Integrand size = 17, antiderivative size = 103 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=a^3 x+\frac {3 a^2 b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+\frac {3 a b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)}+\frac {b^3 \left (F^{g (e+f x)}\right )^{3 n}}{3 f g n \log (F)} \]
a^3*x+3*a^2*b*(F^(g*(f*x+e)))^n/f/g/n/ln(F)+3/2*a*b^2*(F^(g*(f*x+e)))^(2*n )/f/g/n/ln(F)+1/3*b^3*(F^(g*(f*x+e)))^(3*n)/f/g/n/ln(F)
Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=\frac {b \left (F^{g (e+f x)}\right )^n \left (18 a^2+9 a b \left (F^{g (e+f x)}\right )^n+2 b^2 \left (F^{g (e+f x)}\right )^{2 n}\right )+6 a^3 \log \left (\left (F^{g (e+f x)}\right )^n\right )}{6 f g n \log (F)} \]
(b*(F^(g*(e + f*x)))^n*(18*a^2 + 9*a*b*(F^(g*(e + f*x)))^n + 2*b^2*(F^(g*( e + f*x)))^(2*n)) + 6*a^3*Log[(F^(g*(e + f*x)))^n])/(6*f*g*n*Log[F])
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2720, 798, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int F^{-g (e+f x)} \left (b \left (F^{g (e+f x)}\right )^n+a\right )^3dF^{g (e+f x)}}{f g \log (F)}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int F^{-g (e+f x)} \left (b \left (F^{g (e+f x)}\right )^n+a\right )^3d\left (F^{g (e+f x)}\right )^n}{f g n \log (F)}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (3 a b^2 \left (F^{g (e+f x)}\right )^n+a^3 F^{-g (e+f x)}+b^3 F^{2 g (e+f x)}+3 a^2 b\right )d\left (F^{g (e+f x)}\right )^n}{f g n \log (F)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \log \left (\left (F^{g (e+f x)}\right )^n\right )+3 a^2 b \left (F^{g (e+f x)}\right )^n+\frac {3}{2} a b^2 F^{2 g (e+f x)}+\frac {1}{3} b^3 F^{3 g (e+f x)}}{f g n \log (F)}\) |
((3*a*b^2*F^(2*g*(e + f*x)))/2 + (b^3*F^(3*g*(e + f*x)))/3 + 3*a^2*b*(F^(g *(e + f*x)))^n + a^3*Log[(F^(g*(e + f*x)))^n])/(f*g*n*Log[F])
3.1.42.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80
method | result | size |
parallelrisch | \(\frac {6 a^{3} x \ln \left (F \right ) f g n +2 b^{3} \left (F^{g \left (f x +e \right )}\right )^{3 n}+9 a \,b^{2} \left (F^{g \left (f x +e \right )}\right )^{2 n}+18 a^{2} b \left (F^{g \left (f x +e \right )}\right )^{n}}{6 \ln \left (F \right ) f g n}\) | \(82\) |
derivativedivides | \(\frac {\frac {b^{3} \left (F^{g \left (f x +e \right )}\right )^{3 n}}{3}+\frac {3 a \,b^{2} \left (F^{g \left (f x +e \right )}\right )^{2 n}}{2}+3 a^{2} b \left (F^{g \left (f x +e \right )}\right )^{n}+a^{3} \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{g f \ln \left (F \right ) n}\) | \(86\) |
default | \(\frac {\frac {b^{3} \left (F^{g \left (f x +e \right )}\right )^{3 n}}{3}+\frac {3 a \,b^{2} \left (F^{g \left (f x +e \right )}\right )^{2 n}}{2}+3 a^{2} b \left (F^{g \left (f x +e \right )}\right )^{n}+a^{3} \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{g f \ln \left (F \right ) n}\) | \(86\) |
norman | \(a^{3} x +\frac {b^{3} {\mathrm e}^{3 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{3 n g f \ln \left (F \right )}+\frac {3 a \,b^{2} {\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{2 n g f \ln \left (F \right )}+\frac {3 a^{2} b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n g f \ln \left (F \right )}\) | \(109\) |
