3.24 \(\int \coth (x) (a+b \coth ^2(x))^{3/2} \, dx\)
Optimal. Leaf size=63 \[ -(a+b) \sqrt {a+b \coth ^2(x)}-\frac {1}{3} \left (a+b \coth ^2(x)\right )^{3/2}+(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right ) \]
[Out]
(a+b)^(3/2)*arctanh((a+b*coth(x)^2)^(1/2)/(a+b)^(1/2))-1/3*(a+b*coth(x)^2)^(3/2)-(a+b)*(a+b*coth(x)^2)^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.10, antiderivative size = 63, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used =
{3670, 444, 50, 63, 208} \[ -(a+b) \sqrt {a+b \coth ^2(x)}-\frac {1}{3} \left (a+b \coth ^2(x)\right )^{3/2}+(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right ) \]
Antiderivative was successfully verified.
[In]
Int[Coth[x]*(a + b*Coth[x]^2)^(3/2),x]
[Out]
(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Coth[x]^2]/Sqrt[a + b]] - (a + b)*Sqrt[a + b*Coth[x]^2] - (a + b*Coth[x]^2)^(
3/2)/3
Rule 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
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 444
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]
Rule 3670
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))
Rubi steps
\begin {align*} \int \coth (x) \left (a+b \coth ^2(x)\right )^{3/2} \, dx &=\operatorname {Subst}\left (\int \frac {x \left (a+b x^2\right )^{3/2}}{1-x^2} \, dx,x,\coth (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{1-x} \, dx,x,\coth ^2(x)\right )\\ &=-\frac {1}{3} \left (a+b \coth ^2(x)\right )^{3/2}+\frac {1}{2} (a+b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{1-x} \, dx,x,\coth ^2(x)\right )\\ &=-(a+b) \sqrt {a+b \coth ^2(x)}-\frac {1}{3} \left (a+b \coth ^2(x)\right )^{3/2}+\frac {1}{2} (a+b)^2 \operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\coth ^2(x)\right )\\ &=-(a+b) \sqrt {a+b \coth ^2(x)}-\frac {1}{3} \left (a+b \coth ^2(x)\right )^{3/2}+\frac {(a+b)^2 \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \coth ^2(x)}\right )}{b}\\ &=(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right )-(a+b) \sqrt {a+b \coth ^2(x)}-\frac {1}{3} \left (a+b \coth ^2(x)\right )^{3/2}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.16, size = 59, normalized size = 0.94 \[ (a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \coth ^2(x)}}{\sqrt {a+b}}\right )-\frac {1}{3} \sqrt {a+b \coth ^2(x)} \left (4 a+b \coth ^2(x)+3 b\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[x]*(a + b*Coth[x]^2)^(3/2),x]
[Out]
(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Coth[x]^2]/Sqrt[a + b]] - (Sqrt[a + b*Coth[x]^2]*(4*a + 3*b + b*Coth[x]^2))/3
________________________________________________________________________________________
fricas [B] time = 0.58, size = 2362, normalized size = 37.49 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)*(a+b*coth(x)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/12*(3*((a + b)*cosh(x)^6 + 6*(a + b)*cosh(x)*sinh(x)^5 + (a + b)*sinh(x)^6 - 3*(a + b)*cosh(x)^4 + 3*(5*(a
+ b)*cosh(x)^2 - a - b)*sinh(x)^4 + 4*(5*(a + b)*cosh(x)^3 - 3*(a + b)*cosh(x))*sinh(x)^3 + 3*(a + b)*cosh(x)^
2 + 3*(5*(a + b)*cosh(x)^4 - 6*(a + b)*cosh(x)^2 + a + b)*sinh(x)^2 + 6*((a + b)*cosh(x)^5 - 2*(a + b)*cosh(x)
^3 + (a + b)*cosh(x))*sinh(x) - a - b)*sqrt(a + b)*log(-((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)*cosh(x)*sin
