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

Optimal. Leaf size=82 \[ -\frac{x \left (a^2-2 b^2\right )}{2 a^3}+\frac{2 b \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^3}-\frac{\sinh (x) (2 b-a \cosh (x))}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.212117, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3872, 2865, 2735, 2659, 205} \[ -\frac{x \left (a^2-2 b^2\right )}{2 a^3}+\frac{2 b \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^3}-\frac{\sinh (x) (2 b-a \cosh (x))}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Sech[x]),x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 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{\sinh ^2(x)}{a+b \text{sech}(x)} \, dx &=-\int \frac{\cosh (x) \sinh ^2(x)}{-b-a \cosh (x)} \, dx\\ &=-\frac{(2 b-a \cosh (x)) \sinh (x)}{2 a^2}+\frac{\int \frac{-a b+\left (a^2-2 b^2\right ) \cosh (x)}{-b-a \cosh (x)} \, dx}{2 a^2}\\ &=-\frac{\left (a^2-2 b^2\right ) x}{2 a^3}-\frac{(2 b-a \cosh (x)) \sinh (x)}{2 a^2}-\frac{\left (b \left (a^2-b^2\right )\right ) \int \frac{1}{-b-a \cosh (x)} \, dx}{a^3}\\ &=-\frac{\left (a^2-2 b^2\right ) x}{2 a^3}-\frac{(2 b-a \cosh (x)) \sinh (x)}{2 a^2}-\frac{\left (2 b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^3}\\ &=-\frac{\left (a^2-2 b^2\right ) x}{2 a^3}+\frac{2 \sqrt{a-b} b \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^3}-\frac{(2 b-a \cosh (x)) \sinh (x)}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.153419, size = 76, normalized size = 0.93 \[ \frac{-8 b \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )-2 a^2 x+a^2 \sinh (2 x)-4 a b \sinh (x)+4 b^2 x}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*x + 4*b^2*x - 8*b*Sqrt[a^2 - b^2]*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]] - 4*a*b*Sinh[x] + a^2*S
inh[2*x])/(4*a^3)

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Maple [B]  time = 0.027, size = 213, normalized size = 2.6 \begin{align*} -{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{b}{a\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{{b}^{3}}{{a}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/(tanh(1/2*x)+1)^2+1/2/a/(tanh(1/2*x)+1)+1/a^2/(tanh(1/2*x)+1)*b-1/2/a*ln(tanh(1/2*x)+1)+1/a^3*ln(tanh(1
/2*x)+1)*b^2+1/2/a/(tanh(1/2*x)-1)^2+1/2/a/(tanh(1/2*x)-1)+1/a^2/(tanh(1/2*x)-1)*b+1/2/a*ln(tanh(1/2*x)-1)-1/a
^3*ln(tanh(1/2*x)-1)*b^2+2/a*b/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-2*b^3/a^3/((a
+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(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(sinh(x)^2/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.75851, size = 1500, normalized size = 18.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(a^2*cosh(x)^4 + a^2*sinh(x)^4 - 4*a*b*cosh(x)^3 - 4*(a^2 - 2*b^2)*x*cosh(x)^2 + 4*(a^2*cosh(x) - a*b)*si
nh(x)^3 + 4*a*b*cosh(x) + 2*(3*a^2*cosh(x)^2 - 6*a*b*cosh(x) - 2*(a^2 - 2*b^2)*x)*sinh(x)^2 + 8*(b*cosh(x)^2 +
2*b*cosh(x)*sinh(x) + b*sinh(x)^2)*sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2
+ 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sin
h(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) + a)) - a^2 + 4*(a^2*cosh(x)^3 - 3*a*b*cosh(x)^2 - 2*(a^2 - 2
*b^2)*x*cosh(x) + a*b)*sinh(x))/(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2), 1/8*(a^2*cosh(x)^4 +
a^2*sinh(x)^4 - 4*a*b*cosh(x)^3 - 4*(a^2 - 2*b^2)*x*cosh(x)^2 + 4*(a^2*cosh(x) - a*b)*sinh(x)^3 + 4*a*b*cosh(x
) + 2*(3*a^2*cosh(x)^2 - 6*a*b*cosh(x) - 2*(a^2 - 2*b^2)*x)*sinh(x)^2 - 16*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x)
+ b*sinh(x)^2)*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) - a^2 + 4*(a^2*cosh(x)^3 -
3*a*b*cosh(x)^2 - 2*(a^2 - 2*b^2)*x*cosh(x) + a*b)*sinh(x))/(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh
(x)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14827, size = 135, normalized size = 1.65 \begin{align*} \frac{a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} - \frac{{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} + \frac{{\left (4 \, a b e^{x} - a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} + \frac{2 \,{\left (a^{2} b - b^{3}\right )} \arctan \left (\frac{a e^{x} + b}{\sqrt{a^{2} - b^{2}}}\right )}{\sqrt{a^{2} - b^{2}} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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