\(\int f^{a+b x^3} x^8 \, dx\) [99]

Optimal result

Integrand size = 13, antiderivative size = 67 \[ \int f^{a+b x^3} x^8 \, dx=\frac {2 f^{a+b x^3}}{3 b^3 \log ^3(f)}-\frac {2 f^{a+b x^3} x^3}{3 b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^6}{3 b \log (f)} \]

[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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int f^{a+b x^3} x^8 \, dx=\frac {2 f^{a+b x^3}}{3 b^3 \log ^3(f)}-\frac {2 x^3 f^{a+b x^3}}{3 b^2 \log ^2(f)}+\frac {x^6 f^{a+b x^3}}{3 b \log (f)} \]

[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 2240

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 2243

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.61 \[ \int f^{a+b x^3} x^8 \, dx=\frac {f^{a+b x^3} \left (2-2 b x^3 \log (f)+b^2 x^6 \log ^2(f)\right )}{3 b^3 \log ^3(f)} \]

[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)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.60

method	result	size
gosper	\(\frac {\left (b^{2} x^{6} \ln \left (f \right )^{2}-2 b \,x^{3} \ln \left (f \right )+2\right ) f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{3} b^{3}}\)	\(40\)
risch	\(\frac {\left (b^{2} x^{6} \ln \left (f \right )^{2}-2 b \,x^{3} \ln \left (f \right )+2\right ) f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{3} b^{3}}\)	\(40\)
meijerg	\(-\frac {f^{a} \left (2-\frac {\left (3 b^{2} x^{6} \ln \left (f \right )^{2}-6 b \,x^{3} \ln \left (f \right )+6\right ) {\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{3}\right )}{3 b^{3} \ln \left (f \right )^{3}}\)	\(47\)
parallelrisch	\(\frac {f^{b \,x^{3}+a} x^{6} b^{2} \ln \left (f \right )^{2}-2 f^{b \,x^{3}+a} x^{3} b \ln \left (f \right )+2 f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{3} b^{3}}\)	\(59\)
norman	\(\frac {2 \,{\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 \ln \left (f \right )^{3} b^{3}}+\frac {x^{6} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 b \ln \left (f \right )}-\frac {2 x^{3} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 \ln \left (f \right )^{2} b^{2}}\)	\(68\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int f^{a+b x^3} x^8 \, dx=\frac {{\left (b^{2} x^{6} \log \left (f\right )^{2} - 2 \, b x^{3} \log \left (f\right ) + 2\right )} f^{b x^{3} + a}}{3 \, b^{3} \log \left (f\right )^{3}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int f^{a+b x^3} x^8 \, dx=\begin {cases} \frac {f^{a + b x^{3}} \left (b^{2} x^{6} \log {\left (f \right )}^{2} - 2 b x^{3} \log {\left (f \right )} + 2\right )}{3 b^{3} \log {\left (f \right )}^{3}} & \text {for}\: b^{3} \log {\left (f \right )}^{3} \neq 0 \\\frac {x^{9}}{9} & \text {otherwise} \end {cases} \]

[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(b**3*log(f)**3,
0)), (x**9/9, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70 \[ \int f^{a+b x^3} x^8 \, dx=\frac {{\left (b^{2} f^{a} x^{6} \log \left (f\right )^{2} - 2 \, b f^{a} x^{3} \log \left (f\right ) + 2 \, f^{a}\right )} f^{b x^{3}}}{3 \, b^{3} \log \left (f\right )^{3}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int f^{a+b x^3} x^8 \, dx=\frac {b^{2} f^{b x^{3}} f^{a} x^{6} \log \left (f\right )^{2} - 2 \, b f^{b x^{3}} f^{a} x^{3} \log \left (f\right ) + 2 \, f^{b x^{3}} f^{a}}{3 \, b^{3} \log \left (f\right )^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int f^{a+b x^3} x^8 \, dx=\frac {f^{b\,x^3+a}\,\left (\frac {b^2\,x^6\,{\ln \left (f\right )}^2}{3}-\frac {2\,b\,x^3\,\ln \left (f\right )}{3}+\frac {2}{3}\right )}{b^3\,{\ln \left (f\right )}^3} \]

[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)