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

Optimal. Leaf size=65 \[ \frac {f^{a+b x^3} \left (b^4 x^{12} \log ^4(f)-4 b^3 x^9 \log ^3(f)+12 b^2 x^6 \log ^2(f)-24 b x^3 \log (f)+24\right )}{3 b^5 \log ^5(f)} \]

[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

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Rubi [C]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 0.37, 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} \[ \frac {f^a \text {Gamma}\left (5,-b x^3 \log (f)\right )}{3 b^5 \log ^5(f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)*x^14,x]

[Out]

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

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^{14} \, dx &=\frac {f^a \Gamma \left (5,-b x^3 \log (f)\right )}{3 b^5 \log ^5(f)}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 24, normalized size = 0.37 \[ \frac {f^a \Gamma \left (5,-b x^3 \log (f)\right )}{3 b^5 \log ^5(f)} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.42, size = 63, normalized size = 0.97 \[ \frac {{\left (b^{4} x^{12} \log \relax (f)^{4} - 4 \, b^{3} x^{9} \log \relax (f)^{3} + 12 \, b^{2} x^{6} \log \relax (f)^{2} - 24 \, b x^{3} \log \relax (f) + 24\right )} f^{b x^{3} + a}}{3 \, b^{5} \log \relax (f)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 105, normalized size = 1.62 \[ \frac {b^{4} f^{b x^{3}} f^{a} x^{12} \log \relax (f)^{4} - 4 \, b^{3} f^{b x^{3}} f^{a} x^{9} \log \relax (f)^{3} + 12 \, b^{2} f^{b x^{3}} f^{a} x^{6} \log \relax (f)^{2} - 24 \, b f^{b x^{3}} f^{a} x^{3} \log \relax (f) + 24 \, f^{b x^{3}} f^{a}}{3 \, b^{5} \log \relax (f)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 64, normalized size = 0.98 \[ \frac {\left (b^{4} x^{12} \ln \relax (f )^{4}-4 b^{3} x^{9} \ln \relax (f )^{3}+12 b^{2} x^{6} \ln \relax (f )^{2}-24 b \,x^{3} \ln \relax (f )+24\right ) f^{b \,x^{3}+a}}{3 b^{5} \ln \relax (f )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [A]  time = 0.93, size = 77, normalized size = 1.18 \[ \frac {{\left (b^{4} f^{a} x^{12} \log \relax (f)^{4} - 4 \, b^{3} f^{a} x^{9} \log \relax (f)^{3} + 12 \, b^{2} f^{a} x^{6} \log \relax (f)^{2} - 24 \, b f^{a} x^{3} \log \relax (f) + 24 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{5} \log \relax (f)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.54, size = 63, normalized size = 0.97 \[ \frac {f^{b\,x^3+a}\,\left (\frac {b^4\,x^{12}\,{\ln \relax (f)}^4}{3}-\frac {4\,b^3\,x^9\,{\ln \relax (f)}^3}{3}+4\,b^2\,x^6\,{\ln \relax (f)}^2-8\,b\,x^3\,\ln \relax (f)+8\right )}{b^5\,{\ln \relax (f)}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.16, size = 82, normalized size = 1.26 \[ \begin {cases} \frac {f^{a + b x^{3}} \left (b^{4} x^{12} \log {\relax (f )}^{4} - 4 b^{3} x^{9} \log {\relax (f )}^{3} + 12 b^{2} x^{6} \log {\relax (f )}^{2} - 24 b x^{3} \log {\relax (f )} + 24\right )}{3 b^{5} \log {\relax (f )}^{5}} & \text {for}\: 3 b^{5} \log {\relax (f )}^{5} \neq 0 \\\frac {x^{15}}{15} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(3*b**5*log(f)**5, 0)), (x**15/15, True))

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