\(\int f^{a+b x^3} x^{11} \, dx\) [98]

Optimal result

Integrand size = 13, antiderivative size = 84 \[ \int f^{a+b x^3} x^{11} \, dx=-\frac {2 f^{a+b x^3}}{b^4 \log ^4(f)}+\frac {2 f^{a+b x^3} x^3}{b^3 \log ^3(f)}-\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)} \]

[Out]

-2*f^(b*x^3+a)/b^4/ln(f)^4+2*f^(b*x^3+a)*x^3/b^3/ln(f)^3-f^(b*x^3+a)*x^6/b^2/ln(f)^2+1/3*f^(b*x^3+a)*x^9/b/ln(
f)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 84, 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, 2240} \[ \int f^{a+b x^3} x^{11} \, dx=-\frac {2 f^{a+b x^3}}{b^4 \log ^4(f)}+\frac {2 x^3 f^{a+b x^3}}{b^3 \log ^3(f)}-\frac {x^6 f^{a+b x^3}}{b^2 \log ^2(f)}+\frac {x^9 f^{a+b x^3}}{3 b \log (f)} \]

[In]

Int[f^(a + b*x^3)*x^11,x]

[Out]

(-2*f^(a + b*x^3))/(b^4*Log[f]^4) + (2*f^(a + b*x^3)*x^3)/(b^3*Log[f]^3) - (f^(a + b*x^3)*x^6)/(b^2*Log[f]^2)
+ (f^(a + b*x^3)*x^9)/(3*b*Log[f])

Rule 2240

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

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^3} x^9}{3 b \log (f)}-\frac {3 \int f^{a+b x^3} x^8 \, dx}{b \log (f)} \\ & = -\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)}+\frac {6 \int f^{a+b x^3} x^5 \, dx}{b^2 \log ^2(f)} \\ & = \frac {2 f^{a+b x^3} x^3}{b^3 \log ^3(f)}-\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)}-\frac {6 \int f^{a+b x^3} x^2 \, dx}{b^3 \log ^3(f)} \\ & = -\frac {2 f^{a+b x^3}}{b^4 \log ^4(f)}+\frac {2 f^{a+b x^3} x^3}{b^3 \log ^3(f)}-\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.63 \[ \int f^{a+b x^3} x^{11} \, dx=\frac {f^{a+b x^3} \left (-6+6 b x^3 \log (f)-3 b^2 x^6 \log ^2(f)+b^3 x^9 \log ^3(f)\right )}{3 b^4 \log ^4(f)} \]

[In]

Integrate[f^(a + b*x^3)*x^11,x]

[Out]

(f^(a + b*x^3)*(-6 + 6*b*x^3*Log[f] - 3*b^2*x^6*Log[f]^2 + b^3*x^9*Log[f]^3))/(3*b^4*Log[f]^4)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.62

method	result	size
gosper	\(\frac {\left (b^{3} x^{9} \ln \left (f \right )^{3}-3 b^{2} x^{6} \ln \left (f \right )^{2}+6 b \,x^{3} \ln \left (f \right )-6\right ) f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{4} b^{4}}\)	\(52\)
risch	\(\frac {\left (b^{3} x^{9} \ln \left (f \right )^{3}-3 b^{2} x^{6} \ln \left (f \right )^{2}+6 b \,x^{3} \ln \left (f \right )-6\right ) f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{4} b^{4}}\)	\(52\)
meijerg	\(\frac {f^{a} \left (6-\frac {\left (-4 b^{3} x^{9} \ln \left (f \right )^{3}+12 b^{2} x^{6} \ln \left (f \right )^{2}-24 b \,x^{3} \ln \left (f \right )+24\right ) {\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{4}\right )}{3 b^{4} \ln \left (f \right )^{4}}\)	\(59\)
parallelrisch	\(\frac {f^{b \,x^{3}+a} x^{9} b^{3} \ln \left (f \right )^{3}-3 f^{b \,x^{3}+a} x^{6} b^{2} \ln \left (f \right )^{2}+6 f^{b \,x^{3}+a} x^{3} b \ln \left (f \right )-6 f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{4} b^{4}}\)	\(80\)
norman	\(-\frac {2 \,{\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{4} b^{4}}+\frac {x^{9} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 b \ln \left (f \right )}+\frac {2 x^{3} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{3} b^{3}}-\frac {x^{6} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{2} b^{2}}\)	\(91\)

