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

Optimal. Leaf size=34 \[ -\frac {f^a \left (-b x^2 \log (f)\right )^{9/2} \Gamma \left (-\frac {9}{2},-b x^2 \log (f)\right )}{2 x^9} \]

[Out]

-1/2*f^a*(-32/945*Pi^(1/2)*erfc((-b*x^2*ln(f))^(1/2))+32/945/(-b*x^2*ln(f))^(1/2)*exp(b*x^2*ln(f))-16/945/(-b*
x^2*ln(f))^(3/2)*exp(b*x^2*ln(f))+8/315/(-b*x^2*ln(f))^(5/2)*exp(b*x^2*ln(f))-4/63/(-b*x^2*ln(f))^(7/2)*exp(b*
x^2*ln(f))+2/9/(-b*x^2*ln(f))^(9/2)*exp(b*x^2*ln(f)))*(-b*x^2*ln(f))^(9/2)/x^9

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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} \[ -\frac {f^a \left (-b x^2 \log (f)\right )^{9/2} \text {Gamma}\left (-\frac {9}{2},-b x^2 \log (f)\right )}{2 x^9} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^10,x]

[Out]

-(f^a*Gamma[-9/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(9/2))/(2*x^9)

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^{10}} \, dx &=-\frac {f^a \Gamma \left (-\frac {9}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{9/2}}{2 x^9}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ -\frac {f^a \left (-b x^2 \log (f)\right )^{9/2} \Gamma \left (-\frac {9}{2},-b x^2 \log (f)\right )}{2 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^10,x]

[Out]

-1/2*(f^a*Gamma[-9/2, -(b*x^2*Log[f])]*(-(b*x^2*Log[f]))^(9/2))/x^9

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fricas [A]  time = 0.41, size = 97, normalized size = 2.85 \[ -\frac {16 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} b^{4} f^{a} x^{9} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right ) \log \relax (f)^{4} + {\left (16 \, b^{4} x^{8} \log \relax (f)^{4} + 8 \, b^{3} x^{6} \log \relax (f)^{3} + 12 \, b^{2} x^{4} \log \relax (f)^{2} + 30 \, b x^{2} \log \relax (f) + 105\right )} f^{b x^{2} + a}}{945 \, x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/945*(16*sqrt(pi)*sqrt(-b*log(f))*b^4*f^a*x^9*erf(sqrt(-b*log(f))*x)*log(f)^4 + (16*b^4*x^8*log(f)^4 + 8*b^3
*x^6*log(f)^3 + 12*b^2*x^4*log(f)^2 + 30*b*x^2*log(f) + 105)*f^(b*x^2 + a))/x^9

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{2} + a}}{x^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^10, x)

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maple [A]  time = 0.08, size = 133, normalized size = 3.91 \[ \frac {16 \sqrt {\pi }\, b^{5} f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x \right ) \ln \relax (f )^{5}}{945 \sqrt {-b \ln \relax (f )}}-\frac {16 b^{4} f^{a} f^{b \,x^{2}} \ln \relax (f )^{4}}{945 x}-\frac {8 b^{3} f^{a} f^{b \,x^{2}} \ln \relax (f )^{3}}{945 x^{3}}-\frac {4 b^{2} f^{a} f^{b \,x^{2}} \ln \relax (f )^{2}}{315 x^{5}}-\frac {2 b \,f^{a} f^{b \,x^{2}} \ln \relax (f )}{63 x^{7}}-\frac {f^{a} f^{b \,x^{2}}}{9 x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*f^a/x^9*f^(b*x^2)-2/63*f^a*ln(f)*b/x^7*f^(b*x^2)-4/315*f^a*ln(f)^2*b^2/x^5*f^(b*x^2)-8/945*f^a*ln(f)^3*b^
3/x^3*f^(b*x^2)-16/945*f^a*ln(f)^4*b^4/x*f^(b*x^2)+16/945*f^a*ln(f)^5*b^5*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln
(f))^(1/2)*x)

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maxima [A]  time = 1.26, size = 28, normalized size = 0.82 \[ -\frac {\left (-b x^{2} \log \relax (f)\right )^{\frac {9}{2}} f^{a} \Gamma \left (-\frac {9}{2}, -b x^{2} \log \relax (f)\right )}{2 \, x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.51, size = 153, normalized size = 4.50 \[ \frac {16\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \relax (f)}\right )\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{9/2}}{945\,x^9}-\frac {16\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{9/2}}{945\,x^9}-\frac {f^a\,f^{b\,x^2}}{9\,x^9}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^2}{315\,x^5}-\frac {8\,b^3\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^3}{945\,x^3}-\frac {16\,b^4\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^4}{945\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \relax (f)}{63\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)/x^10,x)

[Out]

(16*f^a*pi^(1/2)*erfc((-b*x^2*log(f))^(1/2))*(-b*x^2*log(f))^(9/2))/(945*x^9) - (16*f^a*pi^(1/2)*(-b*x^2*log(f
))^(9/2))/(945*x^9) - (f^a*f^(b*x^2))/(9*x^9) - (4*b^2*f^a*f^(b*x^2)*log(f)^2)/(315*x^5) - (8*b^3*f^a*f^(b*x^2
)*log(f)^3)/(945*x^3) - (16*b^4*f^a*f^(b*x^2)*log(f)^4)/(945*x) - (2*b*f^a*f^(b*x^2)*log(f))/(63*x^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{2}}}{x^{10}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**10, x)

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