Integrand size = 17, antiderivative size = 67 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=a^2 x+\frac {2 a b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)}+\frac {b^2 \left (F^{g (e+f x)}\right )^{2 n}}{2 f g n \log (F)} \] Output:
a^2*x+2*a*b*(F^(g*(f*x+e)))^n/f/g/n/ln(F)+1/2*b^2*(F^(g*(f*x+e)))^(2*n)/f/ g/n/ln(F)
Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=\frac {b \left (F^{g (e+f x)}\right )^n \left (4 a+b \left (F^{g (e+f x)}\right )^n\right )+2 a^2 \log \left (\left (F^{g (e+f x)}\right )^n\right )}{2 f g n \log (F)} \] Input:
Integrate[(a + b*(F^(g*(e + f*x)))^n)^2,x]
Output:
(b*(F^(g*(e + f*x)))^n*(4*a + b*(F^(g*(e + f*x)))^n) + 2*a^2*Log[(F^(g*(e + f*x)))^n])/(2*f*g*n*Log[F])
Time = 0.35 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94, 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.
\(\displaystyle \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int F^{-g (e+f x)} \left (b \left (F^{g (e+f x)}\right )^n+a\right )^2dF^{g (e+f x)}}{f g \log (F)}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int F^{-g (e+f x)} \left (b \left (F^{g (e+f x)}\right )^n+a\right )^2d\left (F^{g (e+f x)}\right )^n}{f g n \log (F)}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (b^2 \left (F^{g (e+f x)}\right )^n+a^2 F^{-g (e+f x)}+2 a b\right )d\left (F^{g (e+f x)}\right )^n}{f g n \log (F)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \log \left (\left (F^{g (e+f x)}\right )^n\right )+2 a b \left (F^{g (e+f x)}\right )^n+\frac {1}{2} b^2 F^{2 g (e+f x)}}{f g n \log (F)}\) |
Input:
Int[(a + b*(F^(g*(e + f*x)))^n)^2,x]
Output:
((b^2*F^(2*g*(e + f*x)))/2 + 2*a*b*(F^(g*(e + f*x)))^n + a^2*Log[(F^(g*(e + f*x)))^n])/(f*g*n*Log[F])
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 = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {2 a^{2} x \ln \left (F \right ) f g n +\left (F^{g \left (f x +e \right )}\right )^{2 n} b^{2}+4 \left (F^{g \left (f x +e \right )}\right )^{n} a b}{2 \ln \left (F \right ) f g n}\) | \(60\) |
derivativedivides | \(\frac {\frac {\left (F^{g \left (f x +e \right )}\right )^{2 n} b^{2}}{2}+2 \left (F^{g \left (f x +e \right )}\right )^{n} a b +a^{2} \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{g f \ln \left (F \right ) n}\) | \(65\) |
default | \(\frac {\frac {\left (F^{g \left (f x +e \right )}\right )^{2 n} b^{2}}{2}+2 \left (F^{g \left (f x +e \right )}\right )^{n} a b +a^{2} \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{g f \ln \left (F \right ) n}\) | \(65\) |
norman | \(a^{2} x +\frac {b^{2} {\mathrm e}^{2 n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{2 n g f \ln \left (F \right )}+\frac {2 a b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n g f \ln \left (F \right )}\) | \(72\) |
orering | \(\frac {\left (2 x n g f \ln \left (F \right )+3\right ) {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}^{2}}{2 \ln \left (F \right ) f g n}-\frac {\left (3 x n g f \ln \left (F \right )+1\right ) \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) b \left (F^{g \left (f x +e \right )}\right )^{n}}{\ln \left (F \right ) f g n}+\frac {x \left (2 b^{2} \left (F^{g \left (f x +e \right )}\right )^{2 n} n^{2} g^{2} f^{2} \ln \left (F \right )^{2}+2 \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) b \left (F^{g \left (f x +e \right )}\right )^{n} n^{2} g^{2} f^{2} \ln \left (F \right )^{2}\right )}{2 \ln \left (F \right )^{2} f^{2} g^{2} n^{2}}\) | \(186\) |
