3.11 \(\int \frac{\text{csch}^3(x)}{a+b \cosh ^2(x)} \, dx\)
Optimal. Leaf size=61 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^2}+\frac{(a+3 b) \tanh ^{-1}(\cosh (x))}{2 (a+b)^2}-\frac{\coth (x) \text{csch}(x)}{2 (a+b)} \]
[Out]
(b^(3/2)*ArcTan[(Sqrt[b]*Cosh[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2) + ((a + 3*b)*ArcTanh[Cosh[x]])/(2*(a + b)^2) -
(Coth[x]*Csch[x])/(2*(a + b))
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Rubi [A] time = 0.104366, antiderivative size = 61, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{3190, 414, 522, 206, 205} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^2}+\frac{(a+3 b) \tanh ^{-1}(\cosh (x))}{2 (a+b)^2}-\frac{\coth (x) \text{csch}(x)}{2 (a+b)} \]
Antiderivative was successfully verified.
[In]
Int[Csch[x]^3/(a + b*Cosh[x]^2),x]
[Out]
(b^(3/2)*ArcTan[(Sqrt[b]*Cosh[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2) + ((a + 3*b)*ArcTanh[Cosh[x]])/(2*(a + b)^2) -
(Coth[x]*Csch[x])/(2*(a + b))
Rule 3190
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 414
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]
Rule 522
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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{\text{csch}^3(x)}{a+b \cosh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\cosh (x)\right )\\ &=-\frac{\coth (x) \text{csch}(x)}{2 (a+b)}+\frac{\operatorname{Subst}\left (\int \frac{a+2 b+b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\cosh (x)\right )}{2 (a+b)}\\ &=-\frac{\coth (x) \text{csch}(x)}{2 (a+b)}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\cosh (x)\right )}{(a+b)^2}+\frac{(a+3 b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cosh (x)\right )}{2 (a+b)^2}\\ &=\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \cosh (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^2}+\frac{(a+3 b) \tanh ^{-1}(\cosh (x))}{2 (a+b)^2}-\frac{\coth (x) \text{csch}(x)}{2 (a+b)}\\ \end{align*}
Mathematica [C] time = 0.278086, size = 154, normalized size = 2.52 \[ \frac{-4 a^{3/2} \log \left (\tanh \left (\frac{x}{2}\right )\right )+8 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a+b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a}}\right )+8 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a+b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a}}\right )-\sqrt{a} (a+b) \text{csch}^2\left (\frac{x}{2}\right )-\sqrt{a} (a+b) \text{sech}^2\left (\frac{x}{2}\right )-12 \sqrt{a} b \log \left (\tanh \left (\frac{x}{2}\right )\right )}{8 \sqrt{a} (a+b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Csch[x]^3/(a + b*Cosh[x]^2),x]
[Out]
(8*b^(3/2)*ArcTan[(Sqrt[b] - I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]] + 8*b^(3/2)*ArcTan[(Sqrt[b] + I*Sqrt[a + b]*Tan
h[x/2])/Sqrt[a]] - Sqrt[a]*(a + b)*Csch[x/2]^2 - 4*a^(3/2)*Log[Tanh[x/2]] - 12*Sqrt[a]*b*Log[Tanh[x/2]] - Sqrt
[a]*(a + b)*Sech[x/2]^2)/(8*Sqrt[a]*(a + b)^2)
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Maple [A] time = 0.029, size = 94, normalized size = 1.5 \begin{align*}{\frac{1}{8\,a+8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{1}{8\,a+8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{a}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{3\,b}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{b}^{2}}{ \left ( a+b \right ) ^{2}}\arctan \left ({\frac{1}{4} \left ( 2\, \left ( a+b \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,a+2\,b \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csch(x)^3/(a+b*cosh(x)^2),x)
[Out]
1/8*tanh(1/2*x)^2/(a+b)-1/8/(a+b)/tanh(1/2*x)^2-1/2/(a+b)^2*ln(tanh(1/2*x))*a-3/2/(a+b)^2*ln(tanh(1/2*x))*b+1/
