Integrand size = 21, antiderivative size = 132 \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \coth ^2(x)}{2 \sqrt {c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \coth ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \]
-1/4*(b+2*c)*arctanh(1/2*(b+2*c*coth(x)^2)/c^(1/2)/(a+b*coth(x)^2+c*coth(x )^4)^(1/2))/c^(1/2)+1/2*arctanh(1/2*(2*a+b+(b+2*c)*coth(x)^2)/(a+b+c)^(1/2 )/(a+b*coth(x)^2+c*coth(x)^4)^(1/2))*(a+b+c)^(1/2)-1/2*(a+b*coth(x)^2+c*co th(x)^4)^(1/2)
Time = 1.19 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.39 \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=-\frac {\sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \tanh ^2(x) \left ((b+2 c) \text {arctanh}\left (\frac {2 c+b \tanh ^2(x)}{2 \sqrt {c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}}\right )-2 \sqrt {c} \sqrt {a+b+c} \text {arctanh}\left (\frac {b+2 c+(2 a+b) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}}\right )+2 \sqrt {c} \coth ^2(x) \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}\right )}{4 \sqrt {c} \sqrt {c+b \tanh ^2(x)+a \tanh ^4(x)}} \]
-1/4*(Sqrt[a + b*Coth[x]^2 + c*Coth[x]^4]*Tanh[x]^2*((b + 2*c)*ArcTanh[(2* c + b*Tanh[x]^2)/(2*Sqrt[c]*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])] - 2*Sqrt [c]*Sqrt[a + b + c]*ArcTanh[(b + 2*c + (2*a + b)*Tanh[x]^2)/(2*Sqrt[a + b + c]*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])] + 2*Sqrt[c]*Coth[x]^2*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4]))/(Sqrt[c]*Sqrt[c + b*Tanh[x]^2 + a*Tanh[x]^4])
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.64, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 26, 4184, 1576, 1162, 25, 1269, 1092, 219, 1154, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int i \cot (i x) \sqrt {a-b \cot (i x)^2+c \cot (i x)^4}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \cot (i x) \sqrt {c \cot (i x)^4-b \cot (i x)^2+a}dx\) |
\(\Big \downarrow \) 4184 |
\(\displaystyle -\int -\frac {i \coth (x) \sqrt {c \coth ^4(x)+b \coth ^2(x)+a}}{1-\coth ^2(x)}d(-i \coth (x))\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle -\frac {1}{2} \int \frac {\sqrt {-c \coth ^2(x)+i b \coth (x)+a}}{1-\coth ^2(x)}d\left (-\coth ^2(x)\right )\) |
\(\Big \downarrow \) 1162 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {2 a+b+i (b+2 c) \coth (x)}{\left (1-\coth ^2(x)\right ) \sqrt {-c \coth ^2(x)+i b \coth (x)+a}}d\left (-\coth ^2(x)\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {2 a+b+i (b+2 c) \coth (x)}{\left (1-\coth ^2(x)\right ) \sqrt {-c \coth ^2(x)+i b \coth (x)+a}}d\left (-\coth ^2(x)\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left ((b+2 c) \int \frac {1}{\sqrt {-c \coth ^2(x)+i b \coth (x)+a}}d\left (-\coth ^2(x)\right )-2 (a+b+c) \int \frac {1}{\left (1-\coth ^2(x)\right ) \sqrt {-c \coth ^2(x)+i b \coth (x)+a}}d\left (-\coth ^2(x)\right )\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (2 (b+2 c) \int \frac {1}{\coth ^2(x)+4 c}d\left (-\frac {b+2 i c \coth (x)}{\sqrt {-c \coth ^2(x)+i b \coth (x)+a}}\right )-2 (a+b+c) \int \frac {1}{\left (1-\coth ^2(x)\right ) \sqrt {-c \coth ^2(x)+i b \coth (x)+a}}d\left (-\coth ^2(x)\right )\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-2 (a+b+c) \int \frac {1}{\left (1-\coth ^2(x)\right ) \sqrt {-c \coth ^2(x)+i b \coth (x)+a}}d\left (-\coth ^2(x)\right )-\frac {i (b+2 c) \arctan \left (\frac {\coth (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (4 (a+b+c) \int \frac {1}{\coth ^2(x)+4 (a+b+c)}d\frac {2 a+b+i (b+2 c) \coth (x)}{\sqrt {-c \coth ^2(x)+i b \coth (x)+a}}-\frac {i (b+2 c) \arctan \left (\frac {\coth (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-2 i \sqrt {a+b+c} \arctan \left (\frac {\coth (x)}{2 \sqrt {a+b+c}}\right )-\frac {i (b+2 c) \arctan \left (\frac {\coth (x)}{2 \sqrt {c}}\right )}{\sqrt {c}}\right )-\sqrt {a+i b \coth (x)-c \coth ^2(x)}\right )\) |
((((-I)*(b + 2*c)*ArcTan[Coth[x]/(2*Sqrt[c])])/Sqrt[c] - (2*I)*Sqrt[a + b + c]*ArcTan[Coth[x]/(2*Sqrt[a + b + c])])/2 - Sqrt[a + I*b*Coth[x] - c*Cot h[x]^2])/2
3.3.10.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p/(e*(m + 2*p + 1)) Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x ] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] , x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
Int[cot[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*(cot[(d_.) + (e_.)*(x_)]*( f_.))^(n_.) + (c_.)*(cot[(d_.) + (e_.)*(x_)]*(f_.))^(n2_.))^(p_), x_Symbol] :> Simp[-f/e Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)), x], x, f*Cot[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[ n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Time = 0.82 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(-\frac {\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\coth \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{\coth \left (x \right )^{2}-1}\right )}{2}\) | \(165\) |
default | \(-\frac {\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\coth \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\coth \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\coth \left (x \right )^{2}-1\right )+a +b +c}}{\coth \left (x \right )^{2}-1}\right )}{2}\) | \(165\) |
-1/2*((coth(x)^2-1)^2*c+(b+2*c)*(coth(x)^2-1)+a+b+c)^(1/2)-1/4*(b+2*c)*ln( (1/2*b+c+c*(coth(x)^2-1))/c^(1/2)+((coth(x)^2-1)^2*c+(b+2*c)*(coth(x)^2-1) +a+b+c)^(1/2))/c^(1/2)+1/2*(a+b+c)^(1/2)*ln((2*a+2*b+2*c+(b+2*c)*(coth(x)^ 2-1)+2*(a+b+c)^(1/2)*((coth(x)^2-1)^2*c+(b+2*c)*(coth(x)^2-1)+a+b+c)^(1/2) )/(coth(x)^2-1))
Leaf count of result is larger than twice the leaf count of optimal. 1840 vs. \(2 (108) = 216\).
Time = 1.44 (sec) , antiderivative size = 7964, normalized size of antiderivative = 60.33 \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\text {Too large to display} \]
\[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int \sqrt {a + b \coth ^{2}{\left (x \right )} + c \coth ^{4}{\left (x \right )}} \coth {\left (x \right )}\, dx \]
\[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int { \sqrt {c \coth \left (x\right )^{4} + b \coth \left (x\right )^{2} + a} \coth \left (x\right ) \,d x } \]
\[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int { \sqrt {c \coth \left (x\right )^{4} + b \coth \left (x\right )^{2} + a} \coth \left (x\right ) \,d x } \]
Timed out. \[ \int \coth (x) \sqrt {a+b \coth ^2(x)+c \coth ^4(x)} \, dx=\int \mathrm {coth}\left (x\right )\,\sqrt {c\,{\mathrm {coth}\left (x\right )}^4+b\,{\mathrm {coth}\left (x\right )}^2+a} \,d x \]