Integrand size = 23, antiderivative size = 259 \[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {b^2 \left (48 a^2-80 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{9/2} (a-b)^{5/2} d}+\frac {\left (8 a^3-4 a^2 b-45 a b^2+35 b^3\right ) \coth (c+d x)}{8 a^4 (a-b)^2 d}-\frac {\left (8 a^2-52 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 (a-b)^2 d}-\frac {b \text {csch}^3(c+d x) \text {sech}^3(c+d x)}{4 a (a-b) d \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac {(10 a-7 b) b \text {csch}^3(c+d x) \text {sech}(c+d x)}{8 a^2 (a-b)^2 d \left (a-(a-b) \tanh ^2(c+d x)\right )} \]
1/8*b^2*(48*a^2-80*a*b+35*b^2)*arctanh((a-b)^(1/2)*tanh(d*x+c)/a^(1/2))/a^ (9/2)/(a-b)^(5/2)/d+1/8*(8*a^3-4*a^2*b-45*a*b^2+35*b^3)*coth(d*x+c)/a^4/(a -b)^2/d-1/24*(8*a^2-52*a*b+35*b^2)*coth(d*x+c)^3/a^3/(a-b)^2/d-1/4*b*csch( d*x+c)^3*sech(d*x+c)^3/a/(a-b)/d/(a-(a-b)*tanh(d*x+c)^2)^2-1/8*(10*a-7*b)* b*csch(d*x+c)^3*sech(d*x+c)/a^2/(a-b)^2/d/(a-(a-b)*tanh(d*x+c)^2)
Time = 7.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.64 \[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {\frac {3 b^2 \left (48 a^2-80 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{(a-b)^{5/2}}+\sqrt {a} \left (-8 \coth (c+d x) \left (-2 a-9 b+a \text {csch}^2(c+d x)\right )+\frac {3 b^3 \left (-32 a^2+40 a b-11 b^2+b (-14 a+11 b) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{(a-b)^2 (2 a-b+b \cosh (2 (c+d x)))^2}\right )}{24 a^{9/2} d} \]
((3*b^2*(48*a^2 - 80*a*b + 35*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqr t[a]])/(a - b)^(5/2) + Sqrt[a]*(-8*Coth[c + d*x]*(-2*a - 9*b + a*Csch[c + d*x]^2) + (3*b^3*(-32*a^2 + 40*a*b - 11*b^2 + b*(-14*a + 11*b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/((a - b)^2*(2*a - b + b*Cosh[2*(c + d*x)])^2))) /(24*a^(9/2)*d)
Time = 0.55 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3666, 370, 439, 437, 2009}
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.
| \(\displaystyle \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (i c+i d x)^4 \left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (1-\tanh ^2(c+d x)\right )^4}{\left (a-(a-b) \tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 370 |
| \(\displaystyle \frac {\frac {\int \frac {\coth ^4(c+d x) \left (1-\tanh ^2(c+d x)\right )^2 \left (-\left ((4 a-b) \tanh ^2(c+d x)\right )+4 a-7 b\right )}{\left (a-(a-b) \tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{4 a (a-b)}-\frac {b \left (1-\tanh ^2(c+d x)\right )^3 \coth ^3(c+d x)}{4 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 439 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\coth ^4(c+d x) \left (1-\tanh ^2(c+d x)\right ) \left (8 a^2-52 b a+35 b^2-\left (8 a^2-12 b a+7 b^2\right ) \tanh ^2(c+d x)\right )}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{2 a (a-b)}-\frac {b (10 a-7 b) \left (1-\tanh ^2(c+d x)\right )^2 \coth ^3(c+d x)}{2 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \left (1-\tanh ^2(c+d x)\right )^3 \coth ^3(c+d x)}{4 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 437 |
| \(\displaystyle \frac {\frac {\frac {\int \left (\frac {\left (8 a^2-52 b a+35 b^2\right ) \coth ^4(c+d x)}{a}+\frac {\left (-8 a^3+4 b a^2+45 b^2 a-35 b^3\right ) \coth ^2(c+d x)}{a^2}+\frac {b^2 \left (48 a^2-80 b a+35 b^2\right )}{a^2 \left (a-(a-b) \tanh ^2(c+d x)\right )}\right )d\tanh (c+d x)}{2 a (a-b)}-\frac {b (10 a-7 b) \left (1-\tanh ^2(c+d x)\right )^2 \coth ^3(c+d x)}{2 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \left (1-\tanh ^2(c+d x)\right )^3 \coth ^3(c+d x)}{4 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\left (8 a^2-52 a b+35 b^2\right ) \coth ^3(c+d x)}{3 a}+\frac {b^2 \left (48 a^2-80 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a-b}}+\frac {\left (8 a^3-4 a^2 b-45 a b^2+35 b^3\right ) \coth (c+d x)}{a^2}}{2 a (a-b)}-\frac {b (10 a-7 b) \left (1-\tanh ^2(c+d x)\right )^2 \coth ^3(c+d x)}{2 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \left (1-\tanh ^2(c+d x)\right )^3 \coth ^3(c+d x)}{4 a (a-b) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
