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]

(a+2*b)*coth(x)/(a+b)^2-1/3*coth(x)^3/(a+b)+b^2*arctanh(a^(1/2)*tanh(x)/(a+b)^(1/2))/(a+b)^(5/2)/a^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.22, size = 59, normalized size = 1.00 \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{\sqrt {a} (a+b)^{5/2}}-\frac {\coth (x) \left ((a+b) \text {csch}^2(x)-2 a-5 b\right )}{3 (a+b)^2} \]

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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fricas [B]  time = 0.54, size = 1875, normalized size = 31.78 \[ \text {result too large to display} \]

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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giac [B]  time = 0.39, size = 107, normalized size = 1.81 \[ \frac {b^{2} \arctan \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt {-a^{2} - a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a^{2} - a b}} + \frac {2 \, {\left (3 \, b e^{\left (4 \, x\right )} - 6 \, a e^{\left (2 \, x\right )} - 12 \, b e^{\left (2 \, x\right )} + 2 \, a + 5 \, b\right )}}{3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

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)

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maple [B]  time = 0.14, size = 177, normalized size = 3.00 \[ -\frac {a \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{24 \left (a +b \right )^{2}}-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b}{24 \left (a +b \right )^{2}}+\frac {3 a \tanh \left (\frac {x}{2}\right )}{8 \left (a +b \right )^{2}}+\frac {7 \tanh \left (\frac {x}{2}\right ) b}{8 \left (a +b \right )^{2}}-\frac {b^{2} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \sqrt {a}\, \tanh \left (\frac {x}{2}\right )+\sqrt {a +b}\right )}{2 \left (a +b \right )^{\frac {5}{2}} \sqrt {a}}+\frac {b^{2} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \sqrt {a}\, \tanh \left (\frac {x}{2}\right )+\sqrt {a +b}\right )}{2 \left (a +b \right )^{\frac {5}{2}} \sqrt {a}}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{3}}+\frac {3 a}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )}+\frac {7 b}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )} \]

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
-1/2*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))-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

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maxima [B]  time = 0.43, size = 161, normalized size = 2.73 \[ -\frac {b^{2} \log \left (\frac {b e^{\left (-2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (-2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{2 \, \sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {2 \, {\left (6 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, x\right )} - 3 \, b e^{\left (-4 \, x\right )} - 2 \, a - 5 \, b\right )}}{3 \, {\left (a^{2} + 2 \, a b + b^{2} - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )} - {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-6 \, x\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*cosh(x)^2),x, algorithm="maxima")

[Out]

-1/2*b^2*log((b*e^(-2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(-2*x) + 2*a + b + 2*sqrt((a + b)*a)))/(sqrt((a +
b)*a)*(a^2 + 2*a*b + b^2)) - 2/3*(6*(a + 2*b)*e^(-2*x) - 3*b*e^(-4*x) - 2*a - 5*b)/(a^2 + 2*a*b + b^2 - 3*(a^
2 + 2*a*b + b^2)*e^(-2*x) + 3*(a^2 + 2*a*b + b^2)*e^(-4*x) - (a^2 + 2*a*b + b^2)*e^(-6*x))

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mupad [B]  time = 1.45, size = 245, normalized size = 4.15 \[ \frac {2\,b}{{\left (a+b\right )}^2\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {4}{\left (a+b\right )\,\left ({\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1\right )}-\frac {8}{3\,\left (a+b\right )\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {b^2\,\ln \left (\frac {4\,b^2\,\left (2\,a\,b+8\,a^2\,{\mathrm {e}}^{2\,x}+b^2\,{\mathrm {e}}^{2\,x}+b^2+8\,a\,b\,{\mathrm {e}}^{2\,x}\right )}{a\,{\left (a+b\right )}^5}-\frac {8\,b^2\,\left (b+4\,a\,{\mathrm {e}}^{2\,x}+2\,b\,{\mathrm {e}}^{2\,x}\right )}{\sqrt {a}\,{\left (a+b\right )}^{9/2}}\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}+\frac {b^2\,\ln \left (\frac {8\,b^2\,\left (b+4\,a\,{\mathrm {e}}^{2\,x}+2\,b\,{\mathrm {e}}^{2\,x}\right )}{\sqrt {a}\,{\left (a+b\right )}^{9/2}}+\frac {4\,b^2\,\left (2\,a\,b+8\,a^2\,{\mathrm {e}}^{2\,x}+b^2\,{\mathrm {e}}^{2\,x}+b^2+8\,a\,b\,{\mathrm {e}}^{2\,x}\right )}{a\,{\left (a+b\right )}^5}\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^4*(a + b*cosh(x)^2)),x)

[Out]

(2*b)/((a + b)^2*(exp(2*x) - 1)) - 4/((a + b)*(exp(4*x) - 2*exp(2*x) + 1)) - 8/(3*(a + b)*(3*exp(2*x) - 3*exp(
4*x) + exp(6*x) - 1)) - (b^2*log((4*b^2*(2*a*b + 8*a^2*exp(2*x) + b^2*exp(2*x) + b^2 + 8*a*b*exp(2*x)))/(a*(a
+ b)^5) - (8*b^2*(b + 4*a*exp(2*x) + 2*b*exp(2*x)))/(a^(1/2)*(a + b)^(9/2))))/(2*a^(1/2)*(a + b)^(5/2)) + (b^2
*log((8*b^2*(b + 4*a*exp(2*x) + 2*b*exp(2*x)))/(a^(1/2)*(a + b)^(9/2)) + (4*b^2*(2*a*b + 8*a^2*exp(2*x) + b^2*
exp(2*x) + b^2 + 8*a*b*exp(2*x)))/(a*(a + b)^5)))/(2*a^(1/2)*(a + b)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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