Integrand size = 17, antiderivative size = 131 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac {b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt {a+b \tanh ^2(x)}}-\frac {(3 a+2 b) (a+4 b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{3 a^3 (a+b)^2} \]
arctanh((a+b)^(1/2)*tanh(x)/(a+b*tanh(x)^2)^(1/2))/(a+b)^(5/2)+1/3*b*(7*a+ 4*b)*coth(x)/a^2/(a+b)^2/(a+b*tanh(x)^2)^(1/2)-1/3*(3*a+2*b)*(a+4*b)*coth( x)*(a+b*tanh(x)^2)^(1/2)/a^3/(a+b)^2+1/3*b*coth(x)/a/(a+b)/(a+b*tanh(x)^2) ^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 8.66 (sec) , antiderivative size = 1375, normalized size of antiderivative = 10.50 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx =\text {Too large to display} \]
-((Cosh[x]^2*Coth[x]*((-20*a*Csch[x]^2)/(3*(a + b)) - (5*a^2*Csch[x]^4)/(a + b)^2 - (40*b*Sech[x]^2)/(a + b) - (30*a*b*Csch[x]^2*Sech[x]^2)/(a + b)^ 2 - (40*b^2*Sech[x]^4)/(a + b)^2 - (92*(a + b)*Hypergeometric2F1[2, 2, 9/2 , -(((a + b)*Sinh[x]^2)/a)]*Sinh[x]^2)/(105*a) - (24*(a + b)*Hypergeometri cPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Sinh[x]^2)/a)]*Sinh[x]^2)/(35*a) - (1 6*(a + b)*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, -(((a + b)*Sinh[x]^ 2)/a)]*Sinh[x]^2)/(105*a) - (160*b^2*Sech[x]^2*Tanh[x]^2)/(3*a*(a + b)) - (124*b*(a + b)*Hypergeometric2F1[2, 2, 9/2, -(((a + b)*Sinh[x]^2)/a)]*Sinh [x]^2*Tanh[x]^2)/(35*a^2) - (16*b*(a + b)*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Sinh[x]^2)/a)]*Sinh[x]^2*Tanh[x]^2)/(7*a^2) - (16*b*(a + b)*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, -(((a + b)*Sinh[x]^2)/a)] *Sinh[x]^2*Tanh[x]^2)/(35*a^2) - (64*b^3*Sech[x]^2*Tanh[x]^4)/(3*a^2*(a + b)) - (152*b^2*(a + b)*Hypergeometric2F1[2, 2, 9/2, -(((a + b)*Sinh[x]^2)/ a)]*Sinh[x]^2*Tanh[x]^4)/(35*a^3) - (88*b^2*(a + b)*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Sinh[x]^2)/a)]*Sinh[x]^2*Tanh[x]^4)/(35*a^3) - (16*b^2*(a + b)*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, -(((a + b)*S inh[x]^2)/a)]*Sinh[x]^2*Tanh[x]^4)/(35*a^3) - (176*b^3*(a + b)*Hypergeomet ric2F1[2, 2, 9/2, -(((a + b)*Sinh[x]^2)/a)]*Sinh[x]^2*Tanh[x]^6)/(105*a^4) - (32*b^3*(a + b)*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, -(((a + b)*Sinh[ x]^2)/a)]*Sinh[x]^2*Tanh[x]^6)/(35*a^4) - (16*b^3*(a + b)*Hypergeometri...
Time = 0.44 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {3042, 25, 4153, 25, 374, 25, 441, 25, 445, 27, 291, 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.
