Integrand size = 17, antiderivative size = 121 \[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=-\frac {\left (8 a^2+4 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )}{8 a^{3/2}}+\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\frac {(4 a+b) \coth ^2(x) \sqrt {a+b \tanh ^2(x)}}{8 a}-\frac {1}{4} \coth ^4(x) \sqrt {a+b \tanh ^2(x)} \]
-1/8*(8*a^2+4*a*b-b^2)*arctanh((a+b*tanh(x)^2)^(1/2)/a^(1/2))/a^(3/2)+arct anh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)-1/8*(4*a+b)*coth(x)^2*( a+b*tanh(x)^2)^(1/2)/a-1/4*coth(x)^4*(a+b*tanh(x)^2)^(1/2)
Time = 0.74 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=\frac {\left (-8 a^2-4 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )+\sqrt {a} \left (8 a \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\coth ^2(x) \left (4 a+b+2 a \coth ^2(x)\right ) \sqrt {a+b \tanh ^2(x)}\right )}{8 a^{3/2}} \]
((-8*a^2 - 4*a*b + b^2)*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a]] + Sqrt[a]*( 8*a*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]] - Coth[x]^2*(4* a + b + 2*a*Coth[x]^2)*Sqrt[a + b*Tanh[x]^2]))/(8*a^(3/2))
Time = 0.37 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {3042, 26, 4153, 26, 354, 110, 27, 168, 27, 174, 73, 221}
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.
\(\displaystyle \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \sqrt {a-b \tan (i x)^2}}{\tan (i x)^5}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\sqrt {a-b \tan (i x)^2}}{\tan (i x)^5}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle i \int -\frac {i \coth ^5(x) \sqrt {b \tanh ^2(x)+a}}{1-\tanh ^2(x)}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\coth ^5(x) \sqrt {a+b \tanh ^2(x)}}{1-\tanh ^2(x)}d\tanh (x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\coth ^3(x) \sqrt {b \tanh ^2(x)+a}}{1-\tanh ^2(x)}d\tanh ^2(x)\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {\coth ^2(x) \left (3 b \tanh ^2(x)+4 a+b\right )}{2 \left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \int \frac {\coth ^2(x) \left (3 b \tanh ^2(x)+4 a+b\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {\int -\frac {\coth (x) \left (8 a^2+4 b a-b^2+b (4 a+b) \tanh ^2(x)\right )}{2 \left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)}{a}-\frac {(4 a+b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}\right )-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\int \frac {\coth (x) \left (8 a^2+4 b a-b^2+b (4 a+b) \tanh ^2(x)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)}{2 a}-\frac {(4 a+b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}\right )-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\left (8 a^2+4 a b-b^2\right ) \int \frac {\coth (x)}{\sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)+8 a (a+b) \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)}{2 a}-\frac {(4 a+b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}\right )-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\frac {2 \left (8 a^2+4 a b-b^2\right ) \int \frac {1}{\frac {\tanh ^4(x)}{b}-\frac {a}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}+\frac {16 a (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}}{2 a}-\frac {(4 a+b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}\right )-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {16 a \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )-\frac {2 \left (8 a^2+4 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 a}-\frac {(4 a+b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}\right )-\frac {1}{2} \coth ^2(x) \sqrt {a+b \tanh ^2(x)}\right )\) |
(-1/2*(Coth[x]^2*Sqrt[a + b*Tanh[x]^2]) + (((-2*(8*a^2 + 4*a*b - b^2)*ArcT anh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a]])/Sqrt[a] + 16*a*Sqrt[a + b]*ArcTanh[Sqr t[a + b*Tanh[x]^2]/Sqrt[a + b]])/(2*a) - ((4*a + b)*Coth[x]*Sqrt[a + b*Tan h[x]^2])/a)/4)/2
3.3.18.3.1 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_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f)) Int[(a + b*x)^(m + 1) *(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
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]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^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] && Ratio nalQ[n]))
\[\int \coth \left (x \right )^{5} \sqrt {a +b \tanh \left (x \right )^{2}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2094 vs. \(2 (99) = 198\).
Time = 0.57 (sec) , antiderivative size = 9642, normalized size of antiderivative = 79.69 \[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=\text {Too large to display} \]
\[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=\int \sqrt {a + b \tanh ^{2}{\left (x \right )}} \coth ^{5}{\left (x \right )}\, dx \]
\[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=\int { \sqrt {b \tanh \left (x\right )^{2} + a} \coth \left (x\right )^{5} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 947 vs. \(2 (99) = 198\).
Time = 1.01 (sec) , antiderivative size = 947, normalized size of antiderivative = 7.83 \[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=\text {Too large to display} \]
-1/2*sqrt(a + b)*log(abs(-(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x ) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))*(a + b) - sqrt(a + b)*(a - b))) + 1/2*sqrt(a + b)*log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) + sqrt(a + b))) - 1/2*sqrt(a + b)*log (abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b *e^(2*x) + a + b) - sqrt(a + b))) + 1/4*(8*a^2 + 4*a*b - b^2)*arctan(-1/2* (sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2 *x) + a + b) - sqrt(a + b))/sqrt(-a))/(sqrt(-a)*a) + 1/2*((16*a^2 + 12*a*b + b^2)*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^7 - (16*a^2 + 52*a*b + 7*b^2)*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^6*sqrt(a + b) - (48*a^3 - 28*a^2*b - 109*a*b^2 - 21*b^3)*(sqrt(a + b)*e^(2*x) - sq rt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^5 + (176*a^ 3 + 156*a^2*b - 115*a*b^2 - 35*b^3)*(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^4*sqrt(a + b) + (304*a^4 - 156*a^3*b - 317*a^2*b^2 + 130*a*b^3 + 35*b^4)*(sqrt(a + b)*e^(2*x) - sq rt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^3 - (48*a^4 + 476*a^3*b - 379*a^2*b^2 + 94*a*b^3 + 21*b^4)*(sqrt(a + b)*e^(2*x) - sqr t(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^2*sqrt(a + b ) - (272*a^5 + 140*a^4*b - 271*a^3*b^2 + 135*a^2*b^3 - 53*a*b^4 - 7*b^5...
Timed out. \[ \int \coth ^5(x) \sqrt {a+b \tanh ^2(x)} \, dx=\int {\mathrm {coth}\left (x\right )}^5\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a} \,d x \]