3.777 \(\int \frac {1}{\sqrt {-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}} \]

[Out]

-arctanh(1/2*(b^2-c^2)^(1/4)*sinh(x+I*arctan(b,-I*c))*2^(1/2)/(-(b^2-c^2)^(1/2)+cosh(x+I*arctan(b,-I*c))*(b^2-
c^2)^(1/2))^(1/2))*2^(1/2)/(b^2-c^2)^(1/4)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3115, 2649, 204} \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]],x]

[Out]

-((Sqrt[2]*ArcTanh[((b^2 - c^2)^(1/4)*Sinh[x + I*ArcTan[b, (-I)*c]])/(Sqrt[2]*Sqrt[-Sqrt[b^2 - c^2] + Sqrt[b^2
- c^2]*Cosh[x + I*ArcTan[b, (-I)*c]]])])/(b^2 - c^2)^(1/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3115

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Int[1/Sqrt[a +
Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-\sqrt {b^2-c^2}+b \cosh (x)+c \sinh (x)}} \, dx &=\int \frac {1}{\sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}} \, dx\\ &=2 i \operatorname {Subst}\left (\int \frac {1}{-2 \sqrt {b^2-c^2}-x^2} \, dx,x,-\frac {i \sqrt {b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt [4]{b^2-c^2} \sinh \left (x+i \tan ^{-1}(b,-i c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2-c^2}+\sqrt {b^2-c^2} \cosh \left (x+i \tan ^{-1}(b,-i c)\right )}}\right )}{\sqrt [4]{b^2-c^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 31.29, size = 52609, normalized size = 515.77 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[-Sqrt[b^2 - c^2] + b*Cosh[x] + c*Sinh[x]],x]

[Out]

Result too large to show

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 680, normalized size = 6.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

[sqrt(2)*log(-((b^2 + 2*b*c + c^2)*cosh(x)^4 + 4*(b^2 + 2*b*c + c^2)*cosh(x)^3*sinh(x) + 6*(b^2 + 2*b*c + c^2)
*cosh(x)^2*sinh(x)^2 + 4*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^3 + (b^2 + 2*b*c + c^2)*sinh(x)^4 - 2*sqrt(2)*sqr
t(1/2)*(2*(b^2 - c^2)*cosh(x)^2 + 4*(b^2 - c^2)*cosh(x)*sinh(x) + 2*(b^2 - c^2)*sinh(x)^2 + ((b + c)*cosh(x)^3
+ 3*(b + c)*cosh(x)*sinh(x)^2 + (b + c)*sinh(x)^3 + (b - c)*cosh(x) + (3*(b + c)*cosh(x)^2 + b - c)*sinh(x))*
sqrt(b^2 - c^2))*sqrt(((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 - 2*sqrt(b^2 - c^2)*(
cosh(x) + sinh(x)) + b - c)/(cosh(x) + sinh(x)))/(b^2 - c^2)^(1/4) - b^2 + 2*b*c - c^2 + 2*((b + c)*cosh(x)^3
+ 3*(b + c)*cosh(x)*sinh(x)^2 + (b + c)*sinh(x)^3 - (b - c)*cosh(x) + (3*(b + c)*cosh(x)^2 - b + c)*sinh(x))*s
qrt(b^2 - c^2))/((b^2 + 2*b*c + c^2)*cosh(x)^4 + 4*(b^2 + 2*b*c + c^2)*cosh(x)*sinh(x)^3 + (b^2 + 2*b*c + c^2)
*sinh(x)^4 - 2*(b^2 - c^2)*cosh(x)^2 + 2*(3*(b^2 + 2*b*c + c^2)*cosh(x)^2 - b^2 + c^2)*sinh(x)^2 + b^2 - 2*b*c
+ c^2 + 4*((b^2 + 2*b*c + c^2)*cosh(x)^3 - (b^2 - c^2)*cosh(x))*sinh(x)))/(b^2 - c^2)^(1/4), 2*sqrt(2)*sqrt(-
1/sqrt(b^2 - c^2))*arctan(sqrt(2)*sqrt(1/2)*(sqrt(b^2 - c^2)*(cosh(x) + sinh(x)) + b - c)*sqrt(((b + c)*cosh(x
)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 - 2*sqrt(b^2 - c^2)*(cosh(x) + sinh(x)) + b - c)/(cosh(x)
+ sinh(x)))*sqrt(-1/sqrt(b^2 - c^2))/((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(x) + (b + c)*sinh(x)^2 - b +
c))]

