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

Optimal. Leaf size=78 \[ -\frac {f^{a+b x^3} \left (-b^5 x^{15} \log ^5(f)+5 b^4 x^{12} \log ^4(f)-20 b^3 x^9 \log ^3(f)+60 b^2 x^6 \log ^2(f)-120 b x^3 \log (f)+120\right )}{3 b^6 \log ^6(f)} \]

[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

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Rubi [C]  time = 0.02, 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} \[ -\frac {f^a \text {Gamma}\left (6,-b x^3 \log (f)\right )}{3 b^6 \log ^6(f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)*x^17,x]

[Out]

-(f^a*Gamma[6, -(b*x^3*Log[f])])/(3*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^3} x^{17} \, dx &=-\frac {f^a \Gamma \left (6,-b x^3 \log (f)\right )}{3 b^6 \log ^6(f)}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 24, normalized size = 0.31 \[ -\frac {f^a \Gamma \left (6,-b x^3 \log (f)\right )}{3 b^6 \log ^6(f)} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.43, size = 75, normalized size = 0.96 \[ \frac {{\left (b^{5} x^{15} \log \relax (f)^{5} - 5 \, b^{4} x^{12} \log \relax (f)^{4} + 20 \, b^{3} x^{9} \log \relax (f)^{3} - 60 \, b^{2} x^{6} \log \relax (f)^{2} + 120 \, b x^{3} \log \relax (f) - 120\right )} f^{b x^{3} + a}}{3 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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,x):;OUTPUT:Poly
nomial exponent overflow. Error: Bad Argument Value

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maple [A]  time = 0.01, size = 76, normalized size = 0.97 \[ \frac {\left (b^{5} x^{15} \ln \relax (f )^{5}-5 b^{4} x^{12} \ln \relax (f )^{4}+20 b^{3} x^{9} \ln \relax (f )^{3}-60 b^{2} x^{6} \ln \relax (f )^{2}+120 b \,x^{3} \ln \relax (f )-120\right ) f^{b \,x^{3}+a}}{3 b^{6} \ln \relax (f )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [A]  time = 0.93, size = 92, normalized size = 1.18 \[ \frac {{\left (b^{5} f^{a} x^{15} \log \relax (f)^{5} - 5 \, b^{4} f^{a} x^{12} \log \relax (f)^{4} + 20 \, b^{3} f^{a} x^{9} \log \relax (f)^{3} - 60 \, b^{2} f^{a} x^{6} \log \relax (f)^{2} + 120 \, b f^{a} x^{3} \log \relax (f) - 120 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{6} \log \relax (f)^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.66, size = 76, normalized size = 0.97 \[ -\frac {f^{b\,x^3+a}\,\left (-\frac {b^5\,x^{15}\,{\ln \relax (f)}^5}{3}+\frac {5\,b^4\,x^{12}\,{\ln \relax (f)}^4}{3}-\frac {20\,b^3\,x^9\,{\ln \relax (f)}^3}{3}+20\,b^2\,x^6\,{\ln \relax (f)}^2-40\,b\,x^3\,\ln \relax (f)+40\right )}{b^6\,{\ln \relax (f)}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.17, size = 95, normalized size = 1.22 \[ \begin {cases} \frac {f^{a + b x^{3}} \left (b^{5} x^{15} \log {\relax (f )}^{5} - 5 b^{4} x^{12} \log {\relax (f )}^{4} + 20 b^{3} x^{9} \log {\relax (f )}^{3} - 60 b^{2} x^{6} \log {\relax (f )}^{2} + 120 b x^{3} \log {\relax (f )} - 120\right )}{3 b^{6} \log {\relax (f )}^{6}} & \text {for}\: 3 b^{6} \log {\relax (f )}^{6} \neq 0 \\\frac {x^{18}}{18} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(3*b**6*log(f)**6, 0)), (x**18/18, True))

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