Integrand size = 17, antiderivative size = 67 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=d^2 x+\frac {2 d e \left (F^{c (a+b x)}\right )^n}{b c n \log (F)}+\frac {e^2 \left (F^{c (a+b x)}\right )^{2 n}}{2 b c n \log (F)} \] Output:
d^2*x+2*d*e*(F^(c*(b*x+a)))^n/b/c/n/ln(F)+1/2*e^2*(F^(c*(b*x+a)))^(2*n)/b/ c/n/ln(F)
Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=\frac {e \left (F^{c (a+b x)}\right )^n \left (4 d+e \left (F^{c (a+b x)}\right )^n\right )+2 d^2 \log \left (\left (F^{c (a+b x)}\right )^n\right )}{2 b c n \log (F)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)^2,x]
Output:
(e*(F^(c*(a + b*x)))^n*(4*d + e*(F^(c*(a + b*x)))^n) + 2*d^2*Log[(F^(c*(a + b*x)))^n])/(2*b*c*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.
\(\displaystyle \int \left (e \left (F^{c (a+b x)}\right )^n+d\right )^2 \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int F^{-c (a+b x)} \left (e \left (F^{c (a+b x)}\right )^n+d\right )^2dF^{c (a+b x)}}{b c \log (F)}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int F^{-c (a+b x)} \left (e \left (F^{c (a+b x)}\right )^n+d\right )^2d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (e^2 \left (F^{c (a+b x)}\right )^n+d^2 F^{-c (a+b x)}+2 d e\right )d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 \log \left (\left (F^{c (a+b x)}\right )^n\right )+2 d e \left (F^{c (a+b x)}\right )^n+\frac {1}{2} e^2 F^{2 c (a+b x)}}{b c n \log (F)}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)^2,x]
Output:
((e^2*F^(2*c*(a + b*x)))/2 + 2*d*e*(F^(c*(a + b*x)))^n + d^2*Log[(F^(c*(a + b*x)))^n])/(b*c*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 = 0.00 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {2 d^{2} x \ln \left (F \right ) b c n +e^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n}+4 d e \left (F^{c \left (b x +a \right )}\right )^{n}}{2 \ln \left (F \right ) b c n}\) | \(60\) |
derivativedivides | \(\frac {\frac {e^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n}}{2}+2 d e \left (F^{c \left (b x +a \right )}\right )^{n}+d^{2} \ln \left (\left (F^{c \left (b x +a \right )}\right )^{n}\right )}{\ln \left (F \right ) b c n}\) | \(65\) |
default | \(\frac {\frac {e^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n}}{2}+2 d e \left (F^{c \left (b x +a \right )}\right )^{n}+d^{2} \ln \left (\left (F^{c \left (b x +a \right )}\right )^{n}\right )}{\ln \left (F \right ) b c n}\) | \(65\) |
norman | \(d^{2} x +\frac {e^{2} {\mathrm e}^{2 n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{2 \ln \left (F \right ) b c n}+\frac {2 d e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) b c n}\) | \(72\) |
orering | \(\frac {\left (2 \ln \left (F \right ) b c n x +3\right ) {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2}}{2 \ln \left (F \right ) b c n}-\frac {\left (3 \ln \left (F \right ) b c n x +1\right ) \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) e \left (F^{c \left (b x +a \right )}\right )^{n}}{\ln \left (F \right ) b c n}+\frac {x \left (2 e^{2} \left (F^{c \left (b x +a \right )}\right )^{2 n} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}+2 \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) e \left (F^{c \left (b x +a \right )}\right )^{n} \ln \left (F \right )^{2} b^{2} c^{2} n^{2}\right )}{2 \ln \left (F \right )^{2} b^{2} c^{2} n^{2}}\) | \(186\) |
