\(\int f^{a+b x^2} x^4 \, dx\) [86]

Optimal result

Integrand size = 13, antiderivative size = 82 \[ \int f^{a+b x^2} x^4 \, dx=\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}-\frac {3 f^{a+b x^2} x}{4 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^3}{2 b \log (f)} \]

[Out]

-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)+3/8*f^a*erfi(x*b^(1/2)*ln(f)^(1/2))*Pi^(1/2)/b^(5/2
)/ln(f)^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2235} \[ \int f^{a+b x^2} x^4 \, dx=\frac {3 \sqrt {\pi } f^a \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}-\frac {3 x f^{a+b x^2}}{4 b^2 \log ^2(f)}+\frac {x^3 f^{a+b x^2}}{2 b \log (f)} \]

[In]

Int[f^(a + b*x^2)*x^4,x]

[Out]

(3*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]])/(8*b^(5/2)*Log[f]^(5/2)) - (3*f^(a + b*x^2)*x)/(4*b^2*Log[f]^2)
+ (f^(a + b*x^2)*x^3)/(2*b*Log[f])

Rule 2235

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]

Rule 2243

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*Log[F])), x] - Dist[(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])

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^2} x^3}{2 b \log (f)}-\frac {3 \int f^{a+b x^2} x^2 \, dx}{2 b \log (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)}+\frac {3 \int f^{a+b x^2} \, dx}{4 b^2 \log ^2(f)} \\ & = \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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int f^{a+b x^2} x^4 \, dx=\frac {f^a \left (3 \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )+2 \sqrt {b} f^{b x^2} x \sqrt {\log (f)} \left (-3+2 b x^2 \log (f)\right )\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)} \]

[In]

Integrate[f^(a + b*x^2)*x^4,x]

[Out]

(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))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91

method	result	size
meijerg	\(\frac {f^{a} \left (-\frac {x \left (-b \right )^{\frac {5}{2}} \sqrt {\ln \left (f \right )}\, \left (-10 b \,x^{2} \ln \left (f \right )+15\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{10 b^{2}}+\frac {3 \left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{4 b^{\frac {5}{2}}}\right )}{2 \ln \left (f \right )^{\frac {5}{2}} b^{2} \sqrt {-b}}\)	\(75\)
risch	\(\frac {f^{a} x^{3} f^{b \,x^{2}}}{2 \ln \left (f \right ) b}-\frac {3 f^{a} x \,f^{b \,x^{2}}}{4 \ln \left (f \right )^{2} b^{2}}+\frac {3 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{8 \ln \left (f \right )^{2} b^{2} \sqrt {-b \ln \left (f \right )}}\)	\(76\)

[In]

int(f^(b*x^2+a)*x^4,x,method=_RETURNVERBOSE)

[Out]

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(x*b^(1/2)*ln(f)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int f^{a+b x^2} x^4 \, dx=-\frac {3 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) - 2 \, {\left (2 \, b^{2} x^{3} \log \left (f\right )^{2} - 3 \, b x \log \left (f\right )\right )} f^{b x^{2} + a}}{8 \, b^{3} \log \left (f\right )^{3}} \]

[In]

integrate(f^(b*x^2+a)*x^4,x, algorithm="fricas")

[Out]

-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)

Sympy [F]

\[ \int f^{a+b x^2} x^4 \, dx=\int f^{a + b x^{2}} x^{4}\, dx \]

[In]

integrate(f**(b*x**2+a)*x**4,x)

[Out]

Integral(f**(a + b*x**2)*x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int f^{a+b x^2} x^4 \, dx=\frac {{\left (2 \, b f^{a} x^{3} \log \left (f\right ) - 3 \, f^{a} x\right )} f^{b x^{2}}}{4 \, b^{2} \log \left (f\right )^{2}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right )}{8 \, \sqrt {-b \log \left (f\right )} b^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(b*x^2+a)*x^4,x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83 \[ \int f^{a+b x^2} x^4 \, dx=-\frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (f\right )} x\right )}{8 \, \sqrt {-b \log \left (f\right )} b^{2} \log \left (f\right )^{2}} + \frac {{\left (2 \, b x^{3} \log \left (f\right ) - 3 \, x\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{4 \, b^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(b*x^2+a)*x^4,x, algorithm="giac")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int f^{a+b x^2} x^4 \, dx=\frac {f^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )-6\,f^{b\,x^2}\,x\,\sqrt {b\,\ln \left (f\right )}\right )}{8\,b^2\,{\ln \left (f\right )}^2\,\sqrt {b\,\ln \left (f\right )}}+\frac {f^a\,f^{b\,x^2}\,x^3}{2\,b\,\ln \left (f\right )} \]

[In]

int(f^(a + b*x^2)*x^4,x)

[Out]

(f^a*(3*pi^(1/2)*erfi((b*x*log(f))/(b*log(f))^(1/2)) - 6*f^(b*x^2)*x*(b*log(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))