\(\int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\) [208]

Optimal result

Integrand size = 22, antiderivative size = 136 \[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=-\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 a \sqrt {\sqrt {a}-\sqrt {b}} d}-\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 a \sqrt {\sqrt {a}+\sqrt {b}} d} \] Output:

-1/2*b^(1/4)*arctan(b^(1/4)*cosh(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/a/(a^(1/2 
)-b^(1/2))^(1/2)/d-arctanh(cosh(d*x+c))/a/d+1/2*b^(1/4)*arctanh(b^(1/4)*co 
sh(d*x+c)/(a^(1/2)+b^(1/2))^(1/2))/a/(a^(1/2)+b^(1/2))^(1/2)/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 9.26 (sec) , antiderivative size = 397, normalized size of antiderivative = 2.92 \[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=-\frac {8 \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )-8 \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )+b \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-c-d x-2 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right )+3 c \text {$\#$1}^2+3 d x \text {$\#$1}^2+6 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^2-3 c \text {$\#$1}^4-3 d x \text {$\#$1}^4-6 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^4+c \text {$\#$1}^6+d x \text {$\#$1}^6+2 \log \left (-\cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}-\sinh \left (\frac {1}{2} (c+d x)\right ) \text {$\#$1}\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{8 a d} \] Input:

Integrate[Csch[c + d*x]/(a - b*Sinh[c + d*x]^4),x]
 

Output:

-1/8*(8*Log[Cosh[(c + d*x)/2]] - 8*Log[Sinh[(c + d*x)/2]] + b*RootSum[b - 
4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-c - d*x - 2*Log[ 
-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + 
d*x)/2]*#1] + 3*c*#1^2 + 3*d*x*#1^2 + 6*Log[-Cosh[(c + d*x)/2] - Sinh[(c + 
 d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^2 - 3*c*#1^4 - 
3*d*x*#1^4 - 6*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x) 
/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^4 + c*#1^6 + d*x*#1^6 + 2*Log[-Cosh[(c + 
 d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1 
]*#1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1^7) & ])/(a*d)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3042, 26, 3694, 1484, 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.

\(\displaystyle \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i}{\sin (i c+i d x) \left (a-b \sin (i c+i d x)^4\right )}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {1}{\sin (i c+i d x) \left (a-b \sin (i c+i d x)^4\right )}dx\)

\(\Big \downarrow \) 3694

\(\displaystyle -\frac {\int \frac {1}{\left (1-\cosh ^2(c+d x)\right ) \left (-b \cosh ^4(c+d x)+2 b \cosh ^2(c+d x)+a-b\right )}d\cosh (c+d x)}{d}\)

\(\Big \downarrow \) 1484

\(\displaystyle -\frac {\int \left (\frac {b-b \cosh ^2(c+d x)}{a \left (-b \cosh ^4(c+d x)+2 b \cosh ^2(c+d x)+a-b\right )}-\frac {1}{a \left (\cosh ^2(c+d x)-1\right )}\right )d\cosh (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 a \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} \cosh (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 a \sqrt {\sqrt {a}+\sqrt {b}}}+\frac {\text {arctanh}(\cosh (c+d x))}{a}}{d}\)

Input:

Int[Csch[c + d*x]/(a - b*Sinh[c + d*x]^4),x]
 

Output:

-(((b^(1/4)*ArcTan[(b^(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]])/(2*a* 
Sqrt[Sqrt[a] - Sqrt[b]]) + ArcTanh[Cosh[c + d*x]]/a - (b^(1/4)*ArcTanh[(b^ 
(1/4)*Cosh[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]])/(2*a*Sqrt[Sqrt[a] + Sqrt[b] 
]))/d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 
 0] && IntegerQ[q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3694

