\(\int \frac {f^{a+b x^2}}{x^{11}} \, dx\) [81]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\frac {1}{2} b^5 f^a \Gamma \left (-5,-b x^2 \log (f)\right ) \log ^5(f) \]

[Out]

-1/2*f^a/x^10*Ei(6,-b*x^2*ln(f))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, 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 = {2250} \[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\frac {1}{2} b^5 f^a \log ^5(f) \Gamma \left (-5,-b x^2 \log (f)\right ) \]

[In]

Int[f^(a + b*x^2)/x^11,x]

[Out]

(b^5*f^a*Gamma[-5, -(b*x^2*Log[f])]*Log[f]^5)/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*} \text {integral}& = \frac {1}{2} b^5 f^a \Gamma \left (-5,-b x^2 \log (f)\right ) \log ^5(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\frac {1}{2} b^5 f^a \Gamma \left (-5,-b x^2 \log (f)\right ) \log ^5(f) \]

[In]

Integrate[f^(a + b*x^2)/x^11,x]

[Out]

(b^5*f^a*Gamma[-5, -(b*x^2*Log[f])]*Log[f]^5)/2

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(109\) vs. \(2(18)=36\).

Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 4.58

method	result	size
risch	\(-\frac {f^{a} \left (\ln \left (f \right )^{5} \operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right ) b^{5} x^{10}+\ln \left (f \right )^{4} f^{b \,x^{2}} b^{4} x^{8}+f^{b \,x^{2}} \ln \left (f \right )^{3} b^{3} x^{6}+2 \ln \left (f \right )^{2} f^{b \,x^{2}} b^{2} x^{4}+6 \ln \left (f \right ) f^{b \,x^{2}} b \,x^{2}+24 f^{b \,x^{2}}\right )}{240 x^{10}}\)	\(110\)
meijerg	\(-\frac {f^{a} b^{5} \ln \left (f \right )^{5} \left (\frac {1}{5 b^{5} x^{10} \ln \left (f \right )^{5}}+\frac {1}{4 b^{4} x^{8} \ln \left (f \right )^{4}}+\frac {1}{6 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {1}{12 b^{2} x^{4} \ln \left (f \right )^{2}}+\frac {1}{24 b \,x^{2} \ln \left (f \right )}+\frac {137}{7200}-\frac {\ln \left (x \right )}{60}-\frac {\ln \left (-b \right )}{120}-\frac {\ln \left (\ln \left (f \right )\right )}{120}-\frac {137 b^{5} x^{10} \ln \left (f \right )^{5}+300 b^{4} x^{8} \ln \left (f \right )^{4}+600 b^{3} x^{6} \ln \left (f \right )^{3}+1200 b^{2} x^{4} \ln \left (f \right )^{2}+1800 b \,x^{2} \ln \left (f \right )+1440}{7200 b^{5} x^{10} \ln \left (f \right )^{5}}+\frac {\left (6 b^{4} x^{8} \ln \left (f \right )^{4}+6 b^{3} x^{6} \ln \left (f \right )^{3}+12 b^{2} x^{4} \ln \left (f \right )^{2}+36 b \,x^{2} \ln \left (f \right )+144\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{720 b^{5} x^{10} \ln \left (f \right )^{5}}+\frac {\ln \left (-b \,x^{2} \ln \left (f \right )\right )}{120}+\frac {\operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right )}{120}\right )}{2}\)	\(249\)

[In]

int(f^(b*x^2+a)/x^11,x,method=_RETURNVERBOSE)

[Out]

-1/240*f^a*(ln(f)^5*Ei(1,-b*x^2*ln(f))*b^5*x^10+ln(f)^4*f^(b*x^2)*b^4*x^8+f^(b*x^2)*ln(f)^3*b^3*x^6+2*ln(f)^2*
f^(b*x^2)*b^2*x^4+6*ln(f)*f^(b*x^2)*b*x^2+24*f^(b*x^2))/x^10

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (18) = 36\).

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.46 \[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\frac {b^{5} f^{a} x^{10} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{5} - {\left (b^{4} x^{8} \log \left (f\right )^{4} + b^{3} x^{6} \log \left (f\right )^{3} + 2 \, b^{2} x^{4} \log \left (f\right )^{2} + 6 \, b x^{2} \log \left (f\right ) + 24\right )} f^{b x^{2} + a}}{240 \, x^{10}} \]

[In]

integrate(f^(b*x^2+a)/x^11,x, algorithm="fricas")

[Out]

1/240*(b^5*f^a*x^10*Ei(b*x^2*log(f))*log(f)^5 - (b^4*x^8*log(f)^4 + b^3*x^6*log(f)^3 + 2*b^2*x^4*log(f)^2 + 6*
b*x^2*log(f) + 24)*f^(b*x^2 + a))/x^10

Sympy [F]

\[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\int \frac {f^{a + b x^{2}}}{x^{11}}\, dx \]

[In]

integrate(f**(b*x**2+a)/x**11,x)

[Out]

Integral(f**(a + b*x**2)/x**11, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\frac {1}{2} \, b^{5} f^{a} \Gamma \left (-5, -b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{5} \]

[In]

integrate(f^(b*x^2+a)/x^11,x, algorithm="maxima")

[Out]

1/2*b^5*f^a*gamma(-5, -b*x^2*log(f))*log(f)^5

Giac [F]

\[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{11}} \,d x } \]

[In]

integrate(f^(b*x^2+a)/x^11,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^11, x)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 4.25 \[ \int \frac {f^{a+b x^2}}{x^{11}} \, dx=-\frac {b^5\,f^a\,{\ln \left (f\right )}^5\,\mathrm {expint}\left (-b\,x^2\,\ln \left (f\right )\right )}{240}-\frac {b^5\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^5\,\left (\frac {1}{120\,b\,x^2\,\ln \left (f\right )}+\frac {1}{120\,b^2\,x^4\,{\ln \left (f\right )}^2}+\frac {1}{60\,b^3\,x^6\,{\ln \left (f\right )}^3}+\frac {1}{20\,b^4\,x^8\,{\ln \left (f\right )}^4}+\frac {1}{5\,b^5\,x^{10}\,{\ln \left (f\right )}^5}\right )}{2} \]

[In]

int(f^(a + b*x^2)/x^11,x)

[Out]

- (b^5*f^a*log(f)^5*expint(-b*x^2*log(f)))/240 - (b^5*f^a*f^(b*x^2)*log(f)^5*(1/(120*b*x^2*log(f)) + 1/(120*b^
2*x^4*log(f)^2) + 1/(60*b^3*x^6*log(f)^3) + 1/(20*b^4*x^8*log(f)^4) + 1/(5*b^5*x^10*log(f)^5)))/2