Integrand size = 23, antiderivative size = 185 \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {(a-5 b) x}{2 (a+b)^4}-\frac {\sqrt {b} \left (15 a^2-10 a b-b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{3/2} (a+b)^4 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {3 b \tanh (c+d x)}{4 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(11 a-b) b \tanh (c+d x)}{8 a (a+b)^3 d \left (a+b \tanh ^2(c+d x)\right )} \] Output:
-1/2*(a-5*b)*x/(a+b)^4-1/8*b^(1/2)*(15*a^2-10*a*b-b^2)*arctan(b^(1/2)*tanh (d*x+c)/a^(1/2))/a^(3/2)/(a+b)^4/d+1/2*cosh(d*x+c)*sinh(d*x+c)/(a+b)/d/(a+ b*tanh(d*x+c)^2)^2-3/4*b*tanh(d*x+c)/(a+b)^2/d/(a+b*tanh(d*x+c)^2)^2-1/8*( 11*a-b)*b*tanh(d*x+c)/a/(a+b)^3/d/(a+b*tanh(d*x+c)^2)
Time = 1.39 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.85 \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {-4 (a-5 b) (c+d x)+\frac {\sqrt {b} \left (-15 a^2+10 a b+b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+2 (a+b) \sinh (2 (c+d x))-\frac {4 b^2 (a+b) \sinh (2 (c+d x))}{(a-b+(a+b) \cosh (2 (c+d x)))^2}-\frac {(9 a-b) b (a+b) \sinh (2 (c+d x))}{a (a-b+(a+b) \cosh (2 (c+d x)))}}{8 (a+b)^4 d} \] Input:
Integrate[Sinh[c + d*x]^2/(a + b*Tanh[c + d*x]^2)^3,x]
Output:
(-4*(a - 5*b)*(c + d*x) + (Sqrt[b]*(-15*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b ]*Tanh[c + d*x])/Sqrt[a]])/a^(3/2) + 2*(a + b)*Sinh[2*(c + d*x)] - (4*b^2* (a + b)*Sinh[2*(c + d*x)])/(a - b + (a + b)*Cosh[2*(c + d*x)])^2 - ((9*a - b)*b*(a + b)*Sinh[2*(c + d*x)])/(a*(a - b + (a + b)*Cosh[2*(c + d*x)])))/ (8*(a + b)^4*d)
Time = 0.47 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.19, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 25, 4146, 373, 402, 27, 402, 25, 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.
| \(\displaystyle \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i c+i d x)^2}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i c+i d x)^2}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2 \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\int \frac {a-5 b \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\int -\frac {2 a \left (-9 b \tanh ^2(c+d x)+2 a-b\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{4 a (a+b)}}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {\int \frac {-9 b \tanh ^2(c+d x)+2 a-b}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{2 (a+b)}+\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {\frac {b (11 a-b) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {\int -\frac {4 a^2-9 b a-b^2-(11 a-b) b \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}}{2 (a+b)}+\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {\frac {\int \frac {4 a^2-9 b a-b^2-(11 a-b) b \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}+\frac {b (11 a-b) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{2 (a+b)}+\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {\frac {\frac {b \left (15 a^2-10 a b-b^2\right ) \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}+\frac {4 a (a-5 b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}}{2 a (a+b)}+\frac {b (11 a-b) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{2 (a+b)}+\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {\frac {\frac {4 a (a-5 b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}+\frac {\sqrt {b} \left (15 a^2-10 a b-b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 a (a+b)}+\frac {b (11 a-b) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{2 (a+b)}+\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{2 (a+b)}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\tanh (c+d x)}{2 (a+b) \left (1-\tanh ^2(c+d x)\right ) \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {\frac {\frac {\frac {\sqrt {b} \left (15 a^2-10 a b-b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}+\frac {4 a (a-5 b) \text {arctanh}(\tanh (c+d x))}{a+b}}{2 a (a+b)}+\frac {b (11 a-b) \tanh (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{2 (a+b)}+\frac {3 b \tanh (c+d x)}{2 (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{2 (a+b)}}{d}\) |
Input:
Int[Sinh[c + d*x]^2/(a + b*Tanh[c + d*x]^2)^3,x]
Output:
(Tanh[c + d*x]/(2*(a + b)*(1 - Tanh[c + d*x]^2)*(a + b*Tanh[c + d*x]^2)^2) - ((3*b*Tanh[c + d*x])/(2*(a + b)*(a + b*Tanh[c + d*x]^2)^2) + (((Sqrt[b] *(15*a^2 - 10*a*b - b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a] *(a + b)) + (4*a*(a - 5*b)*ArcTanh[Tanh[c + d*x]])/(a + b))/(2*a*(a + b)) + ((11*a - b)*b*Tanh[c + d*x])/(2*a*(a + b)*(a + b*Tanh[c + d*x]^2)))/(2*( a + b)))/(2*(a + b)))/d
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_.)*(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[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(487\) vs. \(2(167)=334\).
