Integrand size = 14, antiderivative size = 156 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\frac {x}{a^3}-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{8 a^3 (a-b)^{5/2} d}+\frac {b \coth (c+d x)}{4 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^2}+\frac {(7 a-4 b) b \coth (c+d x)}{8 a^2 (a-b)^2 d \left (a-b+b \coth ^2(c+d x)\right )} \]
x/a^3+1/4*b*coth(d*x+c)/a/(a-b)/d/(a-b+b*coth(d*x+c)^2)^2+1/8*(7*a-4*b)*b* coth(d*x+c)/a^2/(a-b)^2/d/(a-b+b*coth(d*x+c)^2)-1/8*(15*a^2-20*a*b+8*b^2)* arctan((a-b)^(1/2)*tanh(d*x+c)/b^(1/2))*b^(1/2)/a^3/(a-b)^(5/2)/d
Time = 7.02 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\frac {(-a+2 b+a \cosh (2 (c+d x))) \text {csch}^6(c+d x) \left (8 (c+d x) (a-2 b-a \cosh (2 (c+d x)))^2-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right ) (a-2 b-a \cosh (2 (c+d x)))^2}{(a-b)^{5/2}}-\frac {4 a b^2 \sinh (2 (c+d x))}{a-b}+\frac {3 a (3 a-2 b) b (-a+2 b+a \cosh (2 (c+d x))) \sinh (2 (c+d x))}{(a-b)^2}\right )}{64 a^3 d \left (a+b \text {csch}^2(c+d x)\right )^3} \]
((-a + 2*b + a*Cosh[2*(c + d*x)])*Csch[c + d*x]^6*(8*(c + d*x)*(a - 2*b - a*Cosh[2*(c + d*x)])^2 - (Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[b]]*(a - 2*b - a*Cosh[2*(c + d*x)])^2)/(a - b)^( 5/2) - (4*a*b^2*Sinh[2*(c + d*x)])/(a - b) + (3*a*(3*a - 2*b)*b*(-a + 2*b + a*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(a - b)^2))/(64*a^3*d*(a + b*Csc h[c + d*x]^2)^3)
Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4616, 316, 25, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-b \sec \left (i c+i d x+\frac {\pi }{2}\right )^2\right )^3}dx\) |
| \(\Big \downarrow \) 4616 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\coth ^2(c+d x)\right ) \left (b \coth ^2(c+d x)+a-b\right )^3}d\coth (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}-\frac {\int -\frac {-3 b \coth ^2(c+d x)+4 a-b}{\left (1-\coth ^2(c+d x)\right ) \left (b \coth ^2(c+d x)+a-b\right )^2}d\coth (c+d x)}{4 a (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {-3 b \coth ^2(c+d x)+4 a-b}{\left (1-\coth ^2(c+d x)\right ) \left (b \coth ^2(c+d x)+a-b\right )^2}d\coth (c+d x)}{4 a (a-b)}+\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {b (7 a-4 b) \coth (c+d x)}{2 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )}-\frac {\int -\frac {8 a^2-9 b a+4 b^2-(7 a-4 b) b \coth ^2(c+d x)}{\left (1-\coth ^2(c+d x)\right ) \left (b \coth ^2(c+d x)+a-b\right )}d\coth (c+d x)}{2 a (a-b)}}{4 a (a-b)}+\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {8 a^2-9 b a+4 b^2-(7 a-4 b) b \coth ^2(c+d x)}{\left (1-\coth ^2(c+d x)\right ) \left (b \coth ^2(c+d x)+a-b\right )}d\coth (c+d x)}{2 a (a-b)}+\frac {b (7 a-4 b) \coth (c+d x)}{2 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )}}{4 a (a-b)}+\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {b \left (15 a^2-20 a b+8 b^2\right ) \int \frac {1}{b \coth ^2(c+d x)+a-b}d\coth (c+d x)}{a}+\frac {8 (a-b)^2 \int \frac {1}{1-\coth ^2(c+d x)}d\coth (c+d x)}{a}}{2 a (a-b)}+\frac {b (7 a-4 b) \coth (c+d x)}{2 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )}}{4 a (a-b)}+\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a-b)^2 \int \frac {1}{1-\coth ^2(c+d x)}d\coth (c+d x)}{a}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a-b}}\right )}{a \sqrt {a-b}}}{2 a (a-b)}+\frac {b (7 a-4 b) \coth (c+d x)}{2 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )}}{4 a (a-b)}+\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a-b}}\right )}{a \sqrt {a-b}}+\frac {8 (a-b)^2 \text {arctanh}(\coth (c+d x))}{a}}{2 a (a-b)}+\frac {b (7 a-4 b) \coth (c+d x)}{2 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )}}{4 a (a-b)}+\frac {b \coth (c+d x)}{4 a (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2}}{d}\) |
((b*Coth[c + d*x])/(4*a*(a - b)*(a - b + b*Coth[c + d*x]^2)^2) + (((Sqrt[b ]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Coth[c + d*x])/Sqrt[a - b]])/( a*Sqrt[a - b]) + (8*(a - b)^2*ArcTanh[Coth[c + d*x]])/a)/(2*a*(a - b)) + ( (7*a - 4*b)*b*Coth[c + d*x])/(2*a*(a - b)*(a - b + b*Coth[c + d*x]^2)))/(4 *a*(a - b)))/d
3.1.7.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(441\) vs. \(2(142)=284\).
