\(\int f^{a+b x^2} x^9 \, dx\) [71]

Optimal result

Integrand size = 13, antiderivative size = 65 \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^{a+b x^2} \left (24-24 b x^2 \log (f)+12 b^2 x^4 \log ^2(f)-4 b^3 x^6 \log ^3(f)+b^4 x^8 \log ^4(f)\right )}{2 b^5 \log ^5(f)} \]

[Out]

1/2*f^(b*x^2+a)*(24-24*b*x^2*ln(f)+12*b^2*x^4*ln(f)^2-4*b^3*x^6*ln(f)^3+b^4*x^8*ln(f)^4)/b^5/ln(f)^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2249} \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^{a+b x^2} \left (b^4 x^8 \log ^4(f)-4 b^3 x^6 \log ^3(f)+12 b^2 x^4 \log ^2(f)-24 b x^2 \log (f)+24\right )}{2 b^5 \log ^5(f)} \]

[In]

Int[f^(a + b*x^2)*x^9,x]

[Out]

(f^(a + b*x^2)*(24 - 24*b*x^2*Log[f] + 12*b^2*x^4*Log[f]^2 - 4*b^3*x^6*Log[f]^3 + b^4*x^8*Log[f]^4))/(2*b^5*Lo
g[f]^5)

Rule 2249

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{p = Simplify
[(m + 1)/n]}, Simp[(-F^a)*((f/d)^m/(d*n*((-b)*Log[F])^p))*Simplify[FunctionExpand[Gamma[p, (-b)*(c + d*x)^n*Lo
g[F]]]], x] /; IGtQ[p, 0]] /; FreeQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^2} \left (24-24 b x^2 \log (f)+12 b^2 x^4 \log ^2(f)-4 b^3 x^6 \log ^3(f)+b^4 x^8 \log ^4(f)\right )}{2 b^5 \log ^5(f)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.37 \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^a \Gamma \left (5,-b x^2 \log (f)\right )}{2 b^5 \log ^5(f)} \]

[In]

Integrate[f^(a + b*x^2)*x^9,x]

[Out]

(f^a*Gamma[5, -(b*x^2*Log[f])])/(2*b^5*Log[f]^5)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98

method	result	size
gosper	\(\frac {f^{b \,x^{2}+a} \left (24-24 b \,x^{2} \ln \left (f \right )+12 b^{2} x^{4} \ln \left (f \right )^{2}-4 b^{3} x^{6} \ln \left (f \right )^{3}+b^{4} x^{8} \ln \left (f \right )^{4}\right )}{2 b^{5} \ln \left (f \right )^{5}}\)	\(64\)
risch	\(\frac {f^{b \,x^{2}+a} \left (24-24 b \,x^{2} \ln \left (f \right )+12 b^{2} x^{4} \ln \left (f \right )^{2}-4 b^{3} x^{6} \ln \left (f \right )^{3}+b^{4} x^{8} \ln \left (f \right )^{4}\right )}{2 b^{5} \ln \left (f \right )^{5}}\)	\(64\)
meijerg	\(-\frac {f^{a} \left (24-\frac {\left (5 b^{4} x^{8} \ln \left (f \right )^{4}-20 b^{3} x^{6} \ln \left (f \right )^{3}+60 b^{2} x^{4} \ln \left (f \right )^{2}-120 b \,x^{2} \ln \left (f \right )+120\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{5}\right )}{2 b^{5} \ln \left (f \right )^{5}}\)	\(71\)
parallelrisch	\(\frac {f^{b \,x^{2}+a} x^{8} \ln \left (f \right )^{4} b^{4}-4 f^{b \,x^{2}+a} x^{6} \ln \left (f \right )^{3} b^{3}+12 f^{b \,x^{2}+a} x^{4} \ln \left (f \right )^{2} b^{2}-24 f^{b \,x^{2}+a} x^{2} \ln \left (f \right ) b +24 f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{5} b^{5}}\)	\(101\)
norman	\(\frac {12 \,{\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{b^{5} \ln \left (f \right )^{5}}+\frac {x^{8} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}-\frac {12 x^{2} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{4} b^{4}}+\frac {6 x^{4} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{3} b^{3}}-\frac {2 x^{6} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{2} b^{2}}\)	\(114\)