h(x)^7 + (a^3 + a^2*b)*sinh(x)^8 - 2*(2*a^3 + a^2*b)*cosh(x)^6 - 2*(2*a^3 + a^2*b - 14*(a^3 + a^2*b)*cosh(x)^2
)*sinh(x)^6 + 4*(14*(a^3 + a^2*b)*cosh(x)^3 - 3*(2*a^3 + a^2*b)*cosh(x))*sinh(x)^5 + (6*a^3 + 4*a^2*b - a*b^2
+ b^3)*cosh(x)^4 + (70*(a^3 + a^2*b)*cosh(x)^4 + 6*a^3 + 4*a^2*b - a*b^2 + b^3 - 30*(2*a^3 + a^2*b)*cosh(x)^2)
*sinh(x)^4 + 4*(14*(a^3 + a^2*b)*cosh(x)^5 - 10*(2*a^3 + a^2*b)*cosh(x)^3 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*co
sh(x))*sinh(x)^3 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 - 2*(2*a^3 + 3*a^2*b - b^3)*cosh(x)^2 + 2*(14*(a^3 + a^2*b)*c
osh(x)^6 - 15*(2*a^3 + a^2*b)*cosh(x)^4 - 2*a^3 - 3*a^2*b + b^3 + 3*(6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^2)
*sinh(x)^2 + sqrt(2)*(a^2*cosh(x)^6 + 6*a^2*cosh(x)*sinh(x)^5 + a^2*sinh(x)^6 - 3*a^2*cosh(x)^4 + 3*(5*a^2*cos
h(x)^2 - a^2)*sinh(x)^4 + 4*(5*a^2*cosh(x)^3 - 3*a^2*cosh(x))*sinh(x)^3 + (3*a^2 + 2*a*b - b^2)*cosh(x)^2 + (1
5*a^2*cosh(x)^4 - 18*a^2*cosh(x)^2 + 3*a^2 + 2*a*b - b^2)*sinh(x)^2 - a^2 - 2*a*b - b^2 + 2*(3*a^2*cosh(x)^5 -
6*a^2*cosh(x)^3 + (3*a^2 + 2*a*b - b^2)*cosh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(
x)^2 - a + b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a^3 + a^2*b)*cosh(x)^7 - 3*(2*a^3 + a^2*b)*
cosh(x)^5 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^3 - (2*a^3 + 3*a^2*b - b^3)*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)*si
nh(x)^5 + sinh(x)^6)) + 3*((a + b)*cosh(x)^6 + 6*(a + b)*cosh(x)*sinh(x)^5 + (a + b)*sinh(x)^6 - 3*(a + b)*cos
h(x)^4 + 3*(5*(a + b)*cosh(x)^2 - a - b)*sinh(x)^4 + 4*(5*(a + b)*cosh(x)^3 - 3*(a + b)*cosh(x))*sinh(x)^3 + 3
*(a + b)*cosh(x)^2 + 3*(5*(a + b)*cosh(x)^4 - 6*(a + b)*cosh(x)^2 + a + b)*sinh(x)^2 + 6*((a + b)*cosh(x)^5 -
2*(a + b)*cosh(x)^3 + (a + b)*cosh(x))*sinh(x) - a - b)*sqrt(a + b)*log(((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)
*sinh(x)^3 + (a + b)*sinh(x)^4 + 2*b*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 + b)*sinh(x)^2 + sqrt(2)*(cosh(x)^2 +
2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 - a + b)/(cosh(x)^2
- 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*((a + b)*cosh(x)^3 + b*cosh(x))*sinh(x) + a + b)/(cosh(x)^2 + 2*cosh(x)
*sinh(x) + sinh(x)^2)) - 16*sqrt(2)*((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 - (2*
a + b)*cosh(x)^2 + (6*(a + b)*cosh(x)^2 - 2*a - b)*sinh(x)^2 + 2*(2*(a + b)*cosh(x)^3 - (2*a + b)*cosh(x))*sin
h(x) + a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 - a + b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)
))/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 - 1)*sinh(x)^4 - 3*cosh(x)^4 + 4*(5*cosh(x)^3
- 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 - 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 - 2*cosh(x)
^3 + cosh(x))*sinh(x) - 1), -1/6*(3*((a + b)*cosh(x)^6 + 6*(a + b)*cosh(x)*sinh(x)^5 + (a + b)*sinh(x)^6 - 3*(
a + b)*cosh(x)^4 + 3*(5*(a + b)*cosh(x)^2 - a - b)*sinh(x)^4 + 4*(5*(a + b)*cosh(x)^3 - 3*(a + b)*cosh(x))*sin
h(x)^3 + 3*(a + b)*cosh(x)^2 + 3*(5*(a + b)*cosh(x)^4 - 6*(a + b)*cosh(x)^2 + a + b)*sinh(x)^2 + 6*((a + b)*co
sh(x)^5 - 2*(a + b)*cosh(x)^3 + (a + b)*cosh(x))*sinh(x) - a - b)*sqrt(-a - b)*arctan(sqrt(2)*(a*cosh(x)^2 + 2
*a*cosh(x)*sinh(x) + a*sinh(x)^2 - a - b)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 - a + b)/(c
