\(\int \frac {1}{(a+b \text {csch}^2(c+d x))^3} \, dx\) [7]

Optimal result

Integrand size = 14, antiderivative size = 156 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\frac {x}{a^3}-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{8 a^3 (a-b)^{5/2} d}+\frac {b \coth (c+d x)}{4 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^2}+\frac {(7 a-4 b) b \coth (c+d x)}{8 a^2 (a-b)^2 d \left (a-b+b \coth ^2(c+d x)\right )} \]

[Out]

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*cot
h(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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4213, 425, 541, 536, 212, 211} \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\frac {x}{a^3}+\frac {b (7 a-4 b) \coth (c+d x)}{8 a^2 d (a-b)^2 \left (a+b \coth ^2(c+d x)-b\right )}-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{8 a^3 d (a-b)^{5/2}}+\frac {b \coth (c+d x)}{4 a d (a-b) \left (a+b \coth ^2(c+d x)-b\right )^2} \]

[In]

Int[(a + b*Csch[c + d*x]^2)^(-3),x]

[Out]

x/a^3 - (Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[b]])/(8*a^3*(a - b)^(5/2)*d
) + (b*Coth[c + d*x])/(4*a*(a - b)*d*(a - b + b*Coth[c + d*x]^2)^2) + ((7*a - 4*b)*b*Coth[c + d*x])/(8*a^2*(a
- b)^2*d*(a - b + b*Coth[c + d*x]^2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a-b+b x^2\right )^3} \, dx,x,\coth (c+d x)\right )}{d} \\ & = \frac {b \coth (c+d x)}{4 a (a-b) d \left (a-b+b \coth ^2(c+d x)\right )^2}-\frac {\text {Subst}\left (\int \frac {-4 a+b+3 b x^2}{\left (1-x^2\right ) \left (a-b+b x^2\right )^2} \, dx,x,\coth (c+d x)\right )}{4 a (a-b) 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 )}+\frac {\text {Subst}\left (\int \frac {8 a^2-9 a b+4 b^2-(7 a-4 b) b x^2}{\left (1-x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\coth (c+d x)\right )}{8 a^2 (a-b)^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 )}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\coth (c+d x)\right )}{a^3 d}+\frac {\left (b \left (15 a^2-20 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\coth (c+d x)\right )}{8 a^3 (a-b)^2 d} \\ & = \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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.02 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\frac {(-a+2 b+a \cosh (2 (c+d x))) \text {csch}^6(c+d x) \left (8 (c+d x) (a-2 b-a \cosh (2 (c+d x)))^2-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right ) (a-2 b-a \cosh (2 (c+d x)))^2}{(a-b)^{5/2}}-\frac {4 a b^2 \sinh (2 (c+d x))}{a-b}+\frac {3 a (3 a-2 b) b (-a+2 b+a \cosh (2 (c+d x))) \sinh (2 (c+d x))}{(a-b)^2}\right )}{64 a^3 d \left (a+b \text {csch}^2(c+d x)\right )^3} \]

[In]

Integrate[(a + b*Csch[c + d*x]^2)^(-3),x]

[Out]

((-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]*(1
5*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*Csch[c + d*x]^2)^3)

Maple [B] (verified)

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	\(\frac {x}{a^{3}}+\frac {b \left (9 a^{3} {\mathrm e}^{6 d x +6 c}-28 a^{2} b \,{\mathrm e}^{6 d x +6 c}+16 a \,b^{2} {\mathrm e}^{6 d x +6 c}-27 a^{3} {\mathrm e}^{4 d x +4 c}+90 a^{2} b \,{\mathrm e}^{4 d x +4 c}-120 a \,b^{2} {\mathrm e}^{4 d x +4 c}+48 \,{\mathrm e}^{4 d x +4 c} b^{3}+27 a^{3} {\mathrm e}^{2 d x +2 c}-68 a^{2} b \,{\mathrm e}^{2 d x +2 c}+32 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}-9 a^{3}+6 a^{2} b \right )}{4 a^{3} d \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}-2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {15 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a +2 \sqrt {-b \left (a -b \right )}-2 b}{a}\right )}{16 \left (a -b \right )^{3} d a}-\frac {5 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a +2 \sqrt {-b \left (a -b \right )}-2 b}{a}\right ) b}{4 \left (a -b \right )^{3} d \,a^{2}}+\frac {\sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a +2 \sqrt {-b \left (a -b \right )}-2 b}{a}\right ) b^{2}}{2 \left (a -b \right )^{3} d \,a^{3}}-\frac {15 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {-a +2 \sqrt {-b \left (a -b \right )}+2 b}{a}\right )}{16 \left (a -b \right )^{3} d a}+\frac {5 \sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {-a +2 \sqrt {-b \left (a -b \right )}+2 b}{a}\right ) b}{4 \left (a -b \right )^{3} d \,a^{2}}-\frac {\sqrt {-b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {-a +2 \sqrt {-b \left (a -b \right )}+2 b}{a}\right ) b^{2}}{2 \left (a -b \right )^{3} d \,a^{3}}\)	\(579\)

[In]

int(1/(a+b*csch(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

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*(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)^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)))))

Fricas [B] (verification not implemented)

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 \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+b*csch(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\int \frac {1}{\left (a + b \operatorname {csch}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(1/(a+b*csch(d*x+c)**2)**3,x)

[Out]

Integral((a + b*csch(c + d*x)**2)**(-3), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+b*csch(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [F]

\[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\int { \frac {1}{{\left (b \operatorname {csch}\left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]

[In]

integrate(1/(a+b*csch(d*x+c)^2)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \text {csch}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {sinh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \]

[In]

int(1/(a + b/sinh(c + d*x)^2)^3,x)

[Out]

int(1/(a + b/sinh(c + d*x)^2)^3, x)