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, 
 x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 
 1)/2]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.01 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{a d}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{a d}+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4096 a^{5} d^{4}-4096 a^{4} b \,d^{4}\right ) \textit {\_Z}^{4}+128 a^{2} b \,d^{2} \textit {\_Z}^{2}-b \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (\left (\frac {1024 d^{3} a^{4}}{b}-1024 a^{3} d^{3}\right ) \textit {\_R}^{3}+32 a d \textit {\_R} \right ) {\mathrm e}^{d x +c}+1\right )\right )\)	\(130\)
derivativedivides	\(\frac {8 b \left (-\frac {\sqrt {a b}\, \arctan \left (\frac {-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}+2 a}{4 \sqrt {-a b -\sqrt {a b}\, a}}\right )}{16 a b \sqrt {-a b -\sqrt {a b}\, a}}-\frac {\sqrt {a b}\, \arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}-2 a}{4 \sqrt {-a b +\sqrt {a b}\, a}}\right )}{16 a b \sqrt {-a b +\sqrt {a b}\, a}}\right )+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(164\)
default	\(\frac {8 b \left (-\frac {\sqrt {a b}\, \arctan \left (\frac {-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}+2 a}{4 \sqrt {-a b -\sqrt {a b}\, a}}\right )}{16 a b \sqrt {-a b -\sqrt {a b}\, a}}-\frac {\sqrt {a b}\, \arctan \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \sqrt {a b}-2 a}{4 \sqrt {-a b +\sqrt {a b}\, a}}\right )}{16 a b \sqrt {-a b +\sqrt {a b}\, a}}\right )+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\)	\(164\)

Input:

int(csch(d*x+c)/(a-b*sinh(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

-1/a/d*ln(exp(d*x+c)+1)+1/a/d*ln(exp(d*x+c)-1)+2*sum(_R*ln(exp(2*d*x+2*c)+ 
((1024/b*d^3*a^4-1024*a^3*d^3)*_R^3+32*a*d*_R)*exp(d*x+c)+1),_R=RootOf((40 
96*a^5*d^4-4096*a^4*b*d^4)*_Z^4+128*a^2*b*d^2*_Z^2-b))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1067 vs. \(2 (100) = 200\).

Time = 0.17 (sec) , antiderivative size = 1067, normalized size of antiderivative = 7.85 \[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(csch(d*x+c)/(a-b*sinh(d*x+c)^4),x, algorithm="fricas")
 

Output:

1/4*(a*d*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) 
+ b)/((a^3 - a^2*b)*d^2))*log(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d 
*x + c) + b*sinh(d*x + c)^2 + 2*(a*b*d*cosh(d*x + c) + a*b*d*sinh(d*x + c) 
 - ((a^4 - a^3*b)*d^3*cosh(d*x + c) + (a^4 - a^3*b)*d^3*sinh(d*x + c))*sqr 
t(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)))*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b/((a^ 
5 - 2*a^4*b + a^3*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2)) + b) - a*d*sqrt(-(( 
a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + b)/((a^3 - a^2* 
b)*d^2))*log(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh( 
d*x + c)^2 - 2*(a*b*d*cosh(d*x + c) + a*b*d*sinh(d*x + c) - ((a^4 - a^3*b) 
*d^3*cosh(d*x + c) + (a^4 - a^3*b)*d^3*sinh(d*x + c))*sqrt(b/((a^5 - 2*a^4 
*b + a^3*b^2)*d^4)))*sqrt(-((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3 
*b^2)*d^4)) + b)/((a^3 - a^2*b)*d^2)) + b) + a*d*sqrt(((a^3 - a^2*b)*d^2*s 
qrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - b)/((a^3 - a^2*b)*d^2))*log(b*cos 
h(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*(a* 
b*d*cosh(d*x + c) + a*b*d*sinh(d*x + c) + ((a^4 - a^3*b)*d^3*cosh(d*x + c) 
 + (a^4 - a^3*b)*d^3*sinh(d*x + c))*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4) 
))*sqrt(((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - b)/(( 
a^3 - a^2*b)*d^2)) + b) - a*d*sqrt(((a^3 - a^2*b)*d^2*sqrt(b/((a^5 - 2*a^4 
*b + a^3*b^2)*d^4)) - b)/((a^3 - a^2*b)*d^2))*log(b*cosh(d*x + c)^2 + 2*b* 
cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - 2*(a*b*d*cosh(d*x + c...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(csch(d*x+c)/(a-b*sinh(d*x+c)**4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\operatorname {csch}\left (d x + c\right )}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(csch(d*x+c)/(a-b*sinh(d*x+c)^4),x, algorithm="maxima")
 

Output:

-log((e^(d*x + c) + 1)*e^(-c))/(a*d) + log((e^(d*x + c) - 1)*e^(-c))/(a*d) 
 - 2*integrate((b*e^(7*d*x + 7*c) - 3*b*e^(5*d*x + 5*c) + 3*b*e^(3*d*x + 3 
*c) - b*e^(d*x + c))/(a*b*e^(8*d*x + 8*c) - 4*a*b*e^(6*d*x + 6*c) - 4*a*b* 
e^(2*d*x + 2*c) + a*b - 2*(8*a^2*e^(4*c) - 3*a*b*e^(4*c))*e^(4*d*x)), x)
 

