\(\int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx\) [699]

Optimal result

Integrand size = 17, antiderivative size = 34 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {-a+b e^{c+d x}}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[Out]

2*arctan((-a+b*exp(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2320, 65, 211} \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {b e^{c+d x}-a}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[In]

Int[1/Sqrt[-a + b*E^(c + d*x)],x]

[Out]

(2*ArcTan[Sqrt[-a + b*E^(c + d*x)]/Sqrt[a]])/(Sqrt[a]*d)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \sqrt {-a+b x}} \, dx,x,e^{c+d x}\right )}{d} \\ & = \frac {2 \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b e^{c+d x}}\right )}{b d} \\ & = \frac {2 \tan ^{-1}\left (\frac {\sqrt {-a+b e^{c+d x}}}{\sqrt {a}}\right )}{\sqrt {a} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {-a+b e^{c+d x}}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[In]

Integrate[1/Sqrt[-a + b*E^(c + d*x)],x]

[Out]

(2*ArcTan[Sqrt[-a + b*E^(c + d*x)]/Sqrt[a]])/(Sqrt[a]*d)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {2 \arctan \left (\frac {\sqrt {-a +b \,{\mathrm e}^{d x +c}}}{\sqrt {a}}\right )}{d \sqrt {a}}\)	\(28\)
default	\(\frac {2 \arctan \left (\frac {\sqrt {-a +b \,{\mathrm e}^{d x +c}}}{\sqrt {a}}\right )}{d \sqrt {a}}\)	\(28\)

[In]

int(1/(-a+b*exp(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*arctan((-a+b*exp(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\left [-\frac {\sqrt {-a} \log \left ({\left (b e^{\left (d x + c\right )} - 2 \, \sqrt {b e^{\left (d x + c\right )} - a} \sqrt {-a} - 2 \, a\right )} e^{\left (-d x - c\right )}\right )}{a d}, \frac {2 \, \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right )}{\sqrt {a} d}\right ] \]

[In]

integrate(1/(-a+b*exp(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-sqrt(-a)*log((b*e^(d*x + c) - 2*sqrt(b*e^(d*x + c) - a)*sqrt(-a) - 2*a)*e^(-d*x - c))/(a*d), 2*arctan(sqrt(b
*e^(d*x + c) - a)/sqrt(a))/(sqrt(a)*d)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (27) = 54\).

Time = 0.60 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\begin {cases} \frac {\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt {- a + b e^{c} e^{d x}}}{\sqrt {a}} \right )}}{\sqrt {a}} & \text {for}\: b e^{c} \neq 0 \\\frac {\log {\left (e^{d x} \right )}}{\sqrt {- a}} & \text {otherwise} \end {cases}}{d} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {- a + b e^{c}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(-a+b*exp(d*x+c))**(1/2),x)

[Out]

Piecewise((Piecewise((2*atan(sqrt(-a + b*exp(c)*exp(d*x))/sqrt(a))/sqrt(a), Ne(b*exp(c), 0)), (log(exp(d*x))/s
qrt(-a), True))/d, Ne(d, 0)), (x/sqrt(-a + b*exp(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[In]

integrate(1/(-a+b*exp(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2*arctan(sqrt(b*e^(d*x + c) - a)/sqrt(a))/(sqrt(a)*d)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[In]

integrate(1/(-a+b*exp(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2*arctan(sqrt(b*e^(d*x + c) - a)/sqrt(a))/(sqrt(a)*d)

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\sqrt {-a+b e^{c+d x}}} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a}}{\sqrt {-a}}\right )}{\sqrt {-a}\,d} \]

[In]

int(1/(b*exp(c + d*x) - a)^(1/2),x)

[Out]

-(2*atanh((b*exp(d*x)*exp(c) - a)^(1/2)/(-a)^(1/2)))/((-a)^(1/2)*d)