3.70 \(\int f^{a+b x^2} x^{11} \, dx\)

Optimal. Leaf size=78 \[ -\frac {f^{a+b x^2} \left (-b^5 x^{10} \log ^5(f)+5 b^4 x^8 \log ^4(f)-20 b^3 x^6 \log ^3(f)+60 b^2 x^4 \log ^2(f)-120 b x^2 \log (f)+120\right )}{2 b^6 \log ^6(f)} \]

[Out]

-1/2*f^(b*x^2+a)*(120-120*b*x^2*ln(f)+60*b^2*x^4*ln(f)^2-20*b^3*x^6*ln(f)^3+5*b^4*x^8*ln(f)^4-b^5*x^10*ln(f)^5
)/b^6/ln(f)^6

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Rubi [C]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 0.31, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ -\frac {f^a \text {Gamma}\left (6,-b x^2 \log (f)\right )}{2 b^6 \log ^6(f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)*x^11,x]

[Out]

-(f^a*Gamma[6, -(b*x^2*Log[f])])/(2*b^6*Log[f]^6)

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \int f^{a+b x^2} x^{11} \, dx &=-\frac {f^a \Gamma \left (6,-b x^2 \log (f)\right )}{2 b^6 \log ^6(f)}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 24, normalized size = 0.31 \[ -\frac {f^a \Gamma \left (6,-b x^2 \log (f)\right )}{2 b^6 \log ^6(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)*x^11,x]

[Out]

-1/2*(f^a*Gamma[6, -(b*x^2*Log[f])])/(b^6*Log[f]^6)

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fricas [A]  time = 0.42, size = 75, normalized size = 0.96 \[ \frac {{\left (b^{5} x^{10} \log \relax (f)^{5} - 5 \, b^{4} x^{8} \log \relax (f)^{4} + 20 \, b^{3} x^{6} \log \relax (f)^{3} - 60 \, b^{2} x^{4} \log \relax (f)^{2} + 120 \, b x^{2} \log \relax (f) - 120\right )} f^{b x^{2} + a}}{2 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^11,x, algorithm="fricas")

[Out]

1/2*(b^5*x^10*log(f)^5 - 5*b^4*x^8*log(f)^4 + 20*b^3*x^6*log(f)^3 - 60*b^2*x^4*log(f)^2 + 120*b*x^2*log(f) - 1
20)*f^(b*x^2 + a)/(b^6*log(f)^6)

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giac [A]  time = 0.32, size = 79, normalized size = 1.01 \[ \frac {{\left (b^{5} x^{10} \log \relax (f)^{5} - 5 \, b^{4} x^{8} \log \relax (f)^{4} + 20 \, b^{3} x^{6} \log \relax (f)^{3} - 60 \, b^{2} x^{4} \log \relax (f)^{2} + 120 \, b x^{2} \log \relax (f) - 120\right )} e^{\left (b x^{2} \log \relax (f) + a \log \relax (f)\right )}}{2 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^11,x, algorithm="giac")

[Out]

1/2*(b^5*x^10*log(f)^5 - 5*b^4*x^8*log(f)^4 + 20*b^3*x^6*log(f)^3 - 60*b^2*x^4*log(f)^2 + 120*b*x^2*log(f) - 1
20)*e^(b*x^2*log(f) + a*log(f))/(b^6*log(f)^6)

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maple [A]  time = 0.01, size = 76, normalized size = 0.97 \[ \frac {\left (b^{5} x^{10} \ln \relax (f )^{5}-5 b^{4} x^{8} \ln \relax (f )^{4}+20 b^{3} x^{6} \ln \relax (f )^{3}-60 b^{2} x^{4} \ln \relax (f )^{2}+120 b \,x^{2} \ln \relax (f )-120\right ) f^{b \,x^{2}+a}}{2 b^{6} \ln \relax (f )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)*x^11,x)

[Out]

1/2*(b^5*x^10*ln(f)^5-5*b^4*x^8*ln(f)^4+20*b^3*x^6*ln(f)^3-60*b^2*x^4*ln(f)^2+120*b*x^2*ln(f)-120)*f^(b*x^2+a)
/ln(f)^6/b^6

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maxima [A]  time = 0.45, size = 92, normalized size = 1.18 \[ \frac {{\left (b^{5} f^{a} x^{10} \log \relax (f)^{5} - 5 \, b^{4} f^{a} x^{8} \log \relax (f)^{4} + 20 \, b^{3} f^{a} x^{6} \log \relax (f)^{3} - 60 \, b^{2} f^{a} x^{4} \log \relax (f)^{2} + 120 \, b f^{a} x^{2} \log \relax (f) - 120 \, f^{a}\right )} f^{b x^{2}}}{2 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^11,x, algorithm="maxima")

[Out]

1/2*(b^5*f^a*x^10*log(f)^5 - 5*b^4*f^a*x^8*log(f)^4 + 20*b^3*f^a*x^6*log(f)^3 - 60*b^2*f^a*x^4*log(f)^2 + 120*
b*f^a*x^2*log(f) - 120*f^a)*f^(b*x^2)/(b^6*log(f)^6)

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mupad [B]  time = 3.56, size = 76, normalized size = 0.97 \[ -\frac {f^{b\,x^2+a}\,\left (-\frac {b^5\,x^{10}\,{\ln \relax (f)}^5}{2}+\frac {5\,b^4\,x^8\,{\ln \relax (f)}^4}{2}-10\,b^3\,x^6\,{\ln \relax (f)}^3+30\,b^2\,x^4\,{\ln \relax (f)}^2-60\,b\,x^2\,\ln \relax (f)+60\right )}{b^6\,{\ln \relax (f)}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)*x^11,x)

[Out]

-(f^(a + b*x^2)*(30*b^2*x^4*log(f)^2 - 10*b^3*x^6*log(f)^3 + (5*b^4*x^8*log(f)^4)/2 - (b^5*x^10*log(f)^5)/2 -
60*b*x^2*log(f) + 60))/(b^6*log(f)^6)

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sympy [A]  time = 0.16, size = 95, normalized size = 1.22 \[ \begin {cases} \frac {f^{a + b x^{2}} \left (b^{5} x^{10} \log {\relax (f )}^{5} - 5 b^{4} x^{8} \log {\relax (f )}^{4} + 20 b^{3} x^{6} \log {\relax (f )}^{3} - 60 b^{2} x^{4} \log {\relax (f )}^{2} + 120 b x^{2} \log {\relax (f )} - 120\right )}{2 b^{6} \log {\relax (f )}^{6}} & \text {for}\: 2 b^{6} \log {\relax (f )}^{6} \neq 0 \\\frac {x^{12}}{12} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)*x**11,x)

[Out]

Piecewise((f**(a + b*x**2)*(b**5*x**10*log(f)**5 - 5*b**4*x**8*log(f)**4 + 20*b**3*x**6*log(f)**3 - 60*b**2*x*
*4*log(f)**2 + 120*b*x**2*log(f) - 120)/(2*b**6*log(f)**6), Ne(2*b**6*log(f)**6, 0)), (x**12/12, True))

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