Integrand size = 17, antiderivative size = 68 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {a+b}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}-\frac {1}{a^2 \sqrt {a+b \text {sech}^2(x)}} \]
arctanh((a+b*sech(x)^2)^(1/2)/a^(1/2))/a^(5/2)+1/3*(-a-b)/a/b/(a+b*sech(x) ^2)^(3/2)-1/a^2/(a+b*sech(x)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.82 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cosh (2 x)) \left (3 \sqrt {a} (a+2 b)^2 \cosh (x)+a^{3/2} (a+4 b) \cosh (3 x)-3 \sqrt {2} b (a+2 b+a \cosh (2 x))^{3/2} \log \left (\sqrt {2} \sqrt {a} \cosh (x)+\sqrt {a+2 b+a \cosh (2 x)}\right )\right ) \text {sech}^5(x)}{24 a^{5/2} b \left (a+b \text {sech}^2(x)\right )^{5/2}} \]
-1/24*((a + 2*b + a*Cosh[2*x])*(3*Sqrt[a]*(a + 2*b)^2*Cosh[x] + a^(3/2)*(a + 4*b)*Cosh[3*x] - 3*Sqrt[2]*b*(a + 2*b + a*Cosh[2*x])^(3/2)*Log[Sqrt[2]* Sqrt[a]*Cosh[x] + Sqrt[a + 2*b + a*Cosh[2*x]]])*Sech[x]^5)/(a^(5/2)*b*(a + b*Sech[x]^2)^(5/2))
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {3042, 26, 4627, 25, 354, 87, 61, 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.
\(\displaystyle \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \tan (i x)^3}{\left (a+b \sec (i x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\tan (i x)^3}{\left (b \sec (i x)^2+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \int -\frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (a+b \text {sech}^2(x)\right )^{5/2}}d\text {sech}(x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (b \text {sech}^2(x)+a\right )^{5/2}}d\text {sech}(x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {1}{2} \int \frac {\cosh (x) \left (1-\text {sech}^2(x)\right )}{\left (b \text {sech}^2(x)+a\right )^{5/2}}d\text {sech}^2(x)\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\cosh (x)}{\left (b \text {sech}^2(x)+a\right )^{3/2}}d\text {sech}^2(x)}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\int \frac {\cosh (x)}{\sqrt {b \text {sech}^2(x)+a}}d\text {sech}^2(x)}{a}+\frac {2}{a \sqrt {a+b \text {sech}^2(x)}}}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 \int \frac {1}{\frac {\text {sech}^4(x)}{b}-\frac {a}{b}}d\sqrt {b \text {sech}^2(x)+a}}{a b}+\frac {2}{a \sqrt {a+b \text {sech}^2(x)}}}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {\frac {2}{a \sqrt {a+b \text {sech}^2(x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}^2(x)}}{\sqrt {a}}\right )}{a^{3/2}}}{a}-\frac {2 (a+b)}{3 a b \left (a+b \text {sech}^2(x)\right )^{3/2}}\right )\) |
((-2*(a + b))/(3*a*b*(a + b*Sech[x]^2)^(3/2)) - ((-2*ArcTanh[Sqrt[a + b*Se ch[x]^2]/Sqrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b*Sech[x]^2]))/a)/2
3.3.14.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, 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_))*((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])
\[\int \frac {\tanh \left (x \right )^{3}}{\left (a +\operatorname {sech}\left (x \right )^{2} b \right )^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1954 vs. \(2 (56) = 112\).
