\(\int f^{a+b x^2} x^6 \, dx\) [85]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 105 \[ \int f^{a+b x^2} x^6 \, dx=-\frac {15 f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}+\frac {15 f^{a+b x^2} x}{8 b^3 \log ^3(f)}-\frac {5 f^{a+b x^2} x^3}{4 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^5}{2 b \log (f)} \]

[Out]

15/8*f^(b*x^2+a)*x/b^3/ln(f)^3-5/4*f^(b*x^2+a)*x^3/b^2/ln(f)^2+1/2*f^(b*x^2+a)*x^5/b/ln(f)-15/16*f^a*erfi(x*b^
(1/2)*ln(f)^(1/2))*Pi^(1/2)/b^(7/2)/ln(f)^(7/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2235} \[ \int f^{a+b x^2} x^6 \, dx=-\frac {15 \sqrt {\pi } f^a \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}+\frac {15 x f^{a+b x^2}}{8 b^3 \log ^3(f)}-\frac {5 x^3 f^{a+b x^2}}{4 b^2 \log ^2(f)}+\frac {x^5 f^{a+b x^2}}{2 b \log (f)} \]

[In]

Int[f^(a + b*x^2)*x^6,x]

[Out]

(-15*f^a*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]])/(16*b^(7/2)*Log[f]^(7/2)) + (15*f^(a + b*x^2)*x)/(8*b^3*Log[f]
^3) - (5*f^(a + b*x^2)*x^3)/(4*b^2*Log[f]^2) + (f^(a + b*x^2)*x^5)/(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^5}{2 b \log (f)}-\frac {5 \int f^{a+b x^2} x^4 \, dx}{2 b \log (f)} \\ & = -\frac {5 f^{a+b x^2} x^3}{4 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^5}{2 b \log (f)}+\frac {15 \int f^{a+b x^2} x^2 \, dx}{4 b^2 \log ^2(f)} \\ & = \frac {15 f^{a+b x^2} x}{8 b^3 \log ^3(f)}-\frac {5 f^{a+b x^2} x^3}{4 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^5}{2 b \log (f)}-\frac {15 \int f^{a+b x^2} \, dx}{8 b^3 \log ^3(f)} \\ & = -\frac {15 f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}+\frac {15 f^{a+b x^2} x}{8 b^3 \log ^3(f)}-\frac {5 f^{a+b x^2} x^3}{4 b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^5}{2 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79 \[ \int f^{a+b x^2} x^6 \, dx=\frac {f^a \left (-15 \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )+2 \sqrt {b} f^{b x^2} x \sqrt {\log (f)} \left (15-10 b x^2 \log (f)+4 b^2 x^4 \log ^2(f)\right )\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)} \]

[In]

Integrate[f^(a + b*x^2)*x^6,x]

[Out]

(f^a*(-15*Sqrt[Pi]*Erfi[Sqrt[b]*x*Sqrt[Log[f]]] + 2*Sqrt[b]*f^(b*x^2)*x*Sqrt[Log[f]]*(15 - 10*b*x^2*Log[f] + 4
*b^2*x^4*Log[f]^2)))/(16*b^(7/2)*Log[f]^(7/2))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.83

method result size
meijerg \(-\frac {f^{a} \left (\frac {x \left (-b \right )^{\frac {7}{2}} \sqrt {\ln \left (f \right )}\, \left (28 b^{2} x^{4} \ln \left (f \right )^{2}-70 b \,x^{2} \ln \left (f \right )+105\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{28 b^{3}}-\frac {15 \left (-b \right )^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{8 b^{\frac {7}{2}}}\right )}{2 \ln \left (f \right )^{\frac {7}{2}} b^{3} \sqrt {-b}}\) \(87\)
risch \(\frac {f^{a} x^{5} f^{b \,x^{2}}}{2 \ln \left (f \right ) b}-\frac {5 f^{a} x^{3} f^{b \,x^{2}}}{4 \ln \left (f \right )^{2} b^{2}}+\frac {15 f^{a} x \,f^{b \,x^{2}}}{8 \ln \left (f \right )^{3} b^{3}}-\frac {15 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{16 \ln \left (f \right )^{3} b^{3} \sqrt {-b \ln \left (f \right )}}\) \(98\)

[In]

int(f^(b*x^2+a)*x^6,x,method=_RETURNVERBOSE)

[Out]