Giac [F]

\[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\int { -\frac {\operatorname {csch}\left (d x + c\right )}{b \sinh \left (d x + c\right )^{4} - a} \,d x } \] Input:

integrate(csch(d*x+c)/(a-b*sinh(d*x+c)^4),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 1243, normalized size of antiderivative = 9.14 \[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

int(1/(sinh(c + d*x)*(a - b*sinh(c + d*x)^4)),x)
 

Output:

log((((((4294967296*a*d^2*(exp(2*c + 2*d*x) + 1)*(26*a*b - 49*a^2 + 15*b^2 
))/(b^6*(a - b)^3) + (8589934592*a^2*d^3*exp(c + d*x)*(3*a*b + 16*a^2 - 15 
*b^2)*(-(a^2*b + (a^5*b)^(1/2))/(a^4*d^2*(a - b)))^(1/2))/(b^7*(a - b)^2)) 
*(-(a^2*b + (a^5*b)^(1/2))/(a^4*d^2*(a - b)))^(1/2))/4 - (2147483648*d*exp 
(c + d*x)*(17*a - 15*b))/(b^6*(a - b)^2))*(-(a^2*b + (a^5*b)^(1/2))/(a^4*d 
^2*(a - b)))^(1/2))/4 + (268435456*(exp(2*c + 2*d*x) + 1)*(3*a*b + 16*a^2 
- 15*b^2))/(a*b^6*(a - b)^3))*(-(a^2*b + (a^5*b)^(1/2))/(16*(a^5*d^2 - a^4 
*b*d^2)))^(1/2) - log((((((4294967296*a*d^2*(exp(2*c + 2*d*x) + 1)*(26*a*b 
 - 49*a^2 + 15*b^2))/(b^6*(a - b)^3) - (8589934592*a^2*d^3*exp(c + d*x)*(3 
*a*b + 16*a^2 - 15*b^2)*(-(a^2*b + (a^5*b)^(1/2))/(a^4*d^2*(a - b)))^(1/2) 
)/(b^7*(a - b)^2))*(-(a^2*b + (a^5*b)^(1/2))/(a^4*d^2*(a - b)))^(1/2))/4 + 
 (2147483648*d*exp(c + d*x)*(17*a - 15*b))/(b^6*(a - b)^2))*(-(a^2*b + (a^ 
5*b)^(1/2))/(a^4*d^2*(a - b)))^(1/2))/4 + (268435456*(exp(2*c + 2*d*x) + 1 
)*(3*a*b + 16*a^2 - 15*b^2))/(a*b^6*(a - b)^3))*(-(a^2*b + (a^5*b)^(1/2))/ 
(16*(a^5*d^2 - a^4*b*d^2)))^(1/2) - (2*atan((exp(d*x)*exp(c)*(65536*a^2*(- 
a^2*d^2)^(1/2) + 50625*b^2*(-a^2*d^2)^(1/2) - 115200*a*b*(-a^2*d^2)^(1/2)) 
)/(65536*a^3*d + 50625*a*b^2*d - 115200*a^2*b*d)))/(-a^2*d^2)^(1/2) - log( 
(((((4294967296*a*d^2*(exp(2*c + 2*d*x) + 1)*(26*a*b - 49*a^2 + 15*b^2))/( 
b^6*(a - b)^3) - (8589934592*a^2*d^3*exp(c + d*x)*(3*a*b + 16*a^2 - 15*b^2 
)*(-(a^2*b - (a^5*b)^(1/2))/(a^4*d^2*(a - b)))^(1/2))/(b^7*(a - b)^2))*...
 

Reduce [F]

\[ \int \frac {\text {csch}(c+d x)}{a-b \sinh ^4(c+d x)} \, dx=-\left (\int \frac {\mathrm {csch}\left (d x +c \right )}{\sinh \left (d x +c \right )^{4} b -a}d x \right ) \] Input:

int(csch(d*x+c)/(a-b*sinh(d*x+c)^4),x)
 

Output:

 - int(csch(c + d*x)/(sinh(c + d*x)**4*b - a),x)