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]

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

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 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]

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^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*}

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Mathematica [A]  time = 0.04, size = 77, normalized size = 0.80 \[ \frac {f^a \left (4 \sqrt {\pi } b^{5/2} x^5 \log ^{\frac {5}{2}}(f) \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )-f^{b x^2} \left (4 b^2 x^4 \log ^2(f)+2 b x^2 \log (f)+3\right )\right )}{15 x^5} \]

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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fricas [A]  time = 0.41, size = 73, normalized size = 0.76 \[ -\frac {4 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} b^{2} f^{a} x^{5} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right ) \log \relax (f)^{2} + {\left (4 \, b^{2} x^{4} \log \relax (f)^{2} + 2 \, b x^{2} \log \relax (f) + 3\right )} f^{b x^{2} + a}}{15 \, x^{5}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{2} + a}}{x^{6}}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 89, normalized size = 0.93 \[ \frac {4 \sqrt {\pi }\, b^{3} f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x \right ) \ln \relax (f )^{3}}{15 \sqrt {-b \ln \relax (f )}}-\frac {4 b^{2} f^{a} f^{b \,x^{2}} \ln \relax (f )^{2}}{15 x}-\frac {2 b \,f^{a} f^{b \,x^{2}} \ln \relax (f )}{15 x^{3}}-\frac {f^{a} f^{b \,x^{2}}}{5 x^{5}} \]

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 = 1.34, size = 28, normalized size = 0.29 \[ -\frac {\left (-b x^{2} \log \relax (f)\right )^{\frac {5}{2}} f^{a} \Gamma \left (-\frac {5}{2}, -b x^{2} \log \relax (f)\right )}{2 \, x^{5}} \]

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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mupad [B]  time = 3.56, size = 109, normalized size = 1.14 \[ \frac {4\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \relax (f)}\right )\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{5/2}}{15\,x^5}-\frac {4\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{5/2}}{15\,x^5}-\frac {f^a\,f^{b\,x^2}}{5\,x^5}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^2}{15\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \relax (f)}{15\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)/x^6,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{2}}}{x^{6}}\, dx \]

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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