3.33 \(\int \frac{1}{(a+b \cosh ^2(x))^2} \, dx\)

Optimal. Leaf size=65 \[ \frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{2 a^{3/2} (a+b)^{3/2}}-\frac{b \sinh (x) \cosh (x)}{2 a (a+b) \left (a+b \cosh ^2(x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.0574036, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3184, 12, 3181, 208} \[ \frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{2 a^{3/2} (a+b)^{3/2}}-\frac{b \sinh (x) \cosh (x)}{2 a (a+b) \left (a+b \cosh ^2(x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[x]^2)^(-2),x]

[Out]

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

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

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

Mathematica [A]  time = 0.213622, size = 68, normalized size = 1.05 \[ \frac{(2 a+b) \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{2 a^{3/2} (a+b)^{3/2}}-\frac{b \sinh (2 x)}{2 a (a+b) (2 a+b \cosh (2 x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[x]^2)^(-2),x]

[Out]

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

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Maple [B]  time = 0.036, size = 230, normalized size = 3.5 \begin{align*} -2\,{\frac{1}{ \left ( \tanh \left ( x/2 \right ) \right ) ^{4}a+b \left ( \tanh \left ( x/2 \right ) \right ) ^{4}-2\,a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+a+b} \left ( 1/2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}b}{a \left ( a+b \right ) }}+1/2\,{\frac{\tanh \left ( x/2 \right ) b}{a \left ( a+b \right ) }} \right ) }+{\frac{1}{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 ) \left ( a+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}}-{\frac{1}{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 ) \left ( a+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}}+{\frac{b}{4}\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{3}{2}}} \left ( a+b \right ) ^{-{\frac{3}{2}}}}-{\frac{b}{4}\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{3}{2}}} \left ( a+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(1/2*b/a/(a+b)*tanh(1/2*x)^3+1/2*b/a/(a+b)*tanh(1/2*x))/(tanh(1/2*x)^4*a+b*tanh(1/2*x)^4-2*a*tanh(1/2*x)^2+
2*tanh(1/2*x)^2*b+a+b)+1/2/(a+b)^(3/2)/a^(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/(a+b)^(3/2)/a^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2-2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))+1/4/a^(3/2)/(a+b)^(3/
2)*b*ln((a+b)^(1/2)*tanh(1/2*x)^2+2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))-1/4/a^(3/2)/(a+b)^(3/2)*b*ln((a+b)^(1/2)*
tanh(1/2*x)^2-2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.38822, size = 3090, normalized size = 47.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*a^2*b + 4*a*b^2 + 4*(2*a^3 + 3*a^2*b + a*b^2)*cosh(x)^2 + 8*(2*a^3 + 3*a^2*b + a*b^2)*cosh(x)*sinh(x)
+ 4*(2*a^3 + 3*a^2*b + a*b^2)*sinh(x)^2 + ((2*a*b + b^2)*cosh(x)^4 + 4*(2*a*b + b^2)*cosh(x)*sinh(x)^3 + (2*a*
b + b^2)*sinh(x)^4 + 2*(4*a^2 + 4*a*b + b^2)*cosh(x)^2 + 2*(3*(2*a*b + b^2)*cosh(x)^2 + 4*a^2 + 4*a*b + b^2)*s
inh(x)^2 + 2*a*b + b^2 + 4*((2*a*b + b^2)*cosh(x)^3 + (4*a^2 + 4*a*b + 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
))*sinh(x) + b)))/(a^4*b + 2*a^3*b^2 + a^2*b^3 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*cosh(x)^4 + 4*(a^4*b + 2*a^3*b^
2 + a^2*b^3)*cosh(x)*sinh(x)^3 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*sinh(x)^4 + 2*(2*a^5 + 5*a^4*b + 4*a^3*b^2 + a^
2*b^3)*cosh(x)^2 + 2*(2*a^5 + 5*a^4*b + 4*a^3*b^2 + a^2*b^3 + 3*(a^4*b + 2*a^3*b^2 + a^2*b^3)*cosh(x)^2)*sinh(
x)^2 + 4*((a^4*b + 2*a^3*b^2 + a^2*b^3)*cosh(x)^3 + (2*a^5 + 5*a^4*b + 4*a^3*b^2 + a^2*b^3)*cosh(x))*sinh(x)),
1/2*(2*a^2*b + 2*a*b^2 + 2*(2*a^3 + 3*a^2*b + a*b^2)*cosh(x)^2 + 4*(2*a^3 + 3*a^2*b + a*b^2)*cosh(x)*sinh(x)
+ 2*(2*a^3 + 3*a^2*b + a*b^2)*sinh(x)^2 + ((2*a*b + b^2)*cosh(x)^4 + 4*(2*a*b + b^2)*cosh(x)*sinh(x)^3 + (2*a*
b + b^2)*sinh(x)^4 + 2*(4*a^2 + 4*a*b + b^2)*cosh(x)^2 + 2*(3*(2*a*b + b^2)*cosh(x)^2 + 4*a^2 + 4*a*b + b^2)*s
inh(x)^2 + 2*a*b + b^2 + 4*((2*a*b + b^2)*cosh(x)^3 + (4*a^2 + 4*a*b + 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)))/(a^4*b
+ 2*a^3*b^2 + a^2*b^3 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*cosh(x)^4 + 4*(a^4*b + 2*a^3*b^2 + a^2*b^3)*cosh(x)*sinh
(x)^3 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*sinh(x)^4 + 2*(2*a^5 + 5*a^4*b + 4*a^3*b^2 + a^2*b^3)*cosh(x)^2 + 2*(2*a
^5 + 5*a^4*b + 4*a^3*b^2 + a^2*b^3 + 3*(a^4*b + 2*a^3*b^2 + a^2*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^4*b + 2*a^3*
b^2 + a^2*b^3)*cosh(x)^3 + (2*a^5 + 5*a^4*b + 4*a^3*b^2 + a^2*b^3)*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(1/(a+b*cosh(x)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.22918, size = 140, normalized size = 2.15 \begin{align*} \frac{{\left (2 \, a + b\right )} \arctan \left (\frac{b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt{-a^{2} - a b}}\right )}{2 \,{\left (a^{2} + a b\right )} \sqrt{-a^{2} - a b}} + \frac{2 \, a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + b}{{\left (a^{2} + a b\right )}{\left (b e^{\left (4 \, x\right )} + 4 \, a e^{\left (2 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a + b)*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/((a^2 + a*b)*sqrt(-a^2 - a*b)) + (2*a*e^(2*x)
+ b*e^(2*x) + b)/((a^2 + a*b)*(b*e^(4*x) + 4*a*e^(2*x) + 2*b*e^(2*x) + b))