3.84 \(\int \frac {1}{\sqrt {a-a \text {sech}(c+d x)}} \, dx\)
Optimal. Leaf size=87 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a-a \text {sech}(c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {2} \sqrt {a-a \text {sech}(c+d x)}}\right )}{\sqrt {a} d} \]
[Out]
2*arctanh(a^(1/2)*tanh(d*x+c)/(a-a*sech(d*x+c))^(1/2))/d/a^(1/2)-arctanh(1/2*a^(1/2)*tanh(d*x+c)*2^(1/2)/(a-a*
sech(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)
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Rubi [A] time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used =
{3776, 3774, 203, 3795} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a-a \text {sech}(c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {2} \sqrt {a-a \text {sech}(c+d x)}}\right )}{\sqrt {a} d} \]
Antiderivative was successfully verified.
[In]
Int[1/Sqrt[a - a*Sech[c + d*x]],x]
[Out]
(2*ArcTanh[(Sqrt[a]*Tanh[c + d*x])/Sqrt[a - a*Sech[c + d*x]]])/(Sqrt[a]*d) - (Sqrt[2]*ArcTanh[(Sqrt[a]*Tanh[c
+ d*x])/(Sqrt[2]*Sqrt[a - a*Sech[c + d*x]])])/(Sqrt[a]*d)
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3774
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 3776
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[1/a, Int[Sqrt[a + b*Csc[c + d*x]], x], x]
- Dist[b/a, Int[Csc[c + d*x]/Sqrt[a + b*Csc[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 3795
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {a-a \text {sech}(c+d x)}} \, dx &=\frac {\int \sqrt {a-a \text {sech}(c+d x)} \, dx}{a}+\int \frac {\text {sech}(c+d x)}{\sqrt {a-a \text {sech}(c+d x)}} \, dx\\ &=-\frac {(2 i) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\frac {i a \tanh (c+d x)}{\sqrt {a-a \text {sech}(c+d x)}}\right )}{d}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,\frac {i a \tanh (c+d x)}{\sqrt {a-a \text {sech}(c+d x)}}\right )}{d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {a-a \text {sech}(c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {2} \sqrt {a-a \text {sech}(c+d x)}}\right )}{\sqrt {a} d}\\ \end {align*}
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Mathematica [A] time = 2.23, size = 118, normalized size = 1.36 \[ \frac {\left (e^{c+d x}-1\right ) \left (\sqrt {2} \sinh ^{-1}\left (e^{c+d x}\right )-2 \tanh ^{-1}\left (\frac {e^{c+d x}+1}{\sqrt {2} \sqrt {e^{2 (c+d x)}+1}}\right )+\sqrt {2} \tanh ^{-1}\left (\sqrt {e^{2 (c+d x)}+1}\right )\right )}{\sqrt {2} d \sqrt {e^{2 (c+d x)}+1} \sqrt {a-a \text {sech}(c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/Sqrt[a - a*Sech[c + d*x]],x]
[Out]
((-1 + E^(c + d*x))*(Sqrt[2]*ArcSinh[E^(c + d*x)] - 2*ArcTanh[(1 + E^(c + d*x))/(Sqrt[2]*Sqrt[1 + E^(2*(c + d*
x))])] + Sqrt[2]*ArcTanh[Sqrt[1 + E^(2*(c + d*x))]]))/(Sqrt[2]*d*Sqrt[1 + E^(2*(c + d*x))]*Sqrt[a - a*Sech[c +
d*x]])
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fricas [B] time = 0.42, size = 871, normalized size = 10.01 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-a*sech(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/2*(sqrt(2)*sqrt(a)*log(-(3*cosh(d*x + c)^2 + 2*(3*cosh(d*x + c) + 1)*sinh(d*x + c) + 3*sinh(d*x + c)^2 - 2*s
qrt(2)*(cosh(d*x + c)^3 + (3*cosh(d*x + c) + 1)*sinh(d*x + c)^2 + sinh(d*x + c)^3 + cosh(d*x + c)^2 + (3*cosh(
d*x + c)^2 + 2*cosh(d*x + c) + 1)*sinh(d*x + c) + cosh(d*x + c) + 1)*sqrt(a/(cosh(d*x + c)^2 + 2*cosh(d*x + c)
*sinh(d*x + c) + sinh(d*x + c)^2 + 1))/sqrt(a) + 2*cosh(d*x + c) + 3)/(cosh(d*x + c)^2 + 2*(cosh(d*x + c) - 1)
*sinh(d*x + c) + sinh(d*x + c)^2 - 2*cosh(d*x + c) + 1)) + sqrt(a)*log((a*cosh(d*x + c)^4 + a*sinh(d*x + c)^4
+ 3*a*cosh(d*x + c)^3 + (4*a*cosh(d*x + c) + 3*a)*sinh(d*x + c)^3 + 5*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c)^2
+ 9*a*cosh(d*x + c) + 5*a)*sinh(d*x + c)^2 + (cosh(d*x + c)^5 + (5*cosh(d*x + c) + 3)*sinh(d*x + c)^4 + sinh(
d*x + c)^5 + 3*cosh(d*x + c)^4 + (10*cosh(d*x + c)^2 + 12*cosh(d*x + c) + 5)*sinh(d*x + c)^3 + 5*cosh(d*x + c)
^3 + (10*cosh(d*x + c)^3 + 18*cosh(d*x + c)^2 + 15*cosh(d*x + c) + 7)*sinh(d*x + c)^2 + 7*cosh(d*x + c)^2 + (5
*cosh(d*x + c)^4 + 12*cosh(d*x + c)^3 + 15*cosh(d*x + c)^2 + 14*cosh(d*x + c) + 4)*sinh(d*x + c) + 4*cosh(d*x
+ c) + 4)*sqrt(a)*sqrt(a/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)) + 4*a*cosh(d
*x + c) + (4*a*cosh(d*x + c)^3 + 9*a*cosh(d*x + c)^2 + 10*a*cosh(d*x + c) + 4*a)*sinh(d*x + c) + 4*a)/(cosh(d*
x + c)^3 + 3*cosh(d*x + c)^2*sinh(d*x + c) + 3*cosh(d*x + c)*sinh(d*x + c)^2 + sinh(d*x + c)^3)) + sqrt(a)*log
(-(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + (cosh(d*x + c)^3 + (3*cosh(d*x + c) - 1)*sinh(d*x + c)^2 + sinh(d*x
+ c)^3 - cosh(d*x + c)^2 + (3*cosh(d*x + c)^2 - 2*cosh(d*x + c) + 1)*sinh(d*x + c) + cosh(d*x + c) - 1)*sqrt(
a)*sqrt(a/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)) - a*cosh(d*x + c) + (2*a*co
sh(d*x + c) - a)*sinh(d*x + c) + a)/(cosh(d*x + c) + sinh(d*x + c))))/(a*d)
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-a*sech(d*x+c))^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(exp(d*x+c)-1)]Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replac
ing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution v
ariable should perhaps be purged.Error: Bad Argument Type
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maple [F] time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a -a \,\mathrm {sech}\left (d x +c \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a-a*sech(d*x+c))^(1/2),x)
[Out]
int(1/(a-a*sech(d*x+c))^(1/2),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-a \operatorname {sech}\left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-a*sech(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(1/sqrt(-a*sech(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a-\frac {a}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a - a/cosh(c + d*x))^(1/2),x)
[Out]
int(1/(a - a/cosh(c + d*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- a \operatorname {sech}{\left (c + d x \right )} + a}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-a*sech(d*x+c))**(1/2),x)
[Out]
Integral(1/sqrt(-a*sech(c + d*x) + a), x)
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