Time = 28.86 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.64
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b \left (\frac {\left (-\frac {9}{8} a^{2}-\frac {5}{4} a b -\frac {1}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-\frac {\left (27 a^{3}+58 a^{2} b +27 b^{2} a -4 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a}-\frac {\left (27 a^{3}+58 a^{2} b +27 b^{2} a -4 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a}+\left (-\frac {9}{8} a^{2}-\frac {5}{4} a b -\frac {1}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (15 a^{2}-10 a b -b^{2}\right ) \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{8}\right )}{\left (a +b \right )^{4}}+\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a +b \right )^{4}}-\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a +5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a +b \right )^{4}}}{d}\) | \(488\) |
| default | \(\frac {\frac {2 b \left (\frac {\left (-\frac {9}{8} a^{2}-\frac {5}{4} a b -\frac {1}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-\frac {\left (27 a^{3}+58 a^{2} b +27 b^{2} a -4 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a}-\frac {\left (27 a^{3}+58 a^{2} b +27 b^{2} a -4 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a}+\left (-\frac {9}{8} a^{2}-\frac {5}{4} a b -\frac {1}{8} b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (15 a^{2}-10 a b -b^{2}\right ) \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{8}\right )}{\left (a +b \right )^{4}}+\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a +b \right )^{4}}-\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 \left (a +b \right )^{3} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a +5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a +b \right )^{4}}}{d}\) | \(488\) |
| risch | \(-\frac {x a}{2 \left (a +b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {5 x b}{2 \left (a +b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {{\mathrm e}^{2 d x +2 c}}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right ) d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right ) d}+\frac {b \left (9 a^{3} {\mathrm e}^{6 d x +6 c}-5 a^{2} b \,{\mathrm e}^{6 d x +6 c}-13 a \,b^{2} {\mathrm e}^{6 d x +6 c}+b^{3} {\mathrm e}^{6 d x +6 c}+27 \,{\mathrm e}^{4 d x +4 c} a^{3}-21 a^{2} b \,{\mathrm e}^{4 d x +4 c}+29 a \,b^{2} {\mathrm e}^{4 d x +4 c}-3 b^{3} {\mathrm e}^{4 d x +4 c}+27 \,{\mathrm e}^{2 d x +2 c} a^{3}+a^{2} b \,{\mathrm e}^{2 d x +2 c}-23 a \,b^{2} {\mathrm e}^{2 d x +2 c}+3 b^{3} {\mathrm e}^{2 d x +2 c}+9 a^{3}+17 a^{2} b +7 b^{2} a -b^{3}\right )}{4 \left ({\mathrm e}^{4 d x +4 c} a +b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 \,{\mathrm e}^{2 d x +2 c} b +a +b \right )^{2} \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right ) \left (a +b \right ) a d}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{16 \left (a +b \right )^{4} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right ) b}{8 a \left (a +b \right )^{4} d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right ) b^{2}}{16 a^{2} \left (a +b \right )^{4} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{16 \left (a +b \right )^{4} d}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right ) b}{8 a \left (a +b \right )^{4} d}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right ) b^{2}}{16 a^{2} \left (a +b \right )^{4} d}\) | \(712\) |