Input:
int((d+e*(F^(c*(b*x+a)))^n)^2,x,method=_RETURNVERBOSE)
Output:
1/2*(2*d^2*x*ln(F)*b*c*n+e^2*((F^(c*(b*x+a)))^n)^2+4*d*e*(F^(c*(b*x+a)))^n )/ln(F)/b/c/n
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=\frac {2 \, b c d^{2} n x \log \left (F\right ) + 4 \, F^{b c n x + a c n} d e + F^{2 \, b c n x + 2 \, a c n} e^{2}}{2 \, b c n \log \left (F\right )} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^2,x, algorithm="fricas")
Output:
1/2*(2*b*c*d^2*n*x*log(F) + 4*F^(b*c*n*x + a*c*n)*d*e + F^(2*b*c*n*x + 2*a *c*n)*e^2)/(b*c*n*log(F))
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=\begin {cases} x \left (d + e\right )^{2} & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge n = 0 \\x \left (d + e \left (F^{a c}\right )^{n}\right )^{2} & \text {for}\: b = 0 \\x \left (d + e\right )^{2} & \text {for}\: F = 1 \vee c = 0 \vee n = 0 \\d^{2} x + \frac {2 d e \left (F^{a c + b c x}\right )^{n}}{b c n \log {\left (F \right )}} + \frac {e^{2} \left (F^{a c + b c x}\right )^{2 n}}{2 b c n \log {\left (F \right )}} & \text {otherwise} \end {cases} \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)**2,x)
Output:
Piecewise((x*(d + e)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(n, 0)), (x*(d + e*(F**(a*c))**n)**2, Eq(b, 0)), (x*(d + e)**2, Eq(F, 1) | Eq(c, 0) | Eq (n, 0)), (d**2*x + 2*d*e*(F**(a*c + b*c*x))**n/(b*c*n*log(F)) + e**2*(F**( a*c + b*c*x))**(2*n)/(2*b*c*n*log(F)), True))
Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=d^{2} x + \frac {2 \, F^{{\left (b x + a\right )} c n} d e}{b c n \log \left (F\right )} + \frac {F^{2 \, {\left (b x + a\right )} c n} e^{2}}{2 \, b c n \log \left (F\right )} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^2,x, algorithm="maxima")
Output:
d^2*x + 2*F^((b*x + a)*c*n)*d*e/(b*c*n*log(F)) + 1/2*F^(2*(b*x + a)*c*n)*e ^2/(b*c*n*log(F))
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=\frac {4 \, F^{b c n x} F^{a c n} d e + F^{2 \, b c n x} F^{2 \, a c n} e^{2} + 2 \, d^{2} \log \left ({\left | F \right |}^{b c n x} {\left | F \right |}^{a c n}\right )}{2 \, b c n \log \left (F\right )} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^2,x, algorithm="giac")
Output:
1/2*(4*F^(b*c*n*x)*F^(a*c*n)*d*e + F^(2*b*c*n*x)*F^(2*a*c*n)*e^2 + 2*d^2*l og(abs(F)^(b*c*n*x)*abs(F)^(a*c*n)))/(b*c*n*log(F))
Time = 0.00 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=\frac {d^2\,\ln \left (F^{c\,\left (a+b\,x\right )}\right )}{b\,c\,\ln \left (F\right )}+\frac {\frac {e^2\,{\left (F^{a\,c+b\,c\,x}\right )}^{2\,n}}{2}+2\,d\,e\,{\left (F^{a\,c+b\,c\,x}\right )}^n}{b\,c\,n\,\ln \left (F\right )} \] Input:
int((d + e*(F^(c*(a + b*x)))^n)^2,x)
Output:
(d^2*log(F^(c*(a + b*x))))/(b*c*log(F)) + ((e^2*(F^(a*c + b*c*x))^(2*n))/2 + 2*d*e*(F^(a*c + b*c*x))^n)/(b*c*n*log(F))
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 \, dx=\frac {f^{2 b c n x +2 a c n} e^{2}+4 f^{b c n x +a c n} d e +2 \,\mathrm {log}\left (f \right ) b c \,d^{2} n x}{2 \,\mathrm {log}\left (f \right ) b c n} \] Input:
int((d+e*(F^((b*x+a)*c))^n)^2,x)
Output:
(f**(2*a*c*n + 2*b*c*n*x)*e**2 + 4*f**(a*c*n + b*c*n*x)*d*e + 2*log(f)*b*c *d**2*n*x)/(2*log(f)*b*c*n)