3.17 \(\int \frac{\text{csch}^4(x)}{a+b \cosh ^2(x)} \, dx\)
Optimal. Leaf size=59 \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{5/2}}-\frac{\coth ^3(x)}{3 (a+b)}+\frac{(a+2 b) \coth (x)}{(a+b)^2} \]
[Out]
(b^2*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) + ((a + 2*b)*Coth[x])/(a + b)^2 - Coth[x]
^3/(3*(a + b))
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Rubi [A] time = 0.0857252, antiderivative size = 59, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3191, 390, 208} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{5/2}}-\frac{\coth ^3(x)}{3 (a+b)}+\frac{(a+2 b) \coth (x)}{(a+b)^2} \]
Antiderivative was successfully verified.
[In]
Int[Csch[x]^4/(a + b*Cosh[x]^2),x]
[Out]
(b^2*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) + ((a + 2*b)*Coth[x])/(a + b)^2 - Coth[x]
^3/(3*(a + b))
Rule 3191
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Rule 390
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]
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{\text{csch}^4(x)}{a+b \cosh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{a-(a+b) x^2} \, dx,x,\coth (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{a+2 b}{(a+b)^2}-\frac{x^2}{a+b}+\frac{b^2}{(a+b)^2 \left (a-(a+b) x^2\right )}\right ) \, dx,x,\coth (x)\right )\\ &=\frac{(a+2 b) \coth (x)}{(a+b)^2}-\frac{\coth ^3(x)}{3 (a+b)}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a-(a+b) x^2} \, dx,x,\coth (x)\right )}{(a+b)^2}\\ &=\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{5/2}}+\frac{(a+2 b) \coth (x)}{(a+b)^2}-\frac{\coth ^3(x)}{3 (a+b)}\\ \end{align*}
Mathematica [A] time = 0.199082, size = 59, normalized size = 1. \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} (a+b)^{5/2}}-\frac{\coth (x) \left ((a+b) \text{csch}^2(x)-2 a-5 b\right )}{3 (a+b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Csch[x]^4/(a + b*Cosh[x]^2),x]
[Out]
(b^2*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) - (Coth[x]*(-2*a - 5*b + (a + b)*Csch[x]^
2))/(3*(a + b)^2)
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Maple [B] time = 0.042, size = 180, normalized size = 3.1 \begin{align*} -{\frac{a}{24\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{3\,a}{8\, \left ( a+b \right ) ^{2}}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{7\,b}{8\, \left ( a+b \right ) ^{2}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{24\,a+24\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{3\,a}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{7\,b}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{{b}^{2}}{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{5}{2}}}{\frac{1}{\sqrt{a}}}}-{\frac{{b}^{2}}{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{5}{2}}}{\frac{1}{\sqrt{a}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csch(x)^4/(a+b*cosh(x)^2),x)
[Out]
-1/24/(a+b)^2*a*tanh(1/2*x)^3-1/24/(a+b)^2*tanh(1/2*x)^3*b+3/8/(a+b)^2*a*tanh(1/2*x)+7/8/(a+b)^2*tanh(1/2*x)*b
-1/24/(a+b)/tanh(1/2*x)^3+3/8/(a+b)^2/tanh(1/2*x)*a+7/8/(a+b)^2/tanh(1/2*x)*b+1/2*b^2/(a+b)^(5/2)/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*b^2/(a+b)^(5/2)/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))
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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(csch(x)^4/(a+b*cosh(x)^2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.22818, size = 4686, normalized size = 79.42 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(x)^4/(a+b*cosh(x)^2),x, algorithm="fricas")
[Out]
[1/6*(12*(a^2*b + a*b^2)*cosh(x)^4 + 48*(a^2*b + a*b^2)*cosh(x)*sinh(x)^3 + 12*(a^2*b + a*b^2)*sinh(x)^4 + 8*a
^3 + 28*a^2*b + 20*a*b^2 - 24*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(x)^2 - 24*(a^3 + 3*a^2*b + 2*a*b^2 - 3*(a^2*b + a
*b^2)*cosh(x)^2)*sinh(x)^2 + 3*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*
