3.2.18 \(\int f^{a+\frac {b}{x}} x^2 \, dx\) [118]

Optimal. Leaf size=79 \[ \frac {1}{3} f^{a+\frac {b}{x}} x^3+\frac {1}{6} b f^{a+\frac {b}{x}} x^2 \log (f)+\frac {1}{6} b^2 f^{a+\frac {b}{x}} x \log ^2(f)-\frac {1}{6} b^3 f^a \text {Ei}\left (\frac {b \log (f)}{x}\right ) \log ^3(f) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2245, 2237, 2241} \begin {gather*} -\frac {1}{6} b^3 f^a \log ^3(f) \text {Ei}\left (\frac {b \log (f)}{x}\right )+\frac {1}{6} b^2 x \log ^2(f) f^{a+\frac {b}{x}}+\frac {1}{3} x^3 f^{a+\frac {b}{x}}+\frac {1}{6} b x^2 \log (f) f^{a+\frac {b}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x)*x^2,x]

[Out]

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

Rule 2237

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2241

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 2245

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

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Mathematica [A]
time = 0.02, size = 53, normalized size = 0.67 \begin {gather*} \frac {1}{6} f^a \left (-b^3 \text {Ei}\left (\frac {b \log (f)}{x}\right ) \log ^3(f)+f^{b/x} x \left (2 x^2+b x \log (f)+b^2 \log ^2(f)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x)*x^2,x]

[Out]

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

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Maple [A]
time = 0.07, size = 77, normalized size = 0.97

method	result	size
risch	\(\frac {f^{a} f^{\frac {b}{x}} x^{3}}{3}+\frac {b \ln \left (f \right ) f^{a} f^{\frac {b}{x}} x^{2}}{6}+\frac {b^{2} \ln \left (f \right )^{2} f^{a} f^{\frac {b}{x}} x}{6}+\frac {b^{3} \ln \left (f \right )^{3} f^{a} \expIntegral \left (1, -\frac {b \ln \left (f \right )}{x}\right )}{6}\)	\(77\)
meijerg	\(f^{a} \ln \left (f \right )^{3} b^{3} \left (-\frac {x^{3} \left (\frac {22 b^{3} \ln \left (f \right )^{3}}{x^{3}}+\frac {36 b^{2} \ln \left (f \right )^{2}}{x^{2}}+\frac {36 b \ln \left (f \right )}{x}+24\right )}{72 b^{3} \ln \left (f \right )^{3}}+\frac {x^{3} \left (\frac {4 b^{2} \ln \left (f \right )^{2}}{x^{2}}+\frac {4 b \ln \left (f \right )}{x}+8\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x}}}{24 b^{3} \ln \left (f \right )^{3}}+\frac {\ln \left (-\frac {b \ln \left (f \right )}{x}\right )}{6}+\frac {\expIntegral \left (1, -\frac {b \ln \left (f \right )}{x}\right )}{6}+\frac {11}{36}+\frac {\ln \left (x \right )}{6}-\frac {\ln \left (-b \right )}{6}-\frac {\ln \left (\ln \left (f \right )\right )}{6}+\frac {x^{3}}{3 b^{3} \ln \left (f \right )^{3}}+\frac {x^{2}}{2 b^{2} \ln \left (f \right )^{2}}+\frac {x}{2 \ln \left (f \right ) b}\right )\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x)*x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 22, normalized size = 0.28 \begin {gather*} -b^{3} f^{a} \Gamma \left (-3, -\frac {b \log \left (f\right )}{x}\right ) \log \left (f\right )^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x)*x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.46, size = 56, normalized size = 0.71 \begin {gather*} -\frac {1}{6} \, b^{3} f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x}\right ) \log \left (f\right )^{3} + \frac {1}{6} \, {\left (b^{2} x \log \left (f\right )^{2} + b x^{2} \log \left (f\right ) + 2 \, x^{3}\right )} f^{\frac {a x + b}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x)*x^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int f^{a + \frac {b}{x}} x^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x)*x**2,x)

[Out]

Integral(f**(a + b/x)*x**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x)*x^2,x, algorithm="giac")

[Out]

integrate(f^(a + b/x)*x^2, x)

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Mupad [B]
time = 3.60, size = 66, normalized size = 0.84 \begin {gather*} b^3\,f^a\,{\ln \left (f\right )}^3\,\left (f^{b/x}\,\left (\frac {x^2}{6\,b^2\,{\ln \left (f\right )}^2}+\frac {x^3}{3\,b^3\,{\ln \left (f\right )}^3}+\frac {x}{6\,b\,\ln \left (f\right )}\right )+\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x}\right )}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x)*x^2,x)

[Out]

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

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