(a+b)^2*b^2/(a*b)^(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*x)^2-2*a+2*b)/(a*b)^(1/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (a + 3 \, b\right )} \log \left (e^{x} + 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{{\left (a + 3 \, b\right )} \log \left (e^{x} - 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{e^{\left (3 \, x\right )} + e^{x}}{{\left (a + b\right )} e^{\left (4 \, x\right )} - 2 \,{\left (a + b\right )} e^{\left (2 \, x\right )} + a + b} + 8 \, \int \frac{b^{2} e^{\left (3 \, x\right )} - b^{2} e^{x}}{4 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3} +{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} e^{\left (4 \, x\right )} + 2 \,{\left (2 \, a^{3} + 5 \, a^{2} b + 4 \, a b^{2} + b^{3}\right )} e^{\left (2 \, x\right )}\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(x)^3/(a+b*cosh(x)^2),x, algorithm="maxima")
[Out]
1/2*(a + 3*b)*log(e^x + 1)/(a^2 + 2*a*b + b^2) - 1/2*(a + 3*b)*log(e^x - 1)/(a^2 + 2*a*b + b^2) - (e^(3*x) + e
^x)/((a + b)*e^(4*x) - 2*(a + b)*e^(2*x) + a + b) + 8*integrate(1/4*(b^2*e^(3*x) - b^2*e^x)/(a^2*b + 2*a*b^2 +
b^3 + (a^2*b + 2*a*b^2 + b^3)*e^(4*x) + 2*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*e^(2*x)), x)
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Fricas [B] time = 2.52114, size = 3981, normalized size = 65.26 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(x)^3/(a+b*cosh(x)^2),x, algorithm="fricas")
[Out]
[-1/2*(2*(a + b)*cosh(x)^3 + 6*(a + b)*cosh(x)*sinh(x)^2 + 2*(a + b)*sinh(x)^3 - (b*cosh(x)^4 + 4*b*cosh(x)*si
nh(x)^3 + b*sinh(x)^4 - 2*b*cosh(x)^2 + 2*(3*b*cosh(x)^2 - b)*sinh(x)^2 + 4*(b*cosh(x)^3 - b*cosh(x))*sinh(x)
+ b)*sqrt(-b/a)*log((b*cosh(x)^4 + 4*b*cosh(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) + 4*(a*cosh(x)^3 + 3*a*cosh(x)*sinh(x)^
2 + a*sinh(x)^3 + a*cosh(x) + (3*a*cosh(x)^2 + a)*sinh(x))*sqrt(-b/a) + b)/(b*cosh(x)^4 + 4*b*cosh(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)*c
osh(x))*sinh(x) + b)) + 2*(a + b)*cosh(x) - ((a + 3*b)*cosh(x)^4 + 4*(a + 3*b)*cosh(x)*sinh(x)^3 + (a + 3*b)*s
inh(x)^4 - 2*(a + 3*b)*cosh(x)^2 + 2*(3*(a + 3*b)*cosh(x)^2 - a - 3*b)*sinh(x)^2 + 4*((a + 3*b)*cosh(x)^3 - (a
+ 3*b)*cosh(x))*sinh(x) + a + 3*b)*log(cosh(x) + sinh(x) + 1) + ((a + 3*b)*cosh(x)^4 + 4*(a + 3*b)*cosh(x)*si
nh(x)^3 + (a + 3*b)*sinh(x)^4 - 2*(a + 3*b)*cosh(x)^2 + 2*(3*(a + 3*b)*cosh(x)^2 - a - 3*b)*sinh(x)^2 + 4*((a
+ 3*b)*cosh(x)^3 - (a + 3*b)*cosh(x))*sinh(x) + a + 3*b)*log(cosh(x) + sinh(x) - 1) + 2*(3*(a + b)*cosh(x)^2 +
a + b)*sinh(x))/((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2
)*sinh(x)^4 - 2*(a^2 + 2*a*b + b^2)*cosh(x)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(x)^2 - a^2 - 2*a*b - b^2)*sinh(x
)^2 + a^2 + 2*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 - (a^2 + 2*a*b + b^2)*cosh(x))*sinh(x)), -1/2*(2*(a
+ b)*cosh(x)^3 + 6*(a + b)*cosh(x)*sinh(x)^2 + 2*(a + b)*sinh(x)^3 - 2*(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 +
b*sinh(x)^4 - 2*b*cosh(x)^2 + 2*(3*b*cosh(x)^2 - b)*sinh(x)^2 + 4*(b*cosh(x)^3 - b*cosh(x))*sinh(x) + b)*sqrt
(b/a)*arctan(1/2*sqrt(b/a)*(cosh(x) + sinh(x))) + 2*(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 2*b*c
osh(x)^2 + 2*(3*b*cosh(x)^2 - b)*sinh(x)^2 + 4*(b*cosh(x)^3 - b*cosh(x))*sinh(x) + b)*sqrt(b/a)*arctan(1/2*(b*
cosh(x)^3 + 3*b*cosh(x)*sinh(x)^2 + b*sinh(x)^3 + (4*a + b)*cosh(x) + (3*b*cosh(x)^2 + 4*a + b)*sinh(x))*sqrt(
b/a)/b) + 2*(a + b)*cosh(x) - ((a + 3*b)*cosh(x)^4 + 4*(a + 3*b)*cosh(x)*sinh(x)^3 + (a + 3*b)*sinh(x)^4 - 2*(
a + 3*b)*cosh(x)^2 + 2*(3*(a + 3*b)*cosh(x)^2 - a - 3*b)*sinh(x)^2 + 4*((a + 3*b)*cosh(x)^3 - (a + 3*b)*cosh(x
))*sinh(x) + a + 3*b)*log(cosh(x) + sinh(x) + 1) + ((a + 3*b)*cosh(x)^4 + 4*(a + 3*b)*cosh(x)*sinh(x)^3 + (a +
3*b)*sinh(x)^4 - 2*(a + 3*b)*cosh(x)^2 + 2*(3*(a + 3*b)*cosh(x)^2 - a - 3*b)*sinh(x)^2 + 4*((a + 3*b)*cosh(x)
^3 - (a + 3*b)*cosh(x))*sinh(x) + a + 3*b)*log(cosh(x) + sinh(x) - 1) + 2*(3*(a + b)*cosh(x)^2 + a + b)*sinh(x
))/((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2)*sinh(x)^4 -
2*(a^2 + 2*a*b + b^2)*cosh(x)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(x)^2 - a^2 - 2*a*b - b^2)*sinh(x)^2 + a^2 + 2*
a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 - (a^2 + 2*a*b + b^2)*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)**3/(a+b*cosh(x)**2),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csch(x)^3/(a+b*cosh(x)^2),x, algorithm="giac")
[Out]
Exception raised: TypeError