\(\int \frac {f^{a+b x^2}}{x^9} \, dx\) [80]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int \frac {f^{a+b x^2}}{x^9} \, dx=-\frac {1}{2} b^4 f^a \Gamma \left (-4,-b x^2 \log (f)\right ) \log ^4(f) \]

[Out]

-1/2*f^a/x^8*Ei(5,-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} \[ \int \frac {f^{a+b x^2}}{x^9} \, dx=-\frac {1}{2} b^4 f^a \log ^4(f) \Gamma \left (-4,-b x^2 \log (f)\right ) \]

[In]

Int[f^(a + b*x^2)/x^9,x]

[Out]

-1/2*(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)

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^4 f^a \Gamma \left (-4,-b x^2 \log (f)\right ) \log ^4(f) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x^2}}{x^9} \, dx=-\frac {1}{2} b^4 f^a \Gamma \left (-4,-b x^2 \log (f)\right ) \log ^4(f) \]

[In]

Integrate[f^(a + b*x^2)/x^9,x]

[Out]

-1/2*(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(90\) vs. \(2(18)=36\).

Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.79

method	result	size
risch	\(-\frac {f^{a} \left (\operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right ) \ln \left (f \right )^{4} b^{4} x^{8}+f^{b \,x^{2}} \ln \left (f \right )^{3} b^{3} x^{6}+\ln \left (f \right )^{2} f^{b \,x^{2}} b^{2} x^{4}+2 \ln \left (f \right ) f^{b \,x^{2}} b \,x^{2}+6 f^{b \,x^{2}}\right )}{48 x^{8}}\)	\(91\)
meijerg	\(\frac {f^{a} b^{4} \ln \left (f \right )^{4} \left (-\frac {1}{4 b^{4} x^{8} \ln \left (f \right )^{4}}-\frac {1}{3 b^{3} x^{6} \ln \left (f \right )^{3}}-\frac {1}{4 b^{2} x^{4} \ln \left (f \right )^{2}}-\frac {1}{6 b \,x^{2} \ln \left (f \right )}-\frac {25}{288}+\frac {\ln \left (x \right )}{12}+\frac {\ln \left (-b \right )}{24}+\frac {\ln \left (\ln \left (f \right )\right )}{24}+\frac {125 b^{4} x^{8} \ln \left (f \right )^{4}+240 b^{3} x^{6} \ln \left (f \right )^{3}+360 b^{2} x^{4} \ln \left (f \right )^{2}+480 b \,x^{2} \ln \left (f \right )+360}{1440 b^{4} x^{8} \ln \left (f \right )^{4}}-\frac {\left (5 b^{3} x^{6} \ln \left (f \right )^{3}+5 b^{2} x^{4} \ln \left (f \right )^{2}+10 b \,x^{2} \ln \left (f \right )+30\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{120 b^{4} x^{8} \ln \left (f \right )^{4}}-\frac {\ln \left (-b \,x^{2} \ln \left (f \right )\right )}{24}-\frac {\operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right )}{24}\right )}{2}\)	\(213\)

[In]

int(f^(b*x^2+a)/x^9,x,method=_RETURNVERBOSE)

[Out]

-1/48*f^a*(Ei(1,-b*x^2*ln(f))*ln(f)^4*b^4*x^8+f^(b*x^2)*ln(f)^3*b^3*x^6+ln(f)^2*f^(b*x^2)*b^2*x^4+2*ln(f)*f^(b
*x^2)*b*x^2+6*f^(b*x^2))/x^8

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (18) = 36\).

Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96 \[ \int \frac {f^{a+b x^2}}{x^9} \, dx=\frac {b^{4} f^{a} x^{8} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{4} - {\left (b^{3} x^{6} \log \left (f\right )^{3} + b^{2} x^{4} \log \left (f\right )^{2} + 2 \, b x^{2} \log \left (f\right ) + 6\right )} f^{b x^{2} + a}}{48 \, x^{8}} \]

[In]

integrate(f^(b*x^2+a)/x^9,x, algorithm="fricas")

[Out]

1/48*(b^4*f^a*x^8*Ei(b*x^2*log(f))*log(f)^4 - (b^3*x^6*log(f)^3 + b^2*x^4*log(f)^2 + 2*b*x^2*log(f) + 6)*f^(b*
x^2 + a))/x^8

Sympy [F]

\[ \int \frac {f^{a+b x^2}}{x^9} \, dx=\int \frac {f^{a + b x^{2}}}{x^{9}}\, dx \]

[In]

integrate(f**(b*x**2+a)/x**9,x)

[Out]

Integral(f**(a + b*x**2)/x**9, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+b x^2}}{x^9} \, dx=-\frac {1}{2} \, b^{4} f^{a} \Gamma \left (-4, -b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{4} \]

[In]

integrate(f^(b*x^2+a)/x^9,x, algorithm="maxima")

[Out]

-1/2*b^4*f^a*gamma(-4, -b*x^2*log(f))*log(f)^4

Giac [F]

\[ \int \frac {f^{a+b x^2}}{x^9} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{9}} \,d x } \]

[In]

integrate(f^(b*x^2+a)/x^9,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^9, x)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.75 \[ \int \frac {f^{a+b x^2}}{x^9} \, dx=-\frac {b^4\,f^a\,{\ln \left (f\right )}^4\,\mathrm {expint}\left (-b\,x^2\,\ln \left (f\right )\right )}{48}-\frac {b^4\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^4\,\left (\frac {1}{24\,b\,x^2\,\ln \left (f\right )}+\frac {1}{24\,b^2\,x^4\,{\ln \left (f\right )}^2}+\frac {1}{12\,b^3\,x^6\,{\ln \left (f\right )}^3}+\frac {1}{4\,b^4\,x^8\,{\ln \left (f\right )}^4}\right )}{2} \]

[In]

int(f^(a + b*x^2)/x^9,x)

[Out]

- (b^4*f^a*log(f)^4*expint(-b*x^2*log(f)))/48 - (b^4*f^a*f^(b*x^2)*log(f)^4*(1/(24*b*x^2*log(f)) + 1/(24*b^2*x
^4*log(f)^2) + 1/(12*b^3*x^6*log(f)^3) + 1/(4*b^4*x^8*log(f)^4)))/2