\(\int f^{a+b x^3} x^{14} \, dx\) [97]

Optimal result

Integrand size = 13, antiderivative size = 65 \[ \int f^{a+b x^3} x^{14} \, dx=\frac {f^{a+b x^3} \left (24-24 b x^3 \log (f)+12 b^2 x^6 \log ^2(f)-4 b^3 x^9 \log ^3(f)+b^4 x^{12} \log ^4(f)\right )}{3 b^5 \log ^5(f)} \]

[Out]

1/3*f^(b*x^3+a)*(24-24*b*x^3*ln(f)+12*b^2*x^6*ln(f)^2-4*b^3*x^9*ln(f)^3+b^4*x^12*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} \[ \int f^{a+b x^3} x^{14} \, dx=\frac {f^{a+b x^3} \left (b^4 x^{12} \log ^4(f)-4 b^3 x^9 \log ^3(f)+12 b^2 x^6 \log ^2(f)-24 b x^3 \log (f)+24\right )}{3 b^5 \log ^5(f)} \]

[In]

Int[f^(a + b*x^3)*x^14,x]

[Out]

(f^(a + b*x^3)*(24 - 24*b*x^3*Log[f] + 12*b^2*x^6*Log[f]^2 - 4*b^3*x^9*Log[f]^3 + b^4*x^12*Log[f]^4))/(3*b^5*L
og[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^3} \left (24-24 b x^3 \log (f)+12 b^2 x^6 \log ^2(f)-4 b^3 x^9 \log ^3(f)+b^4 x^{12} \log ^4(f)\right )}{3 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 = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.37 \[ \int f^{a+b x^3} x^{14} \, dx=\frac {f^a \Gamma \left (5,-b x^3 \log (f)\right )}{3 b^5 \log ^5(f)} \]

[In]

Integrate[f^(a + b*x^3)*x^14,x]

[Out]

(f^a*Gamma[5, -(b*x^3*Log[f])])/(3*b^5*Log[f]^5)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98

method	result	size
gosper	\(\frac {f^{b \,x^{3}+a} \left (24-24 b \,x^{3} \ln \left (f \right )+12 b^{2} x^{6} \ln \left (f \right )^{2}-4 b^{3} x^{9} \ln \left (f \right )^{3}+b^{4} x^{12} \ln \left (f \right )^{4}\right )}{3 b^{5} \ln \left (f \right )^{5}}\)	\(64\)
risch	\(\frac {f^{b \,x^{3}+a} \left (24-24 b \,x^{3} \ln \left (f \right )+12 b^{2} x^{6} \ln \left (f \right )^{2}-4 b^{3} x^{9} \ln \left (f \right )^{3}+b^{4} x^{12} \ln \left (f \right )^{4}\right )}{3 b^{5} \ln \left (f \right )^{5}}\)	\(64\)
meijerg	\(-\frac {f^{a} \left (24-\frac {\left (5 b^{4} x^{12} \ln \left (f \right )^{4}-20 b^{3} x^{9} \ln \left (f \right )^{3}+60 b^{2} x^{6} \ln \left (f \right )^{2}-120 b \,x^{3} \ln \left (f \right )+120\right ) {\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{5}\right )}{3 b^{5} \ln \left (f \right )^{5}}\)	\(71\)
parallelrisch	\(\frac {f^{b \,x^{3}+a} x^{12} b^{4} \ln \left (f \right )^{4}-4 f^{b \,x^{3}+a} x^{9} b^{3} \ln \left (f \right )^{3}+12 f^{b \,x^{3}+a} x^{6} b^{2} \ln \left (f \right )^{2}-24 f^{b \,x^{3}+a} x^{3} b \ln \left (f \right )+24 f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{5} b^{5}}\)	\(101\)
norman	\(\frac {8 \,{\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{b^{5} \ln \left (f \right )^{5}}+\frac {x^{12} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 b \ln \left (f \right )}-\frac {8 x^{3} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{4} b^{4}}+\frac {4 x^{6} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{3} b^{3}}-\frac {4 x^{9} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 \ln \left (f \right )^{2} b^{2}}\)	\(114\)

[In]

int(f^(b*x^3+a)*x^14,x,method=_RETURNVERBOSE)

[Out]

