Integrand size = 21, antiderivative size = 90 \[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 a^2 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {b \text {sech}(c+d x)}{\left (a^2+b^2\right ) d}-\frac {a \tanh (c+d x)}{\left (a^2+b^2\right ) d} \]
-2*a^2*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)/ d-b*sech(d*x+c)/(a^2+b^2)/d-a*tanh(d*x+c)/(a^2+b^2)/d
Time = 0.38 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18 \[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {-b \sqrt {-a^2-b^2} \text {sech}(c+d x)+a \left (2 a \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )-\sqrt {-a^2-b^2} \tanh (c+d x)\right )}{\left (-a^2-b^2\right )^{3/2} d} \]
-((-(b*Sqrt[-a^2 - b^2]*Sech[c + d*x]) + a*(2*a*ArcTan[(b - a*Tanh[(c + d* x)/2])/Sqrt[-a^2 - b^2]] - Sqrt[-a^2 - b^2]*Tanh[c + d*x]))/((-a^2 - b^2)^ (3/2)*d))
Time = 0.51 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 25, 3206, 26, 3042, 26, 3086, 24, 3139, 1083, 217, 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.
| \(\displaystyle \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\tan (i c+i d x)^2}{a-i b \sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\tan (i c+i d x)^2}{a-i b \sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 3206 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a+b \sinh (c+d x)}dx}{a^2+b^2}-\frac {a \int \text {sech}^2(c+d x)dx}{a^2+b^2}-\frac {i b \int i \text {sech}(c+d x) \tanh (c+d x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a+b \sinh (c+d x)}dx}{a^2+b^2}-\frac {a \int \text {sech}^2(c+d x)dx}{a^2+b^2}+\frac {b \int \text {sech}(c+d x) \tanh (c+d x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}+\frac {b \int -i \sec (i c+i d x) \tan (i c+i d x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {i b \int \sec (i c+i d x) \tan (i c+i d x)dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {b \int 1d\text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {a^2 \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a^2+b^2}-\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {b \text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}-\frac {2 i a^2 \int \frac {1}{-a \tanh ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tanh \left (\frac {1}{2} (c+d x)\right )+a}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (a^2+b^2\right )}-\frac {b \text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}+\frac {4 i a^2 \int \frac {1}{\tanh ^2\left (\frac {1}{2} (c+d x)\right )-4 \left (a^2+b^2\right )}d\left (2 i a \tanh \left (\frac {1}{2} (c+d x)\right )-2 i b\right )}{d \left (a^2+b^2\right )}-\frac {b \text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {a \int \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{a^2+b^2}+\frac {2 a^2 \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {b \text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {i a \int 1d(-i \tanh (c+d x))}{d \left (a^2+b^2\right )}+\frac {2 a^2 \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {b \text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {2 a^2 \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {a \tanh (c+d x)}{d \left (a^2+b^2\right )}-\frac {b \text {sech}(c+d x)}{d \left (a^2+b^2\right )}\) |
(2*a^2*ArcTanh[Tanh[(c + d*x)/2]/(2*Sqrt[a^2 + b^2])])/((a^2 + b^2)^(3/2)* d) - (b*Sech[c + d*x])/((a^2 + b^2)*d) - (a*Tanh[c + d*x])/((a^2 + b^2)*d)
3.4.85.3.1 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, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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 = 0.91 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {\frac {-2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {8 a^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (4 a^{2}+4 b^{2}\right ) \sqrt {a^{2}+b^{2}}}}{d}\) | \(103\) |
| default | \(\frac {\frac {-2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {8 a^{2} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (4 a^{2}+4 b^{2}\right ) \sqrt {a^{2}+b^{2}}}}{d}\) | \(103\) |
| risch | \(\frac {-2 b \,{\mathrm e}^{d x +c}+2 a}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {a^{2} \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\) | \(171\) |
1/d*(2/(a^2+b^2)*(-a*tanh(1/2*d*x+1/2*c)-b)/(1+tanh(1/2*d*x+1/2*c)^2)+8*a^ 2/(4*a^2+4*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/ (a^2+b^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (87) = 174\).
Time = 0.26 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.90 \[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 \, a^{3} + 2 \, a b^{2} + {\left (a^{2} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} \sinh \left (d x + c\right )^{2} + a^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )^{2} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \]
(2*a^3 + 2*a*b^2 + (a^2*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c)*sinh(d*x + c ) + a^2*sinh(d*x + c)^2 + a^2)*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) - 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d* x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) - 2*(a^2*b + b^3)*cosh(d*x + c ) - 2*(a^2*b + b^3)*sinh(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c )^2 + 2*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)*sinh(d*x + c) + (a^4 + 2*a ^2*b^2 + b^4)*d*sinh(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*d)
\[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\tanh ^{2}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.28 \[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a^{2} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}} d} - \frac {2 \, {\left (b e^{\left (-d x - c\right )} + a\right )}}{{\left (a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} \]
a^2*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt( a^2 + b^2)))/((a^2 + b^2)^(3/2)*d) - 2*(b*e^(-d*x - c) + a)/((a^2 + b^2 + (a^2 + b^2)*e^(-2*d*x - 2*c))*d)
Time = 0.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {a^{2} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b e^{\left (d x + c\right )} - a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}}}{d} \]
(a^2*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c ) + 2*a + 2*sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) - 2*(b*e^(d*x + c) - a)/(( a^2 + b^2)*(e^(2*d*x + 2*c) + 1)))/d
Time = 1.51 (sec) , antiderivative size = 422, normalized size of antiderivative = 4.69 \[ \int \frac {\tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2\,a}{d\,\left (a^2+b^2\right )}-\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (\frac {2\,a^2}{b^2\,d\,\sqrt {a^4}\,{\left (a^2+b^2\right )}^2}+\frac {2\,\left (a^3\,d\,\sqrt {a^4}+a\,b^2\,d\,\sqrt {a^4}\right )}{a\,b^2\,\sqrt {-d^2\,{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}\right )-\frac {2\,\left (b^3\,d\,\sqrt {a^4}+a^2\,b\,d\,\sqrt {a^4}\right )}{a\,b^2\,\sqrt {-d^2\,{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}\right )\,\left (\frac {b^3\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}{2}+\frac {a^2\,b\,\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}}{2}\right )\right )\,\sqrt {a^4}}{\sqrt {-a^6\,d^2-3\,a^4\,b^2\,d^2-3\,a^2\,b^4\,d^2-b^6\,d^2}} \]
((2*a)/(d*(a^2 + b^2)) - (2*b*exp(c + d*x))/(d*(a^2 + b^2)))/(exp(2*c + 2* d*x) + 1) - (2*atan((exp(d*x)*exp(c)*((2*a^2)/(b^2*d*(a^4)^(1/2)*(a^2 + b^ 2)^2) + (2*(a^3*d*(a^4)^(1/2) + a*b^2*d*(a^4)^(1/2)))/(a*b^2*(-d^2*(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)*(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b^2 *d^2)^(1/2))) - (2*(b^3*d*(a^4)^(1/2) + a^2*b*d*(a^4)^(1/2)))/(a*b^2*(-d^2 *(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)*(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3 *a^4*b^2*d^2)^(1/2)))*((b^3*(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b ^2*d^2)^(1/2))/2 + (a^2*b*(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4*b^2 *d^2)^(1/2))/2))*(a^4)^(1/2))/(- a^6*d^2 - b^6*d^2 - 3*a^2*b^4*d^2 - 3*a^4 *b^2*d^2)^(1/2)