3.1.31 \(\int \frac {\text {sech}^4(x)}{a+b \cosh ^2(x)} \, dx\) [31]

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(a^(1/2)*tanh(x)/(a+b)^(1/2))/a^(5/2)/(a+b)^(1/2)+(a-b)*tanh(x)/a^2-1/3*tanh(x)^3/a

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Rubi [A]
time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3266, 472, 214} \begin {gather*} \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 {gather*}

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 214

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

Rule 472

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 3266

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]

Rubi steps

\begin {align*} \int \frac {\text {sech}^4(x)}{a+b \cosh ^2(x)} \, dx &=\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^4 \left (a-(a+b) x^2\right )} \, dx,x,\coth (x)\right )\\ &=\text {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 \text {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*}

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Mathematica [A]
time = 0.10, size = 55, normalized size = 1.00 \begin {gather*} \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{a^{5/2} \sqrt {a+b}}+\frac {\left (2 a-3 b+a \text {sech}^2(x)\right ) \tanh (x)}{3 a^2} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(138\) vs. \(2(45)=90\).
time = 0.74, size = 139, normalized size = 2.53

method	result	size
default	\(-\frac {2 b^{2} \left (-\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) \sqrt {a}+\sqrt {a +b}\right )}{4 \sqrt {a}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) \sqrt {a}+\sqrt {a +b}\right )}{4 \sqrt {a}\, \sqrt {a +b}}\right )}{a^{2}}-\frac {2 \left (\left (-a +b \right ) \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )+\left (-\frac {2 a}{3}+2 b \right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+\left (-a +b \right ) \tanh \left (\frac {x}{2}\right )\right )}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}\)	\(139\)
risch	\(-\frac {2 \left (-3 \,{\mathrm e}^{4 x} b +6 \,{\mathrm e}^{2 x} a -6 b \,{\mathrm e}^{2 x}+2 a -3 b \right )}{3 \left (1+{\mathrm e}^{2 x}\right )^{3} a^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {2 a \sqrt {a^{2}+a b}+b \sqrt {a^{2}+a b}-2 a^{2}-2 a b}{b \sqrt {a^{2}+a b}}\right )}{2 \sqrt {a^{2}+a b}\, a^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {2 a \sqrt {a^{2}+a b}+b \sqrt {a^{2}+a b}+2 a^{2}+2 a b}{b \sqrt {a^{2}+a b}}\right )}{2 \sqrt {a^{2}+a b}\, a^{2}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(a+b*cosh(x)^2),x,method=_RETURNVERBOSE)

[Out]

-2*b^2/a^2*(-1/4/a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2+2*tanh(1/2*x)*a^(1/2)+(a+b)^(1/2))+1/4/a^(1/
2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2-2*tanh(1/2*x)*a^(1/2)+(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (45) = 90\).
time = 0.48, size = 119, normalized size = 2.16 \begin {gather*} -\frac {b^{2} \log \left (\frac {b e^{\left (-2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (-2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{2 \, \sqrt {{\left (a + b\right )} a} a^{2}} + \frac {2 \, {\left (6 \, {\left (a - b\right )} e^{\left (-2 \, x\right )} - 3 \, b e^{\left (-4 \, x\right )} + 2 \, a - 3 \, b\right )}}{3 \, {\left (3 \, a^{2} e^{\left (-2 \, x\right )} + 3 \, a^{2} e^{\left (-4 \, x\right )} + a^{2} e^{\left (-6 \, x\right )} + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

-1/2*b^2*log((b*e^(-2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(-2*x) + 2*a + b + 2*sqrt((a + b)*a)))/(sqrt((a +
b)*a)*a^2) + 2/3*(6*(a - b)*e^(-2*x) - 3*b*e^(-4*x) + 2*a - 3*b)/(3*a^2*e^(-2*x) + 3*a^2*e^(-4*x) + a^2*e^(-6
*x) + a^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 608 vs. \(2 (45) = 90\).
time = 0.39, size = 1377, normalized size = 25.04 \begin {gather*} \text {Too large to display} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{4}{\left (x \right )}}{a + b \cosh ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**4/(a+b*cosh(x)**2),x)

[Out]

Integral(sech(x)**4/(a + b*cosh(x)**2), x)

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Giac [A]
time = 0.53, size = 87, normalized size = 1.58 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 1.32, size = 239, normalized size = 4.35 \begin {gather*} \frac {8}{3\,a\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {4}{a\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}+\frac {2\,b}{a^2\,\left ({\mathrm {e}}^{2\,x}+1\right )}-\frac {b^2\,\ln \left (\frac {4\,b^2\,\left (2\,a\,b+8\,a^2\,{\mathrm {e}}^{2\,x}+b^2\,{\mathrm {e}}^{2\,x}+b^2+8\,a\,b\,{\mathrm {e}}^{2\,x}\right )}{a^5\,\left (a+b\right )}-\frac {8\,b^2\,\left (b+4\,a\,{\mathrm {e}}^{2\,x}+2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^{9/2}\,\sqrt {a+b}}\right )}{2\,a^{5/2}\,\sqrt {a+b}}+\frac {b^2\,\ln \left (\frac {8\,b^2\,\left (b+4\,a\,{\mathrm {e}}^{2\,x}+2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^{9/2}\,\sqrt {a+b}}+\frac {4\,b^2\,\left (2\,a\,b+8\,a^2\,{\mathrm {e}}^{2\,x}+b^2\,{\mathrm {e}}^{2\,x}+b^2+8\,a\,b\,{\mathrm {e}}^{2\,x}\right )}{a^5\,\left (a+b\right )}\right )}{2\,a^{5/2}\,\sqrt {a+b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(a + b*cosh(x)^2)),x)

[Out]

8/(3*a*(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1)) - 4/(a*(2*exp(2*x) + exp(4*x) + 1)) + (2*b)/(a^2*(exp(2*x) +
1)) - (b^2*log((4*b^2*(2*a*b + 8*a^2*exp(2*x) + b^2*exp(2*x) + b^2 + 8*a*b*exp(2*x)))/(a^5*(a + b)) - (8*b^2*(
b + 4*a*exp(2*x) + 2*b*exp(2*x)))/(a^(9/2)*(a + b)^(1/2))))/(2*a^(5/2)*(a + b)^(1/2)) + (b^2*log((8*b^2*(b + 4
*a*exp(2*x) + 2*b*exp(2*x)))/(a^(9/2)*(a + b)^(1/2)) + (4*b^2*(2*a*b + 8*a^2*exp(2*x) + b^2*exp(2*x) + b^2 + 8
*a*b*exp(2*x)))/(a^5*(a + b))))/(2*a^(5/2)*(a + b)^(1/2))

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