Integrand size = 21, antiderivative size = 154 \[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{8 a^{5/2} (a+b)^3 d}-\frac {\text {arctanh}(\cosh (c+d x))}{(a+b)^3 d}-\frac {b \cosh ^3(c+d x)}{4 a (a+b) d \left (b+a \cosh ^2(c+d x)\right )^2}-\frac {b (7 a+3 b) \cosh (c+d x)}{8 a^2 (a+b)^2 d \left (b+a \cosh ^2(c+d x)\right )} \]
-arctanh(cosh(d*x+c))/(a+b)^3/d-1/4*b*cosh(d*x+c)^3/a/(a+b)/d/(b+a*cosh(d* x+c)^2)^2-1/8*b*(7*a+3*b)*cosh(d*x+c)/a^2/(a+b)^2/d/(b+a*cosh(d*x+c)^2)+1/ 8*(15*a^2+10*a*b+3*b^2)*arctan(cosh(d*x+c)*a^(1/2)/b^(1/2))*b^(1/2)/a^(5/2 )/(a+b)^3/d
Result contains complex when optimal does not.
Time = 4.49 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.86 \[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^5(c+d x) \left (\frac {8 b^2 (a+b)^2}{a^2}-\frac {2 b (a+b) (9 a+5 b) (a+2 b+a \cosh (2 (c+d x)))}{a^2}+\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \arctan \left (\frac {\left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right ) (a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}(c+d x)}{a^{5/2}}+\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \arctan \left (\frac {\left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right ) (a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}(c+d x)}{a^{5/2}}-8 (a+2 b+a \cosh (2 (c+d x)))^2 \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right ) \text {sech}(c+d x)+8 (a+2 b+a \cosh (2 (c+d x)))^2 \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right ) \text {sech}(c+d x)\right )}{64 (a+b)^3 d \left (a+b \text {sech}^2(c+d x)\right )^3} \]
((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^5*((8*b^2*(a + b)^2)/a^2 - (2*b*(a + b)*(9*a + 5*b)*(a + 2*b + a*Cosh[2*(c + d*x)]))/a^2 + (Sqrt[b]*( 15*a^2 + 10*a*b + 3*b^2)*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt [(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x])/a^(5/2) + (Sqrt[b]*(15*a^2 + 10*a*b + 3*b^2)*ArcTa n[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x )/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[( d*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x])/a^(5/2 ) - 8*(a + 2*b + a*Cosh[2*(c + d*x)])^2*Log[Cosh[(c + d*x)/2]]*Sech[c + d* x] + 8*(a + 2*b + a*Cosh[2*(c + d*x)])^2*Log[Sinh[(c + d*x)/2]]*Sech[c + d *x]))/(64*(a + b)^3*d*(a + b*Sech[c + d*x]^2)^3)
Time = 0.40 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 26, 4621, 372, 440, 397, 218, 219}
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.
