3.3.1 \(\int \frac {\coth ^2(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx\) [201]
Optimal. Leaf size=53 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{\sqrt {a}}-\frac {\coth (x) \sqrt {a+b-b \tanh ^2(x)}}{a+b} \]
[Out]
arctanh(a^(1/2)*tanh(x)/(a+b-b*tanh(x)^2)^(1/2))/a^(1/2)-coth(x)*(a+b-b*tanh(x)^2)^(1/2)/(a+b)
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Rubi [A]
time = 0.13, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {4226, 2000,
491, 12, 385, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )}{\sqrt {a}}-\frac {\coth (x) \sqrt {a-b \tanh ^2(x)+b}}{a+b} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Coth[x]^2/Sqrt[a + b*Sech[x]^2],x]
[Out]
ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]]/Sqrt[a] - (Coth[x]*Sqrt[a + b - b*Tanh[x]^2])/(a + b)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 385
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 491
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]
Rule 2000
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&& !BinomialMatchQ[{u, v}, x]
Rule 4226
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])
Rubi steps
\begin {align*} \int \frac {\coth ^2(x)}{\sqrt {a+b \text {sech}^2(x)}} \, dx &=\text {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \sqrt {a+b \left (1-x^2\right )}} \, dx,x,\tanh (x)\right )\\ &=\text {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=-\frac {\coth (x) \sqrt {a+b-b \tanh ^2(x)}}{a+b}+\frac {\text {Subst}\left (\int \frac {a+b}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )}{a+b}\\ &=-\frac {\coth (x) \sqrt {a+b-b \tanh ^2(x)}}{a+b}+\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=-\frac {\coth (x) \sqrt {a+b-b \tanh ^2(x)}}{a+b}+\text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )}{\sqrt {a}}-\frac {\coth (x) \sqrt {a+b-b \tanh ^2(x)}}{a+b}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 94, normalized size = 1.77 \begin {gather*} \frac {\sqrt {a+2 b+a \cosh (2 x)} \text {sech}(x) \left ((a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {a+b+a \sinh ^2(x)}}\right )-\sqrt {a} \text {csch}(x) \sqrt {a+b+a \sinh ^2(x)}\right )}{\sqrt {2} \sqrt {a} (a+b) \sqrt {a+b \text {sech}^2(x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]^2/Sqrt[a + b*Sech[x]^2],x]
[Out]
(Sqrt[a + 2*b + a*Cosh[2*x]]*Sech[x]*((a + b)*ArcTanh[(Sqrt[a]*Sinh[x])/Sqrt[a + b + a*Sinh[x]^2]] - Sqrt[a]*C
sch[x]*Sqrt[a + b + a*Sinh[x]^2]))/(Sqrt[2]*Sqrt[a]*(a + b)*Sqrt[a + b*Sech[x]^2])
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Maple [F]
time = 1.80, size = 0, normalized size = 0.00 \[\int \frac {\coth ^{2}\left (x \right )}{\sqrt {a +b \mathrm {sech}\left (x \right )^{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)^2/(a+b*sech(x)^2)^(1/2),x)
[Out]
int(coth(x)^2/(a+b*sech(x)^2)^(1/2),x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^2/(a+b*sech(x)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(coth(x)^2/sqrt(b*sech(x)^2 + a), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 403 vs.
\(2 (45) = 90\).
time = 0.45, size = 1365, normalized size = 25.75 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^2/(a+b*sech(x)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/4*(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - a - b)*sqrt(a)*log((a*b^2*cosh(x)^8
+ 8*a*b^2*cosh(x)*sinh(x)^7 + a*b^2*sinh(x)^8 - 2*(a*b^2 - b^3)*cosh(x)^6 + 2*(14*a*b^2*cosh(x)^2 - a*b^2 + b
^3)*sinh(x)^6 + 4*(14*a*b^2*cosh(x)^3 - 3*(a*b^2 - b^3)*cosh(x))*sinh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)
^4 + (70*a*b^2*cosh(x)^4 + a^3 + 4*a^2*b + 9*a*b^2 - 30*(a*b^2 - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*a*b^2*cosh(
x)^5 - 10*(a*b^2 - b^3)*cosh(x)^3 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 + 2*(a^3 + 3*a^2*b)*cos
h(x)^2 + 2*(14*a*b^2*cosh(x)^6 - 15*(a*b^2 - b^3)*cosh(x)^4 + a^3 + 3*a^2*b + 3*(a^3 + 4*a^2*b + 9*a*b^2)*cosh
(x)^2)*sinh(x)^2 + sqrt(2)*(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 + 4*(5*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 + 4*a*b)*cosh(x)^2 + (15*
b^2*cosh(x)^4 - 18*b^2*cosh(x)^2 - a^2 - 4*a*b)*sinh(x)^2 - a^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)^3 - (a^2
+ 4*a*b)*cosh(x))*sinh(x))*sqrt(a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) +
sinh(x)^2)) + 4*(2*a*b^2*cosh(x)^7 - 3*(a*b^2 - b^3)*cosh(x)^5 + (a^3 + 4*a^2*b + 9*a*b^2)*cosh(x)^3 + (a^3 +
3*a^2*b)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3
+ 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6)) + ((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x)
+ (a + b)*sinh(x)^2 - a - b)*sqrt(a)*log(-(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*(a + b)*cosh
(x)^2 + 2*(3*a*cosh(x)^2 + a + b)*sinh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a)*
sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(a*cosh(x)^3 + (a
+ b)*cosh(x))*sinh(x) + a)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 4*sqrt(2)*a*sqrt((a*cosh(x)^2 + a*si
nh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^2 + a*b)*cosh(x)^2 + 2*(a^2 + a*b)*cosh(x
)*sinh(x) + (a^2 + a*b)*sinh(x)^2 - a^2 - a*b), -1/2*(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)
*sinh(x)^2 - a - b)*sqrt(-a)*arctan(sqrt(2)*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + a)*sqrt(-a)*sqr
t((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(a*b*cosh(x)^4 + 4*a*b*co
sh(x)*sinh(x)^3 + a*b*sinh(x)^4 - (a^2 + 3*a*b)*cosh(x)^2 + (6*a*b*cosh(x)^2 - a^2 - 3*a*b)*sinh(x)^2 - a^2 +
2*(2*a*b*cosh(x)^3 - (a^2 + 3*a*b)*cosh(x))*sinh(x))) + ((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a +
b)*sinh(x)^2 - a - b)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 -
2*cosh(x)*sinh(x) + sinh(x)^2))/(a*cosh(x)^2 + 2*a*cosh(x)*sinh(x) + a*sinh(x)^2 + a)) + 2*sqrt(2)*a*sqrt((a*
cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/((a^2 + a*b)*cosh(x)^2 + 2*(a
^2 + a*b)*cosh(x)*sinh(x) + (a^2 + a*b)*sinh(x)^2 - a^2 - a*b)]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth ^{2}{\left (x \right )}}{\sqrt {a + b \operatorname {sech}^{2}{\left (x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)**2/(a+b*sech(x)**2)**(1/2),x)
[Out]
Integral(coth(x)**2/sqrt(a + b*sech(x)**2), x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)^2/(a+b*sech(x)^2)^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {coth}\left (x\right )}^2}{\sqrt {a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)^2/(a + b/cosh(x)^2)^(1/2),x)
[Out]
int(coth(x)^2/(a + b/cosh(x)^2)^(1/2), x)
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