Time = 1.00 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.83
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {2 b \left (\frac {\frac {16 a b \left (7 a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{128 a^{2}-256 a b +128 b^{2}}+\frac {16 \left (36 a^{2}-31 a b +4 b^{2}\right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{128 a^{2}-256 a b +128 b^{2}}+\frac {16 \left (36 a^{2}-31 a b +4 b^{2}\right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128 a^{2}-256 a b +128 b^{2}}+\frac {16 a b \left (7 a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a^{2}-256 a b +128 b^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +b \right )^{2}}+\frac {2 \left (15 a^{2}-20 a b +8 b^{2}\right ) b \left (\frac {\left (\sqrt {a \left (a -b \right )}+a \right ) \arctan \left (\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}\right )}{2 b \sqrt {a \left (a -b \right )}\, \sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}-\frac {\left (\sqrt {a \left (a -b \right )}-a \right ) \operatorname {arctanh}\left (\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}\right )}{2 b \sqrt {a \left (a -b \right )}\, \sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}\right )}{16 a^{2}-32 a b +16 b^{2}}\right )}{a^{3}}}{d}\) | \(442\) |
| default | \(\frac {\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {2 b \left (\frac {\frac {16 a b \left (7 a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{128 a^{2}-256 a b +128 b^{2}}+\frac {16 \left (36 a^{2}-31 a b +4 b^{2}\right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{128 a^{2}-256 a b +128 b^{2}}+\frac {16 \left (36 a^{2}-31 a b +4 b^{2}\right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128 a^{2}-256 a b +128 b^{2}}+\frac {16 a b \left (7 a -4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a^{2}-256 a b +128 b^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +b \right )^{2}}+\frac {2 \left (15 a^{2}-20 a b +8 b^{2}\right ) b \left (\frac {\left (\sqrt {a \left (a -b \right )}+a \right ) \arctan \left (\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}\right )}{2 b \sqrt {a \left (a -b \right )}\, \sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}-\frac {\left (\sqrt {a \left (a -b \right )}-a \right ) \operatorname {arctanh}\left (\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}\right )}{2 b \sqrt {a \left (a -b \right )}\, \sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}\right )}{16 a^{2}-32 a b +16 b^{2}}\right )}{a^{3}}}{d}\) | \(442\) |
| risch | \(\frac {x}{a^{3}}+\frac {b \left (9 a^{3} {\mathrm e}^{6 d x +6 c}-28 a^{2} b \,{\mathrm e}^{6 d x +6 c}+16 a \,b^{2} {\mathrm e}^{6 d x +6 c}-27 a^{3} {\mathrm e}^{4 d x +4 c}+90 a^{2} b \,{\mathrm e}^{4 d x +4 c}-120 a \,b^{2} {\mathrm e}^{4 d x +4 c}+48 \,{\mathrm e}^{4 d x +4 c} b^{3}+27 a^{3} {\mathrm e}^{2 d x +2 c}-68 a^{2} b \,{\mathrm e}^{2 d x +2 c}+32 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}-9 a^{3}+6 a^{2} b \right )}{4 a^{3} d \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}-2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {15 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a +2 \sqrt {-b \left (a -b \right )}-2 b}{a}\right )}{16 \left (a -b \right )^{3} d a}-\frac {5 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a +2 \sqrt {-b \left (a -b \right )}-2 b}{a}\right ) b}{4 \left (a -b \right )^{3} d \,a^{2}}+\frac {\sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a +2 \sqrt {-b \left (a -b \right )}-2 b}{a}\right ) b^{2}}{2 \left (a -b \right )^{3} d \,a^{3}}-\frac {15 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {-a +2 \sqrt {-b \left (a -b \right )}+2 b}{a}\right )}{16 \left (a -b \right )^{3} d a}+\frac {5 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {-a +2 \sqrt {-b \left (a -b \right )}+2 b}{a}\right ) b}{4 \left (a -b \right )^{3} d \,a^{2}}-\frac {\sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {-a +2 \sqrt {-b \left (a -b \right )}+2 b}{a}\right ) b^{2}}{2 \left (a -b \right )^{3} d \,a^{3}}\) | \(579\) |
1/d*(1/a^3*ln(1+tanh(1/2*d*x+1/2*c))-1/a^3*ln(tanh(1/2*d*x+1/2*c)-1)+2/a^3 *b*(16*(1/128*a*b*(7*a-4*b)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7+1/128*(3 6*a^2-31*a*b+4*b^2)*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5+1/128*(36*a^2- 31*a*b+4*b^2)*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+1/128*a*b*(7*a-4*b)/ (a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*b+4*tanh(1/2*d *x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+b)^2+2*(15*a^2-20*a*b+8*b^2)/(16*a ^2-32*a*b+16*b^2)*b*(1/2*((a*(a-b))^(1/2)+a)/b/(a*(a-b))^(1/2)/((2*(a*(a-b ))^(1/2)+2*a-b)*b)^(1/2)*arctan(b*tanh(1/2*d*x+1/2*c)/((2*(a*(a-b))^(1/2)+ 2*a-b)*b)^(1/2))-1/2*((a*(a-b))^(1/2)-a)/b/(a*(a-b))^(1/2)/((2*(a*(a-b))^( 1/2)-2*a+b)*b)^(1/2)*arctanh(b*tanh(1/2*d*x+1/2*c)/((2*(a*(a-b))^(1/2)-2*a +b)*b)^(1/2)))))
Leaf count of result is larger than twice the leaf count of optimal. 3140 vs. \(2 (142) = 284\).
Time = 0.34 (sec) , antiderivative size = 6569, normalized size of antiderivative = 42.11 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\int \frac {1}{\left (a + b \operatorname {csch}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\int { \frac {1}{{\left (b \operatorname {csch}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {sinh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \]