3.106 \(\int \frac {f^{a+b x^3}}{x^{13}} \, dx\)

Optimal. Leaf size=24 \[ -\frac {1}{3} b^4 f^a \log ^4(f) \Gamma \left (-4,-b x^3 \log (f)\right ) \]

[Out]

-1/3*f^a/x^12*Ei(5,-b*x^3*ln(f))

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ -\frac {1}{3} b^4 f^a \log ^4(f) \text {Gamma}\left (-4,-b x^3 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)/x^13,x]

[Out]

-(b^4*f^a*Gamma[-4, -(b*x^3*Log[f])]*Log[f]^4)/3

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \int \frac {f^{a+b x^3}}{x^{13}} \, dx &=-\frac {1}{3} b^4 f^a \Gamma \left (-4,-b x^3 \log (f)\right ) \log ^4(f)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ -\frac {1}{3} b^4 f^a \log ^4(f) \Gamma \left (-4,-b x^3 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3)/x^13,x]

[Out]

-1/3*(b^4*f^a*Gamma[-4, -(b*x^3*Log[f])]*Log[f]^4)

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fricas [B]  time = 0.40, size = 71, normalized size = 2.96 \[ \frac {b^{4} f^{a} x^{12} {\rm Ei}\left (b x^{3} \log \relax (f)\right ) \log \relax (f)^{4} - {\left (b^{3} x^{9} \log \relax (f)^{3} + b^{2} x^{6} \log \relax (f)^{2} + 2 \, b x^{3} \log \relax (f) + 6\right )} f^{b x^{3} + a}}{72 \, x^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^13,x, algorithm="fricas")

[Out]

1/72*(b^4*f^a*x^12*Ei(b*x^3*log(f))*log(f)^4 - (b^3*x^9*log(f)^3 + b^2*x^6*log(f)^2 + 2*b*x^3*log(f) + 6)*f^(b
*x^3 + a))/x^12

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{3} + a}}{x^{13}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^13,x, algorithm="giac")

[Out]

integrate(f^(b*x^3 + a)/x^13, x)

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maple [B]  time = 0.08, size = 213, normalized size = 8.88 \[ \frac {\left (-\frac {\Ei \left (1, -b \,x^{3} \ln \relax (f )\right )}{24}+\frac {\ln \relax (x )}{8}+\frac {\ln \left (-b \right )}{24}-\frac {\ln \left (-b \,x^{3} \ln \relax (f )\right )}{24}+\frac {\ln \left (\ln \relax (f )\right )}{24}-\frac {1}{6 b \,x^{3} \ln \relax (f )}-\frac {1}{4 b^{2} x^{6} \ln \relax (f )^{2}}-\frac {1}{3 b^{3} x^{9} \ln \relax (f )^{3}}-\frac {\left (5 b^{3} x^{9} \ln \relax (f )^{3}+5 b^{2} x^{6} \ln \relax (f )^{2}+10 b \,x^{3} \ln \relax (f )+30\right ) {\mathrm e}^{b \,x^{3} \ln \relax (f )}}{120 b^{4} x^{12} \ln \relax (f )^{4}}+\frac {125 b^{4} x^{12} \ln \relax (f )^{4}+240 b^{3} x^{9} \ln \relax (f )^{3}+360 b^{2} x^{6} \ln \relax (f )^{2}+480 b \,x^{3} \ln \relax (f )+360}{1440 b^{4} x^{12} \ln \relax (f )^{4}}-\frac {1}{4 b^{4} x^{12} \ln \relax (f )^{4}}-\frac {25}{288}\right ) b^{4} f^{a} \ln \relax (f )^{4}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)/x^13,x)

[Out]

1/3*f^a*b^4*ln(f)^4*(1/1440/b^4/x^12/ln(f)^4*(125*b^4*x^12*ln(f)^4+240*b^3*x^9*ln(f)^3+360*b^2*x^6*ln(f)^2+480
*b*x^3*ln(f)+360)-1/120/b^4/x^12/ln(f)^4*(5*b^3*x^9*ln(f)^3+5*b^2*x^6*ln(f)^2+10*b*x^3*ln(f)+30)*exp(b*x^3*ln(
f))-1/24*ln(-b*x^3*ln(f))-1/24*Ei(1,-b*x^3*ln(f))-25/288+1/8*ln(x)+1/24*ln(-b)+1/24*ln(ln(f))-1/4/x^12/b^4/ln(
f)^4-1/3/b^3/x^9/ln(f)^3-1/4/b^2/x^6/ln(f)^2-1/6/b/x^3/ln(f))

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maxima [B]  time = 1.14, size = 22, normalized size = 0.92 \[ -\frac {1}{3} \, b^{4} f^{a} \Gamma \left (-4, -b x^{3} \log \relax (f)\right ) \log \relax (f)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^13,x, algorithm="maxima")

[Out]

-1/3*b^4*f^a*gamma(-4, -b*x^3*log(f))*log(f)^4

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mupad [B]  time = 3.58, size = 90, normalized size = 3.75 \[ -\frac {b^4\,f^a\,{\ln \relax (f)}^4\,\mathrm {expint}\left (-b\,x^3\,\ln \relax (f)\right )}{72}-\frac {b^4\,f^a\,f^{b\,x^3}\,{\ln \relax (f)}^4\,\left (\frac {1}{24\,b\,x^3\,\ln \relax (f)}+\frac {1}{24\,b^2\,x^6\,{\ln \relax (f)}^2}+\frac {1}{12\,b^3\,x^9\,{\ln \relax (f)}^3}+\frac {1}{4\,b^4\,x^{12}\,{\ln \relax (f)}^4}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3)/x^13,x)

[Out]

- (b^4*f^a*log(f)^4*expint(-b*x^3*log(f)))/72 - (b^4*f^a*f^(b*x^3)*log(f)^4*(1/(24*b*x^3*log(f)) + 1/(24*b^2*x
^6*log(f)^2) + 1/(12*b^3*x^9*log(f)^3) + 1/(4*b^4*x^12*log(f)^4)))/3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{3}}}{x^{13}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**3+a)/x**13,x)

[Out]

Integral(f**(a + b*x**3)/x**13, x)

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