3.324 \(\int \frac{\text{sech}^2(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

Optimal. Leaf size=60 \[ \frac{\tanh (c+d x)}{d (a-b)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{3/2}} \]

[Out]

-((b*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(3/2)*d)) + Tanh[c + d*x]/((a - b)*d)

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Rubi [A]  time = 0.0788262, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3191, 388, 208} \[ \frac{\tanh (c+d x)}{d (a-b)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^2/(a + b*Sinh[c + d*x]^2),x]

[Out]

-((b*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(3/2)*d)) + Tanh[c + d*x]/((a - b)*d)

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(c+d x)}{a+b \sinh ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{a-(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\tanh (c+d x)}{(a-b) d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\tanh (c+d x)\right )}{(a-b) d}\\ &=-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} (a-b)^{3/2} d}+\frac{\tanh (c+d x)}{(a-b) d}\\ \end{align*}

Mathematica [A]  time = 0.154095, size = 60, normalized size = 1. \[ \frac{\tanh (c+d x)}{d (a-b)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} d (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^2/(a + b*Sinh[c + d*x]^2),x]

[Out]

-((b*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)^(3/2)*d)) + Tanh[c + d*x]/((a - b)*d)

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Maple [B]  time = 0.06, size = 335, normalized size = 5.6 \begin{align*} -{\frac{b}{d \left ( a-b \right ) }{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}}+{\frac{{b}^{2}}{d \left ( a-b \right ) }{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{-b \left ( a-b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}}+{\frac{b}{d \left ( a-b \right ) }\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}}+{\frac{{b}^{2}}{d \left ( a-b \right ) }\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{-b \left ( a-b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}}+2\,{\frac{\tanh \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*b/(a-b)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)
^(1/2))+1/d*b^2/(a-b)/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*
(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/d*b/(a-b)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c
)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))+1/d*b^2/(a-b)/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*ar
ctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))+2/d/(a-b)*tanh(1/2*d*x+1/2*c)/(tanh(1/2*d*x+1
/2*c)^2+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.63066, size = 1770, normalized size = 29.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)*sqrt(a^2 - a*b)*log((b^2*
cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 + 2*(2*a*b - b^2)*cosh(d*x + c)^2
+ 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b - b^2)*sinh(d*x + c)^2 + 8*a^2 - 8*a*b + b^2 + 4*(b^2*cosh(d*x + c)^3 + (2*
a*b - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x
+ c)^2 + 2*a - b)*sqrt(a^2 - a*b))/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4
+ 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + (2*
a - b)*cosh(d*x + c))*sinh(d*x + c) + b)) + 4*a^2 - 4*a*b)/((a^3 - 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 + 2*(a^3
- 2*a^2*b + a*b^2)*d*cosh(d*x + c)*sinh(d*x + c) + (a^3 - 2*a^2*b + a*b^2)*d*sinh(d*x + c)^2 + (a^3 - 2*a^2*b
+ a*b^2)*d), ((b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)*sqrt(-a^2 + a*b)*
arctan(-1/2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*a - b)*sqrt(-a^2 + a*
b)/(a^2 - a*b)) - 2*a^2 + 2*a*b)/((a^3 - 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 + 2*(a^3 - 2*a^2*b + a*b^2)*d*cosh
(d*x + c)*sinh(d*x + c) + (a^3 - 2*a^2*b + a*b^2)*d*sinh(d*x + c)^2 + (a^3 - 2*a^2*b + a*b^2)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (c + d x \right )}}{a + b \sinh ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(c + d*x)**2/(a + b*sinh(c + d*x)**2), x)

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Giac [A]  time = 1.22527, size = 111, normalized size = 1.85 \begin{align*} -\frac{b \arctan \left (\frac{b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b}{\left (a d - b d\right )}} - \frac{2}{{\left (a d - b d\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2/(a+b*sinh(d*x+c)^2),x, algorithm="giac")

[Out]

-b*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*(a*d - b*d)) - 2/((a*d - b*d)*
(e^(2*d*x + 2*c) + 1))