### 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)*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]])

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Rubi [A]  time = 0.0784631, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, 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 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]

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 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]

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*}

Mathematica [A]  time = 0.134803, 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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Maple [A]  time = 0.362, size = 181, normalized size = 1.9 \begin{align*} -2\,{\frac{ib-2\,a}{b\cosh \left ( 2\,dx+2\,c \right ) \sqrt{4\,a+2\,b\sinh \left ( 2\,dx+2\,c \right ) }d}\sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}}\sqrt{{\frac{ \left ( -\sinh \left ( 2\,dx+2\,c \right ) +i \right ) b}{ib+2\,a}}}\sqrt{{\frac{ \left ( i+\sinh \left ( 2\,dx+2\,c \right ) \right ) b}{ib-2\,a}}}{\it EllipticF} \left ( \sqrt{-{\frac{2\,a+b\sinh \left ( 2\,dx+2\,c \right ) }{ib-2\,a}}},\sqrt{-{\frac{ib-2\,a}{ib+2\,a}}} \right ) } \end{align*}

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)*((I+sinh(2*d*
x+2*c))*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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a}}\,{d x} \end{align*}

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]

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