Input:
int(sinh(d*x+c)^2/(a+tanh(d*x+c)^2*b)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(2*b/(a+b)^4*(((-9/8*a^2-5/4*a*b-1/8*b^2)*tanh(1/2*d*x+1/2*c)^7-1/8*(2 7*a^3+58*a^2*b+27*a*b^2-4*b^3)/a*tanh(1/2*d*x+1/2*c)^5-1/8*(27*a^3+58*a^2* b+27*a*b^2-4*b^3)/a*tanh(1/2*d*x+1/2*c)^3+(-9/8*a^2-5/4*a*b-1/8*b^2)*tanh( 1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tan h(1/2*d*x+1/2*c)^2+a)^2+1/8*(15*a^2-10*a*b-b^2)*(1/2*(a+((a+b)*b)^(1/2)+b) /a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d *x+1/2*c)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/2))-1/2*(-a+((a+b)*b)^(1/2)-b)/ a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d *x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2))))+1/2/(a+b)^3/(tanh(1/2*d*x +1/2*c)-1)^2+1/2/(a+b)^3/(tanh(1/2*d*x+1/2*c)-1)+1/2*(a-5*b)/(a+b)^4*ln(ta nh(1/2*d*x+1/2*c)-1)-1/2/(a+b)^3/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/(a+b)^3/(ta nh(1/2*d*x+1/2*c)+1)+1/2/(a+b)^4*(-a+5*b)*ln(tanh(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 6319 vs. \(2 (167) = 334\).
Time = 0.31 (sec) , antiderivative size = 12965, normalized size of antiderivative = 70.08 \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**2/(a+b*tanh(d*x+c)**2)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1806 vs. \(2 (167) = 334\).
Time = 0.42 (sec) , antiderivative size = 1806, normalized size of antiderivative = 9.76 \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
Output:
3/4*b*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/((a ^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d) - 3/4*b*log(2*(a - b)*e^(-2*d *x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*d) - 3/32*(5*a^3*b - 15*a^2*b^2 - 5*a*b^3 - b^4)*arctan(1 /2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^6 + 4*a^5*b + 6*a^4*b^ 2 + 4*a^3*b^3 + a^2*b^4)*sqrt(a*b)*d) + 3/32*(5*a^3*b - 15*a^2*b^2 - 5*a*b ^3 - b^4)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/((a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4)*sqrt(a*b)*d) + 1/16*(15*a^2*b + 10*a*b^2 + 3*b^3)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b ))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*sqrt(a*b)*d) + 1/16*(9*a^4*b + 4 *a^3*b^2 - 22*a^2*b^3 - 20*a*b^4 - 3*b^5 + 3*(3*a^4*b - 22*a^3*b^2 - 20*a^ 2*b^3 + 6*a*b^4 + b^5)*e^(6*d*x + 6*c) + (27*a^4*b - 156*a^3*b^2 + 110*a^2 *b^3 - 36*a*b^4 - 9*b^5)*e^(4*d*x + 4*c) + (27*a^4*b - 86*a^3*b^2 - 84*a^2 *b^3 + 38*a*b^4 + 9*b^5)*e^(2*d*x + 2*c))/((a^8 + 6*a^7*b + 15*a^6*b^2 + 2 0*a^5*b^3 + 15*a^4*b^4 + 6*a^3*b^5 + a^2*b^6 + (a^8 + 6*a^7*b + 15*a^6*b^2 + 20*a^5*b^3 + 15*a^4*b^4 + 6*a^3*b^5 + a^2*b^6)*e^(8*d*x + 8*c) + 4*(a^8 + 4*a^7*b + 5*a^6*b^2 - 5*a^4*b^4 - 4*a^3*b^5 - a^2*b^6)*e^(6*d*x + 6*c) + 2*(3*a^8 + 10*a^7*b + 13*a^6*b^2 + 12*a^5*b^3 + 13*a^4*b^4 + 10*a^3*b^5 + 3*a^2*b^6)*e^(4*d*x + 4*c) + 4*(a^8 + 4*a^7*b + 5*a^6*b^2 - 5*a^4*b^4 - 4*a^3*b^5 - a^2*b^6)*e^(2*d*x + 2*c))*d) - 1/16*(9*a^4*b + 4*a^3*b^2 - ...
