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

Optimal. Leaf size=24 \[ \frac{f^a \text{Gamma}\left (5,-b x^3 \log (f)\right )}{3 b^5 \log ^5(f)} \]

[Out]

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

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Rubi [A]  time = 0.0392418, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077 \[ \frac{f^a \text{Gamma}\left (5,-b x^3 \log (f)\right )}{3 b^5 \log ^5(f)} \]

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)

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Rubi in Sympy [A]  time = 3.78091, size = 24, normalized size = 1. \[ \frac{f^{a} \Gamma{\left (5,- b x^{3} \log{\left (f \right )} \right )}}{3 b^{5} \log{\left (f \right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(f**(b*x**3+a)*x**14,x)

[Out]

f**a*Gamma(5, -b*x**3*log(f))/(3*b**5*log(f)**5)

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Mathematica [B]  time = 0.0183968, size = 65, normalized size = 2.71 \[ \frac{f^{a+b x^3} \left (b^4 x^{12} \log ^4(f)-4 b^3 x^9 \log ^3(f)+12 b^2 x^6 \log ^2(f)-24 b x^3 \log (f)+24\right )}{3 b^5 \log ^5(f)} \]

Antiderivative was successfully verified.

[In]  Integrate[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*Log[f]^5)

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Maple [A]  time = 0.016, size = 64, normalized size = 2.7 \[{\frac{ \left ({b}^{4}{x}^{12} \left ( \ln \left ( f \right ) \right ) ^{4}-4\,{b}^{3}{x}^{9} \left ( \ln \left ( f \right ) \right ) ^{3}+12\,{b}^{2}{x}^{6} \left ( \ln \left ( f \right ) \right ) ^{2}-24\,b{x}^{3}\ln \left ( f \right ) +24 \right ){f}^{b{x}^{3}+a}}{3\, \left ( \ln \left ( f \right ) \right ) ^{5}{b}^{5}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(f^(b*x^3+a)*x^14,x)

[Out]

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

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Maxima [A]  time = 0.80098, size = 104, normalized size = 4.33 \[ \frac{{\left (b^{4} f^{a} x^{12} \log \left (f\right )^{4} - 4 \, b^{3} f^{a} x^{9} \log \left (f\right )^{3} + 12 \, b^{2} f^{a} x^{6} \log \left (f\right )^{2} - 24 \, b f^{a} x^{3} \log \left (f\right ) + 24 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{5} \log \left (f\right )^{5}} \]

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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Fricas [A]  time = 0.253028, size = 85, normalized size = 3.54 \[ \frac{{\left (b^{4} x^{12} \log \left (f\right )^{4} - 4 \, b^{3} x^{9} \log \left (f\right )^{3} + 12 \, b^{2} x^{6} \log \left (f\right )^{2} - 24 \, b x^{3} \log \left (f\right ) + 24\right )} f^{b x^{3} + a}}{3 \, b^{5} \log \left (f\right )^{5}} \]

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*log(f)^5)

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Sympy [A]  time = 0.309708, size = 82, normalized size = 3.42 \[ \begin{cases} \frac{f^{a + b x^{3}} \left (b^{4} x^{12} \log{\left (f \right )}^{4} - 4 b^{3} x^{9} \log{\left (f \right )}^{3} + 12 b^{2} x^{6} \log{\left (f \right )}^{2} - 24 b x^{3} \log{\left (f \right )} + 24\right )}{3 b^{5} \log{\left (f \right )}^{5}} & \text{for}\: 3 b^{5} \log{\left (f \right )}^{5} \neq 0 \\\frac{x^{15}}{15} & \text{otherwise} \end{cases} \]

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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GIAC/XCAS [A]  time = 0.239167, size = 162, normalized size = 6.75 \[ \frac{b^{4} x^{12} e^{\left (b x^{3}{\rm ln}\left (f\right ) + a{\rm ln}\left (f\right )\right )}{\rm ln}\left (f\right )^{4} - 4 \, b^{3} x^{9} e^{\left (b x^{3}{\rm ln}\left (f\right ) + a{\rm ln}\left (f\right )\right )}{\rm ln}\left (f\right )^{3} + 12 \, b^{2} x^{6} e^{\left (b x^{3}{\rm ln}\left (f\right ) + a{\rm ln}\left (f\right )\right )}{\rm ln}\left (f\right )^{2} - 24 \, b x^{3} e^{\left (b x^{3}{\rm ln}\left (f\right ) + a{\rm ln}\left (f\right )\right )}{\rm ln}\left (f\right ) + 24 \, e^{\left (b x^{3}{\rm ln}\left (f\right ) + a{\rm ln}\left (f\right )\right )}}{3 \, b^{5}{\rm ln}\left (f\right )^{5}} \]

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