Integrand size = 13, antiderivative size = 82 \[ \int f^{a+b x^2} x^4 \, dx=\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}-\frac {3 f^{a+b x^2} x}{4 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^3}{2 b \log (f)} \] Output:
3/8*f^a*Pi^(1/2)*erfi(b^(1/2)*x*ln(f)^(1/2))/b^(5/2)/ln(f)^(5/2)-3/4*f^(b* x^2+a)*x/b^2/ln(f)^2+1/2*f^(b*x^2+a)*x^3/b/ln(f)
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int f^{a+b x^2} x^4 \, dx=\frac {f^a \left (3 \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )+2 \sqrt {b} f^{b x^2} x \sqrt {\log (f)} \left (-3+2 b x^2 \log (f)\right )\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)} \] Input:
Integrate[f^(a + b*x^2)*x^4,x]
Output:
(f^a*(3*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]] + 2*Sqrt[b]*f^(b*x^2)*x*Sqrt [Log[f]]*(-3 + 2*b*x^2*Log[f])))/(8*b^(5/2)*Log[f]^(5/2))
Time = 0.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2641, 2641, 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.
\(\displaystyle \int x^4 f^{a+b x^2} \, dx\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle \frac {x^3 f^{a+b x^2}}{2 b \log (f)}-\frac {3 \int f^{b x^2+a} x^2dx}{2 b \log (f)}\) |
\(\Big \downarrow \) 2641 |
\(\displaystyle \frac {x^3 f^{a+b x^2}}{2 b \log (f)}-\frac {3 \left (\frac {x f^{a+b x^2}}{2 b \log (f)}-\frac {\int f^{b x^2+a}dx}{2 b \log (f)}\right )}{2 b \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {x^3 f^{a+b x^2}}{2 b \log (f)}-\frac {3 \left (\frac {x f^{a+b x^2}}{2 b \log (f)}-\frac {\sqrt {\pi } f^a \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}\right )}{2 b \log (f)}\) |
Input:
Int[f^(a + b*x^2)*x^4,x]
Output:
(f^(a + b*x^2)*x^3)/(2*b*Log[f]) - (3*(-1/4*(f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*S qrt[Log[f]]])/(b^(3/2)*Log[f]^(3/2)) + (f^(a + b*x^2)*x)/(2*b*Log[f])))/(2 *b*Log[f])
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 - n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*L og[F])), x] - Simp[(m - n + 1)/(b*n*Log[F]) 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[0, (m + 1)/n, 5] && IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n , 0])
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91
method | result | size |
meijerg | \(\frac {f^{a} \left (-\frac {x \left (-b \right )^{\frac {5}{2}} \sqrt {\ln \left (f \right )}\, \left (-10 b \,x^{2} \ln \left (f \right )+15\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{10 b^{2}}+\frac {3 \left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, x \sqrt {\ln \left (f \right )}\right )}{4 b^{\frac {5}{2}}}\right )}{2 \ln \left (f \right )^{\frac {5}{2}} b^{2} \sqrt {-b}}\) | \(75\) |
risch | \(\frac {f^{a} x^{3} f^{b \,x^{2}}}{2 \ln \left (f \right ) b}-\frac {3 f^{a} x \,f^{b \,x^{2}}}{4 \ln \left (f \right )^{2} b^{2}}+\frac {3 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{8 \ln \left (f \right )^{2} b^{2} \sqrt {-b \ln \left (f \right )}}\) | \(76\) |
Input:
int(f^(b*x^2+a)*x^4,x,method=_RETURNVERBOSE)
Output:
