\(\int \text {csch}(5 x) \sinh (x) \, dx\) [69]

Optimal result

Integrand size = 7, antiderivative size = 80 \[ \int \text {csch}(5 x) \sinh (x) \, dx=-\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )} \tanh (x)\right )+\sqrt {\frac {2}{5 \left (5+\sqrt {5}\right )}} \arctan \left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )} \tanh (x)\right ) \] Output:

-1/10*(10+2*5^(1/2))^(1/2)*arctan(1/5*(25-10*5^(1/2))^(1/2)*tanh(x))+2^(1/ 
2)/(25+5*5^(1/2))^(1/2)*arctan(1/5*(25+10*5^(1/2))^(1/2)*tanh(x))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \text {csch}(5 x) \sinh (x) \, dx=\frac {\sqrt {5+\sqrt {5}} \arctan \left (\frac {\left (-3+\sqrt {5}\right ) \tanh (x)}{\sqrt {10-2 \sqrt {5}}}\right )+\sqrt {5-\sqrt {5}} \arctan \left (\frac {\left (3+\sqrt {5}\right ) \tanh (x)}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )}{5 \sqrt {2}} \] Input:

Integrate[Csch[5*x]*Sinh[x],x]
 

Output:

(Sqrt[5 + Sqrt[5]]*ArcTan[((-3 + Sqrt[5])*Tanh[x])/Sqrt[10 - 2*Sqrt[5]]] + 
 Sqrt[5 - Sqrt[5]]*ArcTan[((3 + Sqrt[5])*Tanh[x])/Sqrt[2*(5 + Sqrt[5])]])/ 
(5*Sqrt[2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4889, 1480, 216}

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.

\(\displaystyle \int \sinh (x) \text {csch}(5 x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (i x)}{\sin (5 i x)}dx\)

\(\Big \downarrow \) 4889

\(\displaystyle \int \frac {1-\tanh ^2(x)}{\tanh ^4(x)+10 \tanh ^2(x)+5}d\tanh (x)\)

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {1}{10} \left (5-3 \sqrt {5}\right ) \int \frac {1}{\tanh ^2(x)-2 \sqrt {5}+5}d\tanh (x)-\frac {1}{10} \left (5+3 \sqrt {5}\right ) \int \frac {1}{\tanh ^2(x)+2 \sqrt {5}+5}d\tanh (x)\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\left (5-3 \sqrt {5}\right ) \arctan \left (\frac {\tanh (x)}{\sqrt {5-2 \sqrt {5}}}\right )}{10 \sqrt {5-2 \sqrt {5}}}-\frac {\left (5+3 \sqrt {5}\right ) \arctan \left (\frac {\tanh (x)}{\sqrt {5+2 \sqrt {5}}}\right )}{10 \sqrt {5+2 \sqrt {5}}}\)

Input:

Int[Csch[5*x]*Sinh[x],x]
 

Output:

-1/10*((5 - 3*Sqrt[5])*ArcTan[Tanh[x]/Sqrt[5 - 2*Sqrt[5]]])/Sqrt[5 - 2*Sqr 
t[5]] - ((5 + 3*Sqrt[5])*ArcTan[Tanh[x]/Sqrt[5 + 2*Sqrt[5]]])/(10*Sqrt[5 + 
 2*Sqrt[5]])
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 3042

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

rule 4889

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors 
[Tan[v], x]}, Simp[d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1/(1 + d^2*x 
^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /;  !FalseQ[v] && FunctionOfQ[N 
onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] &&  !MatchQ[ 
u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I 
ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
 

Maple [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.51

method	result	size
risch	\(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (32000 \textit {\_Z}^{4}+400 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (4000 \textit {\_R}^{3}-200 \textit {\_R}^{2}+{\mathrm e}^{2 x}+30 \textit {\_R} -1\right )\right )\)	\(41\)

Input:

int(csch(5*x)*sinh(x),x,method=_RETURNVERBOSE)
 

Output:

2*sum(_R*ln(4000*_R^3-200*_R^2+exp(2*x)+30*_R-1),_R=RootOf(32000*_Z^4+400* 
_Z^2+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (53) = 106\).

Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.39 \[ \int \text {csch}(5 x) \sinh (x) \, dx=-\frac {1}{5} \, \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} \arctan \left (\frac {1}{10} \, {\left (4 \, \sqrt {5} \cosh \left (x\right )^{2} + 8 \, \sqrt {5} \cosh \left (x\right ) \sinh \left (x\right ) + 4 \, \sqrt {5} \sinh \left (x\right )^{2} + \sqrt {5} + 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}\right ) + \frac {1}{5} \, \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} \arctan \left (\frac {1}{10} \, {\left (4 \, \sqrt {5} \cosh \left (x\right )^{2} + 8 \, \sqrt {5} \cosh \left (x\right ) \sinh \left (x\right ) + 4 \, \sqrt {5} \sinh \left (x\right )^{2} + \sqrt {5} - 5\right )} \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}\right ) \] Input:

integrate(csch(5*x)*sinh(x),x, algorithm="fricas")
 