Leaf count of result is larger than twice the leaf count of optimal. 535 vs. \(2 (167) = 334\).
Time = 1.09 (sec) , antiderivative size = 535, normalized size of antiderivative = 2.89 \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\frac {4 \, {\left (d x + c\right )} {\left (a - 5 \, b\right )}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {{\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {{\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt {a b}} - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (9 \, a^{3} b e^{\left (6 \, d x + 6 \, c\right )} - 5 \, a^{2} b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 13 \, a b^{3} e^{\left (6 \, d x + 6 \, c\right )} + b^{4} e^{\left (6 \, d x + 6 \, c\right )} + 27 \, a^{3} b e^{\left (4 \, d x + 4 \, c\right )} - 21 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 29 \, a b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 27 \, a^{3} b e^{\left (2 \, d x + 2 \, c\right )} + a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 23 \, a b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a^{3} b + 17 \, a^{2} b^{2} + 7 \, a b^{3} - b^{4}\right )}}{{\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2}}}{8 \, d} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
Output:
-1/8*(4*(d*x + c)*(a - 5*b)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - (2*a*e^(2*d*x + 2*c) - 10*b*e^(2*d*x + 2*c) - a - b)*e^(-2*d*x - 2*c)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) + (15*a^2*b - 10*a*b^2 - b^3)*arct an(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*sqrt(a*b)) - e^(2*d*x + 2*c)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(9*a^3*b*e^(6*d*x + 6*c) - 5*a^2*b^2*e^(6* d*x + 6*c) - 13*a*b^3*e^(6*d*x + 6*c) + b^4*e^(6*d*x + 6*c) + 27*a^3*b*e^( 4*d*x + 4*c) - 21*a^2*b^2*e^(4*d*x + 4*c) + 29*a*b^3*e^(4*d*x + 4*c) - 3*b ^4*e^(4*d*x + 4*c) + 27*a^3*b*e^(2*d*x + 2*c) + a^2*b^2*e^(2*d*x + 2*c) - 23*a*b^3*e^(2*d*x + 2*c) + 3*b^4*e^(2*d*x + 2*c) + 9*a^3*b + 17*a^2*b^2 + 7*a*b^3 - b^4)/((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*(a*e^(4*d* x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)^2))/d
Timed out. \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^2}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:
int(sinh(c + d*x)^2/(a + b*tanh(c + d*x)^2)^3,x)
Output:
int(sinh(c + d*x)^2/(a + b*tanh(c + d*x)^2)^3, x)
Time = 0.36 (sec) , antiderivative size = 4124, normalized size of antiderivative = 22.29 \[ \int \frac {\sinh ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^2)^3,x)
Output:
( - 60*e**(10*c + 10*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**5 - 20*e**(10*c + 10*d*x)*sqrt(b)*sqrt(a)*atan((e**( c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**4*b + 104*e**(10*c + 10*d*x)*s qrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b** 2 + 24*e**(10*c + 10*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**3 - 44*e**(10*c + 10*d*x)*sqrt(b)*sqrt(a)*atan( (e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**4 - 4*e**(10*c + 10*d*x )*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**5 - 240*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sq rt(b))/sqrt(a))*a**5 + 400*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**4*b + 96*e**(8*c + 8*d*x)*sqrt(b)* sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b**2 - 416 *e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b) )/sqrt(a))*a**2*b**3 + 144*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**4 + 16*e**(8*c + 8*d*x)*sqrt(b)* sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**5 - 360*e**( 6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqr t(a))*a**5 + 840*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt( a + b) - sqrt(b))/sqrt(a))*a**4*b - 976*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*a tan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b**2 + 720*e**(6...