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

Optimal. Leaf size=82 \[ \frac {3 \sqrt {\pi } f^a \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}-\frac {3 x f^{a+b x^2}}{4 b^2 \log ^2(f)}+\frac {x^3 f^{a+b x^2}}{2 b \log (f)} \]

[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)

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Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2212, 2204} \[ \frac {3 \sqrt {\pi } f^a \text {Erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}-\frac {3 x f^{a+b x^2}}{4 b^2 \log ^2(f)}+\frac {x^3 f^{a+b x^2}}{2 b \log (f)} \]

Antiderivative was successfully verified.

[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 2204

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 2212

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*} \int f^{a+b x^2} x^4 \, dx &=\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*}

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Mathematica [A]  time = 0.03, size = 71, normalized size = 0.87 \[ \frac {f^a \left (3 \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )+2 \sqrt {b} x \sqrt {\log (f)} f^{b x^2} \left (2 b x^2 \log (f)-3\right )\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.42, size = 65, normalized size = 0.79 \[ -\frac {3 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right ) - 2 \, {\left (2 \, b^{2} x^{3} \log \relax (f)^{2} - 3 \, b x \log \relax (f)\right )} f^{b x^{2} + a}}{8 \, b^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.40, size = 68, normalized size = 0.83 \[ -\frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \relax (f)} x\right )}{8 \, \sqrt {-b \log \relax (f)} b^{2} \log \relax (f)^{2}} + \frac {{\left (2 \, b x^{3} \log \relax (f) - 3 \, x\right )} e^{\left (b x^{2} \log \relax (f) + a \log \relax (f)\right )}}{4 \, b^{2} \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 76, normalized size = 0.93 \[ \frac {x^{3} f^{a} f^{b \,x^{2}}}{2 b \ln \relax (f )}-\frac {3 x \,f^{a} f^{b \,x^{2}}}{4 b^{2} \ln \relax (f )^{2}}+\frac {3 \sqrt {\pi }\, f^{a} \erf \left (\sqrt {-b \ln \relax (f )}\, x \right )}{8 \sqrt {-b \ln \relax (f )}\, b^{2} \ln \relax (f )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f^a/ln(f)/b*x^3*f^(b*x^2)-3/4*f^a/ln(f)^2/b^2*x*f^(b*x^2)+3/8*f^a/ln(f)^2/b^2*Pi^(1/2)/(-b*ln(f))^(1/2)*er
f((-b*ln(f))^(1/2)*x)

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maxima [A]  time = 0.63, size = 67, normalized size = 0.82 \[ \frac {{\left (2 \, b f^{a} x^{3} \log \relax (f) - 3 \, f^{a} x\right )} f^{b x^{2}}}{4 \, b^{2} \log \relax (f)^{2}} + \frac {3 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \relax (f)} x\right )}{8 \, \sqrt {-b \log \relax (f)} b^{2} \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.54, size = 75, normalized size = 0.91 \[ \frac {f^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \relax (f)}{\sqrt {b\,\ln \relax (f)}}\right )-6\,f^{b\,x^2}\,x\,\sqrt {b\,\ln \relax (f)}\right )}{8\,b^2\,{\ln \relax (f)}^2\,\sqrt {b\,\ln \relax (f)}}+\frac {f^a\,f^{b\,x^2}\,x^3}{2\,b\,\ln \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x^{2}} x^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b*x**2)*x**4, x)

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