Integrand size = 19, antiderivative size = 81 \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\frac {2 \sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{b c n \log (F)}-\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b c n \log (F)} \] Output:
2*(d+e*(F^(c*(b*x+a)))^n)^(1/2)/b/c/n/ln(F)-2*d^(1/2)*arctanh((d+e*(F^(c*( b*x+a)))^n)^(1/2)/d^(1/2))/b/c/n/ln(F)
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\frac {2 \left (\sqrt {d+e \left (F^{c (a+b x)}\right )^n}-\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )\right )}{b c n \log (F)} \] Input:
Integrate[Sqrt[d + e*(F^(c*(a + b*x)))^n],x]
Output:
(2*(Sqrt[d + e*(F^(c*(a + b*x)))^n] - Sqrt[d]*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]]))/(b*c*n*Log[F])
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2720, 798, 60, 73, 221}
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.
\(\displaystyle \int \sqrt {e \left (F^{c (a+b x)}\right )^n+d} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int F^{-c (a+b x)} \sqrt {e \left (F^{c (a+b x)}\right )^n+d}dF^{c (a+b x)}}{b c \log (F)}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int F^{-c (a+b x)} \sqrt {e \left (F^{c (a+b x)}\right )^n+d}d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {d \int \frac {F^{-c (a+b x)}}{\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}d\left (F^{c (a+b x)}\right )^n+2 \sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{b c n \log (F)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 d \int \frac {1}{\frac {F^{2 c (a+b x)}}{e}-\frac {d}{e}}d\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{e}+2 \sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{b c n \log (F)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {e \left (F^{c (a+b x)}\right )^n+d}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt {d}}\right )}{b c n \log (F)}\) |
Input:
Int[Sqrt[d + e*(F^(c*(a + b*x)))^n],x]
Output:
(2*Sqrt[d + e*(F^(c*(a + b*x)))^n] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]])/(b*c*n*Log[F])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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 = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {2 \sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}-2 \sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}}{\sqrt {d}}\right )}{\ln \left (F \right ) b c n}\) | \(62\) |
default | \(\frac {2 \sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}-2 \sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}}{\sqrt {d}}\right )}{\ln \left (F \right ) b c n}\) | \(62\) |
risch | \(\frac {2 \sqrt {d +e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}}{n c b \ln \left (F \right )}-\frac {2 \sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {d +e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}}{\sqrt {d}}\right )}{\ln \left (F \right ) b c n}\) | \(80\) |
Input:
int((d+e*(F^(c*(b*x+a)))^n)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/ln(F)/b/c/n*(2*(d+e*(F^(c*(b*x+a)))^n)^(1/2)-2*d^(1/2)*arctanh((d+e*(F^( c*(b*x+a)))^n)^(1/2)/d^(1/2)))
Time = 0.08 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.00 \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\left [\frac {\sqrt {d} \log \left (\frac {F^{b c n x + a c n} e - 2 \, \sqrt {F^{b c n x + a c n} e + d} \sqrt {d} + 2 \, d}{F^{b c n x + a c n}}\right ) + 2 \, \sqrt {F^{b c n x + a c n} e + d}}{b c n \log \left (F\right )}, \frac {2 \, {\left (\sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {F^{b c n x + a c n} e + d}}\right ) + \sqrt {F^{b c n x + a c n} e + d}\right )}}{b c n \log \left (F\right )}\right ] \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^(1/2),x, algorithm="fricas")
Output:
[(sqrt(d)*log((F^(b*c*n*x + a*c*n)*e - 2*sqrt(F^(b*c*n*x + a*c*n)*e + d)*s qrt(d) + 2*d)/F^(b*c*n*x + a*c*n)) + 2*sqrt(F^(b*c*n*x + a*c*n)*e + d))/(b *c*n*log(F)), 2*(sqrt(-d)*arctan(sqrt(-d)/sqrt(F^(b*c*n*x + a*c*n)*e + d)) + sqrt(F^(b*c*n*x + a*c*n)*e + d))/(b*c*n*log(F))]
\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\int \sqrt {d + e \left (F^{c \left (a + b x\right )}\right )^{n}}\, dx \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)**(1/2),x)
Output:
Integral(sqrt(d + e*(F**(c*(a + b*x)))**n), x)
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.25 \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\frac {\sqrt {d} \log \left (\frac {\sqrt {F^{b c n x + a c n} e + d} - \sqrt {d}}{\sqrt {F^{b c n x + a c n} e + d} + \sqrt {d}}\right )}{b c n \log \left (F\right )} + \frac {2 \, \sqrt {F^{b c n x + a c n} e + d}}{b c n \log \left (F\right )} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^(1/2),x, algorithm="maxima")
Output:
sqrt(d)*log((sqrt(F^(b*c*n*x + a*c*n)*e + d) - sqrt(d))/(sqrt(F^(b*c*n*x + a*c*n)*e + d) + sqrt(d)))/(b*c*n*log(F)) + 2*sqrt(F^(b*c*n*x + a*c*n)*e + d)/(b*c*n*log(F))
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.81 \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\frac {2 \, {\left (\frac {d \arctan \left (\frac {\sqrt {F^{b c n x + a c n} e + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + \sqrt {F^{b c n x + a c n} e + d}\right )}}{b c n \log \left (F\right )} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)^(1/2),x, algorithm="giac")
Output:
2*(d*arctan(sqrt(F^(b*c*n*x + a*c*n)*e + d)/sqrt(-d))/sqrt(-d) + sqrt(F^(b *c*n*x + a*c*n)*e + d))/(b*c*n*log(F))
Timed out. \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\int \sqrt {d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n} \,d x \] Input:
int((d + e*(F^(c*(a + b*x)))^n)^(1/2),x)
Output:
int((d + e*(F^(c*(a + b*x)))^n)^(1/2), x)
\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \, dx=\int \sqrt {f^{b c n x +a c n} e +d}d x \] Input:
int((d+e*(F^((b*x+a)*c))^n)^(1/2),x)
Output:
int(sqrt(f**(a*c*n + b*c*n*x)*e + d),x)