Integrand size = 13, antiderivative size = 81 \[ \int \frac {f^{a+b x^2}}{x^7} \, dx=-\frac {f^{a+b x^2}}{6 x^6}-\frac {b f^{a+b x^2} \log (f)}{12 x^4}-\frac {b^2 f^{a+b x^2} \log ^2(f)}{12 x^2}+\frac {1}{12} b^3 f^a \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \log ^3(f) \] Output:
-1/6*f^(b*x^2+a)/x^6-1/12*b*f^(b*x^2+a)*ln(f)/x^4-1/12*b^2*f^(b*x^2+a)*ln( f)^2/x^2+1/12*b^3*f^a*Ei(b*x^2*ln(f))*ln(f)^3
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.73 \[ \int \frac {f^{a+b x^2}}{x^7} \, dx=\frac {f^a \left (b^3 x^6 \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \log ^3(f)-f^{b x^2} \left (2+b x^2 \log (f)+b^2 x^4 \log ^2(f)\right )\right )}{12 x^6} \] Input:
Integrate[f^(a + b*x^2)/x^7,x]
Output:
(f^a*(b^3*x^6*ExpIntegralEi[b*x^2*Log[f]]*Log[f]^3 - f^(b*x^2)*(2 + b*x^2* Log[f] + b^2*x^4*Log[f]^2)))/(12*x^6)
Time = 0.50 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02, 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, 2639}
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.
\(\displaystyle \int \frac {f^{a+b x^2}}{x^7} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (f) \int \frac {f^{b x^2+a}}{x^5}dx-\frac {f^{a+b x^2}}{6 x^6}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (f) \left (\frac {1}{2} b \log (f) \int \frac {f^{b x^2+a}}{x^3}dx-\frac {f^{a+b x^2}}{4 x^4}\right )-\frac {f^{a+b x^2}}{6 x^6}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{3} b \log (f) \left (\frac {1}{2} b \log (f) \left (b \log (f) \int \frac {f^{b x^2+a}}{x}dx-\frac {f^{a+b x^2}}{2 x^2}\right )-\frac {f^{a+b x^2}}{4 x^4}\right )-\frac {f^{a+b x^2}}{6 x^6}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{3} b \log (f) \left (\frac {1}{2} b \log (f) \left (\frac {1}{2} b f^a \log (f) \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right )-\frac {f^{a+b x^2}}{2 x^2}\right )-\frac {f^{a+b x^2}}{4 x^4}\right )-\frac {f^{a+b x^2}}{6 x^6}\) |
Input:
Int[f^(a + b*x^2)/x^7,x]
Output:
-1/6*f^(a + b*x^2)/x^6 + (b*Log[f]*(-1/4*f^(a + b*x^2)/x^4 + (b*Log[f]*(-1 /2*f^(a + b*x^2)/x^2 + (b*f^a*ExpIntegralEi[b*x^2*Log[f]]*Log[f])/2))/2))/ 3
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_ Symbol] :> Simp[F^a*(ExpIntegralEi[b*(c + d*x)^n*Log[F]]/(f*n)), x] /; Free Q[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]
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 = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {f^{a} \left (\ln \left (f \right )^{3} \operatorname {expIntegral}_{1}\left (-b \,x^{2} \ln \left (f \right )\right ) b^{3} x^{6}+\ln \left (f \right )^{2} f^{b \,x^{2}} b^{2} x^{4}+\ln \left (f \right ) f^{b \,x^{2}} b \,x^{2}+2 f^{b \,x^{2}}\right )}{12 x^{6}}\) | \(72\) |
