Integrand size = 17, antiderivative size = 49 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {a+b}{a b \sqrt {a+b \text {sech}^2(x)}} \]
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {a+b}{a b \sqrt {a+b \text {sech}^2(x)}} \]
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 26, 4627, 25, 354, 87, 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.
\(\displaystyle \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \tan (i x)^3}{\left (a+b \sec (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\tan (i x)^3}{\left (b \sec (i x)^2+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \int -\frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (a+b \text {sech}^2(x)\right )^{3/2}}d\text {sech}(x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}(x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {1}{2} \int \frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}^2(x)\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\cosh (x)}{\sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a}-\frac {2 (a+b)}{a b \sqrt {a+b \text {sech}^2(x)}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 \int \frac {1}{\frac {\text {sech}^4(x)}{b}-\frac {a}{b}}d\sqrt {b \text {sech}^2(x)+a}}{a b}-\frac {2 (a+b)}{a b \sqrt {a+b \text {sech}^2(x)}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 (a+b)}{a b \sqrt {a+b \text {sech}^2(x)}}\right )\) |
((2*ArcTanh[Sqrt[a + b*Sech[x]^2]/Sqrt[a]])/a^(3/2) - (2*(a + b))/(a*b*Sqr t[a + b*Sech[x]^2]))/2
3.3.5.3.1 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_), 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] , x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers Q[2*n, p])
\[\int \frac {\tanh \left (x \right )^{3}}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 729 vs. \(2 (41) = 82\).
Time = 0.32 (sec) , antiderivative size = 2194, normalized size of antiderivative = 44.78 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*((a*b*cosh(x)^4 + 4*a*b*cosh(x)*sinh(x)^3 + a*b*sinh(x)^4 + 2*(a*b + 2*b^2)*cosh(x)^2 + 2*(3*a*b*cosh(x)^2 + a*b + 2*b^2)*sinh(x)^2 + a*b + 4*( a*b*cosh(x)^3 + (a*b + 2*b^2)*cosh(x))*sinh(x))*sqrt(a)*log(((a^3 + 2*a^2* b + a*b^2)*cosh(x)^8 + 8*(a^3 + 2*a^2*b + a*b^2)*cosh(x)*sinh(x)^7 + (a^3 + 2*a^2*b + a*b^2)*sinh(x)^8 + 2*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x) ^6 + 2*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3 + 14*(a^3 + 2*a^2*b + a*b^2)*cosh( x)^2)*sinh(x)^6 + 4*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^3 + 3*(2*a^3 + 5*a ^2*b + 4*a*b^2 + b^3)*cosh(x))*sinh(x)^5 + (6*a^3 + 14*a^2*b + 9*a*b^2)*co sh(x)^4 + (70*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^4 + 6*a^3 + 14*a^2*b + 9*a*b ^2 + 30*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a^ 3 + 2*a^2*b + a*b^2)*cosh(x)^5 + 10*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh (x)^3 + (6*a^3 + 14*a^2*b + 9*a*b^2)*cosh(x))*sinh(x)^3 + a^3 + 2*(2*a^3 + 3*a^2*b)*cosh(x)^2 + 2*(14*(a^3 + 2*a^2*b + a*b^2)*cosh(x)^6 + 15*(2*a^3 + 5*a^2*b + 4*a*b^2 + b^3)*cosh(x)^4 + 2*a^3 + 3*a^2*b + 3*(6*a^3 + 14*a^2 *b + 9*a*b^2)*cosh(x)^2)*sinh(x)^2 + sqrt(2)*((a^2 + 2*a*b + b^2)*cosh(x)^ 6 + 6*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^5 + (a^2 + 2*a*b + b^2)*sinh(x)^ 6 + 3*(a^2 + 2*a*b + b^2)*cosh(x)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(x)^2 + a^2 + 2*a*b + b^2)*sinh(x)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^ 2 + 2*a*b + b^2)*cosh(x))*sinh(x)^3 + (3*a^2 + 4*a*b)*cosh(x)^2 + (15*(a^2 + 2*a*b + b^2)*cosh(x)^4 + 18*(a^2 + 2*a*b + b^2)*cosh(x)^2 + 3*a^2 + ...
\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (41) = 82\).
Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.90 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=-\frac {\frac {{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} e^{\left (2 \, x\right )}}{a^{3} b + a^{2} b^{2}} + \frac {a^{3} + 2 \, a^{2} b + a b^{2}}{a^{3} b + a^{2} b^{2}}}{\sqrt {a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + 4 \, b e^{\left (2 \, x\right )} + a}} \]
-((a^3 + 2*a^2*b + a*b^2)*e^(2*x)/(a^3*b + a^2*b^2) + (a^3 + 2*a^2*b + a*b ^2)/(a^3*b + a^2*b^2))/sqrt(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a)
Timed out. \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^3}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{3/2}} \,d x \]