3.53 $$\int \frac{\sqrt{\cosh (a+b x)}}{\sqrt{\sinh (a+b x)}} \, dx$$

Optimal. Leaf size=54 $\frac{\tanh ^{-1}\left (\frac{\sqrt{\cosh (a+b x)}}{\sqrt{\sinh (a+b x)}}\right )}{b}-\frac{\tan ^{-1}\left (\frac{\sqrt{\cosh (a+b x)}}{\sqrt{\sinh (a+b x)}}\right )}{b}$

[Out]

-(ArcTan[Sqrt[Cosh[a + b*x]]/Sqrt[Sinh[a + b*x]]]/b) + ArcTanh[Sqrt[Cosh[a + b*x]]/Sqrt[Sinh[a + b*x]]]/b

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Rubi [A]  time = 0.0381403, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.19, Rules used = {2575, 298, 203, 206} $\frac{\tanh ^{-1}\left (\frac{\sqrt{\cosh (a+b x)}}{\sqrt{\sinh (a+b x)}}\right )}{b}-\frac{\tan ^{-1}\left (\frac{\sqrt{\cosh (a+b x)}}{\sqrt{\sinh (a+b x)}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(ArcTan[Sqrt[Cosh[a + b*x]]/Sqrt[Sinh[a + b*x]]]/b) + ArcTanh[Sqrt[Cosh[a + b*x]]/Sqrt[Sinh[a + b*x]]]/b

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{k = Denomina
tor[m]}, -Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

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 206

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

Rubi steps

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

Mathematica [C]  time = 0.0231303, size = 57, normalized size = 1.06 $\frac{2 \sqrt{\sinh (a+b x)} \sqrt [4]{\cosh ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{4};\frac{5}{4};-\sinh ^2(a+b x)\right )}{b \sqrt{\cosh (a+b x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(Cosh[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, -Sinh[a + b*x]^2]*Sqrt[Sinh[a + b*x]])/(b*Sqrt[Cos
h[a + b*x]])

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Maple [F]  time = 0.09, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{\cosh \left ( bx+a \right ) }{\frac{1}{\sqrt{\sinh \left ( bx+a \right ) }}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.47819, size = 421, normalized size = 7.8 \begin{align*} \frac{2 \, \arctan \left (-\cosh \left (b x + a\right )^{2} + 2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \sqrt{\cosh \left (b x + a\right )} \sqrt{\sinh \left (b x + a\right )} - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - \sinh \left (b x + a\right )^{2}\right ) - \log \left (-\cosh \left (b x + a\right )^{2} + 2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \sqrt{\cosh \left (b x + a\right )} \sqrt{\sinh \left (b x + a\right )} - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - \sinh \left (b x + a\right )^{2}\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*arctan(-cosh(b*x + a)^2 + 2*(cosh(b*x + a) + sinh(b*x + a))*sqrt(cosh(b*x + a))*sqrt(sinh(b*x + a)) - 2
*cosh(b*x + a)*sinh(b*x + a) - sinh(b*x + a)^2) - log(-cosh(b*x + a)^2 + 2*(cosh(b*x + a) + sinh(b*x + a))*sqr
t(cosh(b*x + a))*sqrt(sinh(b*x + a)) - 2*cosh(b*x + a)*sinh(b*x + a) - sinh(b*x + a)^2))/b

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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