3.2.53 \(\int \frac {f^{a+\frac {b}{x^2}}}{x^{14}} \, dx\) [153]

Optimal. Leaf size=34 \[ \frac {f^a \Gamma \left (\frac {13}{2},-\frac {b \log (f)}{x^2}\right )}{2 x^{13} \left (-\frac {b \log (f)}{x^2}\right )^{13/2}} \]

[Out]

1/2*f^a*(524288/5621533568633696205238621875*GAMMA(51/2,-b*ln(f)/x^2)-524288/5621533568633696205238621875*(-b*
ln(f)/x^2)^(49/2)*exp(b*ln(f)/x^2)-262144/114725174870075432759971875*(-b*ln(f)/x^2)^(47/2)*exp(b*ln(f)/x^2)-1
31072/2440961167448413462978125*(-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)-32768/1261478639508224011875*(-b*ln(f)/x^2)^(41/2)*exp(b*ln(f)/x^2)-16384/307677
71695322536875*(-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)-512/595505854125*(-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)^(2
5/2)*exp(b*ln(f)/x^2)-64/30421755*(-b*ln(f)/x^2)^(23/2)*exp(b*ln(f)/x^2)-32/1322685*(-b*ln(f)/x^2)^(21/2)*exp(
b*ln(f)/x^2)-16/62985*(-b*ln(f)/x^2)^(19/2)*exp(b*ln(f)/x^2)-8/3315*(-b*ln(f)/x^2)^(17/2)*exp(b*ln(f)/x^2)-4/1
95*(-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)

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Rubi [A]
time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2250} \begin {gather*} \frac {f^a \text {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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)/x^14,x]

[Out]

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

Rule 2250

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] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \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}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)/x^14,x]

[Out]

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

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Maple [A]
time = 0.08, size = 127, normalized size = 3.74

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^a/b^7/ln(f)^(13/2)*(-b)^(1/2)*(-1/416/x*(-b)^(13/2)*ln(f)^(1/2)*(-416*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))

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Maxima [A]
time = 0.06, size = 28, normalized size = 0.82 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.09, size = 124, normalized size = 3.65 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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^(a+b/x^2)/x^14,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^14, x)

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Mupad [B]
time = 3.71, size = 159, normalized size = 4.68 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)/x^14,x)

[Out]

-((f^a*((10395*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))))/128 - (10395*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) - (1
1*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)

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