3.98 \(\int f^{a+b x^3} x^{11} \, dx\)

Optimal. Leaf size=84 \[ -\frac {2 f^{a+b x^3}}{b^4 \log ^4(f)}+\frac {2 x^3 f^{a+b x^3}}{b^3 \log ^3(f)}-\frac {x^6 f^{a+b x^3}}{b^2 \log ^2(f)}+\frac {x^9 f^{a+b x^3}}{3 b \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2212, 2209} \[ -\frac {x^6 f^{a+b x^3}}{b^2 \log ^2(f)}+\frac {2 x^3 f^{a+b x^3}}{b^3 \log ^3(f)}-\frac {2 f^{a+b x^3}}{b^4 \log ^4(f)}+\frac {x^9 f^{a+b x^3}}{3 b \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)*x^11,x]

[Out]

(-2*f^(a + b*x^3))/(b^4*Log[f]^4) + (2*f^(a + b*x^3)*x^3)/(b^3*Log[f]^3) - (f^(a + b*x^3)*x^6)/(b^2*Log[f]^2)
+ (f^(a + b*x^3)*x^9)/(3*b*Log[f])

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rubi steps

\begin {align*} \int f^{a+b x^3} x^{11} \, dx &=\frac {f^{a+b x^3} x^9}{3 b \log (f)}-\frac {3 \int f^{a+b x^3} x^8 \, dx}{b \log (f)}\\ &=-\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)}+\frac {6 \int f^{a+b x^3} x^5 \, dx}{b^2 \log ^2(f)}\\ &=\frac {2 f^{a+b x^3} x^3}{b^3 \log ^3(f)}-\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)}-\frac {6 \int f^{a+b x^3} x^2 \, dx}{b^3 \log ^3(f)}\\ &=-\frac {2 f^{a+b x^3}}{b^4 \log ^4(f)}+\frac {2 f^{a+b x^3} x^3}{b^3 \log ^3(f)}-\frac {f^{a+b x^3} x^6}{b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^9}{3 b \log (f)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 53, normalized size = 0.63 \[ \frac {f^{a+b x^3} \left (b^3 x^9 \log ^3(f)-3 b^2 x^6 \log ^2(f)+6 b x^3 \log (f)-6\right )}{3 b^4 \log ^4(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3)*x^11,x]

[Out]

(f^(a + b*x^3)*(-6 + 6*b*x^3*Log[f] - 3*b^2*x^6*Log[f]^2 + b^3*x^9*Log[f]^3))/(3*b^4*Log[f]^4)

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fricas [A]  time = 0.43, size = 51, normalized size = 0.61 \[ \frac {{\left (b^{3} x^{9} \log \relax (f)^{3} - 3 \, b^{2} x^{6} \log \relax (f)^{2} + 6 \, b x^{3} \log \relax (f) - 6\right )} f^{b x^{3} + a}}{3 \, b^{4} \log \relax (f)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)*x^11,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 83, normalized size = 0.99 \[ \frac {b^{3} f^{b x^{3}} f^{a} x^{9} \log \relax (f)^{3} - 3 \, b^{2} f^{b x^{3}} f^{a} x^{6} \log \relax (f)^{2} + 6 \, b f^{b x^{3}} f^{a} x^{3} \log \relax (f) - 6 \, f^{b x^{3}} f^{a}}{3 \, b^{4} \log \relax (f)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)*x^11,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 52, normalized size = 0.62 \[ \frac {\left (b^{3} x^{9} \ln \relax (f )^{3}-3 b^{2} x^{6} \ln \relax (f )^{2}+6 b \,x^{3} \ln \relax (f )-6\right ) f^{b \,x^{3}+a}}{3 b^{4} \ln \relax (f )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)*x^11,x)

[Out]

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

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maxima [A]  time = 0.98, size = 62, normalized size = 0.74 \[ \frac {{\left (b^{3} f^{a} x^{9} \log \relax (f)^{3} - 3 \, b^{2} f^{a} x^{6} \log \relax (f)^{2} + 6 \, b f^{a} x^{3} \log \relax (f) - 6 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{4} \log \relax (f)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)*x^11,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.46, size = 51, normalized size = 0.61 \[ -\frac {f^{b\,x^3+a}\,\left (-\frac {b^3\,x^9\,{\ln \relax (f)}^3}{3}+b^2\,x^6\,{\ln \relax (f)}^2-2\,b\,x^3\,\ln \relax (f)+2\right )}{b^4\,{\ln \relax (f)}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3)*x^11,x)

[Out]

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

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sympy [A]  time = 0.15, size = 68, normalized size = 0.81 \[ \begin {cases} \frac {f^{a + b x^{3}} \left (b^{3} x^{9} \log {\relax (f )}^{3} - 3 b^{2} x^{6} \log {\relax (f )}^{2} + 6 b x^{3} \log {\relax (f )} - 6\right )}{3 b^{4} \log {\relax (f )}^{4}} & \text {for}\: 3 b^{4} \log {\relax (f )}^{4} \neq 0 \\\frac {x^{12}}{12} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**3+a)*x**11,x)

[Out]

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

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