1/6*(6*a^3*x*ln(F)*f*g*n+2*b^3*((F^(g*(f*x+e)))^n)^3+9*a*b^2*((F^(g*(f*x+e )))^n)^2+18*a^2*b*(F^(g*(f*x+e)))^n)/ln(F)/f/g/n
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=\frac {6 \, a^{3} f g n x \log \left (F\right ) + 18 \, F^{f g n x + e g n} a^{2} b + 9 \, F^{2 \, f g n x + 2 \, e g n} a b^{2} + 2 \, F^{3 \, f g n x + 3 \, e g n} b^{3}}{6 \, f g n \log \left (F\right )} \]
1/6*(6*a^3*f*g*n*x*log(F) + 18*F^(f*g*n*x + e*g*n)*a^2*b + 9*F^(2*f*g*n*x + 2*e*g*n)*a*b^2 + 2*F^(3*f*g*n*x + 3*e*g*n)*b^3)/(f*g*n*log(F))
Time = 0.61 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.17 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=\begin {cases} x \left (a + b\right )^{3} & \text {for}\: F = 1 \wedge f = 0 \wedge g = 0 \wedge n = 0 \\x \left (a + b \left (F^{e g}\right )^{n}\right )^{3} & \text {for}\: f = 0 \\x \left (a + b\right )^{3} & \text {for}\: F = 1 \vee g = 0 \vee n = 0 \\a^{3} x + \frac {3 a^{2} b \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} + \frac {3 a b^{2} \left (F^{e g + f g x}\right )^{2 n}}{2 f g n \log {\left (F \right )}} + \frac {b^{3} \left (F^{e g + f g x}\right )^{3 n}}{3 f g n \log {\left (F \right )}} & \text {otherwise} \end {cases} \]
Piecewise((x*(a + b)**3, Eq(F, 1) & Eq(f, 0) & Eq(g, 0) & Eq(n, 0)), (x*(a + b*(F**(e*g))**n)**3, Eq(f, 0)), (x*(a + b)**3, Eq(F, 1) | Eq(g, 0) | Eq (n, 0)), (a**3*x + 3*a**2*b*(F**(e*g + f*g*x))**n/(f*g*n*log(F)) + 3*a*b** 2*(F**(e*g + f*g*x))**(2*n)/(2*f*g*n*log(F)) + b**3*(F**(e*g + f*g*x))**(3 *n)/(3*f*g*n*log(F)), True))
Time = 0.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=a^{3} x + \frac {3 \, F^{{\left (f x + e\right )} g n} a^{2} b}{f g n \log \left (F\right )} + \frac {3 \, F^{2 \, {\left (f x + e\right )} g n} a b^{2}}{2 \, f g n \log \left (F\right )} + \frac {F^{3 \, {\left (f x + e\right )} g n} b^{3}}{3 \, f g n \log \left (F\right )} \]
a^3*x + 3*F^((f*x + e)*g*n)*a^2*b/(f*g*n*log(F)) + 3/2*F^(2*(f*x + e)*g*n) *a*b^2/(f*g*n*log(F)) + 1/3*F^(3*(f*x + e)*g*n)*b^3/(f*g*n*log(F))
Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.95 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=\frac {18 \, F^{f g n x} F^{e g n} a^{2} b + 9 \, F^{2 \, f g n x} F^{2 \, e g n} a b^{2} + 2 \, F^{3 \, f g n x} F^{3 \, e g n} b^{3} + 6 \, a^{3} \log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{e g n}\right )}{6 \, f g n \log \left (F\right )} \]
1/6*(18*F^(f*g*n*x)*F^(e*g*n)*a^2*b + 9*F^(2*f*g*n*x)*F^(2*e*g*n)*a*b^2 + 2*F^(3*f*g*n*x)*F^(3*e*g*n)*b^3 + 6*a^3*log(abs(F)^(f*g*n*x)*abs(F)^(e*g*n )))/(f*g*n*log(F))
Time = 0.38 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.20 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^3 \, dx=\frac {a^3\,\ln \left (F^{f\,g\,x}\right )}{f\,g\,\ln \left (F\right )}+\frac {b^3\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{3\,n}}{3\,f\,g\,n\,\ln \left (F\right )}+\frac {3\,a^2\,b\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^n}{f\,g\,n\,\ln \left (F\right )}+\frac {3\,a\,b^2\,{\left (F^{f\,g\,x}\,F^{e\,g}\right )}^{2\,n}}{2\,f\,g\,n\,\ln \left (F\right )} \]