Integrand size = 23, antiderivative size = 52 \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {a (a+b) \text {csch}^2(c+d x)}{d}-\frac {(a+b)^2 \text {csch}^4(c+d x)}{4 d}+\frac {a^2 \log (\sinh (c+d x))}{d} \]
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.48 \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {\left (b+a \cosh ^2(c+d x)\right )^2 \left (4 a (a+b) \text {csch}^2(c+d x)+(a+b)^2 \text {csch}^4(c+d x)-4 a^2 \log (\sinh (c+d x))\right )}{d (a+2 b+a \cosh (2 (c+d x)))^2} \]
-(((b + a*Cosh[c + d*x]^2)^2*(4*a*(a + b)*Csch[c + d*x]^2 + (a + b)^2*Csch [c + d*x]^4 - 4*a^2*Log[Sinh[c + d*x]]))/(d*(a + 2*b + a*Cosh[2*(c + d*x)] )^2))
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4626, 353, 49, 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.
\(\displaystyle \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \left (a+b \sec (i c+i d x)^2\right )^2}{\tan (i c+i d x)^5}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\left (b \sec (i c+i d x)^2+a\right )^2}{\tan (i c+i d x)^5}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\cosh (c+d x) \left (a \cosh ^2(c+d x)+b\right )^2}{\left (1-\cosh ^2(c+d x)\right )^3}d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {\int \frac {\left (a \cosh ^2(c+d x)+b\right )^2}{\left (1-\cosh ^2(c+d x)\right )^3}d\cosh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\int \left (-\frac {a^2}{\cosh ^2(c+d x)-1}-\frac {2 (a+b) a}{\left (\cosh ^2(c+d x)-1\right )^2}-\frac {(a+b)^2}{\left (\cosh ^2(c+d x)-1\right )^3}\right )d\cosh ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \left (-\log \left (1-\cosh ^2(c+d x)\right )\right )-\frac {2 a (a+b)}{1-\cosh ^2(c+d x)}+\frac {(a+b)^2}{2 \left (1-\cosh ^2(c+d x)\right )^2}}{2 d}\) |
-1/2*((a + b)^2/(2*(1 - Cosh[c + d*x]^2)^2) - (2*a*(a + b))/(1 - Cosh[c + d*x]^2) - a^2*Log[1 - Cosh[c + d*x]^2])/d
3.2.21.3.1 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] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 23.42 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.62
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\ln \left (\sinh \left (d x +c \right )\right )-\frac {\coth \left (d x +c \right )^{2}}{2}-\frac {\coth \left (d x +c \right )^{4}}{4}\right )+2 a b \left (-\frac {\cosh \left (d x +c \right )^{2}}{2 \sinh \left (d x +c \right )^{4}}+\frac {1}{4 \sinh \left (d x +c \right )^{4}}\right )-\frac {b^{2}}{4 \sinh \left (d x +c \right )^{4}}}{d}\) | \(84\) |
default | \(\frac {a^{2} \left (\ln \left (\sinh \left (d x +c \right )\right )-\frac {\coth \left (d x +c \right )^{2}}{2}-\frac {\coth \left (d x +c \right )^{4}}{4}\right )+2 a b \left (-\frac {\cosh \left (d x +c \right )^{2}}{2 \sinh \left (d x +c \right )^{4}}+\frac {1}{4 \sinh \left (d x +c \right )^{4}}\right )-\frac {b^{2}}{4 \sinh \left (d x +c \right )^{4}}}{d}\) | \(84\) |
risch | \(-a^{2} x -\frac {2 a^{2} c}{d}-\frac {4 \,{\mathrm e}^{2 d x +2 c} \left (a^{2} {\mathrm e}^{4 d x +4 c}+a b \,{\mathrm e}^{4 d x +4 c}-a^{2} {\mathrm e}^{2 d x +2 c}+{\mathrm e}^{2 d x +2 c} b^{2}+a^{2}+a b \right )}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) a^{2}}{d}\) | \(122\) |
1/d*(a^2*(ln(sinh(d*x+c))-1/2*coth(d*x+c)^2-1/4*coth(d*x+c)^4)+2*a*b*(-1/2 /sinh(d*x+c)^4*cosh(d*x+c)^2+1/4/sinh(d*x+c)^4)-1/4*b^2/sinh(d*x+c)^4)
Leaf count of result is larger than twice the leaf count of optimal. 1252 vs. \(2 (50) = 100\).