| \(\displaystyle \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\tan (i x)^2 \left (a-b \tan (i x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\tan (i x)^2 \left (a-b \tan (i x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\int -\frac {\coth ^2(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{5/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\coth ^2(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{5/2}}d\tanh (x)\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle \frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac {\int -\frac {\coth ^2(x) \left (-4 b \tanh ^2(x)+3 a+4 b\right )}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh (x)}{3 a (a+b)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\coth ^2(x) \left (-4 b \tanh ^2(x)+3 a+4 b\right )}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh (x)}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 441 |
| \(\displaystyle \frac {\frac {b (7 a+4 b) \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\int -\frac {\coth ^2(x) \left ((3 a+2 b) (a+4 b)-2 b (7 a+4 b) \tanh ^2(x)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a (a+b)}}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\coth ^2(x) \left ((3 a+2 b) (a+4 b)-2 b (7 a+4 b) \tanh ^2(x)\right )}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a (a+b)}+\frac {b (7 a+4 b) \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {3 a^3}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{a}-\frac {(3 a+2 b) (a+4 b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}}{a (a+b)}+\frac {b (7 a+4 b) \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 a^2 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)-\frac {(3 a+2 b) (a+4 b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}}{a (a+b)}+\frac {b (7 a+4 b) \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {3 a^2 \int \frac {1}{1-\frac {(a+b) \tanh ^2(x)}{b \tanh ^2(x)+a}}d\frac {\tanh (x)}{\sqrt {b \tanh ^2(x)+a}}-\frac {(3 a+2 b) (a+4 b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}}{a (a+b)}+\frac {b (7 a+4 b) \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{\sqrt {a+b}}-\frac {(3 a+2 b) (a+4 b) \coth (x) \sqrt {a+b \tanh ^2(x)}}{a}}{a (a+b)}+\frac {b (7 a+4 b) \coth (x)}{a (a+b) \sqrt {a+b \tanh ^2(x)}}}{3 a (a+b)}+\frac {b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}\) |
(b*Coth[x])/(3*a*(a + b)*(a + b*Tanh[x]^2)^(3/2)) + ((b*(7*a + 4*b)*Coth[x ])/(a*(a + b)*Sqrt[a + b*Tanh[x]^2]) + ((3*a^2*ArcTanh[(Sqrt[a + b]*Tanh[x ])/Sqrt[a + b*Tanh[x]^2]])/Sqrt[a + b] - ((3*a + 2*b)*(a + 4*b)*Coth[x]*Sq rt[a + b*Tanh[x]^2])/a)/(a*(a + b)))/(3*a*(a + b))
3.3.54.3.1 Defintions of rubi rules used
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_)^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_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si mp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 )^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && LtQ[p, -1]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
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]))
\[\int \frac {\coth \left (x \right )^{2}}{\left (a +b \tanh \left (x \right )^{2}\right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 5021 vs. \(2 (113) = 226\).
Time = 1.34 (sec) , antiderivative size = 10671, normalized size of antiderivative = 81.46 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {\coth ^{2}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int { \frac {\coth \left (x\right )^{2}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 898 vs. \(2 (113) = 226\).
Time = 0.74 (sec) , antiderivative size = 898, normalized size of antiderivative = 6.85 \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
-1/3*((((9*a^13*b^4 + 50*a^12*b^5 + 115*a^11*b^6 + 140*a^10*b^7 + 95*a^9*b ^8 + 34*a^8*b^9 + 5*a^7*b^10)*e^(2*x)/(a^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 20*a^13*b^5 + 15*a^12*b^6 + 6*a^11*b^7 + a^10*b^8) + 3*(3*a^13*b^4 + 6* a^12*b^5 - 11*a^11*b^6 - 44*a^10*b^7 - 51*a^9*b^8 - 26*a^8*b^9 - 5*a^7*b^1 0)/(a^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 20*a^13*b^5 + 15*a^12*b^6 + 6*a^ 11*b^7 + a^10*b^8))*e^(2*x) - 3*(3*a^13*b^4 + 6*a^12*b^5 - 11*a^11*b^6 - 4 4*a^10*b^7 - 51*a^9*b^8 - 26*a^8*b^9 - 5*a^7*b^10)/(a^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 20*a^13*b^5 + 15*a^12*b^6 + 6*a^11*b^7 + a^10*b^8))*e^(2*x ) - (9*a^13*b^4 + 50*a^12*b^5 + 115*a^11*b^6 + 140*a^10*b^7 + 95*a^9*b^8 + 34*a^8*b^9 + 5*a^7*b^10)/(a^16*b^2 + 6*a^15*b^3 + 15*a^14*b^4 + 20*a^13*b ^5 + 15*a^12*b^6 + 6*a^11*b^7 + a^10*b^8))/(a*e^(4*x) + b*e^(4*x) + 2*a*e^ (2*x) - 2*b*e^(2*x) + a + b)^(3/2) - 1/2*log(abs(-(sqrt(a + b)*e^(2*x) - s qrt(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)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) - 1/2*log(abs(-sqr t(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)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) + 1/2*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)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) + 4*(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 + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x)...
Timed out. \[ \int \frac {\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {coth}\left (x\right )}^2}{{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{5/2}} \,d x \]