Time = 0.48 (sec) , antiderivative size = 4644, normalized size of antiderivative = 68.29 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/12*(3*(a^2*b*cosh(x)^8 + 8*a^2*b*cosh(x)*sinh(x)^7 + a^2*b*sinh(x)^8 + 4*(a^2*b + 2*a*b^2)*cosh(x)^6 + 4*(7*a^2*b*cosh(x)^2 + a^2*b + 2*a*b^2)*si nh(x)^6 + 8*(7*a^2*b*cosh(x)^3 + 3*(a^2*b + 2*a*b^2)*cosh(x))*sinh(x)^5 + 2*(3*a^2*b + 8*a*b^2 + 8*b^3)*cosh(x)^4 + 2*(35*a^2*b*cosh(x)^4 + 3*a^2*b + 8*a*b^2 + 8*b^3 + 30*(a^2*b + 2*a*b^2)*cosh(x)^2)*sinh(x)^4 + 8*(7*a^2*b *cosh(x)^5 + 10*(a^2*b + 2*a*b^2)*cosh(x)^3 + (3*a^2*b + 8*a*b^2 + 8*b^3)* cosh(x))*sinh(x)^3 + a^2*b + 4*(a^2*b + 2*a*b^2)*cosh(x)^2 + 4*(7*a^2*b*co sh(x)^6 + 15*(a^2*b + 2*a*b^2)*cosh(x)^4 + a^2*b + 2*a*b^2 + 3*(3*a^2*b + 8*a*b^2 + 8*b^3)*cosh(x)^2)*sinh(x)^2 + 8*(a^2*b*cosh(x)^7 + 3*(a^2*b + 2* a*b^2)*cosh(x)^5 + (3*a^2*b + 8*a*b^2 + 8*b^3)*cosh(x)^3 + (a^2*b + 2*a*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)*cosh(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*...
\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{{\left (b \operatorname {sech}\left (x\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (56) = 112\).
Time = 0.40 (sec) , antiderivative size = 285, normalized size of antiderivative = 4.19 \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=-\frac {{\left ({\left (\frac {{\left (a^{8} b + 6 \, a^{7} b^{2} + 9 \, a^{6} b^{3} + 4 \, a^{5} b^{4}\right )} e^{\left (2 \, x\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}} + \frac {3 \, {\left (a^{8} b + 6 \, a^{7} b^{2} + 13 \, a^{6} b^{3} + 12 \, a^{5} b^{4} + 4 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} + \frac {3 \, {\left (a^{8} b + 6 \, a^{7} b^{2} + 13 \, a^{6} b^{3} + 12 \, a^{5} b^{4} + 4 \, a^{4} b^{5}\right )}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}\right )} e^{\left (2 \, x\right )} + \frac {a^{8} b + 6 \, a^{7} b^{2} + 9 \, a^{6} b^{3} + 4 \, a^{5} b^{4}}{a^{8} b^{2} + 2 \, a^{7} b^{3} + a^{6} b^{4}}}{3 \, {\left (a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + 4 \, b e^{\left (2 \, x\right )} + a\right )}^{\frac {3}{2}}} \]
-1/3*((((a^8*b + 6*a^7*b^2 + 9*a^6*b^3 + 4*a^5*b^4)*e^(2*x)/(a^8*b^2 + 2*a ^7*b^3 + a^6*b^4) + 3*(a^8*b + 6*a^7*b^2 + 13*a^6*b^3 + 12*a^5*b^4 + 4*a^4 *b^5)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^4))*e^(2*x) + 3*(a^8*b + 6*a^7*b^2 + 13 *a^6*b^3 + 12*a^5*b^4 + 4*a^4*b^5)/(a^8*b^2 + 2*a^7*b^3 + a^6*b^4))*e^(2*x ) + (a^8*b + 6*a^7*b^2 + 9*a^6*b^3 + 4*a^5*b^4)/(a^8*b^2 + 2*a^7*b^3 + a^6 *b^4))/(a*e^(4*x) + 2*a*e^(2*x) + 4*b*e^(2*x) + a)^(3/2)
Timed out. \[ \int \frac {\tanh ^3(x)}{\left (a+b \text {sech}^2(x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^3}{{\left (a+\frac {b}{{\mathrm {cosh}\left (x\right )}^2}\right )}^{5/2}} \,d x \]