3.144 \(\int f^{a+\frac{b}{x^2}} x^4 \, dx\)

Optimal. Leaf size=96 \[ -\frac{4}{15} \sqrt{\pi } b^{5/2} f^a \log ^{\frac{5}{2}}(f) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )+\frac{4}{15} b^2 x \log ^2(f) f^{a+\frac{b}{x^2}}+\frac{1}{5} x^5 f^{a+\frac{b}{x^2}}+\frac{2}{15} b x^3 \log (f) f^{a+\frac{b}{x^2}} \]

[Out]

(f^(a + b/x^2)*x^5)/5 + (2*b*f^(a + b/x^2)*x^3*Log[f])/15 + (4*b^2*f^(a + b/x^2)*x*Log[f]^2)/15 - (4*b^(5/2)*f
^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Log[f]^(5/2))/15

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Rubi [A]  time = 0.0881798, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {2214, 2206, 2211, 2204} \[ -\frac{4}{15} \sqrt{\pi } b^{5/2} f^a \log ^{\frac{5}{2}}(f) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )+\frac{4}{15} b^2 x \log ^2(f) f^{a+\frac{b}{x^2}}+\frac{1}{5} x^5 f^{a+\frac{b}{x^2}}+\frac{2}{15} b x^3 \log (f) f^{a+\frac{b}{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^4,x]

[Out]

(f^(a + b/x^2)*x^5)/5 + (2*b*f^(a + b/x^2)*x^3*Log[f])/15 + (4*b^2*f^(a + b/x^2)*x*Log[f]^2)/15 - (4*b^(5/2)*f
^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Log[f]^(5/2))/15

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 2206

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 2211

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rubi steps

\begin{align*} \int f^{a+\frac{b}{x^2}} x^4 \, dx &=\frac{1}{5} f^{a+\frac{b}{x^2}} x^5+\frac{1}{5} (2 b \log (f)) \int f^{a+\frac{b}{x^2}} x^2 \, dx\\ &=\frac{1}{5} f^{a+\frac{b}{x^2}} x^5+\frac{2}{15} b f^{a+\frac{b}{x^2}} x^3 \log (f)+\frac{1}{15} \left (4 b^2 \log ^2(f)\right ) \int f^{a+\frac{b}{x^2}} \, dx\\ &=\frac{1}{5} f^{a+\frac{b}{x^2}} x^5+\frac{2}{15} b f^{a+\frac{b}{x^2}} x^3 \log (f)+\frac{4}{15} b^2 f^{a+\frac{b}{x^2}} x \log ^2(f)+\frac{1}{15} \left (8 b^3 \log ^3(f)\right ) \int \frac{f^{a+\frac{b}{x^2}}}{x^2} \, dx\\ &=\frac{1}{5} f^{a+\frac{b}{x^2}} x^5+\frac{2}{15} b f^{a+\frac{b}{x^2}} x^3 \log (f)+\frac{4}{15} b^2 f^{a+\frac{b}{x^2}} x \log ^2(f)-\frac{1}{15} \left (8 b^3 \log ^3(f)\right ) \operatorname{Subst}\left (\int f^{a+b x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{5} f^{a+\frac{b}{x^2}} x^5+\frac{2}{15} b f^{a+\frac{b}{x^2}} x^3 \log (f)+\frac{4}{15} b^2 f^{a+\frac{b}{x^2}} x \log ^2(f)-\frac{4}{15} b^{5/2} f^a \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right ) \log ^{\frac{5}{2}}(f)\\ \end{align*}

Mathematica [A]  time = 0.032388, size = 74, normalized size = 0.77 \[ \frac{1}{15} f^a \left (x f^{\frac{b}{x^2}} \left (4 b^2 \log ^2(f)+2 b x^2 \log (f)+3 x^4\right )-4 \sqrt{\pi } b^{5/2} \log ^{\frac{5}{2}}(f) \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^4,x]

[Out]

(f^a*(-4*b^(5/2)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Log[f]^(5/2) + f^(b/x^2)*x*(3*x^4 + 2*b*x^2*Log[f] +
4*b^2*Log[f]^2)))/15

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Maple [A]  time = 0.033, size = 89, normalized size = 0.9 \begin{align*}{\frac{{f}^{a}{x}^{5}}{5}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{2\,{f}^{a}\ln \left ( f \right ) b{x}^{3}}{15}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{4\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}x}{15}{f}^{{\frac{b}{{x}^{2}}}}}-{\frac{4\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}\sqrt{\pi }}{15}{\it Erf} \left ({\frac{1}{x}\sqrt{-b\ln \left ( f \right ) }} \right ){\frac{1}{\sqrt{-b\ln \left ( f \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^4,x)

[Out]

1/5*f^a*x^5*f^(b/x^2)+2/15*f^a*ln(f)*b*x^3*f^(b/x^2)+4/15*f^a*ln(f)^2*b^2*x*f^(b/x^2)-4/15*f^a*ln(f)^3*b^3*Pi^
(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)/x)

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Maxima [A]  time = 1.28243, size = 38, normalized size = 0.4 \begin{align*} \frac{1}{2} \, f^{a} x^{5} \left (-\frac{b \log \left (f\right )}{x^{2}}\right )^{\frac{5}{2}} \Gamma \left (-\frac{5}{2}, -\frac{b \log \left (f\right )}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^4,x, algorithm="maxima")

[Out]

1/2*f^a*x^5*(-b*log(f)/x^2)^(5/2)*gamma(-5/2, -b*log(f)/x^2)

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Fricas [A]  time = 1.82148, size = 192, normalized size = 2. \begin{align*} \frac{4}{15} \, \sqrt{\pi } \sqrt{-b \log \left (f\right )} b^{2} f^{a} \operatorname{erf}\left (\frac{\sqrt{-b \log \left (f\right )}}{x}\right ) \log \left (f\right )^{2} + \frac{1}{15} \,{\left (3 \, x^{5} + 2 \, b x^{3} \log \left (f\right ) + 4 \, b^{2} x \log \left (f\right )^{2}\right )} f^{\frac{a x^{2} + b}{x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^4,x, algorithm="fricas")

[Out]

4/15*sqrt(pi)*sqrt(-b*log(f))*b^2*f^a*erf(sqrt(-b*log(f))/x)*log(f)^2 + 1/15*(3*x^5 + 2*b*x^3*log(f) + 4*b^2*x
*log(f)^2)*f^((a*x^2 + b)/x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{2}}} x^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**2)*x**4,x)

[Out]

Integral(f**(a + b/x**2)*x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + \frac{b}{x^{2}}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)*x^4,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)*x^4, x)