Integrand size = 13, antiderivative size = 111 \[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=-\frac {2 a^3 b \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right ) \text {csch}(x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cosh (x)) \text {csch}^3(x)}{3 \left (a^2-b^2\right )} \]
-2*a^3*b*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/ 2)-1/3*(3*a^2*b-a*(2*a^2+b^2)*cosh(x))*csch(x)/(a^2-b^2)^2+1/3*(b-a*cosh(x ))*csch(x)^3/(a^2-b^2)
Time = 0.75 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.41 \[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=\frac {(b+a \cosh (x)) \text {sech}(x) \left (\frac {48 a^3 b \arctan \left (\frac {(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {2 (4 a+b) \coth \left (\frac {x}{2}\right )}{(a+b)^2}+\frac {8 \text {csch}^3(x) \sinh ^4\left (\frac {x}{2}\right )}{a-b}-\frac {\text {csch}^4\left (\frac {x}{2}\right ) \sinh (x)}{2 (a+b)}+\frac {8 a \tanh \left (\frac {x}{2}\right )}{(a-b)^2}-\frac {2 b \tanh \left (\frac {x}{2}\right )}{(a-b)^2}\right )}{24 (a+b \text {sech}(x))} \]
((b + a*Cosh[x])*Sech[x]*((48*a^3*b*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (2*(4*a + b)*Coth[x/2])/(a + b)^2 + (8*Csch[x] ^3*Sinh[x/2]^4)/(a - b) - (Csch[x/2]^4*Sinh[x])/(2*(a + b)) + (8*a*Tanh[x/ 2])/(a - b)^2 - (2*b*Tanh[x/2])/(a - b)^2))/(24*(a + b*Sech[x]))
Time = 0.66 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.22, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.154, Rules used = {3042, 4360, 25, 25, 3042, 25, 3345, 25, 3042, 25, 3345, 27, 3042, 3138, 218}
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.
| \(\displaystyle \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos \left (-\frac {\pi }{2}+i x\right )^4 \left (a-b \csc \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\coth (x) \text {csch}^3(x)}{-a \cosh (x)-b}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\coth (x) \text {csch}^3(x)}{b+a \cosh (x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\coth (x) \text {csch}^3(x)}{a \cosh (x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (-\frac {\pi }{2}+i x\right )}{\cos \left (-\frac {\pi }{2}+i x\right )^4 \left (b-a \sin \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (i x-\frac {\pi }{2}\right )}{\cos \left (i x-\frac {\pi }{2}\right )^4 \left (b-a \sin \left (i x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3345 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\int -\frac {\left (a b-2 a^2 \cosh (x)\right ) \text {csch}^2(x)}{b+a \cosh (x)}dx}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\left (a b-2 a^2 \cosh (x)\right ) \text {csch}^2(x)}{b+a \cosh (x)}dx}{3 \left (a^2-b^2\right )}+\frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}+\frac {\int -\frac {2 \sin \left (i x-\frac {\pi }{2}\right ) a^2+b a}{\cos \left (i x-\frac {\pi }{2}\right )^2 \left (b-a \sin \left (i x-\frac {\pi }{2}\right )\right )}dx}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\int \frac {2 \sin \left (i x-\frac {\pi }{2}\right ) a^2+b a}{\cos \left (i x-\frac {\pi }{2}\right )^2 \left (b-a \sin \left (i x-\frac {\pi }{2}\right )\right )}dx}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3345 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\frac {\int \frac {3 a^3 b}{b+a \cosh (x)}dx}{a^2-b^2}+\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\frac {3 a^3 b \int \frac {1}{b+a \cosh (x)}dx}{a^2-b^2}+\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{a^2-b^2}+\frac {3 a^3 b \int \frac {1}{b+a \sin \left (i x+\frac {\pi }{2}\right )}dx}{a^2-b^2}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\frac {6 a^3 b \int \frac {1}{(a-b) \tanh ^2\left (\frac {x}{2}\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{a^2-b^2}+\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{a^2-b^2}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{a^2-b^2}+\frac {6 a^3 b \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\) |
((b - a*Cosh[x])*Csch[x]^3)/(3*(a^2 - b^2)) - ((6*a^3*b*ArcTan[(Sqrt[a - b ]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)) + ((3*a^2 *b - a*(2*a^2 + b^2)*Cosh[x])*Csch[x])/(a^2 - b^2))/(3*(a^2 - b^2))
3.1.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 1.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.14
| method | result | size |
| default | \(-\frac {\frac {a \tanh \left (\frac {x}{2}\right )^{3}}{3}-\frac {b \tanh \left (\frac {x}{2}\right )^{3}}{3}-3 a \tanh \left (\frac {x}{2}\right )+b \tanh \left (\frac {x}{2}\right )}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{3}}-\frac {-3 a -b}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )}\) | \(127\) |
| risch | \(-\frac {2 \left (3 a^{2} b \,{\mathrm e}^{5 x}-3 a \,b^{2} {\mathrm e}^{4 x}-10 a^{2} b \,{\mathrm e}^{3 x}+4 b^{3} {\mathrm e}^{3 x}+6 a^{3} {\mathrm e}^{2 x}+3 a^{2} b \,{\mathrm e}^{x}-2 a^{3}-a \,b^{2}\right )}{3 \left ({\mathrm e}^{2 x}-1\right )^{3} \left (a^{2}-b^{2}\right )^{2}}-\frac {b \,a^{3} \ln \left ({\mathrm e}^{x}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {b \,a^{3} \ln \left ({\mathrm e}^{x}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\) | \(230\) |
-1/8/(a-b)^2*(1/3*a*tanh(1/2*x)^3-1/3*b*tanh(1/2*x)^3-3*a*tanh(1/2*x)+b*ta nh(1/2*x))-2/(a-b)^2/(a+b)^2*a^3*b/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1 /2*x)/((a+b)*(a-b))^(1/2))-1/24/(a+b)/tanh(1/2*x)^3-1/8/(a+b)^2*(-3*a-b)/t anh(1/2*x)
Leaf count of result is larger than twice the leaf count of optimal. 1129 vs. \(2 (98) = 196\).
