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]

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

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Rubi [A]  time = 0.09, 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} \[ \frac {1}{18} b^3 f^a \log ^3(f) \text {Ei}\left (b x^3 \log (f)\right )-\frac {b^2 \log ^2(f) f^{a+b x^3}}{18 x^3}-\frac {f^{a+b x^3}}{9 x^9}-\frac {b \log (f) f^{a+b x^3}}{18 x^6} \]

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 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 \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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.40, size = 59, normalized size = 0.73 \[ \frac {b^{3} f^{a} x^{9} {\rm Ei}\left (b x^{3} \log \relax (f)\right ) \log \relax (f)^{3} - {\left (b^{2} x^{6} \log \relax (f)^{2} + b x^{3} \log \relax (f) + 2\right )} f^{b x^{3} + a}}{18 \, x^{9}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{3} + a}}{x^{10}}\,{d x} \]

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)

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maple [B]  time = 0.06, size = 177, normalized size = 2.19 \[ -\frac {\left (\frac {\Ei \left (1, -b \,x^{3} \ln \relax (f )\right )}{6}-\frac {\ln \relax (x )}{2}-\frac {\ln \left (-b \right )}{6}+\frac {\ln \left (-b \,x^{3} \ln \relax (f )\right )}{6}-\frac {\ln \left (\ln \relax (f )\right )}{6}+\frac {1}{2 b \,x^{3} \ln \relax (f )}+\frac {1}{2 b^{2} x^{6} \ln \relax (f )^{2}}+\frac {\left (4 b^{2} x^{6} \ln \relax (f )^{2}+4 b \,x^{3} \ln \relax (f )+8\right ) {\mathrm e}^{b \,x^{3} \ln \relax (f )}}{24 b^{3} x^{9} \ln \relax (f )^{3}}-\frac {22 b^{3} x^{9} \ln \relax (f )^{3}+36 b^{2} x^{6} \ln \relax (f )^{2}+36 b \,x^{3} \ln \relax (f )+24}{72 b^{3} x^{9} \ln \relax (f )^{3}}+\frac {1}{3 b^{3} x^{9} \ln \relax (f )^{3}}+\frac {11}{36}\right ) b^{3} f^{a} \ln \relax (f )^{3}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)/x^10,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))+
11/36-1/2*ln(x)-1/6*ln(-b)-1/6*ln(ln(f))+1/3/x^9/b^3/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 = 1.18, size = 22, normalized size = 0.27 \[ \frac {1}{3} \, b^{3} f^{a} \Gamma \left (-3, -b x^{3} \log \relax (f)\right ) \log \relax (f)^{3} \]

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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mupad [B]  time = 3.53, size = 69, normalized size = 0.85 \[ -\frac {b^3\,f^a\,{\ln \relax (f)}^3\,\left (f^{b\,x^3}\,\left (\frac {1}{6\,b\,x^3\,\ln \relax (f)}+\frac {1}{6\,b^2\,x^6\,{\ln \relax (f)}^2}+\frac {1}{3\,b^3\,x^9\,{\ln \relax (f)}^3}\right )+\frac {\mathrm {expint}\left (-b\,x^3\,\ln \relax (f)\right )}{6}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3)/x^10,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{3}}}{x^{10}}\, dx \]

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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