3.30 \(\int \frac{\text{sech}^3(x)}{a+b \cosh ^2(x)} \, dx\)

Optimal. Leaf size=59 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{a^2 \sqrt{a+b}}+\frac{(a-2 b) \tan ^{-1}(\sinh (x))}{2 a^2}+\frac{\tanh (x) \text{sech}(x)}{2 a} \]

[Out]

((a - 2*b)*ArcTan[Sinh[x]])/(2*a^2) + (b^(3/2)*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(a^2*Sqrt[a + b]) + (Sec
h[x]*Tanh[x])/(2*a)

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Rubi [A]  time = 0.090977, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3186, 414, 522, 203, 205} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sinh (x)}{\sqrt{a+b}}\right )}{a^2 \sqrt{a+b}}+\frac{(a-2 b) \tan ^{-1}(\sinh (x))}{2 a^2}+\frac{\tanh (x) \text{sech}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Cosh[x]^2),x]

[Out]

((a - 2*b)*ArcTan[Sinh[x]])/(2*a^2) + (b^(3/2)*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(a^2*Sqrt[a + b]) + (Sec
h[x]*Tanh[x])/(2*a)

Rule 3186

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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 205

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

Rubi steps

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

Mathematica [A]  time = 0.164757, size = 58, normalized size = 0.98 \[ \frac{-\frac{2 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \text{csch}(x)}{\sqrt{b}}\right )}{\sqrt{a+b}}+2 (a-2 b) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+a \tanh (x) \text{sech}(x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(a + b*Cosh[x]^2),x]

[Out]

((-2*b^(3/2)*ArcTan[(Sqrt[a + b]*Csch[x])/Sqrt[b]])/Sqrt[a + b] + 2*(a - 2*b)*ArcTan[Tanh[x/2]] + a*Sech[x]*Ta
nh[x])/(2*a^2)

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Maple [B]  time = 0.045, size = 131, normalized size = 2.2 \begin{align*}{\frac{1}{{a}^{2}}{b}^{{\frac{3}{2}}}\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( x/2 \right ) \sqrt{a+b}+2\,\sqrt{a} \right ){\frac{1}{\sqrt{b}}}} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{{a}^{2}}{b}^{{\frac{3}{2}}}\arctan \left ({\frac{1}{2} \left ( 2\,\tanh \left ( x/2 \right ) \sqrt{a+b}-2\,\sqrt{a} \right ){\frac{1}{\sqrt{b}}}} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{1}{a}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{b\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(a+b*cosh(x)^2),x)

[Out]

b^(3/2)/a^2/(a+b)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*(a+b)^(1/2)+2*a^(1/2))/b^(1/2))+b^(3/2)/a^2/(a+b)^(1/2)*arct
an(1/2*(2*tanh(1/2*x)*(a+b)^(1/2)-2*a^(1/2))/b^(1/2))-1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3+1/a/(tanh(1/2*x)^2
+1)^2*tanh(1/2*x)+1/a*arctan(tanh(1/2*x))-2/a^2*b*arctan(tanh(1/2*x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{e^{\left (3 \, x\right )} - e^{x}}{a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + a} + \frac{{\left (a - 2 \, b\right )} \arctan \left (e^{x}\right )}{a^{2}} + 8 \, \int \frac{b^{2} e^{\left (3 \, x\right )} + b^{2} e^{x}}{4 \,{\left (a^{2} b e^{\left (4 \, x\right )} + a^{2} b + 2 \,{\left (2 \, a^{3} + a^{2} b\right )} e^{\left (2 \, x\right )}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

(e^(3*x) - e^x)/(a*e^(4*x) + 2*a*e^(2*x) + a) + (a - 2*b)*arctan(e^x)/a^2 + 8*integrate(1/4*(b^2*e^(3*x) + b^2
*e^x)/(a^2*b*e^(4*x) + a^2*b + 2*(2*a^3 + a^2*b)*e^(2*x)), x)

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Fricas [B]  time = 2.52415, size = 2944, normalized size = 49.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*cosh(x)^2),x, algorithm="fricas")

[Out]

