3.864 \(\int \frac {1}{\sqrt {a+b \cosh (c+d x) \sinh (c+d x)}} \, dx\)

Optimal. Leaf size=96 \[ -\frac {i \sqrt {2} \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{d \sqrt {2 a+b \sinh (2 c+2 d x)}} \]

[Out]

I*(sin(I*c+1/4*Pi+I*d*x)^2)^(1/2)/sin(I*c+1/4*Pi+I*d*x)*EllipticF(cos(I*c+1/4*Pi+I*d*x),2^(1/2)*(b/(2*I*a+b))^
(1/2))*2^(1/2)*((2*a+b*sinh(2*d*x+2*c))/(2*a-I*b))^(1/2)/d/(2*a+b*sinh(2*d*x+2*c))^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2666, 2663, 2661} \[ -\frac {i \sqrt {2} \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}} F\left (\frac {1}{2} \left (2 i c+2 i d x-\frac {\pi }{2}\right )|\frac {2 b}{2 i a+b}\right )}{d \sqrt {2 a+b \sinh (2 c+2 d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Cosh[c + d*x]*Sinh[c + d*x]],x]

[Out]

((-I)*Sqrt[2]*EllipticF[((2*I)*c - Pi/2 + (2*I)*d*x)/2, (2*b)/((2*I)*a + b)]*Sqrt[(2*a + b*Sinh[2*c + 2*d*x])/
(2*a - I*b)])/(d*Sqrt[2*a + b*Sinh[2*c + 2*d*x]])

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

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

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b \cosh (c+d x) \sinh (c+d x)}} \, dx &=\int \frac {1}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}} \, dx\\ &=\frac {\sqrt {\frac {a+\frac {1}{2} b \sinh (2 c+2 d x)}{a-\frac {i b}{2}}} \int \frac {1}{\sqrt {\frac {a}{a-\frac {i b}{2}}+\frac {b \sinh (2 c+2 d x)}{2 \left (a-\frac {i b}{2}\right )}}} \, dx}{\sqrt {a+\frac {1}{2} b \sinh (2 c+2 d x)}}\\ &=-\frac {i \sqrt {2} F\left (\frac {1}{2} \left (2 i c-\frac {\pi }{2}+2 i d x\right )|\frac {2 b}{2 i a+b}\right ) \sqrt {\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}}}{d \sqrt {2 a+b \sinh (2 c+2 d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 90, normalized size = 0.94 \[ \frac {i \sqrt {\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}} F\left (\frac {1}{4} (-4 i c-4 i d x+\pi )|-\frac {2 i b}{2 a-i b}\right )}{d \sqrt {a+\frac {1}{2} b \sinh (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*Cosh[c + d*x]*Sinh[c + d*x]],x]

[Out]

(I*EllipticF[((-4*I)*c + Pi - (4*I)*d*x)/4, ((-2*I)*b)/(2*a - I*b)]*Sqrt[(2*a + b*Sinh[2*(c + d*x)])/(2*a - I*
b)])/(d*Sqrt[a + (b*Sinh[2*(c + d*x)])/2])

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(b*cosh(d*x + c)*sinh(d*x + c) + a), x)

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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+b*cosh(d*x+c)*sinh(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:inde
x.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [A]  time = 0.90, size = 181, normalized size = 1.89 \[ -\frac {2 \left (i b -2 a \right ) \sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}\, \sqrt {\frac {\left (-\sinh \left (2 d x +2 c \right )+i\right ) b}{i b +2 a}}\, \sqrt {\frac {\left (\sinh \left (2 d x +2 c \right )+i\right ) b}{i b -2 a}}\, \EllipticF \left (\sqrt {-\frac {2 a +b \sinh \left (2 d x +2 c \right )}{i b -2 a}}, \sqrt {-\frac {i b -2 a}{i b +2 a}}\right )}{b \cosh \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sinh \left (2 d x +2 c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(I*b-2*a)*(-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2)*((-sinh(2*d*x+2*c)+I)*b/(I*b+2*a))^(1/2)*((sinh(2*d*x+
2*c)+I)*b/(I*b-2*a))^(1/2)*EllipticF((-(2*a+b*sinh(2*d*x+2*c))/(I*b-2*a))^(1/2),(-(I*b-2*a)/(I*b+2*a))^(1/2))/
b/cosh(2*d*x+2*c)/(4*a+2*b*sinh(2*d*x+2*c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*cosh(d*x + c)*sinh(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a + b*sinh(c + d*x)*cosh(c + d*x)), x)

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