Time = 0.27 (sec) , antiderivative size = 2340, normalized size of antiderivative = 21.08 \[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=\text {Too large to display} \]
[-1/3*(6*(a^4*b - a^2*b^3)*cosh(x)^5 + 6*(a^4*b - a^2*b^3)*sinh(x)^5 - 4*a ^5 + 2*a^3*b^2 + 2*a*b^4 - 6*(a^3*b^2 - a*b^4)*cosh(x)^4 - 6*(a^3*b^2 - a* b^4 - 5*(a^4*b - a^2*b^3)*cosh(x))*sinh(x)^4 - 4*(5*a^4*b - 7*a^2*b^3 + 2* b^5)*cosh(x)^3 - 4*(5*a^4*b - 7*a^2*b^3 + 2*b^5 - 15*(a^4*b - a^2*b^3)*cos h(x)^2 + 6*(a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^3 + 12*(a^5 - a^3*b^2)*cosh( x)^2 + 12*(a^5 - a^3*b^2 + 5*(a^4*b - a^2*b^3)*cosh(x)^3 - 3*(a^3*b^2 - a* b^4)*cosh(x)^2 - (5*a^4*b - 7*a^2*b^3 + 2*b^5)*cosh(x))*sinh(x)^2 + 3*(a^3 *b*cosh(x)^6 + 6*a^3*b*cosh(x)*sinh(x)^5 + a^3*b*sinh(x)^6 - 3*a^3*b*cosh( x)^4 + 3*a^3*b*cosh(x)^2 + 3*(5*a^3*b*cosh(x)^2 - a^3*b)*sinh(x)^4 - a^3*b + 4*(5*a^3*b*cosh(x)^3 - 3*a^3*b*cosh(x))*sinh(x)^3 + 3*(5*a^3*b*cosh(x)^ 4 - 6*a^3*b*cosh(x)^2 + a^3*b)*sinh(x)^2 + 6*(a^3*b*cosh(x)^5 - 2*a^3*b*co sh(x)^3 + a^3*b*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^ 2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x) ^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) + a)) + 6*(a^4*b - a^2*b^3)*c osh(x) + 6*(a^4*b - a^2*b^3 + 5*(a^4*b - a^2*b^3)*cosh(x)^4 - 4*(a^3*b^2 - a*b^4)*cosh(x)^3 - 2*(5*a^4*b - 7*a^2*b^3 + 2*b^5)*cosh(x)^2 + 4*(a^5 - a ^3*b^2)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^6 + 6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^5 + (a^6 - 3*a^4*b^ 2 + 3*a^2*b^4 - b^6)*sinh(x)^6 - a^6 + 3*a^4*b^2 - 3*a^2*b^4 + b^6 - 3*...
\[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\operatorname {csch}^{4}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.34 \[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=-\frac {2 \, a^{3} b \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (3 \, a^{2} b e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} - 10 \, a^{2} b e^{\left (3 \, x\right )} + 4 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} - 2 \, a^{3} - a b^{2}\right )}}{3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]
-2*a^3*b*arctan((a*e^x + b)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt (a^2 - b^2)) - 2/3*(3*a^2*b*e^(5*x) - 3*a*b^2*e^(4*x) - 10*a^2*b*e^(3*x) + 4*b^3*e^(3*x) + 6*a^3*e^(2*x) + 3*a^2*b*e^x - 2*a^3 - a*b^2)/((a^4 - 2*a^ 2*b^2 + b^4)*(e^(2*x) - 1)^3)
Time = 2.33 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.66 \[ \int \frac {\text {csch}^4(x)}{a+b \text {sech}(x)} \, dx=\frac {\frac {4\,\left (a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b-b^3\right )}{3\,{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {\frac {8\,a}{3\,\left (a^2-b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2-b^2\right )}}{3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1}+\frac {\frac {2\,a\,b^2}{{\left (a^2-b^2\right )}^2}-\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}+\frac {a^3\,b\,\ln \left (\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}-\frac {2\,a^2\,b\,\left (a+b\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}-\frac {a^3\,b\,\ln \left (\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}+\frac {2\,a^2\,b\,\left (a+b\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}} \]
((4*(a*b^2 - a^3))/(a^2 - b^2)^2 + (8*exp(x)*(a^2*b - b^3))/(3*(a^2 - b^2) ^2))/(exp(4*x) - 2*exp(2*x) + 1) - ((8*a)/(3*(a^2 - b^2)) - (8*b*exp(x))/( 3*(a^2 - b^2)))/(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1) + ((2*a*b^2)/(a^2 - b^2)^2 - (2*a^2*b*exp(x))/(a^2 - b^2)^2)/(exp(2*x) - 1) + (a^3*b*log((2 *a^2*b*exp(x))/(a^2 - b^2)^2 - (2*a^2*b*(a + b*exp(x)))/((a + b)^(5/2)*(b - a)^(5/2))))/((a + b)^(5/2)*(b - a)^(5/2)) - (a^3*b*log((2*a^2*b*exp(x))/ (a^2 - b^2)^2 + (2*a^2*b*(a + b*exp(x)))/((a + b)^(5/2)*(b - a)^(5/2))))/( (a + b)^(5/2)*(b - a)^(5/2))