1/2*f^a/ln(f)^(5/2)/b^2/(-b)^(1/2)*(-1/10*x*(-b)^(5/2)*ln(f)^(1/2)*(-10*b* x^2*ln(f)+15)/b^2*exp(b*x^2*ln(f))+3/4*(-b)^(5/2)/b^(5/2)*Pi^(1/2)*erfi(b^ (1/2)*x*ln(f)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int f^{a+b x^2} x^4 \, dx=-\frac {3 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) - 2 \, {\left (2 \, b^{2} x^{3} \log \left (f\right )^{2} - 3 \, b x \log \left (f\right )\right )} f^{b x^{2} + a}}{8 \, b^{3} \log \left (f\right )^{3}} \] Input:
integrate(f^(b*x^2+a)*x^4,x, algorithm="fricas")
Output:
-1/8*(3*sqrt(pi)*sqrt(-b*log(f))*f^a*erf(sqrt(-b*log(f))*x) - 2*(2*b^2*x^3 *log(f)^2 - 3*b*x*log(f))*f^(b*x^2 + a))/(b^3*log(f)^3)
\[ \int f^{a+b x^2} x^4 \, dx=\int f^{a + b x^{2}} x^{4}\, dx \] Input:
integrate(f**(b*x**2+a)*x**4,x)
Output:
Integral(f**(a + b*x**2)*x**4, x)
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int f^{a+b x^2} x^4 \, dx=\frac {{\left (2 \, b f^{a} x^{3} \log \left (f\right ) - 3 \, f^{a} x\right )} f^{b x^{2}}}{4 \, b^{2} \log \left (f\right )^{2}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right )}{8 \, \sqrt {-b \log \left (f\right )} b^{2} \log \left (f\right )^{2}} \] Input:
integrate(f^(b*x^2+a)*x^4,x, algorithm="maxima")
Output:
1/4*(2*b*f^a*x^3*log(f) - 3*f^a*x)*f^(b*x^2)/(b^2*log(f)^2) + 3/8*sqrt(pi) *f^a*erf(sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^2*log(f)^2)
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83 \[ \int f^{a+b x^2} x^4 \, dx=-\frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (f\right )} x\right )}{8 \, \sqrt {-b \log \left (f\right )} b^{2} \log \left (f\right )^{2}} + \frac {{\left (2 \, b x^{3} \log \left (f\right ) - 3 \, x\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{4 \, b^{2} \log \left (f\right )^{2}} \] Input:
integrate(f^(b*x^2+a)*x^4,x, algorithm="giac")
Output:
-3/8*sqrt(pi)*f^a*erf(-sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^2*log(f)^2) + 1/4*(2*b*x^3*log(f) - 3*x)*e^(b*x^2*log(f) + a*log(f))/(b^2*log(f)^2)
Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int f^{a+b x^2} x^4 \, dx=\frac {f^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )-6\,f^{b\,x^2}\,x\,\sqrt {b\,\ln \left (f\right )}\right )}{8\,b^2\,{\ln \left (f\right )}^2\,\sqrt {b\,\ln \left (f\right )}}+\frac {f^a\,f^{b\,x^2}\,x^3}{2\,b\,\ln \left (f\right )} \] Input:
int(f^(a + b*x^2)*x^4,x)
Output:
(f^a*(3*pi^(1/2)*erfi((b*x*log(f))/(b*log(f))^(1/2)) - 6*f^(b*x^2)*x*(b*lo g(f))^(1/2)))/(8*b^2*log(f)^2*(b*log(f))^(1/2)) + (f^a*f^(b*x^2)*x^3)/(2*b *log(f))
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int f^{a+b x^2} x^4 \, dx=\frac {f^{a} \left (-3 \sqrt {\pi }\, \mathrm {erf}\left (\sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, i x \right ) i +4 f^{b \,x^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b \,x^{3}-6 f^{b \,x^{2}} \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, x \right )}{8 \sqrt {b}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{2}} \] Input:
int(f^(b*x^2+a)*x^4,x)
Output:
(f**a*( - 3*sqrt(pi)*erf(sqrt(b)*sqrt(log(f))*i*x)*i + 4*f**(b*x**2)*sqrt( b)*sqrt(log(f))*log(f)*b*x**3 - 6*f**(b*x**2)*sqrt(b)*sqrt(log(f))*x))/(8* sqrt(b)*sqrt(log(f))*log(f)**2*b**2)