[In]

int(f^(b*x^3+a)*x^11,x,method=_RETURNVERBOSE)

[Out]

1/3*(b^3*x^9*ln(f)^3-3*b^2*x^6*ln(f)^2+6*b*x^3*ln(f)-6)*f^(b*x^3+a)/ln(f)^4/b^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int f^{a+b x^3} x^{11} \, dx=\frac {{\left (b^{3} x^{9} \log \left (f\right )^{3} - 3 \, b^{2} x^{6} \log \left (f\right )^{2} + 6 \, b x^{3} \log \left (f\right ) - 6\right )} f^{b x^{3} + a}}{3 \, b^{4} \log \left (f\right )^{4}} \]

[In]

integrate(f^(b*x^3+a)*x^11,x, algorithm="fricas")

[Out]

1/3*(b^3*x^9*log(f)^3 - 3*b^2*x^6*log(f)^2 + 6*b*x^3*log(f) - 6)*f^(b*x^3 + a)/(b^4*log(f)^4)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int f^{a+b x^3} x^{11} \, dx=\begin {cases} \frac {f^{a + b x^{3}} \left (b^{3} x^{9} \log {\left (f \right )}^{3} - 3 b^{2} x^{6} \log {\left (f \right )}^{2} + 6 b x^{3} \log {\left (f \right )} - 6\right )}{3 b^{4} \log {\left (f \right )}^{4}} & \text {for}\: b^{4} \log {\left (f \right )}^{4} \neq 0 \\\frac {x^{12}}{12} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(b*x**3+a)*x**11,x)

[Out]

Piecewise((f**(a + b*x**3)*(b**3*x**9*log(f)**3 - 3*b**2*x**6*log(f)**2 + 6*b*x**3*log(f) - 6)/(3*b**4*log(f)*
*4), Ne(b**4*log(f)**4, 0)), (x**12/12, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int f^{a+b x^3} x^{11} \, dx=\frac {{\left (b^{3} f^{a} x^{9} \log \left (f\right )^{3} - 3 \, b^{2} f^{a} x^{6} \log \left (f\right )^{2} + 6 \, b f^{a} x^{3} \log \left (f\right ) - 6 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{4} \log \left (f\right )^{4}} \]

[In]

integrate(f^(b*x^3+a)*x^11,x, algorithm="maxima")

[Out]

1/3*(b^3*f^a*x^9*log(f)^3 - 3*b^2*f^a*x^6*log(f)^2 + 6*b*f^a*x^3*log(f) - 6*f^a)*f^(b*x^3)/(b^4*log(f)^4)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int f^{a+b x^3} x^{11} \, dx=\frac {b^{3} f^{b x^{3}} f^{a} x^{9} \log \left (f\right )^{3} - 3 \, b^{2} f^{b x^{3}} f^{a} x^{6} \log \left (f\right )^{2} + 6 \, b f^{b x^{3}} f^{a} x^{3} \log \left (f\right ) - 6 \, f^{b x^{3}} f^{a}}{3 \, b^{4} \log \left (f\right )^{4}} \]

[In]

integrate(f^(b*x^3+a)*x^11,x, algorithm="giac")

[Out]

1/3*(b^3*f^(b*x^3)*f^a*x^9*log(f)^3 - 3*b^2*f^(b*x^3)*f^a*x^6*log(f)^2 + 6*b*f^(b*x^3)*f^a*x^3*log(f) - 6*f^(b
*x^3)*f^a)/(b^4*log(f)^4)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int f^{a+b x^3} x^{11} \, dx=-\frac {f^{b\,x^3+a}\,\left (-\frac {b^3\,x^9\,{\ln \left (f\right )}^3}{3}+b^2\,x^6\,{\ln \left (f\right )}^2-2\,b\,x^3\,\ln \left (f\right )+2\right )}{b^4\,{\ln \left (f\right )}^4} \]

[In]

int(f^(a + b*x^3)*x^11,x)

[Out]

-(f^(a + b*x^3)*(b^2*x^6*log(f)^2 - (b^3*x^9*log(f)^3)/3 - 2*b*x^3*log(f) + 2))/(b^4*log(f)^4)