\(\int f^{a+b x^3} x^{17} \, dx\) [96]

Optimal result

Integrand size = 13, antiderivative size = 78 \[ \int f^{a+b x^3} x^{17} \, dx=-\frac {f^{a+b x^3} \left (120-120 b x^3 \log (f)+60 b^2 x^6 \log ^2(f)-20 b^3 x^9 \log ^3(f)+5 b^4 x^{12} \log ^4(f)-b^5 x^{15} \log ^5(f)\right )}{3 b^6 \log ^6(f)} \]

[Out]

-1/3*f^(b*x^3+a)*(120-120*b*x^3*ln(f)+60*b^2*x^6*ln(f)^2-20*b^3*x^9*ln(f)^3+5*b^4*x^12*ln(f)^4-b^5*x^15*ln(f)^
5)/b^6/ln(f)^6

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 78, 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} \[ \int f^{a+b x^3} x^{17} \, dx=-\frac {f^{a+b x^3} \left (-b^5 x^{15} \log ^5(f)+5 b^4 x^{12} \log ^4(f)-20 b^3 x^9 \log ^3(f)+60 b^2 x^6 \log ^2(f)-120 b x^3 \log (f)+120\right )}{3 b^6 \log ^6(f)} \]

[In]

Int[f^(a + b*x^3)*x^17,x]

[Out]

-1/3*(f^(a + b*x^3)*(120 - 120*b*x^3*Log[f] + 60*b^2*x^6*Log[f]^2 - 20*b^3*x^9*Log[f]^3 + 5*b^4*x^12*Log[f]^4
- b^5*x^15*Log[f]^5))/(b^6*Log[f]^6)

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 (120-120 b x^3 \log (f)+60 b^2 x^6 \log ^2(f)-20 b^3 x^9 \log ^3(f)+5 b^4 x^{12} \log ^4(f)-b^5 x^{15} \log ^5(f)\right )}{3 b^6 \log ^6(f)} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.31 \[ \int f^{a+b x^3} x^{17} \, dx=-\frac {f^a \Gamma \left (6,-b x^3 \log (f)\right )}{3 b^6 \log ^6(f)} \]

[In]

Integrate[f^(a + b*x^3)*x^17,x]

[Out]

-1/3*(f^a*Gamma[6, -(b*x^3*Log[f])])/(b^6*Log[f]^6)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97

method	result	size
gosper	\(\frac {\left (b^{5} x^{15} \ln \left (f \right )^{5}-5 b^{4} x^{12} \ln \left (f \right )^{4}+20 b^{3} x^{9} \ln \left (f \right )^{3}-60 b^{2} x^{6} \ln \left (f \right )^{2}+120 b \,x^{3} \ln \left (f \right )-120\right ) f^{b \,x^{3}+a}}{3 b^{6} \ln \left (f \right )^{6}}\)	\(76\)
risch	\(\frac {\left (b^{5} x^{15} \ln \left (f \right )^{5}-5 b^{4} x^{12} \ln \left (f \right )^{4}+20 b^{3} x^{9} \ln \left (f \right )^{3}-60 b^{2} x^{6} \ln \left (f \right )^{2}+120 b \,x^{3} \ln \left (f \right )-120\right ) f^{b \,x^{3}+a}}{3 b^{6} \ln \left (f \right )^{6}}\)	\(76\)
meijerg	\(\frac {f^{a} \left (120-\frac {\left (-6 b^{5} x^{15} \ln \left (f \right )^{5}+30 b^{4} x^{12} \ln \left (f \right )^{4}-120 b^{3} x^{9} \ln \left (f \right )^{3}+360 b^{2} x^{6} \ln \left (f \right )^{2}-720 b \,x^{3} \ln \left (f \right )+720\right ) {\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{6}\right )}{3 b^{6} \ln \left (f \right )^{6}}\)	\(83\)
parallelrisch	\(\frac {f^{b \,x^{3}+a} x^{15} b^{5} \ln \left (f \right )^{5}-5 f^{b \,x^{3}+a} x^{12} b^{4} \ln \left (f \right )^{4}+20 f^{b \,x^{3}+a} x^{9} b^{3} \ln \left (f \right )^{3}-60 f^{b \,x^{3}+a} x^{6} b^{2} \ln \left (f \right )^{2}+120 f^{b \,x^{3}+a} x^{3} b \ln \left (f \right )-120 f^{b \,x^{3}+a}}{3 b^{6} \ln \left (f \right )^{6}}\)	\(122\)
norman	\(-\frac {40 \,{\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{b^{6} \ln \left (f \right )^{6}}+\frac {40 x^{3} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{b^{5} \ln \left (f \right )^{5}}+\frac {x^{15} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 b \ln \left (f \right )}-\frac {20 x^{6} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{4} b^{4}}+\frac {20 x^{9} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 \ln \left (f \right )^{3} b^{3}}-\frac {5 x^{12} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 \ln \left (f \right )^{2} b^{2}}\)	\(137\)