1/3*f^(b*x^3+a)*(24-24*b*x^3*ln(f)+12*b^2*x^6*ln(f)^2-4*b^3*x^9*ln(f)^3+b^4*x^12*ln(f)^4)/b^5/ln(f)^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^3} x^{14} \, dx=\frac {{\left (b^{4} x^{12} \log \left (f\right )^{4} - 4 \, b^{3} x^{9} \log \left (f\right )^{3} + 12 \, b^{2} x^{6} \log \left (f\right )^{2} - 24 \, b x^{3} \log \left (f\right ) + 24\right )} f^{b x^{3} + a}}{3 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(b*x^3+a)*x^14,x, algorithm="fricas")

[Out]

1/3*(b^4*x^12*log(f)^4 - 4*b^3*x^9*log(f)^3 + 12*b^2*x^6*log(f)^2 - 24*b*x^3*log(f) + 24)*f^(b*x^3 + a)/(b^5*l
og(f)^5)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int f^{a+b x^3} x^{14} \, dx=\begin {cases} \frac {f^{a + b x^{3}} \left (b^{4} x^{12} \log {\left (f \right )}^{4} - 4 b^{3} x^{9} \log {\left (f \right )}^{3} + 12 b^{2} x^{6} \log {\left (f \right )}^{2} - 24 b x^{3} \log {\left (f \right )} + 24\right )}{3 b^{5} \log {\left (f \right )}^{5}} & \text {for}\: b^{5} \log {\left (f \right )}^{5} \neq 0 \\\frac {x^{15}}{15} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(b*x**3+a)*x**14,x)

[Out]

Piecewise((f**(a + b*x**3)*(b**4*x**12*log(f)**4 - 4*b**3*x**9*log(f)**3 + 12*b**2*x**6*log(f)**2 - 24*b*x**3*
log(f) + 24)/(3*b**5*log(f)**5), Ne(b**5*log(f)**5, 0)), (x**15/15, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int f^{a+b x^3} x^{14} \, dx=\frac {{\left (b^{4} f^{a} x^{12} \log \left (f\right )^{4} - 4 \, b^{3} f^{a} x^{9} \log \left (f\right )^{3} + 12 \, b^{2} f^{a} x^{6} \log \left (f\right )^{2} - 24 \, b f^{a} x^{3} \log \left (f\right ) + 24 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(b*x^3+a)*x^14,x, algorithm="maxima")

[Out]

1/3*(b^4*f^a*x^12*log(f)^4 - 4*b^3*f^a*x^9*log(f)^3 + 12*b^2*f^a*x^6*log(f)^2 - 24*b*f^a*x^3*log(f) + 24*f^a)*
f^(b*x^3)/(b^5*log(f)^5)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.62 \[ \int f^{a+b x^3} x^{14} \, dx=\frac {b^{4} f^{b x^{3}} f^{a} x^{12} \log \left (f\right )^{4} - 4 \, b^{3} f^{b x^{3}} f^{a} x^{9} \log \left (f\right )^{3} + 12 \, b^{2} f^{b x^{3}} f^{a} x^{6} \log \left (f\right )^{2} - 24 \, b f^{b x^{3}} f^{a} x^{3} \log \left (f\right ) + 24 \, f^{b x^{3}} f^{a}}{3 \, b^{5} \log \left (f\right )^{5}} \]

[In]

integrate(f^(b*x^3+a)*x^14,x, algorithm="giac")

[Out]

1/3*(b^4*f^(b*x^3)*f^a*x^12*log(f)^4 - 4*b^3*f^(b*x^3)*f^a*x^9*log(f)^3 + 12*b^2*f^(b*x^3)*f^a*x^6*log(f)^2 -
24*b*f^(b*x^3)*f^a*x^3*log(f) + 24*f^(b*x^3)*f^a)/(b^5*log(f)^5)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^3} x^{14} \, dx=\frac {f^{b\,x^3+a}\,\left (\frac {b^4\,x^{12}\,{\ln \left (f\right )}^4}{3}-\frac {4\,b^3\,x^9\,{\ln \left (f\right )}^3}{3}+4\,b^2\,x^6\,{\ln \left (f\right )}^2-8\,b\,x^3\,\ln \left (f\right )+8\right )}{b^5\,{\ln \left (f\right )}^5} \]

[In]

int(f^(a + b*x^3)*x^14,x)

[Out]

(f^(a + b*x^3)*(4*b^2*x^6*log(f)^2 - (4*b^3*x^9*log(f)^3)/3 + (b^4*x^12*log(f)^4)/3 - 8*b*x^3*log(f) + 8))/(b^
5*log(f)^5)