-1/2*f^a/ln(f)^(7/2)/b^3/(-b)^(1/2)*(1/28*x*(-b)^(7/2)*ln(f)^(1/2)*(28*b^2*x^4*ln(f)^2-70*b*x^2*ln(f)+105)/b^3
*exp(b*x^2*ln(f))-15/8*(-b)^(7/2)/b^(7/2)*Pi^(1/2)*erfi(x*b^(1/2)*ln(f)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int f^{a+b x^2} x^6 \, dx=\frac {15 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) + 2 \, {\left (4 \, b^{3} x^{5} \log \left (f\right )^{3} - 10 \, b^{2} x^{3} \log \left (f\right )^{2} + 15 \, b x \log \left (f\right )\right )} f^{b x^{2} + a}}{16 \, b^{4} \log \left (f\right )^{4}} \]

[In]

integrate(f^(b*x^2+a)*x^6,x, algorithm="fricas")

[Out]

1/16*(15*sqrt(pi)*sqrt(-b*log(f))*f^a*erf(sqrt(-b*log(f))*x) + 2*(4*b^3*x^5*log(f)^3 - 10*b^2*x^3*log(f)^2 + 1
5*b*x*log(f))*f^(b*x^2 + a))/(b^4*log(f)^4)

Sympy [F]

\[ \int f^{a+b x^2} x^6 \, dx=\int f^{a + b x^{2}} x^{6}\, dx \]

[In]

integrate(f**(b*x**2+a)*x**6,x)

[Out]

Integral(f**(a + b*x**2)*x**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.78 \[ \int f^{a+b x^2} x^6 \, dx=\frac {{\left (4 \, b^{2} f^{a} x^{5} \log \left (f\right )^{2} - 10 \, b f^{a} x^{3} \log \left (f\right ) + 15 \, f^{a} x\right )} f^{b x^{2}}}{8 \, b^{3} \log \left (f\right )^{3}} - \frac {15 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right )}{16 \, \sqrt {-b \log \left (f\right )} b^{3} \log \left (f\right )^{3}} \]

[In]

integrate(f^(b*x^2+a)*x^6,x, algorithm="maxima")

[Out]

1/8*(4*b^2*f^a*x^5*log(f)^2 - 10*b*f^a*x^3*log(f) + 15*f^a*x)*f^(b*x^2)/(b^3*log(f)^3) - 15/16*sqrt(pi)*f^a*er
f(sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^3*log(f)^3)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76 \[ \int f^{a+b x^2} x^6 \, dx=\frac {15 \, \sqrt {\pi } f^{a} \operatorname {erf}\left (-\sqrt {-b \log \left (f\right )} x\right )}{16 \, \sqrt {-b \log \left (f\right )} b^{3} \log \left (f\right )^{3}} + \frac {{\left (4 \, b^{2} x^{5} \log \left (f\right )^{2} - 10 \, b x^{3} \log \left (f\right ) + 15 \, x\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{8 \, b^{3} \log \left (f\right )^{3}} \]

[In]

integrate(f^(b*x^2+a)*x^6,x, algorithm="giac")

[Out]

15/16*sqrt(pi)*f^a*erf(-sqrt(-b*log(f))*x)/(sqrt(-b*log(f))*b^3*log(f)^3) + 1/8*(4*b^2*x^5*log(f)^2 - 10*b*x^3
*log(f) + 15*x)*e^(b*x^2*log(f) + a*log(f))/(b^3*log(f)^3)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93 \[ \int f^{a+b x^2} x^6 \, dx=\frac {15\,f^a\,f^{b\,x^2}\,x}{8\,b^3\,{\ln \left (f\right )}^3}+\frac {f^a\,f^{b\,x^2}\,x^5}{2\,b\,\ln \left (f\right )}-\frac {5\,f^a\,f^{b\,x^2}\,x^3}{4\,b^2\,{\ln \left (f\right )}^2}-\frac {15\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,x\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}}\right )}{16\,b^3\,{\ln \left (f\right )}^3\,\sqrt {b\,\ln \left (f\right )}} \]

[In]

int(f^(a + b*x^2)*x^6,x)

[Out]

(15*f^a*f^(b*x^2)*x)/(8*b^3*log(f)^3) + (f^a*f^(b*x^2)*x^5)/(2*b*log(f)) - (5*f^a*f^(b*x^2)*x^3)/(4*b^2*log(f)
^2) - (15*f^a*pi^(1/2)*erfi((b*x*log(f))/(b*log(f))^(1/2)))/(16*b^3*log(f)^3*(b*log(f))^(1/2))