3.92 \(\int \frac{f^{a+b x^2}}{x^8} \, dx\)

Optimal. Leaf size=119 \[ \frac{8}{105} \sqrt{\pi } b^{7/2} f^a \log ^{\frac{7}{2}}(f) \text{Erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right )-\frac{8 b^3 \log ^3(f) f^{a+b x^2}}{105 x}-\frac{4 b^2 \log ^2(f) f^{a+b x^2}}{105 x^3}-\frac{f^{a+b x^2}}{7 x^7}-\frac{2 b \log (f) f^{a+b x^2}}{35 x^5} \]

[Out]

-f^(a + b*x^2)/(7*x^7) - (2*b*f^(a + b*x^2)*Log[f])/(35*x^5) - (4*b^2*f^(a + b*x^2)*Log[f]^2)/(105*x^3) - (8*b
^3*f^(a + b*x^2)*Log[f]^3)/(105*x) + (8*b^(7/2)*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*Log[f]^(7/2))/105

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Rubi [A]  time = 0.107872, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2214, 2204} \[ \frac{8}{105} \sqrt{\pi } b^{7/2} f^a \log ^{\frac{7}{2}}(f) \text{Erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right )-\frac{8 b^3 \log ^3(f) f^{a+b x^2}}{105 x}-\frac{4 b^2 \log ^2(f) f^{a+b x^2}}{105 x^3}-\frac{f^{a+b x^2}}{7 x^7}-\frac{2 b \log (f) f^{a+b x^2}}{35 x^5} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^8,x]

[Out]

-f^(a + b*x^2)/(7*x^7) - (2*b*f^(a + b*x^2)*Log[f])/(35*x^5) - (4*b^2*f^(a + b*x^2)*Log[f]^2)/(105*x^3) - (8*b
^3*f^(a + b*x^2)*Log[f]^3)/(105*x) + (8*b^(7/2)*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*Log[f]^(7/2))/105

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^8} \, dx &=-\frac{f^{a+b x^2}}{7 x^7}+\frac{1}{7} (2 b \log (f)) \int \frac{f^{a+b x^2}}{x^6} \, dx\\ &=-\frac{f^{a+b x^2}}{7 x^7}-\frac{2 b f^{a+b x^2} \log (f)}{35 x^5}+\frac{1}{35} \left (4 b^2 \log ^2(f)\right ) \int \frac{f^{a+b x^2}}{x^4} \, dx\\ &=-\frac{f^{a+b x^2}}{7 x^7}-\frac{2 b f^{a+b x^2} \log (f)}{35 x^5}-\frac{4 b^2 f^{a+b x^2} \log ^2(f)}{105 x^3}+\frac{1}{105} \left (8 b^3 \log ^3(f)\right ) \int \frac{f^{a+b x^2}}{x^2} \, dx\\ &=-\frac{f^{a+b x^2}}{7 x^7}-\frac{2 b f^{a+b x^2} \log (f)}{35 x^5}-\frac{4 b^2 f^{a+b x^2} \log ^2(f)}{105 x^3}-\frac{8 b^3 f^{a+b x^2} \log ^3(f)}{105 x}+\frac{1}{105} \left (16 b^4 \log ^4(f)\right ) \int f^{a+b x^2} \, dx\\ &=-\frac{f^{a+b x^2}}{7 x^7}-\frac{2 b f^{a+b x^2} \log (f)}{35 x^5}-\frac{4 b^2 f^{a+b x^2} \log ^2(f)}{105 x^3}-\frac{8 b^3 f^{a+b x^2} \log ^3(f)}{105 x}+\frac{8}{105} b^{7/2} f^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right ) \log ^{\frac{7}{2}}(f)\\ \end{align*}

Mathematica [A]  time = 0.0414604, size = 89, normalized size = 0.75 \[ \frac{f^a \left (8 \sqrt{\pi } b^{7/2} x^7 \log ^{\frac{7}{2}}(f) \text{Erfi}\left (\sqrt{b} x \sqrt{\log (f)}\right )-f^{b x^2} \left (8 b^3 x^6 \log ^3(f)+4 b^2 x^4 \log ^2(f)+6 b x^2 \log (f)+15\right )\right )}{105 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^8,x]

[Out]

(f^a*(8*b^(7/2)*Sqrt[Pi]*x^7*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*Log[f]^(7/2) - f^(b*x^2)*(15 + 6*b*x^2*Log[f] + 4*b^
2*x^4*Log[f]^2 + 8*b^3*x^6*Log[f]^3)))/(105*x^7)

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Maple [A]  time = 0.044, size = 111, normalized size = 0.9 \begin{align*} -{\frac{{f}^{a}{f}^{b{x}^{2}}}{7\,{x}^{7}}}-{\frac{2\,{f}^{a}\ln \left ( f \right ) b{f}^{b{x}^{2}}}{35\,{x}^{5}}}-{\frac{4\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{f}^{b{x}^{2}}}{105\,{x}^{3}}}-{\frac{8\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}{f}^{b{x}^{2}}}{105\,x}}+{\frac{8\,{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{4}{b}^{4}\sqrt{\pi }}{105}{\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^8,x)

[Out]

-1/7*f^a/x^7*f^(b*x^2)-2/35*f^a*ln(f)*b/x^5*f^(b*x^2)-4/105*f^a*ln(f)^2*b^2/x^3*f^(b*x^2)-8/105*f^a*ln(f)^3*b^
3/x*f^(b*x^2)+8/105*f^a*ln(f)^4*b^4*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)*x)

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Maxima [A]  time = 1.17596, size = 38, normalized size = 0.32 \begin{align*} -\frac{\left (-b x^{2} \log \left (f\right )\right )^{\frac{7}{2}} f^{a} \Gamma \left (-\frac{7}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^8,x, algorithm="maxima")

[Out]

-1/2*(-b*x^2*log(f))^(7/2)*f^a*gamma(-7/2, -b*x^2*log(f))/x^7

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Fricas [A]  time = 1.74106, size = 223, normalized size = 1.87 \begin{align*} -\frac{8 \, \sqrt{\pi } \sqrt{-b \log \left (f\right )} b^{3} f^{a} x^{7} \operatorname{erf}\left (\sqrt{-b \log \left (f\right )} x\right ) \log \left (f\right )^{3} +{\left (8 \, b^{3} x^{6} \log \left (f\right )^{3} + 4 \, b^{2} x^{4} \log \left (f\right )^{2} + 6 \, b x^{2} \log \left (f\right ) + 15\right )} f^{b x^{2} + a}}{105 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^8,x, algorithm="fricas")

[Out]

-1/105*(8*sqrt(pi)*sqrt(-b*log(f))*b^3*f^a*x^7*erf(sqrt(-b*log(f))*x)*log(f)^3 + (8*b^3*x^6*log(f)^3 + 4*b^2*x
^4*log(f)^2 + 6*b*x^2*log(f) + 15)*f^(b*x^2 + a))/x^7

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x^{2}}}{x^{8}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)/x**8,x)

[Out]

Integral(f**(a + b*x**2)/x**8, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{b x^{2} + a}}{x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^8,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^8, x)