meijerg | \(-\frac {f^{a} b^{3} \ln \left (f \right )^{3} \left (\frac {1}{3 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {1}{2 b^{2} x^{4} \ln \left (f \right )^{2}}+\frac {1}{2 b \,x^{2} \ln \left (f \right )}+\frac {11}{36}-\frac {\ln \left (x \right )}{3}-\frac {\ln \left (-b \right )}{6}-\frac {\ln \left (\ln \left (f \right )\right )}{6}-\frac {22 b^{3} x^{6} \ln \left (f \right )^{3}+36 b^{2} x^{4} \ln \left (f \right )^{2}+36 b \,x^{2} \ln \left (f \right )+24}{72 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {\left (4 b^{2} x^{4} \ln \left (f \right )^{2}+4 b \,x^{2} \ln \left (f \right )+8\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{24 b^{3} x^{6} \ln \left (f \right )^{3}}+\frac {\ln \left (-b \,x^{2} \ln \left (f \right )\right )}{6}+\frac {\operatorname {expIntegral}_{1}\left (-b \,x^{2} \ln \left (f \right )\right )}{6}\right )}{2}\) | \(177\) |
Input:
int(f^(b*x^2+a)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/12*f^a*(ln(f)^3*Ei(1,-b*x^2*ln(f))*b^3*x^6+ln(f)^2*f^(b*x^2)*b^2*x^4+ln (f)*f^(b*x^2)*b*x^2+2*f^(b*x^2))/x^6
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.73 \[ \int \frac {f^{a+b x^2}}{x^7} \, dx=\frac {b^{3} f^{a} x^{6} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{3} - {\left (b^{2} x^{4} \log \left (f\right )^{2} + b x^{2} \log \left (f\right ) + 2\right )} f^{b x^{2} + a}}{12 \, x^{6}} \] Input:
integrate(f^(b*x^2+a)/x^7,x, algorithm="fricas")
Output:
1/12*(b^3*f^a*x^6*Ei(b*x^2*log(f))*log(f)^3 - (b^2*x^4*log(f)^2 + b*x^2*lo g(f) + 2)*f^(b*x^2 + a))/x^6
\[ \int \frac {f^{a+b x^2}}{x^7} \, dx=\int \frac {f^{a + b x^{2}}}{x^{7}}\, dx \] Input:
integrate(f**(b*x**2+a)/x**7,x)
Output:
Integral(f**(a + b*x**2)/x**7, x)
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.27 \[ \int \frac {f^{a+b x^2}}{x^7} \, dx=\frac {1}{2} \, b^{3} f^{a} \Gamma \left (-3, -b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{3} \] Input:
integrate(f^(b*x^2+a)/x^7,x, algorithm="maxima")
Output:
1/2*b^3*f^a*gamma(-3, -b*x^2*log(f))*log(f)^3
\[ \int \frac {f^{a+b x^2}}{x^7} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{7}} \,d x } \] Input:
integrate(f^(b*x^2+a)/x^7,x, algorithm="giac")
Output:
integrate(f^(b*x^2 + a)/x^7, x)
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int \frac {f^{a+b x^2}}{x^7} \, dx=-\frac {b^3\,f^a\,{\ln \left (f\right )}^3\,\left (f^{b\,x^2}\,\left (\frac {1}{6\,b\,x^2\,\ln \left (f\right )}+\frac {1}{6\,b^2\,x^4\,{\ln \left (f\right )}^2}+\frac {1}{3\,b^3\,x^6\,{\ln \left (f\right )}^3}\right )+\frac {\mathrm {expint}\left (-b\,x^2\,\ln \left (f\right )\right )}{6}\right )}{2} \] Input:
int(f^(a + b*x^2)/x^7,x)
Output:
-(b^3*f^a*log(f)^3*(f^(b*x^2)*(1/(6*b*x^2*log(f)) + 1/(6*b^2*x^4*log(f)^2) + 1/(3*b^3*x^6*log(f)^3)) + expint(-b*x^2*log(f))/6))/2
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {f^{a+b x^2}}{x^7} \, dx=\frac {f^{a} \left (\mathit {ei} \left (\mathrm {log}\left (f \right ) b \,x^{2}\right ) \mathrm {log}\left (f \right )^{3} b^{3} x^{6}-f^{b \,x^{2}} \mathrm {log}\left (f \right )^{2} b^{2} x^{4}-f^{b \,x^{2}} \mathrm {log}\left (f \right ) b \,x^{2}-2 f^{b \,x^{2}}\right )}{12 x^{6}} \] Input:
int(f^(b*x^2+a)/x^7,x)
Output:
(f**a*(ei(log(f)*b*x**2)*log(f)**3*b**3*x**6 - f**(b*x**2)*log(f)**2*b**2* x**4 - f**(b*x**2)*log(f)*b*x**2 - 2*f**(b*x**2)))/(12*x**6)