Input:
int((a+b*(F^(g*(f*x+e)))^n)^2,x,method=_RETURNVERBOSE)
Output:
1/2*(2*a^2*x*ln(F)*f*g*n+b^2*((F^(g*(f*x+e)))^n)^2+4*(F^(g*(f*x+e)))^n*a*b )/ln(F)/f/g/n
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=\frac {2 \, a^{2} f g n x \log \left (F\right ) + 4 \, F^{f g n x + e g n} a b + F^{2 \, f g n x + 2 \, e g n} b^{2}}{2 \, f g n \log \left (F\right )} \] Input:
integrate((a+b*(F^(g*(f*x+e)))^n)^2,x, algorithm="fricas")
Output:
1/2*(2*a^2*f*g*n*x*log(F) + 4*F^(f*g*n*x + e*g*n)*a*b + F^(2*f*g*n*x + 2*e *g*n)*b^2)/(f*g*n*log(F))
Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=\begin {cases} x \left (a + b\right )^{2} & \text {for}\: F = 1 \wedge f = 0 \wedge g = 0 \wedge n = 0 \\x \left (a + b \left (F^{e g}\right )^{n}\right )^{2} & \text {for}\: f = 0 \\x \left (a + b\right )^{2} & \text {for}\: F = 1 \vee g = 0 \vee n = 0 \\a^{2} x + \frac {2 a b \left (F^{e g + f g x}\right )^{n}}{f g n \log {\left (F \right )}} + \frac {b^{2} \left (F^{e g + f g x}\right )^{2 n}}{2 f g n \log {\left (F \right )}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*(F**(g*(f*x+e)))**n)**2,x)
Output:
Piecewise((x*(a + b)**2, Eq(F, 1) & Eq(f, 0) & Eq(g, 0) & Eq(n, 0)), (x*(a + b*(F**(e*g))**n)**2, Eq(f, 0)), (x*(a + b)**2, Eq(F, 1) | Eq(g, 0) | Eq (n, 0)), (a**2*x + 2*a*b*(F**(e*g + f*g*x))**n/(f*g*n*log(F)) + b**2*(F**( e*g + f*g*x))**(2*n)/(2*f*g*n*log(F)), True))
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=a^{2} x + \frac {2 \, F^{{\left (f x + e\right )} g n} a b}{f g n \log \left (F\right )} + \frac {F^{2 \, {\left (f x + e\right )} g n} b^{2}}{2 \, f g n \log \left (F\right )} \] Input:
integrate((a+b*(F^(g*(f*x+e)))^n)^2,x, algorithm="maxima")
Output:
a^2*x + 2*F^((f*x + e)*g*n)*a*b/(f*g*n*log(F)) + 1/2*F^(2*(f*x + e)*g*n)*b ^2/(f*g*n*log(F))
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=\frac {4 \, F^{f g n x} F^{e g n} a b + F^{2 \, f g n x} F^{2 \, e g n} b^{2} + 2 \, a^{2} \log \left ({\left | F \right |}^{f g n x} {\left | F \right |}^{e g n}\right )}{2 \, f g n \log \left (F\right )} \] Input:
integrate((a+b*(F^(g*(f*x+e)))^n)^2,x, algorithm="giac")
Output:
1/2*(4*F^(f*g*n*x)*F^(e*g*n)*a*b + F^(2*f*g*n*x)*F^(2*e*g*n)*b^2 + 2*a^2*l og(abs(F)^(f*g*n*x)*abs(F)^(e*g*n)))/(f*g*n*log(F))
Time = 23.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=\frac {\frac {b^2\,{\left (F^{e\,g+f\,g\,x}\right )}^{2\,n}}{2}+2\,a\,b\,{\left (F^{e\,g+f\,g\,x}\right )}^n}{f\,g\,n\,\ln \left (F\right )}+\frac {a^2\,\ln \left (F^{g\,\left (e+f\,x\right )}\right )}{f\,g\,\ln \left (F\right )} \] Input:
int((a + b*(F^(g*(e + f*x)))^n)^2,x)
Output:
((b^2*(F^(e*g + f*g*x))^(2*n))/2 + 2*a*b*(F^(e*g + f*g*x))^n)/(f*g*n*log(F )) + (a^2*log(F^(g*(e + f*x))))/(f*g*log(F))
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2 \, dx=\frac {f^{2 f g n x +2 e g n} b^{2}+4 f^{f g n x +e g n} a b +2 \,\mathrm {log}\left (f \right ) a^{2} f g n x}{2 \,\mathrm {log}\left (f \right ) f g n} \] Input:
int((a+b*(F^(g*(f*x+e)))^n)^2,x)
Output:
(f**(2*e*g*n + 2*f*g*n*x)*b**2 + 4*f**(e*g*n + f*g*n*x)*a*b + 2*log(f)*a** 2*f*g*n*x)/(2*log(f)*f*g*n)