\(\int f^{a+b x^3} x^5 \, dx\) [100]

Optimal result

Integrand size = 13, antiderivative size = 44 \[ \int f^{a+b x^3} x^5 \, dx=-\frac {f^{a+b x^3}}{3 b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^3}{3 b \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int f^{a+b x^3} x^5 \, dx=\frac {x^3 f^{a+b x^3}}{3 b \log (f)}-\frac {f^{a+b x^3}}{3 b^2 \log ^2(f)} \]

[In]

Int[f^(a + b*x^3)*x^5,x]

[Out]

-1/3*f^(a + b*x^3)/(b^2*Log[f]^2) + (f^(a + b*x^3)*x^3)/(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^3}{3 b \log (f)}-\frac {\int f^{a+b x^3} x^2 \, dx}{b \log (f)} \\ & = -\frac {f^{a+b x^3}}{3 b^2 \log ^2(f)}+\frac {f^{a+b x^3} x^3}{3 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int f^{a+b x^3} x^5 \, dx=\frac {f^{a+b x^3} \left (-1+b x^3 \log (f)\right )}{3 b^2 \log ^2(f)} \]

[In]

Integrate[f^(a + b*x^3)*x^5,x]

[Out]

(f^(a + b*x^3)*(-1 + b*x^3*Log[f]))/(3*b^2*Log[f]^2)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {\left (b \,x^{3} \ln \left (f \right )-1\right ) f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{2} b^{2}}\)	\(28\)
risch	\(\frac {\left (b \,x^{3} \ln \left (f \right )-1\right ) f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{2} b^{2}}\)	\(28\)
meijerg	\(\frac {f^{a} \left (1-\frac {\left (2-2 b \,x^{3} \ln \left (f \right )\right ) {\mathrm e}^{b \,x^{3} \ln \left (f \right )}}{2}\right )}{3 b^{2} \ln \left (f \right )^{2}}\)	\(35\)
parallelrisch	\(\frac {f^{b \,x^{3}+a} x^{3} b \ln \left (f \right )-f^{b \,x^{3}+a}}{3 \ln \left (f \right )^{2} b^{2}}\)	\(38\)
norman	\(-\frac {{\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 \ln \left (f \right )^{2} b^{2}}+\frac {x^{3} {\mathrm e}^{\left (b \,x^{3}+a \right ) \ln \left (f \right )}}{3 b \ln \left (f \right )}\)	\(45\)

[In]

int(f^(b*x^3+a)*x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.61 \[ \int f^{a+b x^3} x^5 \, dx=\frac {{\left (b x^{3} \log \left (f\right ) - 1\right )} f^{b x^{3} + a}}{3 \, b^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(b*x^3+a)*x^5,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int f^{a+b x^3} x^5 \, dx=\begin {cases} \frac {f^{a + b x^{3}} \left (b x^{3} \log {\left (f \right )} - 1\right )}{3 b^{2} \log {\left (f \right )}^{2}} & \text {for}\: b^{2} \log {\left (f \right )}^{2} \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(f**(b*x**3+a)*x**5,x)

[Out]

Piecewise((f**(a + b*x**3)*(b*x**3*log(f) - 1)/(3*b**2*log(f)**2), Ne(b**2*log(f)**2, 0)), (x**6/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int f^{a+b x^3} x^5 \, dx=\frac {{\left (b f^{a} x^{3} \log \left (f\right ) - f^{a}\right )} f^{b x^{3}}}{3 \, b^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(b*x^3+a)*x^5,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 689, normalized size of antiderivative = 15.66 \[ \int f^{a+b x^3} x^5 \, dx=\text {Too large to display} \]

[In]

integrate(f^(b*x^3+a)*x^5,x, algorithm="giac")

[Out]

1/3*(2*((b*x^3*log(abs(f)) - 1)*(pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)/((pi^2*b^2*sgn(f) - pi^2*b^
2 + 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2) + (pi*b*x^3*sgn(f) - pi*b*x
^3)*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))/((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2 + 4
*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2))*cos(-1/2*pi*b*x^3*sgn(f) + 1/2*pi*b*x^3 - 1/2*pi*a*sgn(f
) + 1/2*pi*a) + ((pi*b*x^3*sgn(f) - pi*b*x^3)*(pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)/((pi^2*b^2*sg
n(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2) - 4*(b*x^3*lo
g(abs(f)) - 1)*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))/((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f
))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2))*sin(-1/2*pi*b*x^3*sgn(f) + 1/2*pi*b*x^3 - 1/2
*pi*a*sgn(f) + 1/2*pi*a))*e^(b*x^3*log(abs(f)) + a*log(abs(f))) - 1/6*I*((pi*b*x^3*sgn(f) - pi*b*x^3 - 2*I*b*x
^3*log(abs(f)) + 2*I)*e^(1/2*I*pi*b*x^3*sgn(f) - 1/2*I*pi*b*x^3 + 1/2*I*pi*a*sgn(f) - 1/2*I*pi*a)/(pi^2*b^2*sg
n(f) + 2*I*pi*b^2*log(abs(f))*sgn(f) - pi^2*b^2 - 2*I*pi*b^2*log(abs(f)) + 2*b^2*log(abs(f))^2) + (pi*b*x^3*sg
n(f) - pi*b*x^3 + 2*I*b*x^3*log(abs(f)) - 2*I)*e^(-1/2*I*pi*b*x^3*sgn(f) + 1/2*I*pi*b*x^3 - 1/2*I*pi*a*sgn(f)
+ 1/2*I*pi*a)/(pi^2*b^2*sgn(f) - 2*I*pi*b^2*log(abs(f))*sgn(f) - pi^2*b^2 + 2*I*pi*b^2*log(abs(f)) + 2*b^2*log
(abs(f))^2))*e^(b*x^3*log(abs(f)) + a*log(abs(f)))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.61 \[ \int f^{a+b x^3} x^5 \, dx=\frac {f^{b\,x^3+a}\,\left (\frac {b\,x^3\,\ln \left (f\right )}{3}-\frac {1}{3}\right )}{b^2\,{\ln \left (f\right )}^2} \]

[In]

int(f^(a + b*x^3)*x^5,x)

[Out]

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