________________________________________________________________________________________

giac [B]  time = 1.43, size = 299, normalized size = 2.93 \[ -\frac {2 \, \sqrt {2} {\left (b^{2} - c^{2} - b + c\right )} \sqrt {b - c} \arctan \left (\frac {b^{3} e^{\left (-\frac {1}{2} \, x\right )} - b^{2} c e^{\left (-\frac {1}{2} \, x\right )} - b c^{2} e^{\left (-\frac {1}{2} \, x\right )} + c^{3} e^{\left (-\frac {1}{2} \, x\right )} - b^{2} e^{\left (-\frac {1}{2} \, x\right )} + 2 \, b c e^{\left (-\frac {1}{2} \, x\right )} - c^{2} e^{\left (-\frac {1}{2} \, x\right )}}{\sqrt {-{\left (b^{5} - b^{4} c - 2 \, b^{3} c^{2} + 2 \, b^{2} c^{3} + b c^{4} - c^{5} - 2 \, b^{4} + 4 \, b^{3} c - 4 \, b c^{3} + 2 \, c^{4} + b^{3} - 3 \, b^{2} c + 3 \, b c^{2} - c^{3}\right )} \sqrt {b^{2} - c^{2}}}}\right )}{\sqrt {-{\left (b^{5} - b^{4} c - 2 \, b^{3} c^{2} + 2 \, b^{2} c^{3} + b c^{4} - c^{5} - 2 \, b^{4} + 4 \, b^{3} c - 4 \, b c^{3} + 2 \, c^{4} + b^{3} - 3 \, b^{2} c + 3 \, b c^{2} - c^{3}\right )} \sqrt {b^{2} - c^{2}}} \mathrm {sgn}\left (-\sqrt {b^{2} - c^{2}} e^{x} + b - c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

-2*sqrt(2)*(b^2 - c^2 - b + c)*sqrt(b - c)*arctan((b^3*e^(-1/2*x) - b^2*c*e^(-1/2*x) - b*c^2*e^(-1/2*x) + c^3*
e^(-1/2*x) - b^2*e^(-1/2*x) + 2*b*c*e^(-1/2*x) - c^2*e^(-1/2*x))/sqrt(-(b^5 - b^4*c - 2*b^3*c^2 + 2*b^2*c^3 +
b*c^4 - c^5 - 2*b^4 + 4*b^3*c - 4*b*c^3 + 2*c^4 + b^3 - 3*b^2*c + 3*b*c^2 - c^3)*sqrt(b^2 - c^2)))/(sqrt(-(b^5
- b^4*c - 2*b^3*c^2 + 2*b^2*c^3 + b*c^4 - c^5 - 2*b^4 + 4*b^3*c - 4*b*c^3 + 2*c^4 + b^3 - 3*b^2*c + 3*b*c^2 -
c^3)*sqrt(b^2 - c^2))*sgn(-sqrt(b^2 - c^2)*e^x + b - c))

________________________________________________________________________________________

maple [A]  time = 0.58, size = 129, normalized size = 1.26 \[ \frac {\sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \relax (x )+1\right ) \left (\sinh ^{2}\relax (x )\right )}\, \arctan \left (\frac {\sqrt {\sqrt {b^{2}-c^{2}}\, \left (\sinh \relax (x )+1\right )}\, \cosh \relax (x )}{\sqrt {-\sqrt {b^{2}-c^{2}}\, \left (\sinh \relax (x )+1\right ) \left (\sinh ^{2}\relax (x )\right )}}\right )}{\sqrt {\sqrt {b^{2}-c^{2}}\, \left (\sinh \relax (x )+1\right )}\, \sinh \relax (x ) \sqrt {-\frac {\sinh \relax (x ) b^{2}-\sinh \relax (x ) c^{2}+b^{2}-c^{2}}{\sqrt {b^{2}-c^{2}}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(1/2),x)

[Out]

(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2)/((b^2-c^2)^(1/2)*(sinh(x)+1))^(1/2)*arctan(((b^2-c^2)^(1/2)*(si
nh(x)+1))^(1/2)*cosh(x)/(-(b^2-c^2)^(1/2)*(sinh(x)+1)*sinh(x)^2)^(1/2))/sinh(x)/(-(sinh(x)*b^2-sinh(x)*c^2+b^2
-c^2)/(b^2-c^2)^(1/2))^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cosh \relax (x) + c \sinh \relax (x) - \sqrt {b^{2} - c^{2}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)-(b^2-c^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*cosh(x) + c*sinh(x) - sqrt(b^2 - c^2)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {b\,\mathrm {cosh}\relax (x)-\sqrt {b^2-c^2}+c\,\mathrm {sinh}\relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*cosh(x) - (b^2 - c^2)^(1/2) + c*sinh(x))^(1/2),x)

[Out]

int(1/(b*cosh(x) - (b^2 - c^2)^(1/2) + c*sinh(x))^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cosh {\relax (x )} + c \sinh {\relax (x )} - \sqrt {b^{2} - c^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*cosh(x)+c*sinh(x)-(b**2-c**2)**(1/2))**(1/2),x)

[Out]

Integral(1/sqrt(b*cosh(x) + c*sinh(x) - sqrt(b**2 - c**2)), x)

________________________________________________________________________________________