Integrand size = 13, antiderivative size = 119 \[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\frac {1}{7} f^{a+\frac {b}{x^2}} x^7+\frac {2}{35} b f^{a+\frac {b}{x^2}} x^5 \log (f)+\frac {4}{105} b^2 f^{a+\frac {b}{x^2}} x^3 \log ^2(f)+\frac {8}{105} b^3 f^{a+\frac {b}{x^2}} x \log ^3(f)-\frac {8}{105} b^{7/2} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {7}{2}}(f) \] Output:
1/7*f^(a+b/x^2)*x^7+2/35*b*f^(a+b/x^2)*x^5*ln(f)+4/105*b^2*f^(a+b/x^2)*x^3 *ln(f)^2+8/105*b^3*f^(a+b/x^2)*x*ln(f)^3-8/105*b^(7/2)*f^a*Pi^(1/2)*erfi(b ^(1/2)*ln(f)^(1/2)/x)*ln(f)^(7/2)
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.72 \[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\frac {1}{105} f^a \left (-8 b^{7/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \log ^{\frac {7}{2}}(f)+f^{\frac {b}{x^2}} x \left (15 x^6+6 b x^4 \log (f)+4 b^2 x^2 \log ^2(f)+8 b^3 \log ^3(f)\right )\right ) \] Input:
Integrate[f^(a + b/x^2)*x^6,x]
Output:
(f^a*(-8*b^(7/2)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Log[f]^(7/2) + f^ (b/x^2)*x*(15*x^6 + 6*b*x^4*Log[f] + 4*b^2*x^2*Log[f]^2 + 8*b^3*Log[f]^3)) )/105
Time = 0.64 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2643, 2643, 2643, 2635, 2640, 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.
| \(\displaystyle \int x^6 f^{a+\frac {b}{x^2}} \, dx\) |
| \(\Big \downarrow \) 2643 |
| \(\displaystyle \frac {2}{7} b \log (f) \int f^{a+\frac {b}{x^2}} x^4dx+\frac {1}{7} x^7 f^{a+\frac {b}{x^2}}\) |
| \(\Big \downarrow \) 2643 |
| \(\displaystyle \frac {2}{7} b \log (f) \left (\frac {2}{5} b \log (f) \int f^{a+\frac {b}{x^2}} x^2dx+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\right )+\frac {1}{7} x^7 f^{a+\frac {b}{x^2}}\) |
| \(\Big \downarrow \) 2643 |
| \(\displaystyle \frac {2}{7} b \log (f) \left (\frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \int f^{a+\frac {b}{x^2}}dx+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\right )+\frac {1}{7} x^7 f^{a+\frac {b}{x^2}}\) |
| \(\Big \downarrow \) 2635 |
| \(\displaystyle \frac {2}{7} b \log (f) \left (\frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (2 b \log (f) \int \frac {f^{a+\frac {b}{x^2}}}{x^2}dx+x f^{a+\frac {b}{x^2}}\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\right )+\frac {1}{7} x^7 f^{a+\frac {b}{x^2}}\) |
| \(\Big \downarrow \) 2640 |
| \(\displaystyle \frac {2}{7} b \log (f) \left (\frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (x f^{a+\frac {b}{x^2}}-2 b \log (f) \int f^{a+\frac {b}{x^2}}d\frac {1}{x}\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\right )+\frac {1}{7} x^7 f^{a+\frac {b}{x^2}}\) |
| \(\Big \downarrow \) 2633 |
| \(\displaystyle \frac {2}{7} b \log (f) \left (\frac {2}{5} b \log (f) \left (\frac {2}{3} b \log (f) \left (x f^{a+\frac {b}{x^2}}-\sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )\right )+\frac {1}{3} x^3 f^{a+\frac {b}{x^2}}\right )+\frac {1}{5} x^5 f^{a+\frac {b}{x^2}}\right )+\frac {1}{7} x^7 f^{a+\frac {b}{x^2}}\) |
Input:
Int[f^(a + b/x^2)*x^6,x]
Output:
(f^(a + b/x^2)*x^7)/7 + (2*b*Log[f]*((f^(a + b/x^2)*x^5)/5 + (2*b*Log[f]*( (f^(a + b/x^2)*x^3)/3 + (2*b*(f^(a + b/x^2)*x - Sqrt[b]*f^a*Sqrt[Pi]*Erfi[ (Sqrt[b]*Sqrt[Log[f]])/x]*Sqrt[Log[f]])*Log[f])/3))/5))/7
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_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x] - Simp[b*n*Log[F] Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] && ILtQ[n, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[1/(d*(m + 1)) Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1)]
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 = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84
| method | result | size |
| meijerg | \(\frac {f^{a} \ln \left (f \right )^{\frac {7}{2}} b^{3} \sqrt {-b}\, \left (-\frac {2 x^{7} \left (\frac {8 b^{3} \ln \left (f \right )^{3}}{15 x^{6}}+\frac {4 b^{2} \ln \left (f \right )^{2}}{15 x^{4}}+\frac {2 b \ln \left (f \right )}{5 x^{2}}+1\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{7 \left (-b \right )^{\frac {7}{2}} \ln \left (f \right )^{\frac {7}{2}}}+\frac {16 b^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{105 \left (-b \right )^{\frac {7}{2}}}\right )}{2}\) | \(100\) |