[1/2*(2*a*cosh(x)^3 + 6*a*cosh(x)*sinh(x)^2 + 2*a*sinh(x)^3 + (b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)
^4 + 2*b*cosh(x)^2 + 2*(3*b*cosh(x)^2 + b)*sinh(x)^2 + 4*(b*cosh(x)^3 + b*cosh(x))*sinh(x) + b)*sqrt(-b/(a + b
))*log((b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 2*(2*a + 3*b)*cosh(x)^2 + 2*(3*b*cosh(x)^2 - 2*a -
3*b)*sinh(x)^2 + 4*(b*cosh(x)^3 - (2*a + 3*b)*cosh(x))*sinh(x) + 4*((a + b)*cosh(x)^3 + 3*(a + b)*cosh(x)*sin
h(x)^2 + (a + b)*sinh(x)^3 - (a + b)*cosh(x) + (3*(a + b)*cosh(x)^2 - a - b)*sinh(x))*sqrt(-b/(a + b)) + 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)) + 2*((a - 2*b)*cosh(x)^4 + 4*(a - 2*b)*cosh(x)*sinh(x)^
3 + (a - 2*b)*sinh(x)^4 + 2*(a - 2*b)*cosh(x)^2 + 2*(3*(a - 2*b)*cosh(x)^2 + a - 2*b)*sinh(x)^2 + 4*((a - 2*b)
*cosh(x)^3 + (a - 2*b)*cosh(x))*sinh(x) + a - 2*b)*arctan(cosh(x) + sinh(x)) - 2*a*cosh(x) + 2*(3*a*cosh(x)^2
- a)*sinh(x))/(a^2*cosh(x)^4 + 4*a^2*cosh(x)*sinh(x)^3 + a^2*sinh(x)^4 + 2*a^2*cosh(x)^2 + 2*(3*a^2*cosh(x)^2
+ a^2)*sinh(x)^2 + a^2 + 4*(a^2*cosh(x)^3 + a^2*cosh(x))*sinh(x)), (a*cosh(x)^3 + 3*a*cosh(x)*sinh(x)^2 + a*si
nh(x)^3 + (b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 + 2*b*cosh(x)^2 + 2*(3*b*cosh(x)^2 + b)*sinh(x)^2
+ 4*(b*cosh(x)^3 + b*cosh(x))*sinh(x) + b)*sqrt(b/(a + b))*arctan(1/2*sqrt(b/(a + b))*(cosh(x) + sinh(x))) +
(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 + 2*b*cosh(x)^2 + 2*(3*b*cosh(x)^2 + b)*sinh(x)^2 + 4*(b*co
sh(x)^3 + b*cosh(x))*sinh(x) + b)*sqrt(b/(a + b))*arctan(1/2*(b*cosh(x)^3 + 3*b*cosh(x)*sinh(x)^2 + b*sinh(x)^
3 + (4*a + 3*b)*cosh(x) + (3*b*cosh(x)^2 + 4*a + 3*b)*sinh(x))*sqrt(b/(a + b))/b) + ((a - 2*b)*cosh(x)^4 + 4*(
a - 2*b)*cosh(x)*sinh(x)^3 + (a - 2*b)*sinh(x)^4 + 2*(a - 2*b)*cosh(x)^2 + 2*(3*(a - 2*b)*cosh(x)^2 + a - 2*b)
*sinh(x)^2 + 4*((a - 2*b)*cosh(x)^3 + (a - 2*b)*cosh(x))*sinh(x) + a - 2*b)*arctan(cosh(x) + sinh(x)) - a*cosh
(x) + (3*a*cosh(x)^2 - a)*sinh(x))/(a^2*cosh(x)^4 + 4*a^2*cosh(x)*sinh(x)^3 + a^2*sinh(x)^4 + 2*a^2*cosh(x)^2
+ 2*(3*a^2*cosh(x)^2 + a^2)*sinh(x)^2 + a^2 + 4*(a^2*cosh(x)^3 + a^2*cosh(x))*sinh(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{3}{\left (x \right )}}{a + b \cosh ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**3/(a+b*cosh(x)**2),x)

[Out]

Integral(sech(x)**3/(a + b*cosh(x)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*cosh(x)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError