3.105 \(\int \frac{f^{a+b x^3}}{x^{10}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0897108, 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} \[ \frac{1}{18} b^3 f^a \log ^3(f) \text{Ei}\left (b x^3 \log (f)\right )-\frac{b^2 \log ^2(f) f^{a+b x^3}}{18 x^3}-\frac{f^{a+b x^3}}{9 x^9}-\frac{b \log (f) f^{a+b x^3}}{18 x^6} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)/x^10,x]

[Out]

-f^(a + b*x^3)/(9*x^9) - (b*f^(a + b*x^3)*Log[f])/(18*x^6) - (b^2*f^(a + b*x^3)*Log[f]^2)/(18*x^3) + (b^3*f^a*
ExpIntegralEi[b*x^3*Log[f]]*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 \frac{f^{a+b x^3}}{x^{10}} \, dx &=-\frac{f^{a+b x^3}}{9 x^9}+\frac{1}{3} (b \log (f)) \int \frac{f^{a+b x^3}}{x^7} \, dx\\ &=-\frac{f^{a+b x^3}}{9 x^9}-\frac{b f^{a+b x^3} \log (f)}{18 x^6}+\frac{1}{6} \left (b^2 \log ^2(f)\right ) \int \frac{f^{a+b x^3}}{x^4} \, dx\\ &=-\frac{f^{a+b x^3}}{9 x^9}-\frac{b f^{a+b x^3} \log (f)}{18 x^6}-\frac{b^2 f^{a+b x^3} \log ^2(f)}{18 x^3}+\frac{1}{6} \left (b^3 \log ^3(f)\right ) \int \frac{f^{a+b x^3}}{x} \, dx\\ &=-\frac{f^{a+b x^3}}{9 x^9}-\frac{b f^{a+b x^3} \log (f)}{18 x^6}-\frac{b^2 f^{a+b x^3} \log ^2(f)}{18 x^3}+\frac{1}{18} b^3 f^a \text{Ei}\left (b x^3 \log (f)\right ) \log ^3(f)\\ \end{align*}

Mathematica [A]  time = 0.023417, size = 59, normalized size = 0.73 \[ \frac{f^a \left (b^3 x^9 \log ^3(f) \text{Ei}\left (b x^3 \log (f)\right )-f^{b x^3} \left (b^2 x^6 \log ^2(f)+b x^3 \log (f)+2\right )\right )}{18 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3)/x^10,x]

[Out]

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

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Maple [B]  time = 0.041, size = 177, normalized size = 2.2 \begin{align*} -{\frac{{f}^{a}{b}^{3} \left ( \ln \left ( f \right ) \right ) ^{3}}{3} \left ({\frac{1}{3\,{b}^{3}{x}^{9} \left ( \ln \left ( f \right ) \right ) ^{3}}}+{\frac{1}{2\,{b}^{2}{x}^{6} \left ( \ln \left ( f \right ) \right ) ^{2}}}+{\frac{1}{2\,b{x}^{3}\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{22\,{b}^{3}{x}^{9} \left ( \ln \left ( f \right ) \right ) ^{3}+36\,{b}^{2}{x}^{6} \left ( \ln \left ( f \right ) \right ) ^{2}+36\,b{x}^{3}\ln \left ( f \right ) +24}{72\,{b}^{3}{x}^{9} \left ( \ln \left ( f \right ) \right ) ^{3}}}+{\frac{ \left ( 4\,{b}^{2}{x}^{6} \left ( \ln \left ( f \right ) \right ) ^{2}+4\,b{x}^{3}\ln \left ( f \right ) +8 \right ){{\rm e}^{b{x}^{3}\ln \left ( f \right ) }}}{24\,{b}^{3}{x}^{9} \left ( \ln \left ( f \right ) \right ) ^{3}}}+{\frac{\ln \left ( -b{x}^{3}\ln \left ( f \right ) \right ) }{6}}+{\frac{{\it Ei} \left ( 1,-b{x}^{3}\ln \left ( f \right ) \right ) }{6}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)/x^10,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*l
n(ln(f))-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)))

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Maxima [A]  time = 1.18032, size = 30, normalized size = 0.37 \begin{align*} \frac{1}{3} \, b^{3} f^{a} \Gamma \left (-3, -b x^{3} \log \left (f\right )\right ) \log \left (f\right )^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^10,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73534, size = 140, normalized size = 1.73 \begin{align*} \frac{b^{3} f^{a} x^{9}{\rm Ei}\left (b x^{3} \log \left (f\right )\right ) \log \left (f\right )^{3} -{\left (b^{2} x^{6} \log \left (f\right )^{2} + b x^{3} \log \left (f\right ) + 2\right )} f^{b x^{3} + a}}{18 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^10,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x^{3}}}{x^{10}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**3+a)/x**10,x)

[Out]

Integral(f**(a + b*x**3)/x**10, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{b x^{3} + a}}{x^{10}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^10,x, algorithm="giac")

[Out]

integrate(f^(b*x^3 + a)/x^10, x)