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

Optimal. Leaf size=57 \[ \frac {2 \sqrt {b e^{c+d x}-a}}{d}-\frac {2 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b e^{c+d x}-a}}{\sqrt {a}}\right )}{d} \]

[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

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2282, 50, 63, 205} \[ \frac {2 \sqrt {b e^{c+d x}-a}}{d}-\frac {2 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b e^{c+d x}-a}}{\sqrt {a}}\right )}{d} \]

Antiderivative was successfully verified.

[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 50

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 63

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 205

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

Rule 2282

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*} \int \sqrt {-a+b e^{c+d x}} \, dx &=\frac {\operatorname {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 \operatorname {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) \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.96 \[ \frac {2 \sqrt {b e^{c+d x}-a}-2 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b e^{c+d x}-a}}{\sqrt {a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-a + b*E^(c + d*x)],x]

[Out]

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

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fricas [A]  time = 0.41, size = 117, normalized size = 2.05 \[ \left [\frac {\sqrt {-a} \log \left ({\left (b e^{\left (d x + c\right )} - 2 \, \sqrt {b e^{\left (d x + c\right )} - a} \sqrt {-a} - 2 \, a\right )} e^{\left (-d x - c\right )}\right ) + 2 \, \sqrt {b e^{\left (d x + c\right )} - a}}{d}, -\frac {2 \, {\left (\sqrt {a} \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right ) - \sqrt {b e^{\left (d x + c\right )} - a}\right )}}{d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.22, size = 45, normalized size = 0.79 \[ -\frac {2 \, {\left (\sqrt {a} \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right ) - \sqrt {b e^{\left (d x + c\right )} - a}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 48, normalized size = 0.84 \[ -\frac {2 \sqrt {a}\, \arctan \left (\frac {\sqrt {b \,{\mathrm e}^{d x +c}-a}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {b \,{\mathrm e}^{d x +c}-a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.19, size = 47, normalized size = 0.82 \[ -\frac {2 \, \sqrt {a} \arctan \left (\frac {\sqrt {b e^{\left (d x + c\right )} - a}}{\sqrt {a}}\right )}{d} + \frac {2 \, \sqrt {b e^{\left (d x + c\right )} - a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.60, size = 47, normalized size = 0.82 \[ \frac {2\,\sqrt {b\,{\mathrm {e}}^{c+d\,x}-a}}{d}-\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a}}{\sqrt {a}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- a + b e^{c + d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-a + b*exp(c + d*x)), x)

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