3.183 \(\int \coth ^2(x) \sqrt {a+b \text {sech}^2(x)} \, dx\)
Optimal. Leaf size=48 \[ \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )-\coth (x) \sqrt {a-b \tanh ^2(x)+b} \]
[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)
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Rubi [A] time = 0.18, antiderivative size = 48, 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 =
{4141, 1975, 475, 12, 377, 206} \[ \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a-b \tanh ^2(x)+b}}\right )-\coth (x) \sqrt {a-b \tanh ^2(x)+b} \]
Antiderivative was successfully verified.
[In]
Int[Coth[x]^2*Sqrt[a + b*Sech[x]^2],x]
[Out]
Sqrt[a]*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b - b*Tanh[x]^2]] - Coth[x]*Sqrt[a + b - b*Tanh[x]^2]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 377
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 475
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)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 1975
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 4141
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 \coth ^2(x) \sqrt {a+b \text {sech}^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b \left (1-x^2\right )}}{x^2 \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b-b x^2}}{x^2 \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=-\coth (x) \sqrt {a+b-b \tanh ^2(x)}+\operatorname {Subst}\left (\int \frac {a}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=-\coth (x) \sqrt {a+b-b \tanh ^2(x)}+a \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b-b x^2}} \, dx,x,\tanh (x)\right )\\ &=-\coth (x) \sqrt {a+b-b \tanh ^2(x)}+a \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )\\ &=\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b-b \tanh ^2(x)}}\right )-\coth (x) \sqrt {a+b-b \tanh ^2(x)}\\ \end {align*}
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Mathematica [A] time = 0.52, size = 75, normalized size = 1.56 \[ \sqrt {a+b \text {sech}^2(x)} \left (\frac {\sqrt {2} \sqrt {a} \cosh (x) \sinh ^{-1}\left (\frac {\sqrt {a} \sinh (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b} \sqrt {\frac {a \cosh (2 x)+a+2 b}{a+b}}}-\coth (x)\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]^2*Sqrt[a + b*Sech[x]^2],x]
[Out]
((Sqrt[2]*Sqrt[a]*ArcSinh[(Sqrt[a]*Sinh[x])/Sqrt[a + b]]*Cosh[x])/(Sqrt[a + b]*Sqrt[(a + 2*b + a*Cosh[2*x])/(a
+ b)]) - Coth[x])*Sqrt[a + b*Sech[x]^2]
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fricas [B] time = 0.52, size = 1303, normalized size = 27.15 \[ \text {result too large to display} \]
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*((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*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)*cosh(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))*s
qrt(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*co
sh(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)) + (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*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)*(c
osh(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)*sqrt((a*cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x
)^2)))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1), -1/2*((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*
sqrt(-a)*arctan(sqrt(2)*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + a)*sqrt(-a)*sqrt((a*cosh(x)^2 + a*s
inh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(a*b*cosh(x)^4 + 4*a*b*cosh(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))) + (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-a)*arctan(sqrt(2)*(c
osh(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))/(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*(a + 2*b)*cosh(x)^2 +
2*(3*a*cosh(x)^2 + a + 2*b)*sinh(x)^2 + 4*(a*cosh(x)^3 + (a + 2*b)*cosh(x))*sinh(x) + a)) + 2*sqrt(2)*sqrt((a*
cosh(x)^2 + a*sinh(x)^2 + a + 2*b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(cosh(x)^2 + 2*cosh(x)*sinh(x
) + sinh(x)^2 - 1)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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,x):;OUTPUT:Erro
r: Bad Argument Type
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maple [F] time = 0.39, size = 0, normalized size = 0.00 \[ \int \left (\coth ^{2}\relax (x )\right ) \sqrt {a +b \mathrm {sech}\relax (x )^{2}}\, dx \]
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 \[ \int \sqrt {b \operatorname {sech}\relax (x)^{2} + a} \coth \relax (x)^{2}\,{d x} \]
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(sqrt(b*sech(x)^2 + a)*coth(x)^2, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {coth}\relax (x)}^2\,\sqrt {a+\frac {b}{{\mathrm {cosh}\relax (x)}^2}} \,d x \]
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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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \operatorname {sech}^{2}{\relax (x )}} \coth ^{2}{\relax (x )}\, dx \]
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(sqrt(a + b*sech(x)**2)*coth(x)**2, x)
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