3.104 \(\int \frac {f^{a+b x^3}}{x^7} \, dx\)

Optimal. Leaf size=58 \[ \frac {1}{6} b^2 f^a \log ^2(f) \text {Ei}\left (b x^3 \log (f)\right )-\frac {b \log (f) f^{a+b x^3}}{6 x^3}-\frac {f^{a+b x^3}}{6 x^6} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2214, 2210} \[ \frac {1}{6} b^2 f^a \log ^2(f) \text {Ei}\left (b x^3 \log (f)\right )-\frac {f^{a+b x^3}}{6 x^6}-\frac {b \log (f) f^{a+b x^3}}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)/x^7,x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.83 \[ \frac {f^a \left (b^2 x^6 \log ^2(f) \text {Ei}\left (b x^3 \log (f)\right )-f^{b x^3} \left (b x^3 \log (f)+1\right )\right )}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3)/x^7,x]

[Out]

(f^a*(b^2*x^6*ExpIntegralEi[b*x^3*Log[f]]*Log[f]^2 - f^(b*x^3)*(1 + b*x^3*Log[f])))/(6*x^6)

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fricas [A]  time = 0.41, size = 48, normalized size = 0.83 \[ \frac {b^{2} f^{a} x^{6} {\rm Ei}\left (b x^{3} \log \relax (f)\right ) \log \relax (f)^{2} - {\left (b x^{3} \log \relax (f) + 1\right )} f^{b x^{3} + a}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^7,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{3} + a}}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^7,x, algorithm="giac")

[Out]

integrate(f^(b*x^3 + a)/x^7, x)

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maple [B]  time = 0.06, size = 141, normalized size = 2.43 \[ \frac {\left (-\frac {\Ei \left (1, -b \,x^{3} \ln \relax (f )\right )}{2}+\frac {3 \ln \relax (x )}{2}+\frac {\ln \left (-b \right )}{2}-\frac {\ln \left (-b \,x^{3} \ln \relax (f )\right )}{2}+\frac {\ln \left (\ln \relax (f )\right )}{2}-\frac {1}{b \,x^{3} \ln \relax (f )}-\frac {\left (3 b \,x^{3} \ln \relax (f )+3\right ) {\mathrm e}^{b \,x^{3} \ln \relax (f )}}{6 b^{2} x^{6} \ln \relax (f )^{2}}+\frac {9 b^{2} x^{6} \ln \relax (f )^{2}+12 b \,x^{3} \ln \relax (f )+6}{12 b^{2} x^{6} \ln \relax (f )^{2}}-\frac {1}{2 b^{2} x^{6} \ln \relax (f )^{2}}-\frac {3}{4}\right ) b^{2} f^{a} \ln \relax (f )^{2}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)/x^7,x)

[Out]

1/3*f^a*b^2*ln(f)^2*(1/12/b^2/x^6/ln(f)^2*(9*b^2*x^6*ln(f)^2+12*b*x^3*ln(f)+6)-1/6/b^2/x^6/ln(f)^2*(3+3*b*x^3*
ln(f))*exp(b*x^3*ln(f))-1/2*ln(-b*x^3*ln(f))-1/2*Ei(1,-b*x^3*ln(f))-3/4+3/2*ln(x)+1/2*ln(-b)+1/2*ln(ln(f))-1/2
/x^6/b^2/ln(f)^2-1/b/x^3/ln(f))

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maxima [A]  time = 1.19, size = 22, normalized size = 0.38 \[ -\frac {1}{3} \, b^{2} f^{a} \Gamma \left (-2, -b x^{3} \log \relax (f)\right ) \log \relax (f)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)/x^7,x, algorithm="maxima")

[Out]

-1/3*b^2*f^a*gamma(-2, -b*x^3*log(f))*log(f)^2

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mupad [B]  time = 3.32, size = 57, normalized size = 0.98 \[ -\frac {b^2\,f^a\,{\ln \relax (f)}^2\,\left (f^{b\,x^3}\,\left (\frac {1}{2\,b\,x^3\,\ln \relax (f)}+\frac {1}{2\,b^2\,x^6\,{\ln \relax (f)}^2}\right )+\frac {\mathrm {expint}\left (-b\,x^3\,\ln \relax (f)\right )}{2}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3)/x^7,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{3}}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**3+a)/x**7,x)

[Out]

Integral(f**(a + b*x**3)/x**7, x)

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