3.91 \(\int \frac{f^{a+b x^2}}{x^6} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0805698, antiderivative size = 96, normalized size of antiderivative = 1., 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, 2204} \[ \frac{4}{15} \sqrt{\pi } b^{5/2} f^a \log ^{\frac{5}{2}}(f) \text{Erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right )-\frac{4 b^2 \log ^2(f) f^{a+b x^2}}{15 x}-\frac{f^{a+b x^2}}{5 x^5}-\frac{2 b \log (f) f^{a+b x^2}}{15 x^3} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^6,x]

[Out]

-f^(a + b*x^2)/(5*x^5) - (2*b*f^(a + b*x^2)*Log[f])/(15*x^3) - (4*b^2*f^(a + b*x^2)*Log[f]^2)/(15*x) + (4*b^(5
/2)*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*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 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 \frac{f^{a+b x^2}}{x^6} \, dx &=-\frac{f^{a+b x^2}}{5 x^5}+\frac{1}{5} (2 b \log (f)) \int \frac{f^{a+b x^2}}{x^4} \, dx\\ &=-\frac{f^{a+b x^2}}{5 x^5}-\frac{2 b f^{a+b x^2} \log (f)}{15 x^3}+\frac{1}{15} \left (4 b^2 \log ^2(f)\right ) \int \frac{f^{a+b x^2}}{x^2} \, dx\\ &=-\frac{f^{a+b x^2}}{5 x^5}-\frac{2 b f^{a+b x^2} \log (f)}{15 x^3}-\frac{4 b^2 f^{a+b x^2} \log ^2(f)}{15 x}+\frac{1}{15} \left (8 b^3 \log ^3(f)\right ) \int f^{a+b x^2} \, dx\\ &=-\frac{f^{a+b x^2}}{5 x^5}-\frac{2 b f^{a+b x^2} \log (f)}{15 x^3}-\frac{4 b^2 f^{a+b x^2} \log ^2(f)}{15 x}+\frac{4}{15} b^{5/2} f^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right ) \log ^{\frac{5}{2}}(f)\\ \end{align*}

Mathematica [A]  time = 0.0343455, size = 77, normalized size = 0.8 \[ \frac{f^a \left (4 \sqrt{\pi } b^{5/2} x^5 \log ^{\frac{5}{2}}(f) \text{Erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right )-f^{b x^2} \left (4 b^2 x^4 \log ^2(f)+2 b x^2 \log (f)+3\right )\right )}{15 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^6,x]

[Out]

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

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Maple [A]  time = 0.034, size = 89, normalized size = 0.9 \begin{align*} -{\frac{{f}^{a}{f}^{b{x}^{2}}}{5\,{x}^{5}}}-{\frac{2\,{f}^{a}\ln \left ( f \right ) b{f}^{b{x}^{2}}}{15\,{x}^{3}}}-{\frac{4\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{f}^{b{x}^{2}}}{15\,x}}+{\frac{4\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}\sqrt{\pi }}{15}{\it Erf} \left ( \sqrt{-b\ln \left ( f \right ) }x \right ){\frac{1}{\sqrt{-b\ln \left ( f \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)/x^6,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 = 1.23129, size = 38, normalized size = 0.4 \begin{align*} -\frac{\left (-b x^{2} \log \left (f\right )\right )^{\frac{5}{2}} f^{a} \Gamma \left (-\frac{5}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^6,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73238, size = 192, normalized size = 2. \begin{align*} -\frac{4 \, \sqrt{\pi } \sqrt{-b \log \left (f\right )} b^{2} f^{a} x^{5} \operatorname{erf}\left (\sqrt{-b \log \left (f\right )} x\right ) \log \left (f\right )^{2} +{\left (4 \, b^{2} x^{4} \log \left (f\right )^{2} + 2 \, b x^{2} \log \left (f\right ) + 3\right )} f^{b x^{2} + a}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^6,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x^{2}}}{x^{6}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)/x**6,x)

[Out]

Integral(f**(a + b*x**2)/x**6, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{b x^{2} + a}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^6,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^6, x)