[In]

int(f^(b*x^3+a)*x^17,x,method=_RETURNVERBOSE)

[Out]

1/3*(b^5*x^15*ln(f)^5-5*b^4*x^12*ln(f)^4+20*b^3*x^9*ln(f)^3-60*b^2*x^6*ln(f)^2+120*b*x^3*ln(f)-120)*f^(b*x^3+a
)/b^6/ln(f)^6

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int f^{a+b x^3} x^{17} \, dx=\frac {{\left (b^{5} x^{15} \log \left (f\right )^{5} - 5 \, b^{4} x^{12} \log \left (f\right )^{4} + 20 \, b^{3} x^{9} \log \left (f\right )^{3} - 60 \, b^{2} x^{6} \log \left (f\right )^{2} + 120 \, b x^{3} \log \left (f\right ) - 120\right )} f^{b x^{3} + a}}{3 \, b^{6} \log \left (f\right )^{6}} \]

[In]

integrate(f^(b*x^3+a)*x^17,x, algorithm="fricas")

[Out]

1/3*(b^5*x^15*log(f)^5 - 5*b^4*x^12*log(f)^4 + 20*b^3*x^9*log(f)^3 - 60*b^2*x^6*log(f)^2 + 120*b*x^3*log(f) -
120)*f^(b*x^3 + a)/(b^6*log(f)^6)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int f^{a+b x^3} x^{17} \, dx=\begin {cases} \frac {f^{a + b x^{3}} \left (b^{5} x^{15} \log {\left (f \right )}^{5} - 5 b^{4} x^{12} \log {\left (f \right )}^{4} + 20 b^{3} x^{9} \log {\left (f \right )}^{3} - 60 b^{2} x^{6} \log {\left (f \right )}^{2} + 120 b x^{3} \log {\left (f \right )} - 120\right )}{3 b^{6} \log {\left (f \right )}^{6}} & \text {for}\: b^{6} \log {\left (f \right )}^{6} \neq 0 \\\frac {x^{18}}{18} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(b*x**3+a)*x**17,x)

[Out]

Piecewise((f**(a + b*x**3)*(b**5*x**15*log(f)**5 - 5*b**4*x**12*log(f)**4 + 20*b**3*x**9*log(f)**3 - 60*b**2*x
**6*log(f)**2 + 120*b*x**3*log(f) - 120)/(3*b**6*log(f)**6), Ne(b**6*log(f)**6, 0)), (x**18/18, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.18 \[ \int f^{a+b x^3} x^{17} \, dx=\frac {{\left (b^{5} f^{a} x^{15} \log \left (f\right )^{5} - 5 \, b^{4} f^{a} x^{12} \log \left (f\right )^{4} + 20 \, b^{3} f^{a} x^{9} \log \left (f\right )^{3} - 60 \, b^{2} f^{a} x^{6} \log \left (f\right )^{2} + 120 \, b f^{a} x^{3} \log \left (f\right ) - 120 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{6} \log \left (f\right )^{6}} \]

[In]

integrate(f^(b*x^3+a)*x^17,x, algorithm="maxima")

[Out]

1/3*(b^5*f^a*x^15*log(f)^5 - 5*b^4*f^a*x^12*log(f)^4 + 20*b^3*f^a*x^9*log(f)^3 - 60*b^2*f^a*x^6*log(f)^2 + 120
*b*f^a*x^3*log(f) - 120*f^a)*f^(b*x^3)/(b^6*log(f)^6)

Giac [F(-2)]

Exception generated. \[ \int f^{a+b x^3} x^{17} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(f^(b*x^3+a)*x^17,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Polynomial exponent overflow. Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^3} x^{17} \, dx=-\frac {f^{b\,x^3+a}\,\left (-\frac {b^5\,x^{15}\,{\ln \left (f\right )}^5}{3}+\frac {5\,b^4\,x^{12}\,{\ln \left (f\right )}^4}{3}-\frac {20\,b^3\,x^9\,{\ln \left (f\right )}^3}{3}+20\,b^2\,x^6\,{\ln \left (f\right )}^2-40\,b\,x^3\,\ln \left (f\right )+40\right )}{b^6\,{\ln \left (f\right )}^6} \]

[In]

int(f^(a + b*x^3)*x^17,x)

[Out]

-(f^(a + b*x^3)*(20*b^2*x^6*log(f)^2 - (20*b^3*x^9*log(f)^3)/3 + (5*b^4*x^12*log(f)^4)/3 - (b^5*x^15*log(f)^5)
/3 - 40*b*x^3*log(f) + 40))/(b^6*log(f)^6)