(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 3*b^2*cosh(x)^2 + 4*(5*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 + 3*(5*b^2
*cosh(x)^4 - 6*b^2*cosh(x)^2 + b^2)*sinh(x)^2 - b^2 + 6*(b^2*cosh(x)^5 - 2*b^2*cosh(x)^3 + 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*co
sh(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)) + 48*((a^2*b + a*b^2)*cosh(x)^3 - (a^3 + 3*a^2*b + 2*a*b^2)*cosh(x))*sinh(
x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^6 + 6*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)*sinh(x)^5 +
(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sinh(x)^6 - 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^4 - 3*(a^4 + 3*
a^3*b + 3*a^2*b^2 + a*b^3 - 5*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2)*sinh(x)^4 - a^4 - 3*a^3*b - 3*a^2
*b^2 - a*b^3 + 4*(5*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^3 - 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh
(x))*sinh(x)^3 + 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2 + 3*(5*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*co
sh(x)^4 + a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - 6*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2)*sinh(x)^2 + 6*(
(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^5 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^3 + (a^4 + 3*a^3
*b + 3*a^2*b^2 + a*b^3)*cosh(x))*sinh(x)), 1/3*(6*(a^2*b + a*b^2)*cosh(x)^4 + 24*(a^2*b + a*b^2)*cosh(x)*sinh(
x)^3 + 6*(a^2*b + a*b^2)*sinh(x)^4 + 4*a^3 + 14*a^2*b + 10*a*b^2 - 12*(a^3 + 3*a^2*b + 2*a*b^2)*cosh(x)^2 - 12
*(a^3 + 3*a^2*b + 2*a*b^2 - 3*(a^2*b + a*b^2)*cosh(x)^2)*sinh(x)^2 + 3*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^
5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 - b^2)*sinh(x)^4 + 3*b^2*cosh(x)^2 + 4*(5*b^2*cosh(x)
^3 - 3*b^2*cosh(x))*sinh(x)^3 + 3*(5*b^2*cosh(x)^4 - 6*b^2*cosh(x)^2 + b^2)*sinh(x)^2 - b^2 + 6*(b^2*cosh(x)^5
- 2*b^2*cosh(x)^3 + 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)) + 24*((a^2*b + a*b^2)*cosh(x)^3 - (a^3 + 3*a^2*b + 2*a*b^2)
*cosh(x))*sinh(x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^6 + 6*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh
(x)*sinh(x)^5 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sinh(x)^6 - 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^
4 - 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - 5*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2)*sinh(x)^4 - a^4 -
3*a^3*b - 3*a^2*b^2 - a*b^3 + 4*(5*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^3 - 3*(a^4 + 3*a^3*b + 3*a^2*b^
2 + a*b^3)*cosh(x))*sinh(x)^3 + 3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2 + 3*(5*(a^4 + 3*a^3*b + 3*a^2*
b^2 + a*b^3)*cosh(x)^4 + a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - 6*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^2)*
sinh(x)^2 + 6*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^5 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cosh(x)^3
+ (a^4 + 3*a^3*b + 3*a^2*b^2 + a*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(csch(x)**4/(a+b*cosh(x)**2),x)
[Out]
Timed out
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Giac [B] time = 1.40293, size = 144, normalized size = 2.44 \begin{align*} \frac{b^{2} \arctan \left (\frac{b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt{-a^{2} - a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-a^{2} - a b}} + \frac{2 \,{\left (3 \, b e^{\left (4 \, x\right )} - 6 \, a e^{\left (2 \, x\right )} - 12 \, b e^{\left (2 \, x\right )} + 2 \, a + 5 \, b\right )}}{3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(x)^4/(a+b*cosh(x)^2),x, algorithm="giac")
[Out]
b^2*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/((a^2 + 2*a*b + b^2)*sqrt(-a^2 - a*b)) + 2/3*(3*b*e^(4*
x) - 6*a*e^(2*x) - 12*b*e^(2*x) + 2*a + 5*b)/((a^2 + 2*a*b + b^2)*(e^(2*x) - 1)^3)