\(\int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx\) [87]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\frac {f^a \Gamma \left (\frac {13}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{13} \left (-\frac {b \log (f)}{x^2}\right )^{13/2}} \] Output:

1/2*f^a*(524288/5621533568633696205238621875*GAMMA(51/2,-b*ln(f)/x^2)-5242 
88/5621533568633696205238621875*(-b*ln(f)/x^2)^(49/2)*exp(b*ln(f)/x^2)-262 
144/114725174870075432759971875*(-b*ln(f)/x^2)^(47/2)*exp(b*ln(f)/x^2)-131 
072/2440961167448413462978125*(-b*ln(f)/x^2)^(45/2)*exp(b*ln(f)/x^2)-65536 
/54243581498853632510625*(-b*ln(f)/x^2)^(43/2)*exp(b*ln(f)/x^2)-32768/1261 
478639508224011875*(-b*ln(f)/x^2)^(41/2)*exp(b*ln(f)/x^2)-16384/3076777169 
5322536875*(-b*ln(f)/x^2)^(39/2)*exp(b*ln(f)/x^2)-8192/788917222956988125* 
(-b*ln(f)/x^2)^(37/2)*exp(b*ln(f)/x^2)-4096/21322087106945625*(-b*ln(f)/x^ 
2)^(35/2)*exp(b*ln(f)/x^2)-2048/609202488769875*(-b*ln(f)/x^2)^(33/2)*exp( 
b*ln(f)/x^2)-1024/18460681477875*(-b*ln(f)/x^2)^(31/2)*exp(b*ln(f)/x^2)-51 
2/595505854125*(-b*ln(f)/x^2)^(29/2)*exp(b*ln(f)/x^2)-256/20534684625*(-b* 
ln(f)/x^2)^(27/2)*exp(b*ln(f)/x^2)-128/760543875*(-b*ln(f)/x^2)^(25/2)*exp 
(b*ln(f)/x^2)-64/30421755*(-b*ln(f)/x^2)^(23/2)*exp(b*ln(f)/x^2)-32/132268 
5*(-b*ln(f)/x^2)^(21/2)*exp(b*ln(f)/x^2)-16/62985*(-b*ln(f)/x^2)^(19/2)*ex 
p(b*ln(f)/x^2)-8/3315*(-b*ln(f)/x^2)^(17/2)*exp(b*ln(f)/x^2)-4/195*(-b*ln( 
f)/x^2)^(15/2)*exp(b*ln(f)/x^2)-2/13*(-b*ln(f)/x^2)^(13/2)*exp(b*ln(f)/x^2 
))/x^13/(-b*ln(f)/x^2)^(13/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\frac {f^a \Gamma \left (\frac {13}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{13} \left (-\frac {b \log (f)}{x^2}\right )^{13/2}} \] Input:

Integrate[f^(a + b/x^2)/x^14,x]
 

Output:

(f^a*Gamma[13/2, -((b*Log[f])/x^2)])/(2*x^13*(-((b*Log[f])/x^2))^(13/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2648}

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.

\(\displaystyle \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx\)

\(\Big \downarrow \) 2648

\(\displaystyle \frac {f^a \Gamma \left (\frac {13}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{13} \left (-\frac {b \log (f)}{x^2}\right )^{13/2}}\)

Input:

Int[f^(a + b/x^2)/x^14,x]
 

Output:

(f^a*Gamma[13/2, -((b*Log[f])/x^2)])/(2*x^13*(-((b*Log[f])/x^2))^(13/2))
 

Defintions of rubi rules used

rule 2648

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(-F^a)*((e + f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[ 
F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; FreeQ[{F 
, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 127, normalized size of antiderivative = 3.74

method	result	size
meijerg	\(\frac {f^{a} \sqrt {-b}\, \left (-\frac {\left (-b \right )^{\frac {13}{2}} \sqrt {\ln \left (f \right )}\, \left (-\frac {416 b^{5} \ln \left (f \right )^{5}}{x^{10}}+\frac {2288 b^{4} \ln \left (f \right )^{4}}{x^{8}}-\frac {10296 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {36036 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {90090 b \ln \left (f \right )}{x^{2}}+135135\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{416 x \,b^{6}}+\frac {10395 \left (-b \right )^{\frac {13}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{64 b^{\frac {13}{2}}}\right )}{2 b^{7} \ln \left (f \right )^{\frac {13}{2}}}\)	\(127\)
risch	\(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{11} b \ln \left (f \right )}+\frac {11 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x^{9}}-\frac {99 f^{a} f^{\frac {b}{x^{2}}}}{8 \ln \left (f \right )^{3} b^{3} x^{7}}+\frac {693 f^{a} f^{\frac {b}{x^{2}}}}{16 \ln \left (f \right )^{4} b^{4} x^{5}}-\frac {3465 f^{a} f^{\frac {b}{x^{2}}}}{32 \ln \left (f \right )^{5} b^{5} x^{3}}+\frac {10395 f^{a} f^{\frac {b}{x^{2}}}}{64 \ln \left (f \right )^{6} b^{6} x}-\frac {10395 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{128 \ln \left (f \right )^{6} b^{6} \sqrt {-b \ln \left (f \right )}}\)	\(168\)

Input:

int(f^(a+b/x^2)/x^14,x,method=_RETURNVERBOSE)
 

Output:

1/2*f^a/b^7/ln(f)^(13/2)*(-b)^(1/2)*(-1/416/x*(-b)^(13/2)*ln(f)^(1/2)*(-41 
6*b^5*ln(f)^5/x^10+2288*b^4*ln(f)^4/x^8-10296*b^3*ln(f)^3/x^6+36036*b^2*ln 
(f)^2/x^4-90090*b*ln(f)/x^2+135135)/b^6*exp(b*ln(f)/x^2)+10395/64*(-b)^(13 
/2)/b^(13/2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.65 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\frac {10395 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{11} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (10395 \, b x^{10} \log \left (f\right ) - 6930 \, b^{2} x^{8} \log \left (f\right )^{2} + 2772 \, b^{3} x^{6} \log \left (f\right )^{3} - 792 \, b^{4} x^{4} \log \left (f\right )^{4} + 176 \, b^{5} x^{2} \log \left (f\right )^{5} - 32 \, b^{6} \log \left (f\right )^{6}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{128 \, b^{7} x^{11} \log \left (f\right )^{7}} \] Input:

integrate(f^(a+b/x^2)/x^14,x, algorithm="fricas")
 

Output:

1/128*(10395*sqrt(pi)*sqrt(-b*log(f))*f^a*x^11*erf(sqrt(-b*log(f))/x) + 2* 
(10395*b*x^10*log(f) - 6930*b^2*x^8*log(f)^2 + 2772*b^3*x^6*log(f)^3 - 792 
*b^4*x^4*log(f)^4 + 176*b^5*x^2*log(f)^5 - 32*b^6*log(f)^6)*f^((a*x^2 + b) 
/x^2))/(b^7*x^11*log(f)^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\text {Timed out} \] Input:

integrate(f**(a+b/x**2)/x**14,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\frac {f^{a} \Gamma \left (\frac {13}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{13} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {13}{2}}} \] Input:

integrate(f^(a+b/x^2)/x^14,x, algorithm="maxima")
 

Output:

1/2*f^a*gamma(13/2, -b*log(f)/x^2)/(x^13*(-b*log(f)/x^2)^(13/2))
 

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{14}} \,d x } \] Input:

integrate(f^(a+b/x^2)/x^14,x, algorithm="giac")
 

Output:

integrate(f^(a + b/x^2)/x^14, x)
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 4.68 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=-\frac {\frac {f^a\,\left (\frac {10395\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )}{128}-\frac {10395\,f^{\frac {b}{x^2}}\,\sqrt {b\,\ln \left (f\right )}}{64\,x}\right )}{\sqrt {b\,\ln \left (f\right )}}-\frac {693\,b^2\,f^{a+\frac {b}{x^2}}\,{\ln \left (f\right )}^2}{16\,x^5}+\frac {99\,b^3\,f^{a+\frac {b}{x^2}}\,{\ln \left (f\right )}^3}{8\,x^7}-\frac {11\,b^4\,f^{a+\frac {b}{x^2}}\,{\ln \left (f\right )}^4}{4\,x^9}+\frac {b^5\,f^{a+\frac {b}{x^2}}\,{\ln \left (f\right )}^5}{2\,x^{11}}+\frac {3465\,b\,f^{a+\frac {b}{x^2}}\,\ln \left (f\right )}{32\,x^3}}{b^6\,{\ln \left (f\right )}^6} \] Input:

int(f^(a + b/x^2)/x^14,x)
 

Output:

-((f^a*((10395*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))))/128 - (1039 
5*f^(b/x^2)*(b*log(f))^(1/2))/(64*x)))/(b*log(f))^(1/2) - (693*b^2*f^(a + 
b/x^2)*log(f)^2)/(16*x^5) + (99*b^3*f^(a + b/x^2)*log(f)^3)/(8*x^7) - (11* 
b^4*f^(a + b/x^2)*log(f)^4)/(4*x^9) + (b^5*f^(a + b/x^2)*log(f)^5)/(2*x^11 
) + (3465*b*f^(a + b/x^2)*log(f))/(32*x^3))/(b^6*log(f)^6)
 

Reduce [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx=\int \frac {f^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{14}}d x \] Input:

int(f^(a+b/x^2)/x^14,x)
 

Output:

int(f**((a*x**2 + b)/x**2)/x**14,x)