| risch | \(\frac {f^{a} f^{\frac {b}{x^{2}}} x^{7}}{7}+\frac {2 f^{a} \ln \left (f \right ) b \,x^{5} f^{\frac {b}{x^{2}}}}{35}+\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} x^{3} f^{\frac {b}{x^{2}}}}{105}+\frac {8 f^{a} \ln \left (f \right )^{3} b^{3} x \,f^{\frac {b}{x^{2}}}}{105}-\frac {8 f^{a} \ln \left (f \right )^{4} b^{4} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{105 \sqrt {-b \ln \left (f \right )}}\) | \(111\) |
Input:
int(f^(a+b/x^2)*x^6,x,method=_RETURNVERBOSE)
Output:
1/2*f^a*ln(f)^(7/2)*b^3*(-b)^(1/2)*(-2/7*x^7/(-b)^(7/2)/ln(f)^(7/2)*(8/15* b^3*ln(f)^3/x^6+4/15*b^2*ln(f)^2/x^4+2/5*b*ln(f)/x^2+1)*exp(b*ln(f)/x^2)+1 6/105/(-b)^(7/2)*b^(7/2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.72 \[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\frac {8}{105} \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{3} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) \log \left (f\right )^{3} + \frac {1}{105} \, {\left (15 \, x^{7} + 6 \, b x^{5} \log \left (f\right ) + 4 \, b^{2} x^{3} \log \left (f\right )^{2} + 8 \, b^{3} x \log \left (f\right )^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}} \] Input:
integrate(f^(a+b/x^2)*x^6,x, algorithm="fricas")
Output:
8/105*sqrt(pi)*sqrt(-b*log(f))*b^3*f^a*erf(sqrt(-b*log(f))/x)*log(f)^3 + 1 /105*(15*x^7 + 6*b*x^5*log(f) + 4*b^2*x^3*log(f)^2 + 8*b^3*x*log(f)^3)*f^( (a*x^2 + b)/x^2)
\[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\int f^{a + \frac {b}{x^{2}}} x^{6}\, dx \] Input:
integrate(f**(a+b/x**2)*x**6,x)
Output:
Integral(f**(a + b/x**2)*x**6, x)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.24 \[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\frac {1}{2} \, f^{a} x^{7} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {7}{2}} \Gamma \left (-\frac {7}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right ) \] Input:
integrate(f^(a+b/x^2)*x^6,x, algorithm="maxima")
Output:
1/2*f^a*x^7*(-b*log(f)/x^2)^(7/2)*gamma(-7/2, -b*log(f)/x^2)
\[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\int { f^{a + \frac {b}{x^{2}}} x^{6} \,d x } \] Input:
integrate(f^(a+b/x^2)*x^6,x, algorithm="giac")
Output:
integrate(f^(a + b/x^2)*x^6, x)
Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08 \[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\frac {f^a\,f^{\frac {b}{x^2}}\,x^7}{7}-\frac {8\,f^a\,x^7\,\sqrt {\pi }\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{7/2}}{105}+\frac {8\,f^a\,x^7\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (f\right )}{x^2}}\right )\,{\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}^{7/2}}{105}+\frac {8\,b^3\,f^a\,f^{\frac {b}{x^2}}\,x\,{\ln \left (f\right )}^3}{105}+\frac {4\,b^2\,f^a\,f^{\frac {b}{x^2}}\,x^3\,{\ln \left (f\right )}^2}{105}+\frac {2\,b\,f^a\,f^{\frac {b}{x^2}}\,x^5\,\ln \left (f\right )}{35} \] Input:
int(f^(a + b/x^2)*x^6,x)
Output:
(f^a*f^(b/x^2)*x^7)/7 - (8*f^a*x^7*pi^(1/2)*(-(b*log(f))/x^2)^(7/2))/105 + (8*f^a*x^7*pi^(1/2)*erfc((-(b*log(f))/x^2)^(1/2))*(-(b*log(f))/x^2)^(7/2) )/105 + (8*b^3*f^a*f^(b/x^2)*x*log(f)^3)/105 + (4*b^2*f^a*f^(b/x^2)*x^3*lo g(f)^2)/105 + (2*b*f^a*f^(b/x^2)*x^5*log(f))/35
\[ \int f^{a+\frac {b}{x^2}} x^6 \, dx=\frac {-16 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{4} b^{4}+8 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{3} b^{3} x^{2}+4 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right )^{2} b^{2} x^{4}+6 f^{\frac {a \,x^{2}+b}{x^{2}}} \mathrm {log}\left (f \right ) b \,x^{6}+15 f^{\frac {a \,x^{2}+b}{x^{2}}} x^{8}-32 \left (\int \frac {f^{\frac {a \,x^{2}+b}{x^{2}}}}{x^{4}}d x \right ) \mathrm {log}\left (f \right )^{5} b^{5} x}{105 x} \] Input:
int(f^(a+b/x^2)*x^6,x)
Output:
( - 16*f**((a*x**2 + b)/x**2)*log(f)**4*b**4 + 8*f**((a*x**2 + b)/x**2)*lo g(f)**3*b**3*x**2 + 4*f**((a*x**2 + b)/x**2)*log(f)**2*b**2*x**4 + 6*f**(( a*x**2 + b)/x**2)*log(f)*b*x**6 + 15*f**((a*x**2 + b)/x**2)*x**8 - 32*int( f**((a*x**2 + b)/x**2)/x**4,x)*log(f)**5*b**5*x)/(105*x)