osh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a^2 + a*b)*cosh(x)^4 + 4*(a^2 + a*b)*cosh(x)*sinh(x)^3 + (a^2 + a
*b)*sinh(x)^4 - (2*a^2 + a*b - b^2)*cosh(x)^2 + (6*(a^2 + a*b)*cosh(x)^2 - 2*a^2 - a*b + b^2)*sinh(x)^2 + a^2
+ 2*a*b + b^2 + 2*(2*(a^2 + a*b)*cosh(x)^3 - (2*a^2 + a*b - b^2)*cosh(x))*sinh(x))) + 3*((a + b)*cosh(x)^6 + 6
*(a + b)*cosh(x)*sinh(x)^5 + (a + b)*sinh(x)^6 - 3*(a + b)*cosh(x)^4 + 3*(5*(a + b)*cosh(x)^2 - a - b)*sinh(x)
^4 + 4*(5*(a + b)*cosh(x)^3 - 3*(a + b)*cosh(x))*sinh(x)^3 + 3*(a + b)*cosh(x)^2 + 3*(5*(a + b)*cosh(x)^4 - 6*
(a + b)*cosh(x)^2 + a + b)*sinh(x)^2 + 6*((a + b)*cosh(x)^5 - 2*(a + b)*cosh(x)^3 + (a + b)*cosh(x))*sinh(x) -
a - b)*sqrt(-a - b)*arctan(sqrt(2)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 - a + b)/(cosh(x)
^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a +
b)) + 8*sqrt(2)*((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 - (2*a + b)*cosh(x)^2 + (
6*(a + b)*cosh(x)^2 - 2*a - b)*sinh(x)^2 + 2*(2*(a + b)*cosh(x)^3 - (2*a + b)*cosh(x))*sinh(x) + a + b)*sqrt((
(a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 - a + b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(cosh(x)^6 + 6*co
sh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 - 1)*sinh(x)^4 - 3*cosh(x)^4 + 4*(5*cosh(x)^3 - 3*cosh(x))*sinh(x
)^3 + 3*(5*cosh(x)^4 - 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 - 2*cosh(x)^3 + cosh(x))*sinh(x
) - 1)]
________________________________________________________________________________________
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)*(a+b*coth(x)^2)^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(exp(2*x)-1)]Evaluation time: 2.32Error: Bad Argument Type
________________________________________________________________________________________
maple [B] time = 0.07, size = 578, normalized size = 9.17 \[ -\frac {\left (\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b \right )^{\frac {3}{2}}}{6}-\frac {b \sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}\, \coth \relax (x )}{4}-\frac {3 \sqrt {b}\, \ln \left (\frac {\left (\coth \relax (x )-1\right ) b +b}{\sqrt {b}}+\sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}\right ) a}{4}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b +2 \left (\coth \relax (x )-1\right ) b +2 \sqrt {a +b}\, \sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}}{\coth \relax (x )-1}\right ) a}{2}-\frac {\sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}\, a}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b +2 \left (\coth \relax (x )-1\right ) b +2 \sqrt {a +b}\, \sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}}{\coth \relax (x )-1}\right ) b}{2}-\frac {b^{\frac {3}{2}} \ln \left (\frac {\left (\coth \relax (x )-1\right ) b +b}{\sqrt {b}}+\sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}\right )}{2}-\frac {\sqrt {\left (\coth \relax (x )-1\right )^{2} b +2 \left (\coth \relax (x )-1\right ) b +a +b}\, b}{2}-\frac {\left (\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b \right )^{\frac {3}{2}}}{6}+\frac {b \sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}\, \coth \relax (x )}{4}+\frac {3 \sqrt {b}\, \ln \left (\frac {\left (1+\coth \relax (x )\right ) b -b}{\sqrt {b}}+\sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}\right ) a}{4}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b -2 \left (1+\coth \relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}}{1+\coth \relax (x )}\right ) a}{2}-\frac {\sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}\, a}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 a +2 b -2 \left (1+\coth \relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}}{1+\coth \relax (x )}\right ) b}{2}+\frac {b^{\frac {3}{2}} \ln \left (\frac {\left (1+\coth \relax (x )\right ) b -b}{\sqrt {b}}+\sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}\right )}{2}-\frac {\sqrt {\left (1+\coth \relax (x )\right )^{2} b -2 \left (1+\coth \relax (x )\right ) b +a +b}\, b}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)*(a+b*coth(x)^2)^(3/2),x)
[Out]
-1/6*((coth(x)-1)^2*b+2*(coth(x)-1)*b+a+b)^(3/2)-1/4*b*((coth(x)-1)^2*b+2*(coth(x)-1)*b+a+b)^(1/2)*coth(x)-3/4
*b^(1/2)*ln(((coth(x)-1)*b+b)/b^(1/2)+((coth(x)-1)^2*b+2*(coth(x)-1)*b+a+b)^(1/2))*a+1/2*(a+b)^(1/2)*ln((2*a+2
*b+2*(coth(x)-1)*b+2*(a+b)^(1/2)*((coth(x)-1)^2*b+2*(coth(x)-1)*b+a+b)^(1/2))/(coth(x)-1))*a-1/2*((coth(x)-1)^
2*b+2*(coth(x)-1)*b+a+b)^(1/2)*a+1/2*(a+b)^(1/2)*ln((2*a+2*b+2*(coth(x)-1)*b+2*(a+b)^(1/2)*((coth(x)-1)^2*b+2*
(coth(x)-1)*b+a+b)^(1/2))/(coth(x)-1))*b-1/2*b^(3/2)*ln(((coth(x)-1)*b+b)/b^(1/2)+((coth(x)-1)^2*b+2*(coth(x)-
1)*b+a+b)^(1/2))-1/2*((coth(x)-1)^2*b+2*(coth(x)-1)*b+a+b)^(1/2)*b-1/6*((1+coth(x))^2*b-2*(1+coth(x))*b+a+b)^(
3/2)+1/4*b*((1+coth(x))^2*b-2*(1+coth(x))*b+a+b)^(1/2)*coth(x)+3/4*b^(1/2)*ln(((1+coth(x))*b-b)/b^(1/2)+((1+co
th(x))^2*b-2*(1+coth(x))*b+a+b)^(1/2))*a+1/2*(a+b)^(1/2)*ln((2*a+2*b-2*(1+coth(x))*b+2*(a+b)^(1/2)*((1+coth(x)
)^2*b-2*(1+coth(x))*b+a+b)^(1/2))/(1+coth(x)))*a-1/2*((1+coth(x))^2*b-2*(1+coth(x))*b+a+b)^(1/2)*a+1/2*(a+b)^(
1/2)*ln((2*a+2*b-2*(1+coth(x))*b+2*(a+b)^(1/2)*((1+coth(x))^2*b-2*(1+coth(x))*b+a+b)^(1/2))/(1+coth(x)))*b+1/2
*b^(3/2)*ln(((1+coth(x))*b-b)/b^(1/2)+((1+coth(x))^2*b-2*(1+coth(x))*b+a+b)^(1/2))-1/2*((1+coth(x))^2*b-2*(1+c
oth(x))*b+a+b)^(1/2)*b
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \coth \relax (x)^{2} + a\right )}^{\frac {3}{2}} \coth \relax (x)\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)*(a+b*coth(x)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*coth(x)^2 + a)^(3/2)*coth(x), x)
________________________________________________________________________________________
mupad [B] time = 3.40, size = 64, normalized size = 1.02 \[ \mathrm {atanh}\left (\frac {{\left (a+b\right )}^{3/2}\,\sqrt {b\,{\mathrm {coth}\relax (x)}^2+a}}{a^2+2\,a\,b+b^2}\right )\,{\left (a+b\right )}^{3/2}-\left (a+b\right )\,\sqrt {b\,{\mathrm {coth}\relax (x)}^2+a}-\frac {{\left (b\,{\mathrm {coth}\relax (x)}^2+a\right )}^{3/2}}{3} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(x)*(a + b*coth(x)^2)^(3/2),x)
[Out]
atanh(((a + b)^(3/2)*(a + b*coth(x)^2)^(1/2))/(2*a*b + a^2 + b^2))*(a + b)^(3/2) - (a + b)*(a + b*coth(x)^2)^(
1/2) - (a + b*coth(x)^2)^(3/2)/3
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \coth ^{2}{\relax (x )}\right )^{\frac {3}{2}} \coth {\relax (x )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(x)*(a+b*coth(x)**2)**(3/2),x)
[Out]
Integral((a + b*coth(x)**2)**(3/2)*coth(x), x)
________________________________________________________________________________________