Integrand size = 13, antiderivative size = 96 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=-\frac {f^{a+b x^2}}{5 x^5}-\frac {2 b f^{a+b x^2} \log (f)}{15 x^3}-\frac {4 b^2 f^{a+b x^2} \log ^2(f)}{15 x}+\frac {4}{15} b^{5/2} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {5}{2}}(f) \] Output:
-1/5*f^(b*x^2+a)/x^5-2/15*b*f^(b*x^2+a)*ln(f)/x^3-4/15*b^2*f^(b*x^2+a)*ln( f)^2/x+4/15*b^(5/2)*f^a*Pi^(1/2)*erfi(b^(1/2)*x*ln(f)^(1/2))*ln(f)^(5/2)
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\frac {f^a \left (4 b^{5/2} \sqrt {\pi } x^5 \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {5}{2}}(f)-f^{b x^2} \left (3+2 b x^2 \log (f)+4 b^2 x^4 \log ^2(f)\right )\right )}{15 x^5} \] Input:
Integrate[f^(a + b*x^2)/x^6,x]
Output:
(f^a*(4*b^(5/2)*Sqrt[Pi]*x^5*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*Log[f]^(5/2) - f ^(b*x^2)*(3 + 2*b*x^2*Log[f] + 4*b^2*x^4*Log[f]^2)))/(15*x^5)
Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2643, 2643, 2643, 2633}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {f^{a+b x^2}}{x^6} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {2}{5} b \log (f) \int \frac {f^{b x^2+a}}{x^4}dx-\frac {f^{a+b x^2}}{5 x^5}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \int \frac {f^{b x^2+a}}{x^2}dx-\frac {f^{a+b x^2}}{3 x^3}\right )-\frac {f^{a+b x^2}}{5 x^5}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (2 b \log (f) \int f^{b x^2+a}dx-\frac {f^{a+b x^2}}{x}\right )-\frac {f^{a+b x^2}}{3 x^3}\right )-\frac {f^{a+b x^2}}{5 x^5}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (\sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )-\frac {f^{a+b x^2}}{x}\right )-\frac {f^{a+b x^2}}{3 x^3}\right )-\frac {f^{a+b x^2}}{5 x^5}\) |
Input:
Int[f^(a + b*x^2)/x^6,x]
Output:
-1/5*f^(a + b*x^2)/x^5 + (2*b*Log[f]*(-1/3*f^(a + b*x^2)/x^3 + (2*b*(-(f^( a + b*x^2)/x) + Sqrt[b]*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]]*Sqrt[Log [f]])*Log[f])/3))/5
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) , x] - Simp[b*n*(Log[F]/(m + 1)) Int[(c + d*x)^(m + n)*F^(a + b*(c + d*x) ^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[ -4, (m + 1)/n, 5] && IntegerQ[n] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90
method | result | size |
meijerg | \(-\frac {f^{a} \ln \left (f \right )^{\frac {5}{2}} b^{3} \left (-\frac {2 \left (\frac {4 b^{2} x^{4} \ln \left (f \right )^{2}}{3}+\frac {2 b \,x^{2} \ln \left (f \right )}{3}+1\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{5 x^{5} \left (-b \right )^{\frac {5}{2}} \ln \left (f \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, x \sqrt {\ln \left (f \right )}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{2 \sqrt {-b}}\) | \(86\) |
risch | \(-\frac {f^{a} f^{b \,x^{2}}}{5 x^{5}}-\frac {2 f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{15 x^{3}}-\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{15 x}+\frac {4 f^{a} \ln \left (f \right )^{3} b^{3} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{15 \sqrt {-b \ln \left (f \right )}}\) | \(89\) |
Input:
int(f^(b*x^2+a)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/2*f^a*ln(f)^(5/2)*b^3/(-b)^(1/2)*(-2/5/x^5/(-b)^(5/2)/ln(f)^(5/2)*(4/3* b^2*x^4*ln(f)^2+2/3*b*x^2*ln(f)+1)*exp(b*x^2*ln(f))+8/15/(-b)^(5/2)*b^(5/2 )*Pi^(1/2)*erfi(b^(1/2)*x*ln(f)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=-\frac {4 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{2} f^{a} x^{5} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) \log \left (f\right )^{2} + {\left (4 \, b^{2} x^{4} \log \left (f\right )^{2} + 2 \, b x^{2} \log \left (f\right ) + 3\right )} f^{b x^{2} + a}}{15 \, x^{5}} \] Input:
integrate(f^(b*x^2+a)/x^6,x, algorithm="fricas")
Output:
-1/15*(4*sqrt(pi)*sqrt(-b*log(f))*b^2*f^a*x^5*erf(sqrt(-b*log(f))*x)*log(f )^2 + (4*b^2*x^4*log(f)^2 + 2*b*x^2*log(f) + 3)*f^(b*x^2 + a))/x^5
\[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\int \frac {f^{a + b x^{2}}}{x^{6}}\, dx \] Input:
integrate(f**(b*x**2+a)/x**6,x)
Output:
Integral(f**(a + b*x**2)/x**6, x)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.29 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=-\frac {\left (-b x^{2} \log \left (f\right )\right )^{\frac {5}{2}} f^{a} \Gamma \left (-\frac {5}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{5}} \] Input:
integrate(f^(b*x^2+a)/x^6,x, algorithm="maxima")
Output:
-1/2*(-b*x^2*log(f))^(5/2)*f^a*gamma(-5/2, -b*x^2*log(f))/x^5
\[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{6}} \,d x } \] Input:
integrate(f^(b*x^2+a)/x^6,x, algorithm="giac")
Output:
integrate(f^(b*x^2 + a)/x^6, x)
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\frac {4\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \left (f\right )}\right )\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{5/2}}{15\,x^5}-\frac {4\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{5/2}}{15\,x^5}-\frac {f^a\,f^{b\,x^2}}{5\,x^5}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^2}{15\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \left (f\right )}{15\,x^3} \] Input:
int(f^(a + b*x^2)/x^6,x)
Output:
(4*f^a*pi^(1/2)*erfc((-b*x^2*log(f))^(1/2))*(-b*x^2*log(f))^(5/2))/(15*x^5 ) - (4*f^a*pi^(1/2)*(-b*x^2*log(f))^(5/2))/(15*x^5) - (f^a*f^(b*x^2))/(5*x ^5) - (4*b^2*f^a*f^(b*x^2)*log(f)^2)/(15*x) - (2*b*f^a*f^(b*x^2)*log(f))/( 15*x^3)
\[ \int \frac {f^{a+b x^2}}{x^6} \, dx=f^{a} \left (\int \frac {f^{b \,x^{2}}}{x^{6}}d x \right ) \] Input:
int(f^(b*x^2+a)/x^6,x)
Output:
f**a*int(f**(b*x**2)/x**6,x)