Time = 0.26 (sec) , antiderivative size = 1252, normalized size of antiderivative = 24.08 \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\text {Too large to display} \]
-(a^2*d*x*cosh(d*x + c)^8 + 8*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + a^2* d*x*sinh(d*x + c)^8 - 4*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^6 + 4*(7*a^2*d *x*cosh(d*x + c)^2 - a^2*d*x + a^2 + a*b)*sinh(d*x + c)^6 + 8*(7*a^2*d*x*c osh(d*x + c)^3 - 3*(a^2*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a^2*d*x - 2*a^2 + 2*b^2)*cosh(d*x + c)^4 + 2*(35*a^2*d*x*cosh(d*x + c )^4 + 3*a^2*d*x - 30*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^2 - 2*a^2 + 2*b^2 )*sinh(d*x + c)^4 + a^2*d*x + 8*(7*a^2*d*x*cosh(d*x + c)^5 - 10*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^3 + (3*a^2*d*x - 2*a^2 + 2*b^2)*cosh(d*x + c))*s inh(d*x + c)^3 - 4*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^2 + 4*(7*a^2*d*x*co sh(d*x + c)^6 - 15*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^4 - a^2*d*x + 3*(3* a^2*d*x - 2*a^2 + 2*b^2)*cosh(d*x + c)^2 + a^2 + a*b)*sinh(d*x + c)^2 - (a ^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 - 4*a^2*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 - a^2)*sinh(d*x + c)^6 + 6*a^2*cosh(d*x + c)^4 + 8*(7*a^2*cosh(d*x + c)^3 - 3*a^2*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 - 30*a^2*cosh(d*x + c)^2 + 3*a^2)*sinh(d*x + c)^4 - 4*a^2*cosh(d*x + c)^2 + 8*(7*a^2*cosh(d*x + c)^ 5 - 10*a^2*cosh(d*x + c)^3 + 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*a ^2*cosh(d*x + c)^6 - 15*a^2*cosh(d*x + c)^4 + 9*a^2*cosh(d*x + c)^2 - a^2) *sinh(d*x + c)^2 + a^2 + 8*(a^2*cosh(d*x + c)^7 - 3*a^2*cosh(d*x + c)^5 + 3*a^2*cosh(d*x + c)^3 - a^2*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*...
Timed out. \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (50) = 100\).
Time = 0.20 (sec) , antiderivative size = 282, normalized size of antiderivative = 5.42 \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 4 \, a b {\left (\frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} + \frac {e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} - \frac {4 \, b^{2}}{d {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4}} \]
a^2*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - c) - 1)/d + 4*(e^(- 2*d*x - 2*c) - e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) - 1))) + 4*a *b*(e^(-2*d*x - 2*c)/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6 *d*x - 6*c) - e^(-8*d*x - 8*c) - 1)) + e^(-6*d*x - 6*c)/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) - 1))) - 4*b^2/(d*(e^(d*x + c) - e^(-d*x - c))^4)
Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (50) = 100\).
Time = 0.37 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.88 \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {12 \, {\left (d x + c\right )} a^{2} - 12 \, a^{2} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {25 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} - 52 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 102 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 52 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 48 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}}}{12 \, d} \]
-1/12*(12*(d*x + c)*a^2 - 12*a^2*log(abs(e^(2*d*x + 2*c) - 1)) + (25*a^2*e ^(8*d*x + 8*c) - 52*a^2*e^(6*d*x + 6*c) + 48*a*b*e^(6*d*x + 6*c) + 102*a^2 *e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) - 52*a^2*e^(2*d*x + 2*c) + 48*a* b*e^(2*d*x + 2*c) + 25*a^2)/(e^(2*d*x + 2*c) - 1)^4)/d
Time = 2.38 (sec) , antiderivative size = 207, normalized size of antiderivative = 3.98 \[ \int \coth ^5(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {a^2\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )}{d}-\frac {4\,\left (a^2+2\,a\,b+b^2\right )}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {4\,\left (2\,a^2+3\,a\,b+b^2\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-a^2\,x-\frac {4\,\left (a^2+b\,a\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {8\,\left (a^2+2\,a\,b+b^2\right )}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )} \]
(a^2*log(exp(2*c)*exp(2*d*x) - 1))/d - (4*(2*a*b + a^2 + b^2))/(d*(6*exp(4 *c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) - (4*(3*a*b + 2*a^2 + b^2))/(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - a^2*x - (4*(a*b + a^2))/(d*(exp(2*c + 2*d*x) - 1)) - (8*(2*a*b + a^2 + b^2))/(d*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1))