Integrand size = 13, antiderivative size = 69 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=-\frac {2 a^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {b \text {sech}(x)}{a^2+b^2}-\frac {a \tanh (x)}{a^2+b^2} \] Output:
-2*a^2*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)-b*sech(x )/(a^2+b^2)-a*tanh(x)/(a^2+b^2)
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\frac {-b \text {sech}(x)+a \left (\frac {2 a \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\tanh (x)\right )}{a^2+b^2} \] Input:
Integrate[Tanh[x]^2/(a + b*Sinh[x]),x]
Output:
(-(b*Sech[x]) + a*((2*a*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[- a^2 - b^2] - Tanh[x]))/(a^2 + b^2)
Time = 0.47 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 25, 3206, 26, 3042, 26, 3086, 24, 3139, 1083, 219, 4254, 24}
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.
| \(\displaystyle \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\tan (i x)^2}{a-i b \sin (i x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\tan (i x)^2}{a-i b \sin (i x)}dx\) |
| \(\Big \downarrow \) 3206 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a+b \sinh (x)}dx}{a^2+b^2}-\frac {a \int \text {sech}^2(x)dx}{a^2+b^2}-\frac {i b \int i \text {sech}(x) \tanh (x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a+b \sinh (x)}dx}{a^2+b^2}-\frac {a \int \text {sech}^2(x)dx}{a^2+b^2}+\frac {b \int \text {sech}(x) \tanh (x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}+\frac {b \int -i \sec (i x) \tan (i x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {i b \int \sec (i x) \tan (i x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {b \int 1d\text {sech}(x)}{a^2+b^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {b \text {sech}(x)}{a^2+b^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}+\frac {2 a^2 \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{a^2+b^2}-\frac {b \text {sech}(x)}{a^2+b^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {4 a^2 \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{a^2+b^2}-\frac {b \text {sech}(x)}{a^2+b^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {a \int \csc \left (i x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {2 a^2 \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {b \text {sech}(x)}{a^2+b^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {i a \int 1d(-i \tanh (x))}{a^2+b^2}-\frac {2 a^2 \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {b \text {sech}(x)}{a^2+b^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {2 a^2 \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {a \tanh (x)}{a^2+b^2}-\frac {b \text {sech}(x)}{a^2+b^2}\) |
Input:
Int[Tanh[x]^2/(a + b*Sinh[x]),x]
Output:
(-2*a^2*ArcTanh[(2*b - 2*a*Tanh[x/2])/(2*Sqrt[a^2 + b^2])])/(a^2 + b^2)^(3 /2) - (b*Sech[x])/(a^2 + b^2) - (a*Tanh[x])/(a^2 + b^2)
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[(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[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[a/(a^2 - b^2) Int[(g*Tan[e + f*x])^p/Sin[e + f*x] ^2, x], x] + (-Simp[b*(g/(a^2 - b^2)) Int[(g*Tan[e + f*x])^(p - 1)/Cos[e + f*x], x], x] - Simp[a^2*(g^2/(a^2 - b^2)) Int[(g*Tan[e + f*x])^(p - 2)/ (a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2 , 0] && IntegersQ[2*p] && GtQ[p, 1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.48 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(\frac {8 a^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (4 a^{2}+4 b^{2}\right ) \sqrt {a^{2}+b^{2}}}+\frac {-2 a \tanh \left (\frac {x}{2}\right )-2 b}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )}\) | \(84\) |
| risch | \(\frac {-2 \,{\mathrm e}^{x} b +2 a}{\left ({\mathrm e}^{2 x}+1\right ) \left (a^{2}+b^{2}\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {a^{2} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\) | \(145\) |
Input:
int(tanh(x)^2/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
8*a^2/(4*a^2+4*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2 +b^2)^(1/2))+2/(a^2+b^2)*(-a*tanh(1/2*x)-b)/(1+tanh(1/2*x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (65) = 130\).
Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.72 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\frac {2 \, a^{3} + 2 \, a b^{2} + {\left (a^{2} \cosh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} \sinh \left (x\right )^{2} + a^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}} \] Input:
integrate(tanh(x)^2/(a+b*sinh(x)),x, algorithm="fricas")
Output:
(2*a^3 + 2*a*b^2 + (a^2*cosh(x)^2 + 2*a^2*cosh(x)*sinh(x) + a^2*sinh(x)^2 + a^2)*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh( x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh( x) + a)*sinh(x) - b)) - 2*(a^2*b + b^3)*cosh(x) - 2*(a^2*b + b^3)*sinh(x)) /(a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 + 2*a ^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^2)
\[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\int \frac {\tanh ^{2}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:
integrate(tanh(x)**2/(a+b*sinh(x)),x)
Output:
Integral(tanh(x)**2/(a + b*sinh(x)), x)
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\frac {a^{2} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{\left (-x\right )} + a\right )}}{a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )}} \] Input:
integrate(tanh(x)^2/(a+b*sinh(x)),x, algorithm="maxima")
Output:
a^2*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2))) /(a^2 + b^2)^(3/2) - 2*(b*e^(-x) + a)/(a^2 + b^2 + (a^2 + b^2)*e^(-2*x))
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\frac {a^{2} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}} \] Input:
integrate(tanh(x)^2/(a+b*sinh(x)),x, algorithm="giac")
Output:
a^2*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt( a^2 + b^2)))/(a^2 + b^2)^(3/2) - 2*(b*e^x - a)/((a^2 + b^2)*(e^(2*x) + 1))
Time = 1.86 (sec) , antiderivative size = 330, normalized size of antiderivative = 4.78 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\frac {\frac {2\,a}{a^2+b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {2\,\mathrm {atan}\left (\left (\frac {b^3\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}{2}+\frac {a^2\,b\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}{2}\right )\,\left ({\mathrm {e}}^x\,\left (\frac {2\,a^2}{b^2\,\sqrt {a^4}\,{\left (a^2+b^2\right )}^2}+\frac {2\,\left (a^3\,\sqrt {a^4}+a\,b^2\,\sqrt {a^4}\right )}{a\,b^2\,\sqrt {-{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}\right )-\frac {2\,\left (b^3\,\sqrt {a^4}+a^2\,b\,\sqrt {a^4}\right )}{a\,b^2\,\sqrt {-{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}\right )\right )\,\sqrt {a^4}}{\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}} \] Input:
int(tanh(x)^2/(a + b*sinh(x)),x)
Output:
((2*a)/(a^2 + b^2) - (2*b*exp(x))/(a^2 + b^2))/(exp(2*x) + 1) - (2*atan((( b^3*(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/2 + (a^2*b*(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/2)*(exp(x)*((2*a^2)/(b^2*(a^4)^(1/2)*(a^2 + b^2)^2) + (2*(a^3*(a^4)^(1/2) + a*b^2*(a^4)^(1/2)))/(a*b^2*(-(a^2 + b^2)^ 3)^(1/2)*(a^2 + b^2)*(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2))) - (2*(b ^3*(a^4)^(1/2) + a^2*b*(a^4)^(1/2)))/(a*b^2*(-(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)*(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2))))*(a^4)^(1/2))/(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.38 \[ \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx=\frac {2 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i +2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i -2 e^{2 x} a^{3}-2 e^{2 x} a \,b^{2}-2 e^{x} a^{2} b -2 e^{x} b^{3}}{e^{2 x} a^{4}+2 e^{2 x} a^{2} b^{2}+e^{2 x} b^{4}+a^{4}+2 a^{2} b^{2}+b^{4}} \] Input:
int(tanh(x)^2/(a+b*sinh(x)),x)
Output:
(2*(e**(2*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a* *2*i + sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a**2*i - e**(2*x)*a**3 - e**(2*x)*a*b**2 - e**x*a**2*b - e**x*b**3))/(e**(2*x)*a** 4 + 2*e**(2*x)*a**2*b**2 + e**(2*x)*b**4 + a**4 + 2*a**2*b**2 + b**4)