Integrand size = 17, antiderivative size = 70 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+\frac {(a-b) \sqrt {a+b \tanh ^2(x)}}{b^2}-\frac {\left (a+b \tanh ^2(x)\right )^{3/2}}{3 b^2} \]
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(1/2)+(a-b)*(a+b*tanh(x)^ 2)^(1/2)/b^2-1/3*(a+b*tanh(x)^2)^(3/2)/b^2
Time = 0.57 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+\frac {(a-b+(a-2 b) \cosh (2 x)) \text {sech}^2(x) \sqrt {a+b \tanh ^2(x)}}{3 b^2} \]
ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]]/Sqrt[a + b] + ((a - b + (a - 2* b)*Cosh[2*x])*Sech[x]^2*Sqrt[a + b*Tanh[x]^2])/(3*b^2)
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 26, 4153, 26, 354, 99, 2009}
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.
\(\displaystyle \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)^5}{\sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)^5}{\sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -i \int \frac {i \tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {a+b \tanh ^2(x)}}d\tanh (x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\tanh ^4(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a-b}{b \sqrt {b \tanh ^2(x)+a}}-\frac {\sqrt {b \tanh ^2(x)+a}}{b}+\frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}\right )d\tanh ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 \left (a+b \tanh ^2(x)\right )^{3/2}}{3 b^2}+\frac {2 (a-b) \sqrt {a+b \tanh ^2(x)}}{b^2}\right )\) |
((2*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]])/Sqrt[a + b] + (2*(a - b)*S qrt[a + b*Tanh[x]^2])/b^2 - (2*(a + b*Tanh[x]^2)^(3/2))/(3*b^2))/2
3.3.29.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]))
Leaf count of result is larger than twice the leaf count of optimal. \(163\) vs. \(2(58)=116\).
Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.34
method | result | size |
derivativedivides | \(-\frac {\sqrt {a +b \tanh \left (x \right )^{2}}}{b}-\frac {\tanh \left (x \right )^{2} \sqrt {a +b \tanh \left (x \right )^{2}}}{3 b}+\frac {2 a \sqrt {a +b \tanh \left (x \right )^{2}}}{3 b^{2}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \sqrt {a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2 \sqrt {a +b}}\) | \(164\) |
default | \(-\frac {\sqrt {a +b \tanh \left (x \right )^{2}}}{b}-\frac {\tanh \left (x \right )^{2} \sqrt {a +b \tanh \left (x \right )^{2}}}{3 b}+\frac {2 a \sqrt {a +b \tanh \left (x \right )^{2}}}{3 b^{2}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \sqrt {a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2 \sqrt {a +b}}\) | \(164\) |
-(a+b*tanh(x)^2)^(1/2)/b-1/3*tanh(x)^2/b*(a+b*tanh(x)^2)^(1/2)+2/3*a/b^2*( a+b*tanh(x)^2)^(1/2)+1/2/(a+b)^(1/2)*ln((2*a+2*b+2*b*(tanh(x)-1)+2*(a+b)^( 1/2)*(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2))/(tanh(x)-1))+1/2/(a+b)^( 1/2)*ln((2*a+2*b-2*b*(1+tanh(x))+2*(a+b)^(1/2)*(b*(1+tanh(x))^2-2*b*(1+tan h(x))+a+b)^(1/2))/(1+tanh(x)))
Leaf count of result is larger than twice the leaf count of optimal. 1131 vs. \(2 (58) = 116\).
Time = 0.43 (sec) , antiderivative size = 2827, normalized size of antiderivative = 40.39 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\text {Too large to display} \]
[1/12*(3*(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 + 3*b^2*cosh(x)^2 + 4*(5*b ^2*cosh(x)^3 + 3*b^2*cosh(x))*sinh(x)^3 + 3*(5*b^2*cosh(x)^4 + 6*b^2*cosh( x)^2 + b^2)*sinh(x)^2 + b^2 + 6*(b^2*cosh(x)^5 + 2*b^2*cosh(x)^3 + b^2*cos h(x))*sinh(x))*sqrt(a + b)*log(((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)* cosh(x)*sinh(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)*cosh(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)*cosh(x)^6 + 15*(2*a^3 + a^2*b)*c osh(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*s inh(x)^6 + 3*a^2*cosh(x)^4 + 3*(5*a^2*cosh(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 + (15*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)*s...
\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\tanh ^{5}{\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )}}}\, dx \]
\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{\sqrt {b \tanh \left (x\right )^{2} + a}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (58) = 116\).
Time = 0.67 (sec) , antiderivative size = 592, normalized size of antiderivative = 8.46 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=-\frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, \sqrt {a + b}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, \sqrt {a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, \sqrt {a + b}} - \frac {8 \, {\left (3 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{5} + 3 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{4} \sqrt {a + b} + 2 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{3} {\left (3 \, a - 5 \, b\right )} + 6 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{2} \sqrt {a + b} {\left (a - 3 \, b\right )} - 3 \, {\left (3 \, a^{2} + 6 \, a b - 13 \, b^{2}\right )} {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} - {\left (9 \, a^{2} - 22 \, a b + 17 \, b^{2}\right )} \sqrt {a + b}\right )}}{3 \, {\left ({\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )}^{2} + 2 \, {\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} \sqrt {a + b} + a - 3 \, b\right )}^{3}} \]
-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))*(a + b) - sqrt(a + b)*(a - b)))/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)))/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)))/sqrt(a + b) - 8/3*(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))^5 + 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) + 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))^3*(3*a - 5*b) + 6*(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))^2*sqrt( a + b)*(a - 3*b) - 3*(3*a^2 + 6*a*b - 13*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)) - (9*a^2 - 22* a*b + 17*b^2)*sqrt(a + b))/((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))^2 + 2*(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))*sqrt(a + b) + a - 3*b)^3
Time = 2.68 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {{\left (b\,{\mathrm {tanh}\left (x\right )}^2+a\right )}^{3/2}}{3\,b^2}-\left (\frac {a+b}{b^2}-\frac {2\,a}{b^2}\right )\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a} \]