3.99 \(\int \frac{\sinh ^2(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=92 \[ -\frac{b^3 \log (a+b \coth (x))}{\left (a^2-b^2\right )^2}-\frac{\sinh ^2(x) (b-a \coth (x))}{2 \left (a^2-b^2\right )}+\frac{(a+2 b) \log (1-\coth (x))}{4 (a+b)^2}-\frac{(a-2 b) \log (\coth (x)+1)}{4 (a-b)^2} \]

[Out]

((a + 2*b)*Log[1 - Coth[x]])/(4*(a + b)^2) - ((a - 2*b)*Log[1 + Coth[x]])/(4*(a - b)^2) - (b^3*Log[a + b*Coth[
x]])/(a^2 - b^2)^2 - ((b - a*Coth[x])*Sinh[x]^2)/(2*(a^2 - b^2))

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Rubi [A]  time = 0.141564, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3506, 741, 801} \[ -\frac{b^3 \log (a+b \coth (x))}{\left (a^2-b^2\right )^2}-\frac{\sinh ^2(x) (b-a \coth (x))}{2 \left (a^2-b^2\right )}+\frac{(a+2 b) \log (1-\coth (x))}{4 (a+b)^2}-\frac{(a-2 b) \log (\coth (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Coth[x]),x]

[Out]

((a + 2*b)*Log[1 - Coth[x]])/(4*(a + b)^2) - ((a - 2*b)*Log[1 + Coth[x]])/(4*(a - b)^2) - (b^3*Log[a + b*Coth[
x]])/(a^2 - b^2)^2 - ((b - a*Coth[x])*Sinh[x]^2)/(2*(a^2 - b^2))

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{\sinh ^2(x)}{a+b \coth (x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{(a+x) \left (1-\frac{x^2}{b^2}\right )^2} \, dx,x,b \coth (x)\right )}{b}\\ &=-\frac{(b-a \coth (x)) \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac{b \operatorname{Subst}\left (\int \frac{-2+\frac{a^2}{b^2}+\frac{a x}{b^2}}{(a+x) \left (1-\frac{x^2}{b^2}\right )} \, dx,x,b \coth (x)\right )}{2 \left (a^2-b^2\right )}\\ &=-\frac{(b-a \coth (x)) \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac{b \operatorname{Subst}\left (\int \left (\frac{(a-b) (a+2 b)}{2 b (a+b) (b-x)}+\frac{2 b^2}{(a-b) (a+b) (a+x)}+\frac{(a-2 b) (a+b)}{2 (a-b) b (b+x)}\right ) \, dx,x,b \coth (x)\right )}{2 \left (a^2-b^2\right )}\\ &=\frac{(a+2 b) \log (1-\coth (x))}{4 (a+b)^2}-\frac{(a-2 b) \log (1+\coth (x))}{4 (a-b)^2}-\frac{b^3 \log (a+b \coth (x))}{\left (a^2-b^2\right )^2}-\frac{(b-a \coth (x)) \sinh ^2(x)}{2 \left (a^2-b^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.151654, size = 75, normalized size = 0.82 \[ \frac{a \left (a^2-b^2\right ) \sinh (2 x)+\left (b^3-a^2 b\right ) \cosh (2 x)-2 a^3 x+6 a b^2 x-4 b^3 \log (a \sinh (x)+b \cosh (x))}{4 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*x + 6*a*b^2*x + (-(a^2*b) + b^3)*Cosh[2*x] - 4*b^3*Log[b*Cosh[x] + a*Sinh[x]] + a*(a^2 - b^2)*Sinh[2*x
])/(4*(a - b)^2*(a + b)^2)

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Maple [B]  time = 0.039, size = 175, normalized size = 1.9 \begin{align*} -8\,{\frac{1}{ \left ( 16\,a-16\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{2}}}+16\,{\frac{1}{ \left ( 32\,a-32\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) }}-{\frac{a}{2\, \left ( a-b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{b}{ \left ( a-b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+8\,{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{2}}}+16\,{\frac{1}{ \left ( 32\,a+32\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) }}+{\frac{a}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{b}{ \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{{b}^{3}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+2\,a\tanh \left ( x/2 \right ) +b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/(16*a-16*b)/(tanh(1/2*x)+1)^2+16/(32*a-32*b)/(tanh(1/2*x)+1)-1/2*a/(a-b)^2*ln(tanh(1/2*x)+1)+1/(a-b)^2*ln(t
anh(1/2*x)+1)*b+8/(16*a+16*b)/(tanh(1/2*x)-1)^2+16/(32*a+32*b)/(tanh(1/2*x)-1)+1/2*a/(a+b)^2*ln(tanh(1/2*x)-1)
+1/(a+b)^2*ln(tanh(1/2*x)-1)*b-b^3/(a-b)^2/(a+b)^2*ln(tanh(1/2*x)^2*b+2*a*tanh(1/2*x)+b)

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Maxima [A]  time = 1.17738, size = 112, normalized size = 1.22 \begin{align*} -\frac{b^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a + 2 \, b\right )} x}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{e^{\left (2 \, x\right )}}{8 \,{\left (a + b\right )}} - \frac{e^{\left (-2 \, x\right )}}{8 \,{\left (a - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.69496, size = 818, normalized size = 8.89 \begin{align*} \frac{{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{4} - 4 \,{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x \cosh \left (x\right )^{2} - a^{3} - a^{2} b + a b^{2} + b^{3} + 2 \,{\left (3 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{2} - 2 \,{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x\right )} \sinh \left (x\right )^{2} - 8 \,{\left (b^{3} \cosh \left (x\right )^{2} + 2 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right ) + b^{3} \sinh \left (x\right )^{2}\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \,{\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{3} - 2 \,{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right )}{8 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (a^3 - a^2*b -
a*b^2 + b^3)*sinh(x)^4 - 4*(a^3 - 3*a*b^2 - 2*b^3)*x*cosh(x)^2 - a^3 - a^2*b + a*b^2 + b^3 + 2*(3*(a^3 - a^2*b
- a*b^2 + b^3)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 - 2*b^3)*x)*sinh(x)^2 - 8*(b^3*cosh(x)^2 + 2*b^3*cosh(x)*sinh(x)
+ b^3*sinh(x)^2)*log(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^3
- 2*(a^3 - 3*a*b^2 - 2*b^3)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4
)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 + b^4)*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 \coth{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17309, size = 154, normalized size = 1.67 \begin{align*} -\frac{b^{3} \log \left ({\left | -a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a - 2 \, b\right )} x}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a e^{\left (2 \, x\right )} - 4 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-2 \, x\right )}}{8 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{e^{\left (2 \, x\right )}}{8 \,{\left (a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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