,b\,\ln \relax (f)}+\frac {x^6}{6\,b^2\,{\ln \relax (f)}^2}+\frac {x^9}{3\,b^3\,{\ln \relax (f)}^3}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \relax (f)}{x^3}\right )}{6}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{3}}} x^{8}\, dx \]

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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