3.99 \(\int f^{a+b x^3} x^8 \, dx\)

Optimal. Leaf size=67 \[ \frac {2 f^{a+b x^3}}{3 b^3 \log ^3(f)}-\frac {2 x^3 f^{a+b x^3}}{3 b^2 \log ^2(f)}+\frac {x^6 f^{a+b x^3}}{3 b \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2212, 2209} \[ -\frac {2 x^3 f^{a+b x^3}}{3 b^2 \log ^2(f)}+\frac {2 f^{a+b x^3}}{3 b^3 \log ^3(f)}+\frac {x^6 f^{a+b x^3}}{3 b \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3)*x^8,x]

[Out]

(2*f^(a + b*x^3))/(3*b^3*Log[f]^3) - (2*f^(a + b*x^3)*x^3)/(3*b^2*Log[f]^2) + (f^(a + b*x^3)*x^6)/(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^8 \, dx &=\frac {f^{a+b x^3} x^6}{3 b \log (f)}-\frac {2 \int f^{a+b x^3} x^5 \, dx}{b \log (f)}\\ &=-\frac {2 f^{a+b x^3} x^3}{3 b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^6}{3 b \log (f)}+\frac {2 \int f^{a+b x^3} x^2 \, dx}{b^2 \log ^2(f)}\\ &=\frac {2 f^{a+b x^3}}{3 b^3 \log ^3(f)}-\frac {2 f^{a+b x^3} x^3}{3 b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^6}{3 b \log (f)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.61 \[ \frac {f^{a+b x^3} \left (b^2 x^6 \log ^2(f)-2 b x^3 \log (f)+2\right )}{3 b^3 \log ^3(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3)*x^8,x]

[Out]

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

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fricas [A]  time = 0.42, size = 39, normalized size = 0.58 \[ \frac {{\left (b^{2} x^{6} \log \relax (f)^{2} - 2 \, b x^{3} \log \relax (f) + 2\right )} f^{b x^{3} + a}}{3 \, b^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)*x^8,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 61, normalized size = 0.91 \[ \frac {b^{2} f^{b x^{3}} f^{a} x^{6} \log \relax (f)^{2} - 2 \, b f^{b x^{3}} f^{a} x^{3} \log \relax (f) + 2 \, f^{b x^{3}} f^{a}}{3 \, b^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)*x^8,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 40, normalized size = 0.60 \[ \frac {\left (b^{2} x^{6} \ln \relax (f )^{2}-2 b \,x^{3} \ln \relax (f )+2\right ) f^{b \,x^{3}+a}}{3 b^{3} \ln \relax (f )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a)*x^8,x)

[Out]

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

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maxima [A]  time = 0.95, size = 47, normalized size = 0.70 \[ \frac {{\left (b^{2} f^{a} x^{6} \log \relax (f)^{2} - 2 \, b f^{a} x^{3} \log \relax (f) + 2 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^3+a)*x^8,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.47, size = 39, normalized size = 0.58 \[ \frac {f^{b\,x^3+a}\,\left (\frac {b^2\,x^6\,{\ln \relax (f)}^2}{3}-\frac {2\,b\,x^3\,\ln \relax (f)}{3}+\frac {2}{3}\right )}{b^3\,{\ln \relax (f)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3)*x^8,x)

[Out]

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

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sympy [A]  time = 0.14, size = 54, normalized size = 0.81 \[ \begin {cases} \frac {f^{a + b x^{3}} \left (b^{2} x^{6} \log {\relax (f )}^{2} - 2 b x^{3} \log {\relax (f )} + 2\right )}{3 b^{3} \log {\relax (f )}^{3}} & \text {for}\: 3 b^{3} \log {\relax (f )}^{3} \neq 0 \\\frac {x^{9}}{9} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**3+a)*x**8,x)

[Out]

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

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