\(\displaystyle \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\sin (i c+i d x) \left (a+b \sec (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\left (b \sec (i c+i d x)^2+a\right )^3 \sin (i c+i d x)}dx\) |
\(\Big \downarrow \) 4621 |
\(\displaystyle -\frac {\int \frac {\cosh ^6(c+d x)}{\left (1-\cosh ^2(c+d x)\right ) \left (a \cosh ^2(c+d x)+b\right )^3}d\cosh (c+d x)}{d}\) |
\(\Big \downarrow \) 372 |
\(\displaystyle -\frac {\frac {b \cosh ^3(c+d x)}{4 a (a+b) \left (a \cosh ^2(c+d x)+b\right )^2}-\frac {\int \frac {\cosh ^2(c+d x) \left (3 b-(4 a+3 b) \cosh ^2(c+d x)\right )}{\left (1-\cosh ^2(c+d x)\right ) \left (a \cosh ^2(c+d x)+b\right )^2}d\cosh (c+d x)}{4 a (a+b)}}{d}\) |
\(\Big \downarrow \) 440 |
\(\displaystyle -\frac {\frac {b \cosh ^3(c+d x)}{4 a (a+b) \left (a \cosh ^2(c+d x)+b\right )^2}-\frac {\frac {\int \frac {b (7 a+3 b)-\left (8 a^2+7 b a+3 b^2\right ) \cosh ^2(c+d x)}{\left (1-\cosh ^2(c+d x)\right ) \left (a \cosh ^2(c+d x)+b\right )}d\cosh (c+d x)}{2 a (a+b)}-\frac {b (7 a+3 b) \cosh (c+d x)}{2 a (a+b) \left (a \cosh ^2(c+d x)+b\right )}}{4 a (a+b)}}{d}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle -\frac {\frac {b \cosh ^3(c+d x)}{4 a (a+b) \left (a \cosh ^2(c+d x)+b\right )^2}-\frac {\frac {\frac {b \left (15 a^2+10 a b+3 b^2\right ) \int \frac {1}{a \cosh ^2(c+d x)+b}d\cosh (c+d x)}{a+b}-\frac {8 a^2 \int \frac {1}{1-\cosh ^2(c+d x)}d\cosh (c+d x)}{a+b}}{2 a (a+b)}-\frac {b (7 a+3 b) \cosh (c+d x)}{2 a (a+b) \left (a \cosh ^2(c+d x)+b\right )}}{4 a (a+b)}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\frac {b \cosh ^3(c+d x)}{4 a (a+b) \left (a \cosh ^2(c+d x)+b\right )^2}-\frac {\frac {\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{\sqrt {a} (a+b)}-\frac {8 a^2 \int \frac {1}{1-\cosh ^2(c+d x)}d\cosh (c+d x)}{a+b}}{2 a (a+b)}-\frac {b (7 a+3 b) \cosh (c+d x)}{2 a (a+b) \left (a \cosh ^2(c+d x)+b\right )}}{4 a (a+b)}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {b \cosh ^3(c+d x)}{4 a (a+b) \left (a \cosh ^2(c+d x)+b\right )^2}-\frac {\frac {\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{\sqrt {a} (a+b)}-\frac {8 a^2 \text {arctanh}(\cosh (c+d x))}{a+b}}{2 a (a+b)}-\frac {b (7 a+3 b) \cosh (c+d x)}{2 a (a+b) \left (a \cosh ^2(c+d x)+b\right )}}{4 a (a+b)}}{d}\) |
-(((b*Cosh[c + d*x]^3)/(4*a*(a + b)*(b + a*Cosh[c + d*x]^2)^2) - (((Sqrt[b ]*(15*a^2 + 10*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/(Sqrt [a]*(a + b)) - (8*a^2*ArcTanh[Cosh[c + d*x]])/(a + b))/(2*a*(a + b)) - (b* (7*a + 3*b)*Cosh[c + d*x])/(2*a*(a + b)*(b + a*Cosh[c + d*x]^2)))/(4*a*(a + b)))/d)
3.1.45.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(290\) vs. \(2(140)=280\).
