Integrand size = 23, antiderivative size = 169 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}+\frac {2 a \left (a^2-2 b^2\right ) (a+b \text {sech}(c+d x))^{3/2}}{3 b^4 d}-\frac {2 \left (3 a^2-2 b^2\right ) (a+b \text {sech}(c+d x))^{5/2}}{5 b^4 d}+\frac {6 a (a+b \text {sech}(c+d x))^{7/2}}{7 b^4 d}-\frac {2 (a+b \text {sech}(c+d x))^{9/2}}{9 b^4 d} \]
2/3*a*(a^2-2*b^2)*(a+b*sech(d*x+c))^(3/2)/b^4/d-2/5*(3*a^2-2*b^2)*(a+b*sec h(d*x+c))^(5/2)/b^4/d+6/7*a*(a+b*sech(d*x+c))^(7/2)/b^4/d-2/9*(a+b*sech(d* x+c))^(9/2)/b^4/d+2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-2*( a+b*sech(d*x+c))^(1/2)/d
Time = 2.65 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.91 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=-\frac {-2 \sqrt {a} b^4 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )+2 b^4 \sqrt {a+b \text {sech}(c+d x)}-\frac {2}{3} a \left (a^2-2 b^2\right ) (a+b \text {sech}(c+d x))^{3/2}+\frac {2}{5} \left (3 a^2-2 b^2\right ) (a+b \text {sech}(c+d x))^{5/2}-\frac {6}{7} a (a+b \text {sech}(c+d x))^{7/2}+\frac {2}{9} (a+b \text {sech}(c+d x))^{9/2}}{b^4 d} \]
-((-2*Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]] + 2*b^4*Sqrt[ a + b*Sech[c + d*x]] - (2*a*(a^2 - 2*b^2)*(a + b*Sech[c + d*x])^(3/2))/3 + (2*(3*a^2 - 2*b^2)*(a + b*Sech[c + d*x])^(5/2))/5 - (6*a*(a + b*Sech[c + d*x])^(7/2))/7 + (2*(a + b*Sech[c + d*x])^(9/2))/9)/(b^4*d))
Time = 0.39 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78, 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.
\(\displaystyle \int \tanh ^5(c+d x) \sqrt {a+b \text {sech}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int i \cot \left (i c+i d x+\frac {\pi }{2}\right )^5 \sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^5 \sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -\frac {\int \frac {\cosh (c+d x) \sqrt {a+b \text {sech}(c+d x)} \left (b^2-b^2 \text {sech}^2(c+d x)\right )^2}{b}d(b \text {sech}(c+d x))}{b^4 d}\) |
\(\Big \downarrow \) 517 |
\(\displaystyle -\frac {2 \int -\frac {b^2 \text {sech}^2(c+d x) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )^2}{a-b^2 \text {sech}^2(c+d x)}d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \int \frac {b^2 \text {sech}^2(c+d x) \left (b^4 \text {sech}^4(c+d x)-2 a b^2 \text {sech}^2(c+d x)+a^2-b^2\right )^2}{a-b^2 \text {sech}^2(c+d x)}d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {2 \int \left (-b^8 \text {sech}^8(c+d x)+3 a b^6 \text {sech}^6(c+d x)-b^4 \left (3 a^2-2 b^2\right ) \text {sech}^4(c+d x)+a b^2 \left (a^2-2 b^2\right ) \text {sech}^2(c+d x)-b^4+\frac {a b^4}{a-b^2 \text {sech}^2(c+d x)}\right )d\sqrt {a+b \text {sech}(c+d x)}}{b^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (\frac {1}{5} b^5 \left (3 a^2-2 b^2\right ) \text {sech}^5(c+d x)-\frac {1}{3} a b^3 \left (a^2-2 b^2\right ) \text {sech}^3(c+d x)-\sqrt {a} b^4 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )-\frac {3}{7} a b^7 \text {sech}^7(c+d x)+\frac {1}{9} b^9 \text {sech}^9(c+d x)+b^5 \text {sech}(c+d x)\right )}{b^4 d}\) |
(-2*(-(Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]]) + b^5*Sech[ c + d*x] - (a*b^3*(a^2 - 2*b^2)*Sech[c + d*x]^3)/3 + (b^5*(3*a^2 - 2*b^2)* Sech[c + d*x]^5)/5 - (3*a*b^7*Sech[c + d*x]^7)/7 + (b^9*Sech[c + d*x]^9)/9 ))/(b^4*d)
3.2.25.3.1 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]
\[\int \sqrt {a +b \,\operatorname {sech}\left (d x +c \right )}\, \tanh \left (d x +c \right )^{5}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2052 vs. \(2 (145) = 290\).
Time = 0.68 (sec) , antiderivative size = 4363, normalized size of antiderivative = 25.82 \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=\text {Too large to display} \]
[1/630*(315*(b^4*cosh(d*x + c)^8 + 8*b^4*cosh(d*x + c)*sinh(d*x + c)^7 + b ^4*sinh(d*x + c)^8 + 4*b^4*cosh(d*x + c)^6 + 6*b^4*cosh(d*x + c)^4 + 4*(7* b^4*cosh(d*x + c)^2 + b^4)*sinh(d*x + c)^6 + 4*b^4*cosh(d*x + c)^2 + 8*(7* b^4*cosh(d*x + c)^3 + 3*b^4*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*b^4*cos h(d*x + c)^4 + 30*b^4*cosh(d*x + c)^2 + 3*b^4)*sinh(d*x + c)^4 + b^4 + 8*( 7*b^4*cosh(d*x + c)^5 + 10*b^4*cosh(d*x + c)^3 + 3*b^4*cosh(d*x + c))*sinh (d*x + c)^3 + 4*(7*b^4*cosh(d*x + c)^6 + 15*b^4*cosh(d*x + c)^4 + 9*b^4*co sh(d*x + c)^2 + b^4)*sinh(d*x + c)^2 + 8*(b^4*cosh(d*x + c)^7 + 3*b^4*cosh (d*x + c)^5 + 3*b^4*cosh(d*x + c)^3 + b^4*cosh(d*x + c))*sinh(d*x + c))*sq rt(a)*log(-(2*a^2*cosh(d*x + c)^4 + 2*a^2*sinh(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*cos h(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(d*x + c)^2)) + 4*((16*a^4 - ...
\[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=\int \sqrt {a + b \operatorname {sech}{\left (c + d x \right )}} \tanh ^{5}{\left (c + d x \right )}\, dx \]
\[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=\int { \sqrt {b \operatorname {sech}\left (d x + c\right ) + a} \tanh \left (d x + c\right )^{5} \,d x } \]
\[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=\int { \sqrt {b \operatorname {sech}\left (d x + c\right ) + a} \tanh \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \sqrt {a+b \text {sech}(c+d x)} \tanh ^5(c+d x) \, dx=\int {\mathrm {tanh}\left (c+d\,x\right )}^5\,\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}} \,d x \]