3.31 \(\int \frac{\text{sech}^4(x)}{a+b \cosh ^2(x)} \, dx\)
Optimal. Leaf size=55 \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{a^{5/2} \sqrt{a+b}}+\frac{(a-b) \tanh (x)}{a^2}-\frac{\tanh ^3(x)}{3 a} \]
[Out]
(b^2*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(a^(5/2)*Sqrt[a + b]) + ((a - b)*Tanh[x])/a^2 - Tanh[x]^3/(3*a)
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Rubi [A] time = 0.0991718, antiderivative size = 55, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3187, 461, 208} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{a^{5/2} \sqrt{a+b}}+\frac{(a-b) \tanh (x)}{a^2}-\frac{\tanh ^3(x)}{3 a} \]
Antiderivative was successfully verified.
[In]
Int[Sech[x]^4/(a + b*Cosh[x]^2),x]
[Out]
(b^2*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(a^(5/2)*Sqrt[a + b]) + ((a - b)*Tanh[x])/a^2 - Tanh[x]^3/(3*a)
Rule 3187
Int[sin[(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^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Rule 461
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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}^4(x)}{a+b \cosh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^4 \left (a-(a+b) x^2\right )} \, dx,x,\coth (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{a x^4}+\frac{-a+b}{a^2 x^2}+\frac{b^2}{a^2 \left (a-(a+b) x^2\right )}\right ) \, dx,x,\coth (x)\right )\\ &=\frac{(a-b) \tanh (x)}{a^2}-\frac{\tanh ^3(x)}{3 a}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a-(a+b) x^2} \, dx,x,\coth (x)\right )}{a^2}\\ &=\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{a^{5/2} \sqrt{a+b}}+\frac{(a-b) \tanh (x)}{a^2}-\frac{\tanh ^3(x)}{3 a}\\ \end{align*}
Mathematica [A] time = 0.143477, size = 55, normalized size = 1. \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{a^{5/2} \sqrt{a+b}}+\frac{\tanh (x) \left (a \text{sech}^2(x)+2 a-3 b\right )}{3 a^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Sech[x]^4/(a + b*Cosh[x]^2),x]
[Out]
(b^2*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(a^(5/2)*Sqrt[a + b]) + ((2*a - 3*b + a*Sech[x]^2)*Tanh[x])/(3*a^
2)
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Maple [B] time = 0.046, size = 139, normalized size = 2.5 \begin{align*}{\frac{{b}^{2}}{2}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{{b}^{2}}{2}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\sqrt{a}\tanh \left ( x/2 \right ) -\sqrt{a+b} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b}}}}-2\,{\frac{ \left ( -a+b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{5}+ \left ( -2/3\,a+2\,b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}+ \left ( -a+b \right ) \tanh \left ( x/2 \right ) }{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sech(x)^4/(a+b*cosh(x)^2),x)
[Out]
1/2*b^2/a^(5/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2+2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))-1/2*b^2/a^(5/2)/(a
+b)^(1/2)*ln(-(a+b)^(1/2)*tanh(1/2*x)^2+2*a^(1/2)*tanh(1/2*x)-(a+b)^(1/2))-2/a^2*((-a+b)*tanh(1/2*x)^5+(-2/3*a
+2*b)*tanh(1/2*x)^3+(-a+b)*tanh(1/2*x))/(tanh(1/2*x)^2+1)^3
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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(x)^4/(a+b*cosh(x)^2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.40596, size = 3634, normalized size = 66.07 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)^4/(a+b*cosh(x)^2),x, algorithm="fricas")
[Out]
[1/6*(12*(a^2*b + a*b^2)*cosh(x)^4 + 48*(a^2*b + a*b^2)*cosh(x)*sinh(x)^3 + 12*(a^2*b + a*b^2)*sinh(x)^4 - 8*a
^3 + 4*a^2*b + 12*a*b^2 - 24*(a^3 - a*b^2)*cosh(x)^2 - 24*(a^3 - a*b^2 - 3*(a^2*b + a*b^2)*cosh(x)^2)*sinh(x)^
2 + 3*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 + 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 + b^2)*s
inh(x)^4 + 3*b^2*cosh(x)^2 + 4*(5*b^2*cosh(x)^3 + 3*b^2*cosh(x))*sinh(x)^3 + 3*(5*b^2*cosh(x)^4 + 6*b^2*cosh(x
)^2 + b^2)*sinh(x)^2 + b^2 + 6*(b^2*cosh(x)^5 + 2*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x))*sqrt(a^2 + a*b)*log((b
^2*cosh(x)^4 + 4*b^2*cosh(x)*sinh(x)^3 + b^2*sinh(x)^4 + 2*(2*a*b + b^2)*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + 2*a*
b + b^2)*sinh(x)^2 + 8*a^2 + 8*a*b + b^2 + 4*(b^2*cosh(x)^3 + (2*a*b + b^2)*cosh(x))*sinh(x) - 4*(b*cosh(x)^2
+ 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sqrt(a^2 + a*b))/(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(
x)^4 + 2*(2*a + b)*cosh(x)^2 + 2*(3*b*cosh(x)^2 + 2*a + b)*sinh(x)^2 + 4*(b*cosh(x)^3 + (2*a + b)*cosh(x))*sin
h(x) + b)) + 48*((a^2*b + a*b^2)*cosh(x)^3 - (a^3 - a*b^2)*cosh(x))*sinh(x))/((a^4 + a^3*b)*cosh(x)^6 + 6*(a^4
+ a^3*b)*cosh(x)*sinh(x)^5 + (a^4 + a^3*b)*sinh(x)^6 + 3*(a^4 + a^3*b)*cosh(x)^4 + 3*(a^4 + a^3*b + 5*(a^4 +
a^3*b)*cosh(x)^2)*sinh(x)^4 + a^4 + a^3*b + 4*(5*(a^4 + a^3*b)*cosh(x)^3 + 3*(a^4 + a^3*b)*cosh(x))*sinh(x)^3
+ 3*(a^4 + a^3*b)*cosh(x)^2 + 3*(5*(a^4 + a^3*b)*cosh(x)^4 + a^4 + a^3*b + 6*(a^4 + a^3*b)*cosh(x)^2)*sinh(x)^
2 + 6*((a^4 + a^3*b)*cosh(x)^5 + 2*(a^4 + a^3*b)*cosh(x)^3 + (a^4 + a^3*b)*cosh(x))*sinh(x)), 1/3*(6*(a^2*b +
a*b^2)*cosh(x)^4 + 24*(a^2*b + a*b^2)*cosh(x)*sinh(x)^3 + 6*(a^2*b + a*b^2)*sinh(x)^4 - 4*a^3 + 2*a^2*b + 6*a*
b^2 - 12*(a^3 - a*b^2)*cosh(x)^2 - 12*(a^3 - a*b^2 - 3*(a^2*b + a*b^2)*cosh(x)^2)*sinh(x)^2 + 3*(b^2*cosh(x)^6
+ 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 + 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 + b^2)*sinh(x)^4 + 3*b^2*cos
h(x)^2 + 4*(5*b^2*cosh(x)^3 + 3*b^2*cosh(x))*sinh(x)^3 + 3*(5*b^2*cosh(x)^4 + 6*b^2*cosh(x)^2 + b^2)*sinh(x)^2
+ b^2 + 6*(b^2*cosh(x)^5 + 2*b^2*cosh(x)^3 + b^2*cosh(x))*sinh(x))*sqrt(-a^2 - a*b)*arctan(1/2*(b*cosh(x)^2 +
2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sqrt(-a^2 - a*b)/(a^2 + a*b)) + 24*((a^2*b + a*b^2)*cosh(x)^3 -
(a^3 - a*b^2)*cosh(x))*sinh(x))/((a^4 + a^3*b)*cosh(x)^6 + 6*(a^4 + a^3*b)*cosh(x)*sinh(x)^5 + (a^4 + a^3*b)*s
inh(x)^6 + 3*(a^4 + a^3*b)*cosh(x)^4 + 3*(a^4 + a^3*b + 5*(a^4 + a^3*b)*cosh(x)^2)*sinh(x)^4 + a^4 + a^3*b + 4
*(5*(a^4 + a^3*b)*cosh(x)^3 + 3*(a^4 + a^3*b)*cosh(x))*sinh(x)^3 + 3*(a^4 + a^3*b)*cosh(x)^2 + 3*(5*(a^4 + a^3
*b)*cosh(x)^4 + a^4 + a^3*b + 6*(a^4 + a^3*b)*cosh(x)^2)*sinh(x)^2 + 6*((a^4 + a^3*b)*cosh(x)^5 + 2*(a^4 + a^3
*b)*cosh(x)^3 + (a^4 + a^3*b)*cosh(x))*sinh(x))]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)**4/(a+b*cosh(x)**2),x)
[Out]
Timed out
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Giac [A] time = 1.38746, size = 117, normalized size = 2.13 \begin{align*} \frac{b^{2} \arctan \left (\frac{b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt{-a^{2} - a b}}\right )}{\sqrt{-a^{2} - a b} a^{2}} + \frac{2 \,{\left (3 \, b e^{\left (4 \, x\right )} - 6 \, a e^{\left (2 \, x\right )} + 6 \, b e^{\left (2 \, x\right )} - 2 \, a + 3 \, b\right )}}{3 \, a^{2}{\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)^4/(a+b*cosh(x)^2),x, algorithm="giac")
[Out]
b^2*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*a^2) + 2/3*(3*b*e^(4*x) - 6*a*e^(2*x)
+ 6*b*e^(2*x) - 2*a + 3*b)/(a^2*(e^(2*x) + 1)^3)