Time = 0.26 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.89
\[\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a +b \right )^{3}}+\frac {2 b \left (\frac {-\frac {\left (9 a^{3}-a^{2} b -13 a \,b^{2}-3 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 a^{2}}-\frac {3 \left (9 a^{3}-3 a^{2} b +7 a \,b^{2}+3 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 a^{2}}-\frac {\left (27 a^{3}+13 a^{2} b -23 a \,b^{2}-9 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 a^{2}}-\frac {3 \left (3 a^{3}+7 a^{2} b +5 a \,b^{2}+b^{3}\right )}{8 a^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (15 a^{2}+10 a b +3 b^{2}\right ) \arctan \left (\frac {2 \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 a -2 b}{4 \sqrt {a b}}\right )}{16 a^{2} \sqrt {a b}}\right )}{\left (a +b \right )^{3}}}{d}\]
1/d*(1/(a+b)^3*ln(tanh(1/2*d*x+1/2*c))+2*b/(a+b)^3*((-1/8*(9*a^3-a^2*b-13* a*b^2-3*b^3)/a^2*tanh(1/2*d*x+1/2*c)^6-3/8*(9*a^3-3*a^2*b+7*a*b^2+3*b^3)/a ^2*tanh(1/2*d*x+1/2*c)^4-1/8*(27*a^3+13*a^2*b-23*a*b^2-9*b^3)/a^2*tanh(1/2 *d*x+1/2*c)^2-3/8*(3*a^3+7*a^2*b+5*a*b^2+b^3)/a^2)/(tanh(1/2*d*x+1/2*c)^4* a+tanh(1/2*d*x+1/2*c)^4*b+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^ 2*b+a+b)^2+1/16*(15*a^2+10*a*b+3*b^2)/a^2/(a*b)^(1/2)*arctan(1/4*(2*(a+b)* tanh(1/2*d*x+1/2*c)^2+2*a-2*b)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 4721 vs. \(2 (140) = 280\).
Time = 0.39 (sec) , antiderivative size = 8742, normalized size of antiderivative = 56.77 \[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {csch}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
\[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {csch}\left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
-1/4*((9*a^2*b*e^(7*c) + 5*a*b^2*e^(7*c))*e^(7*d*x) + (27*a^2*b*e^(5*c) + 43*a*b^2*e^(5*c) + 12*b^3*e^(5*c))*e^(5*d*x) + (27*a^2*b*e^(3*c) + 43*a*b^ 2*e^(3*c) + 12*b^3*e^(3*c))*e^(3*d*x) + (9*a^2*b*e^c + 5*a*b^2*e^c)*e^(d*x ))/(a^6*d + 2*a^5*b*d + a^4*b^2*d + (a^6*d*e^(8*c) + 2*a^5*b*d*e^(8*c) + a ^4*b^2*d*e^(8*c))*e^(8*d*x) + 4*(a^6*d*e^(6*c) + 4*a^5*b*d*e^(6*c) + 5*a^4 *b^2*d*e^(6*c) + 2*a^3*b^3*d*e^(6*c))*e^(6*d*x) + 2*(3*a^6*d*e^(4*c) + 14* a^5*b*d*e^(4*c) + 27*a^4*b^2*d*e^(4*c) + 24*a^3*b^3*d*e^(4*c) + 8*a^2*b^4* d*e^(4*c))*e^(4*d*x) + 4*(a^6*d*e^(2*c) + 4*a^5*b*d*e^(2*c) + 5*a^4*b^2*d* e^(2*c) + 2*a^3*b^3*d*e^(2*c))*e^(2*d*x)) - log((e^(d*x + c) + 1)*e^(-c))/ (a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + log((e^(d*x + c) - 1)*e^(-c))/(a ^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + 2*integrate(1/8*((15*a^2*b*e^(3*c) + 10*a*b^2*e^(3*c) + 3*b^3*e^(3*c))*e^(3*d*x) - (15*a^2*b*e^c + 10*a*b^2* e^c + 3*b^3*e^c)*e^(d*x))/(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3 + (a^6*e^(4 *c) + 3*a^5*b*e^(4*c) + 3*a^4*b^2*e^(4*c) + a^3*b^3*e^(4*c))*e^(4*d*x) + 2 *(a^6*e^(2*c) + 5*a^5*b*e^(2*c) + 9*a^4*b^2*e^(2*c) + 7*a^3*b^3*e^(2*c) + 2*a^2*b^4*e^(2*c))*e^(2*d*x)), x)
\[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {csch}\left (d x + c\right )}{{\left (b \operatorname {sech}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {csch}(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6}{\mathrm {sinh}\left (c+d\,x\right )\,{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^3} \,d x \]