[In]

int(f^(b*x^2+a)*x^9,x,method=_RETURNVERBOSE)

[Out]

1/2*f^(b*x^2+a)*(24-24*b*x^2*ln(f)+12*b^2*x^4*ln(f)^2-4*b^3*x^6*ln(f)^3+b^4*x^8*ln(f)^4)/b^5/ln(f)^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^2} x^9 \, dx=\frac {{\left (b^{4} x^{8} \log \left (f\right )^{4} - 4 \, b^{3} x^{6} \log \left (f\right )^{3} + 12 \, b^{2} x^{4} \log \left (f\right )^{2} - 24 \, b x^{2} \log \left (f\right ) + 24\right )} f^{b x^{2} + a}}{2 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(b*x^2+a)*x^9,x, algorithm="fricas")

[Out]

1/2*(b^4*x^8*log(f)^4 - 4*b^3*x^6*log(f)^3 + 12*b^2*x^4*log(f)^2 - 24*b*x^2*log(f) + 24)*f^(b*x^2 + a)/(b^5*lo
g(f)^5)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int f^{a+b x^2} x^9 \, dx=\begin {cases} \frac {f^{a + b x^{2}} \left (b^{4} x^{8} \log {\left (f \right )}^{4} - 4 b^{3} x^{6} \log {\left (f \right )}^{3} + 12 b^{2} x^{4} \log {\left (f \right )}^{2} - 24 b x^{2} \log {\left (f \right )} + 24\right )}{2 b^{5} \log {\left (f \right )}^{5}} & \text {for}\: b^{5} \log {\left (f \right )}^{5} \neq 0 \\\frac {x^{10}}{10} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(b*x**2+a)*x**9,x)

[Out]

Piecewise((f**(a + b*x**2)*(b**4*x**8*log(f)**4 - 4*b**3*x**6*log(f)**3 + 12*b**2*x**4*log(f)**2 - 24*b*x**2*l
og(f) + 24)/(2*b**5*log(f)**5), Ne(b**5*log(f)**5, 0)), (x**10/10, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int f^{a+b x^2} x^9 \, dx=\frac {{\left (b^{4} f^{a} x^{8} \log \left (f\right )^{4} - 4 \, b^{3} f^{a} x^{6} \log \left (f\right )^{3} + 12 \, b^{2} f^{a} x^{4} \log \left (f\right )^{2} - 24 \, b f^{a} x^{2} \log \left (f\right ) + 24 \, f^{a}\right )} f^{b x^{2}}}{2 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(b*x^2+a)*x^9,x, algorithm="maxima")

[Out]

1/2*(b^4*f^a*x^8*log(f)^4 - 4*b^3*f^a*x^6*log(f)^3 + 12*b^2*f^a*x^4*log(f)^2 - 24*b*f^a*x^2*log(f) + 24*f^a)*f
^(b*x^2)/(b^5*log(f)^5)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int f^{a+b x^2} x^9 \, dx=\frac {{\left (b^{4} x^{8} \log \left (f\right )^{4} - 4 \, b^{3} x^{6} \log \left (f\right )^{3} + 12 \, b^{2} x^{4} \log \left (f\right )^{2} - 24 \, b x^{2} \log \left (f\right ) + 24\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{2 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(b*x^2+a)*x^9,x, algorithm="giac")

[Out]

1/2*(b^4*x^8*log(f)^4 - 4*b^3*x^6*log(f)^3 + 12*b^2*x^4*log(f)^2 - 24*b*x^2*log(f) + 24)*e^(b*x^2*log(f) + a*l
og(f))/(b^5*log(f)^5)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^{b\,x^2+a}\,\left (\frac {b^4\,x^8\,{\ln \left (f\right )}^4}{2}-2\,b^3\,x^6\,{\ln \left (f\right )}^3+6\,b^2\,x^4\,{\ln \left (f\right )}^2-12\,b\,x^2\,\ln \left (f\right )+12\right )}{b^5\,{\ln \left (f\right )}^5} \]

[In]

int(f^(a + b*x^2)*x^9,x)

[Out]

(f^(a + b*x^2)*(6*b^2*x^4*log(f)^2 - 2*b^3*x^6*log(f)^3 + (b^4*x^8*log(f)^4)/2 - 12*b*x^2*log(f) + 12))/(b^5*l
og(f)^5)