Integrand size = 19, antiderivative size = 96 \[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=-\frac {2 f^{a+b x^n} x^{-3 n/2}}{3 n}-\frac {4 b f^{a+b x^n} x^{-n/2} \log (f)}{3 n}+\frac {4 b^{3/2} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x^{n/2} \sqrt {\log (f)}\right ) \log ^{\frac {3}{2}}(f)}{3 n} \] Output:
-2/3*f^(a+b*x^n)/n/(x^(3/2*n))-4/3*b*f^(a+b*x^n)*ln(f)/n/(x^(1/2*n))+4/3*b ^(3/2)*f^a*Pi^(1/2)*erfi(b^(1/2)*x^(1/2*n)*ln(f)^(1/2))*ln(f)^(3/2)/n
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.41 \[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=-\frac {f^a x^{-3 n/2} \Gamma \left (-\frac {3}{2},-b x^n \log (f)\right ) \left (-b x^n \log (f)\right )^{3/2}}{n} \] Input:
Integrate[f^(a + b*x^n)*x^(-1 - (3*n)/2),x]
Output:
-((f^a*Gamma[-3/2, -(b*x^n*Log[f])]*(-(b*x^n*Log[f]))^(3/2))/(n*x^((3*n)/2 )))
Time = 0.54 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2644, 2644, 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.
\(\displaystyle \int x^{-\frac {3 n}{2}-1} f^{a+b x^n} \, dx\) |
\(\Big \downarrow \) 2644 |
\(\displaystyle \frac {2}{3} b \log (f) \int f^{b x^n+a} x^{-\frac {n}{2}-1}dx-\frac {2 x^{-3 n/2} f^{a+b x^n}}{3 n}\) |
\(\Big \downarrow \) 2644 |
\(\displaystyle \frac {2}{3} b \log (f) \left (2 b \log (f) \int f^{b x^n+a} x^{\frac {n-2}{2}}dx-\frac {2 x^{-n/2} f^{a+b x^n}}{n}\right )-\frac {2 x^{-3 n/2} f^{a+b x^n}}{3 n}\) |
\(\Big \downarrow \) 2640 |
\(\displaystyle \frac {2}{3} b \log (f) \left (\frac {4 b \log (f) \int f^{b x^n+a}dx^{n/2}}{n}-\frac {2 x^{-n/2} f^{a+b x^n}}{n}\right )-\frac {2 x^{-3 n/2} f^{a+b x^n}}{3 n}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {2}{3} b \log (f) \left (\frac {2 \sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\sqrt {b} \sqrt {\log (f)} x^{n/2}\right )}{n}-\frac {2 x^{-n/2} f^{a+b x^n}}{n}\right )-\frac {2 x^{-3 n/2} f^{a+b x^n}}{3 n}\) |
Input:
Int[f^(a + b*x^n)*x^(-1 - (3*n)/2),x]
Output:
(-2*f^(a + b*x^n))/(3*n*x^((3*n)/2)) + (2*b*((-2*f^(a + b*x^n))/(n*x^(n/2) ) + (2*Sqrt[b]*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x^(n/2)*Sqrt[Log[f]]]*Sqrt[Log[f] ])/n)*Log[f])/3
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[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)^Simplify[m + n]*F^(a + b*( c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && IntegerQ[2*Simpli fy[(m + 1)/n]] && LtQ[-4, Simplify[(m + 1)/n], 5] && !RationalQ[m] && SumS implerQ[m, n]
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82
method | result | size |
meijerg | \(\frac {f^{a} \left (-b \right )^{\frac {3}{2}} \ln \left (f \right )^{\frac {3}{2}} \left (-\frac {2 x^{-\frac {3 n}{2}} \left (2 b \,x^{n} \ln \left (f \right )+1\right ) {\mathrm e}^{b \,x^{n} \ln \left (f \right )}}{3 \left (-b \right )^{\frac {3}{2}} \ln \left (f \right )^{\frac {3}{2}}}+\frac {4 b^{\frac {3}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, x^{\frac {n}{2}} \sqrt {\ln \left (f \right )}\right )}{3 \left (-b \right )^{\frac {3}{2}}}\right )}{n}\) | \(79\) |
risch | \(-\frac {2 f^{a} x^{-\frac {3 n}{2}} f^{b \,x^{n}}}{3 n}-\frac {4 f^{a} \ln \left (f \right ) b \,x^{-\frac {n}{2}} f^{b \,x^{n}}}{3 n}+\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x^{\frac {n}{2}}\right )}{3 n \sqrt {-b \ln \left (f \right )}}\) | \(88\) |
Input:
int(f^(a+b*x^n)*x^(-1-3/2*n),x,method=_RETURNVERBOSE)
Output:
f^a*(-b)^(3/2)*ln(f)^(3/2)/n*(-2/3*x^(-3/2*n)/(-b)^(3/2)/ln(f)^(3/2)*(2*b* x^n*ln(f)+1)*exp(b*x^n*ln(f))+4/3/(-b)^(3/2)*b^(3/2)*Pi^(1/2)*erfi(b^(1/2) *x^(1/2*n)*ln(f)^(1/2)))
\[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=\int { f^{b x^{n} + a} x^{-\frac {3}{2} \, n - 1} \,d x } \] Input:
integrate(f^(a+b*x^n)*x^(-1-3/2*n),x, algorithm="fricas")
Output:
integral(f^(b*x^n + a)*x^(-3/2*n - 1), x)
\[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=\int f^{a + b x^{n}} x^{- \frac {3 n}{2} - 1}\, dx \] Input:
integrate(f**(a+b*x**n)*x**(-1-3/2*n),x)
Output:
Integral(f**(a + b*x**n)*x**(-3*n/2 - 1), x)
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.36 \[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=-\frac {\left (-b x^{n} \log \left (f\right )\right )^{\frac {3}{2}} f^{a} \Gamma \left (-\frac {3}{2}, -b x^{n} \log \left (f\right )\right )}{n x^{\frac {3}{2} \, n}} \] Input:
integrate(f^(a+b*x^n)*x^(-1-3/2*n),x, algorithm="maxima")
Output:
-(-b*x^n*log(f))^(3/2)*f^a*gamma(-3/2, -b*x^n*log(f))/(n*x^(3/2*n))
\[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=\int { f^{b x^{n} + a} x^{-\frac {3}{2} \, n - 1} \,d x } \] Input:
integrate(f^(a+b*x^n)*x^(-1-3/2*n),x, algorithm="giac")
Output:
integrate(f^(b*x^n + a)*x^(-3/2*n - 1), x)
Timed out. \[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=\int \frac {f^{a+b\,x^n}}{x^{\frac {3\,n}{2}+1}} \,d x \] Input:
int(f^(a + b*x^n)/x^((3*n)/2 + 1),x)
Output:
int(f^(a + b*x^n)/x^((3*n)/2 + 1), x)
\[ \int f^{a+b x^n} x^{-1-\frac {3 n}{2}} \, dx=f^{a} \left (\int \frac {f^{x^{n} b}}{x^{\frac {3 n}{2}} x}d x \right ) \] Input:
int(f^(a+b*x^n)*x^(-1-3/2*n),x)
Output:
f**a*int(f**(x**n*b)/(x**((3*n)/2)*x),x)