\(\int \frac {f^{a+b x^2}}{x^8} \, dx\) [92]

Optimal result

Integrand size = 13, antiderivative size = 119 \[ \int \frac {f^{a+b x^2}}{x^8} \, dx=-\frac {f^{a+b x^2}}{7 x^7}-\frac {2 b f^{a+b x^2} \log (f)}{35 x^5}-\frac {4 b^2 f^{a+b x^2} \log ^2(f)}{105 x^3}-\frac {8 b^3 f^{a+b x^2} \log ^3(f)}{105 x}+\frac {8}{105} b^{7/2} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {7}{2}}(f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2235} \[ \int \frac {f^{a+b x^2}}{x^8} \, dx=\frac {8}{105} \sqrt {\pi } b^{7/2} f^a \log ^{\frac {7}{2}}(f) \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )-\frac {8 b^3 \log ^3(f) f^{a+b x^2}}{105 x}-\frac {4 b^2 \log ^2(f) f^{a+b x^2}}{105 x^3}-\frac {f^{a+b x^2}}{7 x^7}-\frac {2 b \log (f) f^{a+b x^2}}{35 x^5} \]

[In]

Int[f^(a + b*x^2)/x^8,x]

[Out]

-1/7*f^(a + b*x^2)/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 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*} \text {integral}& = -\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] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \frac {f^{a+b x^2}}{x^8} \, dx=\frac {f^a \left (8 b^{7/2} \sqrt {\pi } x^7 \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {7}{2}}(f)-f^{b x^2} \left (15+6 b x^2 \log (f)+4 b^2 x^4 \log ^2(f)+8 b^3 x^6 \log ^3(f)\right )\right )}{105 x^7} \]

[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)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82

method	result	size
meijerg	\(\frac {f^{a} b^{4} \ln \left (f \right )^{\frac {7}{2}} \left (-\frac {2 \left (\frac {8 b^{3} x^{6} \ln \left (f \right )^{3}}{15}+\frac {4 b^{2} x^{4} \ln \left (f \right )^{2}}{15}+\frac {2 b \,x^{2} \ln \left (f \right )}{5}+1\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{7 x^{7} \left (-b \right )^{\frac {7}{2}} \ln \left (f \right )^{\frac {7}{2}}}+\frac {16 b^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{105 \left (-b \right )^{\frac {7}{2}}}\right )}{2 \sqrt {-b}}\)	\(98\)
risch	\(-\frac {f^{a} f^{b \,x^{2}}}{7 x^{7}}-\frac {2 f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{35 x^{5}}-\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{105 x^{3}}-\frac {8 f^{a} \ln \left (f \right )^{3} b^{3} f^{b \,x^{2}}}{105 x}+\frac {8 f^{a} \ln \left (f \right )^{4} b^{4} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{105 \sqrt {-b \ln \left (f \right )}}\)	\(111\)

[In]

int(f^(b*x^2+a)/x^8,x,method=_RETURNVERBOSE)

[Out]

1/2*f^a*b^4*ln(f)^(7/2)/(-b)^(1/2)*(-2/7/x^7/(-b)^(7/2)/ln(f)^(7/2)*(8/15*b^3*x^6*ln(f)^3+4/15*b^2*x^4*ln(f)^2
+2/5*b*x^2*ln(f)+1)*exp(b*x^2*ln(f))+16/105/(-b)^(7/2)*b^(7/2)*Pi^(1/2)*erfi(x*b^(1/2)*ln(f)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71 \[ \int \frac {f^{a+b x^2}}{x^8} \, dx=-\frac {8 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{3} f^{a} x^{7} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) \log \left (f\right )^{3} + {\left (8 \, b^{3} x^{6} \log \left (f\right )^{3} + 4 \, b^{2} x^{4} \log \left (f\right )^{2} + 6 \, b x^{2} \log \left (f\right ) + 15\right )} f^{b x^{2} + a}}{105 \, x^{7}} \]

[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

Sympy [F]

\[ \int \frac {f^{a+b x^2}}{x^8} \, dx=\int \frac {f^{a + b x^{2}}}{x^{8}}\, dx \]

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**8, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.24 \[ \int \frac {f^{a+b x^2}}{x^8} \, dx=-\frac {\left (-b x^{2} \log \left (f\right )\right )^{\frac {7}{2}} f^{a} \Gamma \left (-\frac {7}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{7}} \]

[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

Giac [F]

\[ \int \frac {f^{a+b x^2}}{x^8} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{8}} \,d x } \]

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^8, x)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10 \[ \int \frac {f^{a+b x^2}}{x^8} \, dx=\frac {8\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{7/2}}{105\,x^7}-\frac {f^a\,f^{b\,x^2}}{7\,x^7}-\frac {8\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \left (f\right )}\right )\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{7/2}}{105\,x^7}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^2}{105\,x^3}-\frac {8\,b^3\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^3}{105\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \left (f\right )}{35\,x^5} \]

[In]

int(f^(a + b*x^2)/x^8,x)

[Out]

(8*f^a*pi^(1/2)*(-b*x^2*log(f))^(7/2))/(105*x^7) - (f^a*f^(b*x^2))/(7*x^7) - (8*f^a*pi^(1/2)*erfc((-b*x^2*log(
f))^(1/2))*(-b*x^2*log(f))^(7/2))/(105*x^7) - (4*b^2*f^a*f^(b*x^2)*log(f)^2)/(105*x^3) - (8*b^3*f^a*f^(b*x^2)*
log(f)^3)/(105*x) - (2*b*f^a*f^(b*x^2)*log(f))/(35*x^5)