(-1/4*(b*Coth[c + d*x]^3*(1 - Tanh[c + d*x]^2)^3)/(a*(a - b)*(a - (a - b)* Tanh[c + d*x]^2)^2) + (((b^2*(48*a^2 - 80*a*b + 35*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(a^(5/2)*Sqrt[a - b]) + ((8*a^3 - 4*a^2*b - 45 *a*b^2 + 35*b^3)*Coth[c + d*x])/a^2 - ((8*a^2 - 52*a*b + 35*b^2)*Coth[c + d*x]^3)/(3*a))/(2*a*(a - b)) - ((10*a - 7*b)*b*Coth[c + d*x]^3*(1 - Tanh[c + d*x]^2)^2)/(2*a*(a - b)*(a - (a - b)*Tanh[c + d*x]^2)))/(4*a*(a - b)))/ d
3.1.59.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(a*b*e*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1)) Int[(e*x) ^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*Simp[c*(b*c*2*(p + 1) + (b*c - a *d)*(m + 1)) + d*(b*c*2*(p + 1) + (b*c - a*d)*(m + 2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*( a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2)^r, x], x] /; FreeQ[{a, b, c, d, e, f , g, m}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 ))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G tQ[q, 0] && !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(496\) vs. \(2(241)=482\).
Time = 1.18 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a -12 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{4}}-\frac {1}{24 a^{3} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-3 a -12 b}{8 a^{4} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {4 b^{2} \left (\frac {\frac {a b \left (16 a -13 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16 a^{2}-32 a b +16 b^{2}}-\frac {\left (16 a^{2}-69 a b +44 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16 \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (16 a^{2}-69 a b +44 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16 \left (a^{2}-2 a b +b^{2}\right )}+\frac {a b \left (16 a -13 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a^{2}-32 a b +16 b^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (48 a^{2}-80 a b +35 b^{2}\right ) a \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{16 a^{2}-32 a b +16 b^{2}}\right )}{a^{4}}}{d}\) | \(497\) |
| default | \(\frac {-\frac {\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a -12 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{4}}-\frac {1}{24 a^{3} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-3 a -12 b}{8 a^{4} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {4 b^{2} \left (\frac {\frac {a b \left (16 a -13 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16 a^{2}-32 a b +16 b^{2}}-\frac {\left (16 a^{2}-69 a b +44 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16 \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (16 a^{2}-69 a b +44 b^{2}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16 \left (a^{2}-2 a b +b^{2}\right )}+\frac {a b \left (16 a -13 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a^{2}-32 a b +16 b^{2}}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (48 a^{2}-80 a b +35 b^{2}\right ) a \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{16 a^{2}-32 a b +16 b^{2}}\right )}{a^{4}}}{d}\) | \(497\) |
| risch | \(\text {Expression too large to display}\) | \(1065\) |
1/d*(-1/8/a^4*(1/3*a*tanh(1/2*d*x+1/2*c)^3-3*tanh(1/2*d*x+1/2*c)*a-12*b*ta nh(1/2*d*x+1/2*c))-1/24/a^3/tanh(1/2*d*x+1/2*c)^3-1/8/a^4*(-3*a-12*b)/tanh (1/2*d*x+1/2*c)-4*b^2/a^4*((1/16*a*b*(16*a-13*b)/(a^2-2*a*b+b^2)*tanh(1/2* d*x+1/2*c)^7-1/16*(16*a^2-69*a*b+44*b^2)*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/ 2*c)^5-1/16*(16*a^2-69*a*b+44*b^2)*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3 +1/16*a*b*(16*a-13*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*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tanh(1/2*d*x+1/2*c)^2+a)^2+1/16*(4 8*a^2-80*a*b+35*b^2)/(a^2-2*a*b+b^2)*a*(1/2*((-b*(a-b))^(1/2)+b)/a/(-b*(a- b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c )/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))-1/2*((-b*(a-b))^(1/2)-b)/a/(-b*(a- b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2* c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)))))
Leaf count of result is larger than twice the leaf count of optimal. 8519 vs. \(2 (243) = 486\).
Time = 0.49 (sec) , antiderivative size = 17294, normalized size of antiderivative = 66.77 \[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {csch}\left (d x + c\right )^{4}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {csch}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {sinh}\left (c+d\,x\right )}^4\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]