\(\int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx\) [85]

Optimal result

Integrand size = 13, antiderivative size = 132 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=-\frac {105 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{32 b^{9/2} \log ^{\frac {9}{2}}(f)}+\frac {105 f^{a+\frac {b}{x^2}}}{16 b^4 x \log ^4(f)}-\frac {35 f^{a+\frac {b}{x^2}}}{8 b^3 x^3 \log ^3(f)}+\frac {7 f^{a+\frac {b}{x^2}}}{4 b^2 x^5 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)} \] Output:

-105/32*f^a*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x)/b^(9/2)/ln(f)^(9/2)+105/1 
6*f^(a+b/x^2)/b^4/x/ln(f)^4-35/8*f^(a+b/x^2)/b^3/x^3/ln(f)^3+7/4*f^(a+b/x^ 
2)/b^2/x^5/ln(f)^2-1/2*f^(a+b/x^2)/b/x^7/ln(f)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\frac {f^a \left (-105 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )+\frac {2 \sqrt {b} f^{\frac {b}{x^2}} \sqrt {\log (f)} \left (105 x^6-70 b x^4 \log (f)+28 b^2 x^2 \log ^2(f)-8 b^3 \log ^3(f)\right )}{x^7}\right )}{32 b^{9/2} \log ^{\frac {9}{2}}(f)} \] Input:

Integrate[f^(a + b/x^2)/x^10,x]
 

Output:

(f^a*(-105*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x] + (2*Sqrt[b]*f^(b/x^2)* 
Sqrt[Log[f]]*(105*x^6 - 70*b*x^4*Log[f] + 28*b^2*x^2*Log[f]^2 - 8*b^3*Log[ 
f]^3))/x^7))/(32*b^(9/2)*Log[f]^(9/2))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2641, 2641, 2641, 2641, 2640, 2633}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {7 \int \frac {f^{a+\frac {b}{x^2}}}{x^8}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {7 \left (-\frac {5 \int \frac {f^{a+\frac {b}{x^2}}}{x^6}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \int \frac {f^{a+\frac {b}{x^2}}}{x^4}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\)

\(\Big \downarrow \) 2641

\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (-\frac {\int \frac {f^{a+\frac {b}{x^2}}}{x^2}dx}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\)

\(\Big \downarrow \) 2640

\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\int f^{a+\frac {b}{x^2}}d\frac {1}{x}}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\)

\(\Big \downarrow \) 2633

\(\displaystyle -\frac {7 \left (-\frac {5 \left (-\frac {3 \left (\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}\right )}{2 b \log (f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^7 \log (f)}\)

Input:

Int[f^(a + b/x^2)/x^10,x]
 

Output:

-1/2*f^(a + b/x^2)/(b*x^7*Log[f]) - (7*(-1/2*f^(a + b/x^2)/(b*x^5*Log[f]) 
- (5*(-1/2*f^(a + b/x^2)/(b*x^3*Log[f]) - (3*((f^a*Sqrt[Pi]*Erfi[(Sqrt[b]* 
Sqrt[Log[f]])/x])/(4*b^(3/2)*Log[f]^(3/2)) - f^(a + b/x^2)/(2*b*x*Log[f])) 
)/(2*b*Log[f])))/(2*b*Log[f])))/(2*b*Log[f])
 

Defintions of rubi rules used

rule 2633

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 2640

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[1/(d*(m + 1))   Subst[Int[F^(a + b*x^2), x], x, (c + 
d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1)]
 

rule 2641

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(c + d*x)^(m - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L 
og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F])   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[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n 
, 0])
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.78

method	result	size
meijerg	\(\frac {f^{a} \sqrt {-b}\, \left (-\frac {\left (-b \right )^{\frac {9}{2}} \sqrt {\ln \left (f \right )}\, \left (-\frac {72 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {252 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {630 b \ln \left (f \right )}{x^{2}}+945\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{72 x \,b^{4}}+\frac {105 \left (-b \right )^{\frac {9}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{16 b^{\frac {9}{2}}}\right )}{2 b^{5} \ln \left (f \right )^{\frac {9}{2}}}\)	\(103\)
risch	\(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{7} b \ln \left (f \right )}+\frac {7 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x^{5}}-\frac {35 f^{a} f^{\frac {b}{x^{2}}}}{8 \ln \left (f \right )^{3} b^{3} x^{3}}+\frac {105 f^{a} f^{\frac {b}{x^{2}}}}{16 \ln \left (f \right )^{4} b^{4} x}-\frac {105 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{32 \ln \left (f \right )^{4} b^{4} \sqrt {-b \ln \left (f \right )}}\)	\(124\)

Input:

int(f^(a+b/x^2)/x^10,x,method=_RETURNVERBOSE)
 

Output:

1/2*f^a/b^5/ln(f)^(9/2)*(-b)^(1/2)*(-1/72/x*(-b)^(9/2)*ln(f)^(1/2)*(-72*b^ 
3*ln(f)^3/x^6+252*b^2*ln(f)^2/x^4-630*b*ln(f)/x^2+945)/b^4*exp(b*ln(f)/x^2 
)+105/16*(-b)^(9/2)/b^(9/2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\frac {105 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{7} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (105 \, b x^{6} \log \left (f\right ) - 70 \, b^{2} x^{4} \log \left (f\right )^{2} + 28 \, b^{3} x^{2} \log \left (f\right )^{3} - 8 \, b^{4} \log \left (f\right )^{4}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{32 \, b^{5} x^{7} \log \left (f\right )^{5}} \] Input:

integrate(f^(a+b/x^2)/x^10,x, algorithm="fricas")
 

Output:

1/32*(105*sqrt(pi)*sqrt(-b*log(f))*f^a*x^7*erf(sqrt(-b*log(f))/x) + 2*(105 
*b*x^6*log(f) - 70*b^2*x^4*log(f)^2 + 28*b^3*x^2*log(f)^3 - 8*b^4*log(f)^4 
)*f^((a*x^2 + b)/x^2))/(b^5*x^7*log(f)^5)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\text {Timed out} \] Input:

integrate(f**(a+b/x**2)/x**10,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.21 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\frac {f^{a} \Gamma \left (\frac {9}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{9} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {9}{2}}} \] Input:

integrate(f^(a+b/x^2)/x^10,x, algorithm="maxima")
 

Output:

1/2*f^a*gamma(9/2, -b*log(f)/x^2)/(x^9*(-b*log(f)/x^2)^(9/2))
 

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{10}} \,d x } \] Input:

integrate(f^(a+b/x^2)/x^10,x, algorithm="giac")
 

Output:

integrate(f^(a + b/x^2)/x^10, x)
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=-\frac {\frac {f^a\,\left (105\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )-\frac {210\,f^{\frac {b}{x^2}}\,\sqrt {b\,\ln \left (f\right )}}{x}\right )}{32\,\sqrt {b\,\ln \left (f\right )}}-\frac {7\,b^2\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^2}{4\,x^5}+\frac {b^3\,f^a\,f^{\frac {b}{x^2}}\,{\ln \left (f\right )}^3}{2\,x^7}+\frac {35\,b\,f^a\,f^{\frac {b}{x^2}}\,\ln \left (f\right )}{8\,x^3}}{b^4\,{\ln \left (f\right )}^4} \] Input:

int(f^(a + b/x^2)/x^10,x)
 

Output:

-((f^a*(105*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))) - (210*f^(b/x^2 
)*(b*log(f))^(1/2))/x))/(32*(b*log(f))^(1/2)) - (7*b^2*f^a*f^(b/x^2)*log(f 
)^2)/(4*x^5) + (b^3*f^a*f^(b/x^2)*log(f)^3)/(2*x^7) + (35*b*f^a*f^(b/x^2)* 
log(f))/(8*x^3))/(b^4*log(f)^4)
 

Reduce [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{10}} \, dx=\int \frac {f^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{10}}d x \] Input:

int(f^(a+b/x^2)/x^10,x)
 

Output:

int(f**((a*x**2 + b)/x**2)/x**10,x)