3.1.91 \(\int \frac {f^{a+b x^2}}{x^6} \, dx\) [91]

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

[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 = 0.06, 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 = {2245, 2235} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^6,x]

[Out]

-1/5*f^(a + b*x^2)/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 2235

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 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 \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 = 0.06, size = 77, normalized size = 0.80 \begin {gather*} \frac {f^a \left (4 b^{5/2} \sqrt {\pi } x^5 \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {5}{2}}(f)-f^{b x^2} \left (3+2 b x^2 \log (f)+4 b^2 x^4 \log ^2(f)\right )\right )}{15 x^5} \end {gather*}

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 = 0.03, size = 86, normalized size = 0.90

method	result	size
meijerg	\(-\frac {f^{a} \ln \left (f \right )^{\frac {5}{2}} b^{3} \left (-\frac {2 \left (\frac {4 b^{2} x^{4} \ln \left (f \right )^{2}}{3}+\frac {2 b \,x^{2} \ln \left (f \right )}{3}+1\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{5 x^{5} \left (-b \right )^{\frac {5}{2}} \ln \left (f \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \erfi \left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{2 \sqrt {-b}}\)	\(86\)
risch	\(-\frac {f^{a} f^{b \,x^{2}}}{5 x^{5}}-\frac {2 f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{15 x^{3}}-\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{15 x}+\frac {4 f^{a} \ln \left (f \right )^{3} b^{3} \sqrt {\pi }\, \erf \left (\sqrt {-b \ln \left (f \right )}\, x \right )}{15 \sqrt {-b \ln \left (f \right )}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/2*f^a*ln(f)^(5/2)*b^3/(-b)^(1/2)*(-2/5/x^5/(-b)^(5/2)/ln(f)^(5/2)*(4/3*b^2*x^4*ln(f)^2+2/3*b*x^2*ln(f)+1)*e
xp(b*x^2*ln(f))+8/15/(-b)^(5/2)*b^(5/2)*Pi^(1/2)*erfi(x*b^(1/2)*ln(f)^(1/2)))

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Maxima [A]
time = 0.33, size = 28, normalized size = 0.29 \begin {gather*} -\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 {gather*}

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 = 0.38, size = 73, normalized size = 0.76 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {f^{a + b x^{2}}}{x^{6}}\, dx \end {gather*}

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 = 0.00, 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^(b*x^2+a)/x^6,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^6, x)

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

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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