Integrand size = 23, antiderivative size = 148 \[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}{b^4 d}+\frac {2 a (a+b \text {sech}(c+d x))^{3/2}}{b^4 d}-\frac {2 (a+b \text {sech}(c+d x))^{5/2}}{5 b^4 d} \]
2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2*a*(a+b*sech(d*x+c)) ^(3/2)/b^4/d-2/5*(a+b*sech(d*x+c))^(5/2)/b^4/d-2*(a^2-b^2)^2/a/b^4/d/(a+b* sech(d*x+c))^(1/2)-2*(3*a^2-2*b^2)*(a+b*sech(d*x+c))^(1/2)/b^4/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.89 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=-\frac {2 \left (5 b^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \text {sech}(c+d x)}{a}\right )+a \left (4 a \left (4 a^2-5 b^2\right )+2 b \left (4 a^2-5 b^2\right ) \text {sech}(c+d x)-2 a b^2 \text {sech}^2(c+d x)+b^3 \text {sech}^3(c+d x)\right )\right )}{5 a b^4 d \sqrt {a+b \text {sech}(c+d x)}} \]
(-2*(5*b^4*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sech[c + d*x])/a] + a*(4 *a*(4*a^2 - 5*b^2) + 2*b*(4*a^2 - 5*b^2)*Sech[c + d*x] - 2*a*b^2*Sech[c + d*x]^2 + b^3*Sech[c + d*x]^3)))/(5*a*b^4*d*Sqrt[a + b*Sech[c + d*x]])
Time = 0.38 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4373, 517, 25, 1584, 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.
| \(\displaystyle \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \cot \left (i c+i d x+\frac {\pi }{2}\right )^5}{\left (a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^5}{\left (a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle -\frac {\int \frac {\cosh (c+d x) \left (b^2-b^2 \text {sech}^2(c+d x)\right )^2}{b (a+b \text {sech}(c+d x))^{3/2}}d(b \text {sech}(c+d x))}{b^4 d}\) |
| \(\Big \downarrow \) 517 |
| \(\displaystyle -\frac {2 \int -\frac {\cosh ^2(c+d x) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )^2}{b^2 \left (a-b^2 \text {sech}^2(c+d x)\right )}d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \int \frac {\cosh ^2(c+d x) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )^2}{b^2 \left (a-b^2 \text {sech}^2(c+d x)\right )}d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {2 \int \left (-\text {sech}^4(c+d x) b^4+\frac {b^4}{a \left (a-b^2 \text {sech}^2(c+d x)\right )}+3 a \text {sech}^2(c+d x) b^2-3 a^2 \left (1-\frac {2 b^2}{3 a^2}\right )+\frac {\left (a^2-b^2\right )^2 \cosh ^2(c+d x)}{a b^2}\right )d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \left (-\frac {b^4 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\left (a^2-b^2\right )^2 \cosh (c+d x)}{a b}+\left (3 a^2-2 b^2\right ) \sqrt {a+b \text {sech}(c+d x)}-a b^3 \text {sech}^3(c+d x)+\frac {1}{5} b^5 \text {sech}^5(c+d x)\right )}{b^4 d}\) |
(-2*(-((b^4*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/a^(3/2)) + ((a^2 - b^2)^2*Cosh[c + d*x])/(a*b) - a*b^3*Sech[c + d*x]^3 + (b^5*Sech[c + d*x]^ 5)/5 + (3*a^2 - 2*b^2)*Sqrt[a + b*Sech[c + d*x]]))/(b^4*d)
3.2.42.3.1 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[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
\[\int \frac {\tanh \left (d x +c \right )^{5}}{\left (a +b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1743 vs. \(2 (132) = 264\).
Time = 0.69 (sec) , antiderivative size = 3745, normalized size of antiderivative = 25.30 \[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\text {Too large to display} \]
[1/10*(5*(a*b^4*cosh(d*x + c)^6 + a*b^4*sinh(d*x + c)^6 + 2*b^5*cosh(d*x + c)^5 + 3*a*b^4*cosh(d*x + c)^4 + 4*b^5*cosh(d*x + c)^3 + 3*a*b^4*cosh(d*x + c)^2 + 2*b^5*cosh(d*x + c) + 2*(3*a*b^4*cosh(d*x + c) + b^5)*sinh(d*x + c)^5 + a*b^4 + (15*a*b^4*cosh(d*x + c)^2 + 10*b^5*cosh(d*x + c) + 3*a*b^4 )*sinh(d*x + c)^4 + 4*(5*a*b^4*cosh(d*x + c)^3 + 5*b^5*cosh(d*x + c)^2 + 3 *a*b^4*cosh(d*x + c) + b^5)*sinh(d*x + c)^3 + (15*a*b^4*cosh(d*x + c)^4 + 20*b^5*cosh(d*x + c)^3 + 18*a*b^4*cosh(d*x + c)^2 + 12*b^5*cosh(d*x + c) + 3*a*b^4)*sinh(d*x + c)^2 + 2*(3*a*b^4*cosh(d*x + c)^5 + 5*b^5*cosh(d*x + c)^4 + 6*a*b^4*cosh(d*x + c)^3 + 6*b^5*cosh(d*x + c)^2 + 3*a*b^4*cosh(d*x + c) + b^5)*sinh(d*x + c))*sqrt(a)*log(-(2*a^2*cosh(d*x + c)^4 + 2*a^2*sin h(d*x + c)^4 + 4*a*b*cosh(d*x + c)^3 + 4*(2*a^2*cosh(d*x + c) + a*b)*sinh( d*x + c)^3 + 4*a*b*cosh(d*x + c) + (4*a^2 + b^2)*cosh(d*x + c)^2 + (12*a^2 *cosh(d*x + c)^2 + 12*a*b*cosh(d*x + c) + 4*a^2 + b^2)*sinh(d*x + c)^2 + 2 *a^2 + 2*(a*cosh(d*x + c)^4 + a*sinh(d*x + c)^4 + b*cosh(d*x + c)^3 + (4*a *cosh(d*x + c) + b)*sinh(d*x + c)^3 + 2*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c)^2 + 3*b*cosh(d*x + c) + 2*a)*sinh(d*x + c)^2 + b*cosh(d*x + c) + (4*a *cosh(d*x + c)^3 + 3*b*cosh(d*x + c)^2 + 4*a*cosh(d*x + c) + b)*sinh(d*x + c) + a)*sqrt(a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)) + 2*(4*a^2*cosh (d*x + c)^3 + 6*a*b*cosh(d*x + c)^2 + 2*a*b + (4*a^2 + b^2)*cosh(d*x + c)) *sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh...
\[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {\tanh ^{5}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{5}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int { \frac {\tanh \left (d x + c\right )^{5}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^5}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \]