\(\int \sqrt {-a+b e^{c+d x}} \, dx\) [701]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 17, antiderivative size = 57 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=\frac {2 \sqrt {-a+b e^{c+d x}}}{d}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {-a+b e^{c+d x}}}{\sqrt {a}}\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2320, 52, 65, 211} \[ \int \sqrt {-a+b e^{c+d x}} \, dx=\frac {2 \sqrt {b e^{c+d x}-a}}{d}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b e^{c+d x}-a}}{\sqrt {a}}\right )}{d} \]

[In]

Int[Sqrt[-a + b*E^(c + d*x)],x]

[Out]

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

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=\frac {2 \left (\sqrt {-a+b e^{c+d x}}-\sqrt {a} \arctan \left (\frac {\sqrt {-a+b e^{c+d x}}}{\sqrt {a}}\right )\right )}{d} \]

[In]

Integrate[Sqrt[-a + b*E^(c + d*x)],x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81

method result size
derivativedivides \(\frac {2 \sqrt {-a +b \,{\mathrm e}^{d x +c}}-2 \sqrt {a}\, \arctan \left (\frac {\sqrt {-a +b \,{\mathrm e}^{d x +c}}}{\sqrt {a}}\right )}{d}\) \(46\)
default \(\frac {2 \sqrt {-a +b \,{\mathrm e}^{d x +c}}-2 \sqrt {a}\, \arctan \left (\frac {\sqrt {-a +b \,{\mathrm e}^{d x +c}}}{\sqrt {a}}\right )}{d}\) \(46\)
risch \(-\frac {2 \left (a -b \,{\mathrm e}^{d x +c}\right )}{d \sqrt {-a +b \,{\mathrm e}^{d x +c}}}-\frac {2 \arctan \left (\frac {\sqrt {-a +b \,{\mathrm e}^{d x +c}}}{\sqrt {a}}\right ) \sqrt {a}}{d}\) \(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.05 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=\left [\frac {\sqrt {-a} \log \left ({\left (b e^{\left (d x + c\right )} - 2 \, \sqrt {b e^{\left (d x + c\right )} - a} \sqrt {-a} - 2 \, a\right )} e^{\left (-d x - c\right )}\right ) + 2 \, \sqrt {b e^{\left (d x + c\right )} - a}}{d}, -\frac {2 \, {\left (\sqrt {a} \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right ) - \sqrt {b e^{\left (d x + c\right )} - a}\right )}}{d}\right ] \]

[In]

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

Sympy [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=\begin {cases} \frac {\begin {cases} - 2 \sqrt {a} \operatorname {atan}{\left (\frac {\sqrt {- a + b e^{c} e^{d x}}}{\sqrt {a}} \right )} + 2 \sqrt {- a + b e^{c} e^{d x}} & \text {for}\: b e^{c} \neq 0 \\\sqrt {- a} \log {\left (e^{d x} \right )} & \text {otherwise} \end {cases}}{d} & \text {for}\: d \neq 0 \\x \sqrt {- a + b e^{c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=-\frac {2 \, \sqrt {a} \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right )}{d} + \frac {2 \, \sqrt {b e^{\left (d x + c\right )} - a}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=-\frac {2 \, {\left (\sqrt {a} \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right ) - \sqrt {b e^{\left (d x + c\right )} - a}\right )}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \sqrt {-a+b e^{c+d x}} \, dx=\frac {2\,\sqrt {b\,{\mathrm {e}}^{c+d\,x}-a}}{d}-\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a}}{\sqrt {a}}\right )}{d} \]

[In]

int((b*exp(c + d*x) - a)^(1/2),x)

[Out]

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