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

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

[Out]

-1/2*(a^2-2*b^2)*x/a^3-1/2*(2*b-a*cosh(x))*sinh(x)/a^2+2*b*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))*(a-b)^(
1/2)*(a+b)^(1/2)/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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 = 0.20, 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)

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 536, normalized size = 6.54 \[ \left [\frac {a^{2} \cosh \relax (x)^{4} + a^{2} \sinh \relax (x)^{4} - 4 \, a b \cosh \relax (x)^{3} - 4 \, {\left (a^{2} - 2 \, b^{2}\right )} x \cosh \relax (x)^{2} + 4 \, {\left (a^{2} \cosh \relax (x) - a b\right )} \sinh \relax (x)^{3} + 4 \, a b \cosh \relax (x) + 2 \, {\left (3 \, a^{2} \cosh \relax (x)^{2} - 6 \, a b \cosh \relax (x) - 2 \, {\left (a^{2} - 2 \, b^{2}\right )} x\right )} \sinh \relax (x)^{2} + 8 \, {\left (b \cosh \relax (x)^{2} + 2 \, b \cosh \relax (x) \sinh \relax (x) + b \sinh \relax (x)^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) - a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) + a}\right ) - a^{2} + 4 \, {\left (a^{2} \cosh \relax (x)^{3} - 3 \, a b \cosh \relax (x)^{2} - 2 \, {\left (a^{2} - 2 \, b^{2}\right )} x \cosh \relax (x) + a b\right )} \sinh \relax (x)}{8 \, {\left (a^{3} \cosh \relax (x)^{2} + 2 \, a^{3} \cosh \relax (x) \sinh \relax (x) + a^{3} \sinh \relax (x)^{2}\right )}}, \frac {a^{2} \cosh \relax (x)^{4} + a^{2} \sinh \relax (x)^{4} - 4 \, a b \cosh \relax (x)^{3} - 4 \, {\left (a^{2} - 2 \, b^{2}\right )} x \cosh \relax (x)^{2} + 4 \, {\left (a^{2} \cosh \relax (x) - a b\right )} \sinh \relax (x)^{3} + 4 \, a b \cosh \relax (x) + 2 \, {\left (3 \, a^{2} \cosh \relax (x)^{2} - 6 \, a b \cosh \relax (x) - 2 \, {\left (a^{2} - 2 \, b^{2}\right )} x\right )} \sinh \relax (x)^{2} - 16 \, {\left (b \cosh \relax (x)^{2} + 2 \, b \cosh \relax (x) \sinh \relax (x) + b \sinh \relax (x)^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cosh \relax (x) + a \sinh \relax (x) + b}{\sqrt {a^{2} - b^{2}}}\right ) - a^{2} + 4 \, {\left (a^{2} \cosh \relax (x)^{3} - 3 \, a b \cosh \relax (x)^{2} - 2 \, {\left (a^{2} - 2 \, b^{2}\right )} x \cosh \relax (x) + a b\right )} \sinh \relax (x)}{8 \, {\left (a^{3} \cosh \relax (x)^{2} + 2 \, a^{3} \cosh \relax (x) \sinh \relax (x) + a^{3} \sinh \relax (x)^{2}\right )}}\right ] \]

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)]

________________________________________________________________________________________

giac [A]  time = 0.14, size = 100, normalized size = 1.22 \[ \frac {a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} + \frac {{\left (4 \, a b e^{x} - a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} + \frac {2 \, {\left (a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a^{3}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.12, size = 213, normalized size = 2.60 \[ \frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{a^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{a^{3}}+\frac {2 b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

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*b/a/((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))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 1.67, size = 173, normalized size = 2.11 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {b\,{\mathrm {e}}^x}{2\,a^2}+\frac {b\,{\mathrm {e}}^{-x}}{2\,a^2}-\frac {x\,\left (a^2-2\,b^2\right )}{2\,a^3}+\frac {b\,\ln \left (-\frac {2\,b\,{\mathrm {e}}^x\,\left (a^2-b^2\right )}{a^4}-\frac {2\,b\,\sqrt {a+b}\,\left (a+b\,{\mathrm {e}}^x\right )\,\sqrt {b-a}}{a^4}\right )\,\sqrt {a+b}\,\sqrt {b-a}}{a^3}-\frac {b\,\ln \left (\frac {2\,b\,\sqrt {a+b}\,\left (a+b\,{\mathrm {e}}^x\right )\,\sqrt {b-a}}{a^4}-\frac {2\,b\,{\mathrm {e}}^x\,\left (a^2-b^2\right )}{a^4}\right )\,\sqrt {a+b}\,\sqrt {b-a}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a + b/cosh(x)),x)

[Out]

exp(2*x)/(8*a) - exp(-2*x)/(8*a) - (b*exp(x))/(2*a^2) + (b*exp(-x))/(2*a^2) - (x*(a^2 - 2*b^2))/(2*a^3) + (b*l
og(- (2*b*exp(x)*(a^2 - b^2))/a^4 - (2*b*(a + b)^(1/2)*(a + b*exp(x))*(b - a)^(1/2))/a^4)*(a + b)^(1/2)*(b - a
)^(1/2))/a^3 - (b*log((2*b*(a + b)^(1/2)*(a + b*exp(x))*(b - a)^(1/2))/a^4 - (2*b*exp(x)*(a^2 - b^2))/a^4)*(a
+ b)^(1/2)*(b - a)^(1/2))/a^3

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

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)

________________________________________________________________________________________