3.94 \(\int \frac {f^{a+b x^2}}{x^{12}} \, dx\)

Optimal. Leaf size=34 \[ -\frac {f^a \left (-b x^2 \log (f)\right )^{11/2} \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right )}{2 x^{11}} \]

[Out]

-1/2*f^a*(64/10395*Pi^(1/2)*erfc((-b*x^2*ln(f))^(1/2))-64/10395/(-b*x^2*ln(f))^(1/2)*exp(b*x^2*ln(f))+32/10395
/(-b*x^2*ln(f))^(3/2)*exp(b*x^2*ln(f))-16/3465/(-b*x^2*ln(f))^(5/2)*exp(b*x^2*ln(f))+8/693/(-b*x^2*ln(f))^(7/2
)*exp(b*x^2*ln(f))-4/99/(-b*x^2*ln(f))^(9/2)*exp(b*x^2*ln(f))+2/11/(-b*x^2*ln(f))^(11/2)*exp(b*x^2*ln(f)))*(-b
*x^2*ln(f))^(11/2)/x^11

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Rubi [A]  time = 0.02, 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 = {2218} \[ -\frac {f^a \left (-b x^2 \log (f)\right )^{11/2} \text {Gamma}\left (-\frac {11}{2},-b x^2 \log (f)\right )}{2 x^{11}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^12,x]

[Out]

-(f^a*Gamma[-11/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(11/2))/(2*x^11)

Rule 2218

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

Rubi steps

\begin {align*} \int \frac {f^{a+b x^2}}{x^{12}} \, dx &=-\frac {f^a \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{11/2}}{2 x^{11}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ -\frac {f^a \left (-b x^2 \log (f)\right )^{11/2} \Gamma \left (-\frac {11}{2},-b x^2 \log (f)\right )}{2 x^{11}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^12,x]

[Out]

-1/2*(f^a*Gamma[-11/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(11/2))/x^11

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fricas [A]  time = 0.48, size = 109, normalized size = 3.21 \[ -\frac {32 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} b^{5} f^{a} x^{11} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right ) \log \relax (f)^{5} + {\left (32 \, b^{5} x^{10} \log \relax (f)^{5} + 16 \, b^{4} x^{8} \log \relax (f)^{4} + 24 \, b^{3} x^{6} \log \relax (f)^{3} + 60 \, b^{2} x^{4} \log \relax (f)^{2} + 210 \, b x^{2} \log \relax (f) + 945\right )} f^{b x^{2} + a}}{10395 \, x^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^12,x, algorithm="fricas")

[Out]

-1/10395*(32*sqrt(pi)*sqrt(-b*log(f))*b^5*f^a*x^11*erf(sqrt(-b*log(f))*x)*log(f)^5 + (32*b^5*x^10*log(f)^5 + 1
6*b^4*x^8*log(f)^4 + 24*b^3*x^6*log(f)^3 + 60*b^2*x^4*log(f)^2 + 210*b*x^2*log(f) + 945)*f^(b*x^2 + a))/x^11

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{2} + a}}{x^{12}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^12,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^12, x)

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maple [A]  time = 0.11, size = 155, normalized size = 4.56 \[ \frac {32 \sqrt {\pi }\, b^{6} f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x \right ) \ln \relax (f )^{6}}{10395 \sqrt {-b \ln \relax (f )}}-\frac {32 b^{5} f^{a} f^{b \,x^{2}} \ln \relax (f )^{5}}{10395 x}-\frac {16 b^{4} f^{a} f^{b \,x^{2}} \ln \relax (f )^{4}}{10395 x^{3}}-\frac {8 b^{3} f^{a} f^{b \,x^{2}} \ln \relax (f )^{3}}{3465 x^{5}}-\frac {4 b^{2} f^{a} f^{b \,x^{2}} \ln \relax (f )^{2}}{693 x^{7}}-\frac {2 b \,f^{a} f^{b \,x^{2}} \ln \relax (f )}{99 x^{9}}-\frac {f^{a} f^{b \,x^{2}}}{11 x^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)/x^12,x)

[Out]

-1/11*f^a/x^11*f^(b*x^2)-2/99*f^a*ln(f)*b/x^9*f^(b*x^2)-4/693*f^a*ln(f)^2*b^2/x^7*f^(b*x^2)-8/3465*f^a*ln(f)^3
*b^3/x^5*f^(b*x^2)-16/10395*f^a*ln(f)^4*b^4/x^3*f^(b*x^2)-32/10395*f^a*ln(f)^5*b^5/x*f^(b*x^2)+32/10395*f^a*ln
(f)^6*b^6*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)*x)

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maxima [A]  time = 1.37, size = 28, normalized size = 0.82 \[ -\frac {\left (-b x^{2} \log \relax (f)\right )^{\frac {11}{2}} f^{a} \Gamma \left (-\frac {11}{2}, -b x^{2} \log \relax (f)\right )}{2 \, x^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^12,x, algorithm="maxima")

[Out]

-1/2*(-b*x^2*log(f))^(11/2)*f^a*gamma(-11/2, -b*x^2*log(f))/x^11

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mupad [B]  time = 3.63, size = 175, normalized size = 5.15 \[ \frac {32\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{11/2}}{10395\,x^{11}}-\frac {f^a\,f^{b\,x^2}}{11\,x^{11}}-\frac {32\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \relax (f)}\right )\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{11/2}}{10395\,x^{11}}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^2}{693\,x^7}-\frac {8\,b^3\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^3}{3465\,x^5}-\frac {16\,b^4\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^4}{10395\,x^3}-\frac {32\,b^5\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^5}{10395\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \relax (f)}{99\,x^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)/x^12,x)

[Out]

(32*f^a*pi^(1/2)*(-b*x^2*log(f))^(11/2))/(10395*x^11) - (f^a*f^(b*x^2))/(11*x^11) - (32*f^a*pi^(1/2)*erfc((-b*
x^2*log(f))^(1/2))*(-b*x^2*log(f))^(11/2))/(10395*x^11) - (4*b^2*f^a*f^(b*x^2)*log(f)^2)/(693*x^7) - (8*b^3*f^
a*f^(b*x^2)*log(f)^3)/(3465*x^5) - (16*b^4*f^a*f^(b*x^2)*log(f)^4)/(10395*x^3) - (32*b^5*f^a*f^(b*x^2)*log(f)^
5)/(10395*x) - (2*b*f^a*f^(b*x^2)*log(f))/(99*x^9)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{2}}}{x^{12}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)/x**12,x)

[Out]

Integral(f**(a + b*x**2)/x**12, x)

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