Output:

-1/5*sqrt(1/2*sqrt(5) + 5/2)*arctan(1/10*(4*sqrt(5)*cosh(x)^2 + 8*sqrt(5)* 
cosh(x)*sinh(x) + 4*sqrt(5)*sinh(x)^2 + sqrt(5) + 5)*sqrt(1/2*sqrt(5) + 5/ 
2)) + 1/5*sqrt(-1/2*sqrt(5) + 5/2)*arctan(1/10*(4*sqrt(5)*cosh(x)^2 + 8*sq 
rt(5)*cosh(x)*sinh(x) + 4*sqrt(5)*sinh(x)^2 + sqrt(5) - 5)*sqrt(-1/2*sqrt( 
5) + 5/2))
 

Sympy [F]

\[ \int \text {csch}(5 x) \sinh (x) \, dx=\int \sinh {\left (x \right )} \operatorname {csch}{\left (5 x \right )}\, dx \] Input:

integrate(csch(5*x)*sinh(x),x)
 

Output:

Integral(sinh(x)*csch(5*x), x)
 

Maxima [F]

\[ \int \text {csch}(5 x) \sinh (x) \, dx=\int { \operatorname {csch}\left (5 \, x\right ) \sinh \left (x\right ) \,d x } \] Input:

integrate(csch(5*x)*sinh(x),x, algorithm="maxima")
 

Output:

1/10*(-1)^(3/5)*log((-1)^(1/5) + e^(-2*x)) + 1/10*sqrt(5)*(-1)^(3/5)*log(( 
sqrt(5)*(-1)^(1/5) + (-1)^(1/5)*sqrt(2*sqrt(5) - 10) + (-1)^(1/5) - 4*e^(- 
2*x))/(sqrt(5)*(-1)^(1/5) - (-1)^(1/5)*sqrt(2*sqrt(5) - 10) + (-1)^(1/5) - 
 4*e^(-2*x)))/sqrt(2*sqrt(5) - 10) - 1/10*sqrt(5)*(-1)^(3/5)*log((sqrt(5)* 
(-1)^(1/5) - (-1)^(1/5)*sqrt(-2*sqrt(5) - 10) - (-1)^(1/5) + 4*e^(-2*x))/( 
sqrt(5)*(-1)^(1/5) + (-1)^(1/5)*sqrt(-2*sqrt(5) - 10) - (-1)^(1/5) + 4*e^( 
-2*x)))/sqrt(-2*sqrt(5) - 10) - 1/10*log(-(sqrt(5)*(-1)^(1/5) + (-1)^(1/5) 
)*e^(-2*x) + 2*(-1)^(2/5) + 2*e^(-4*x))/(sqrt(5)*(-1)^(2/5) + (-1)^(2/5)) 
+ 1/10*log((sqrt(5)*(-1)^(1/5) - (-1)^(1/5))*e^(-2*x) + 2*(-1)^(2/5) + 2*e 
^(-4*x))/(sqrt(5)*(-1)^(2/5) - (-1)^(2/5)) - 1/10*integrate((e^(3*x) + 2*e 
^(2*x) + 3*e^x + 4)*e^x/(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1), x) - 1/10 
*integrate((e^(3*x) - 2*e^(2*x) + 3*e^x - 4)*e^x/(e^(4*x) - e^(3*x) + e^(2 
*x) - e^x + 1), x) + 1/10*log(e^x + 1) + 1/10*log(e^x - 1)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \text {csch}(5 x) \sinh (x) \, dx=\frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (-\frac {\sqrt {5} - 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) - \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {\sqrt {5} + 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) \] Input:

integrate(csch(5*x)*sinh(x),x, algorithm="giac")
 

Output:

1/10*sqrt(-2*sqrt(5) + 10)*arctan(-(sqrt(5) - 4*e^(2*x) - 1)/sqrt(2*sqrt(5 
) + 10)) - 1/10*sqrt(2*sqrt(5) + 10)*arctan((sqrt(5) + 4*e^(2*x) + 1)/sqrt 
(-2*sqrt(5) + 10))
 

Mupad [B] (verification not implemented)

Time = 3.73 (sec) , antiderivative size = 282, normalized size of antiderivative = 3.52 \[ \int \text {csch}(5 x) \sinh (x) \, dx=2\,\mathrm {atan}\left (\frac {\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {9\,\sqrt {5}}{25}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}+\frac {4}{5}}{5\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\frac {9\,\sqrt {5}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}}{5}+\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\frac {9\,\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}}{5}}\right )\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,\left (\ln \left (\frac {9\,\sqrt {5}}{25}-\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}-\frac {4}{5}-{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,5{}\mathrm {i}+\frac {\sqrt {5}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}-\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,1{}\mathrm {i}+\frac {\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}\right )\,1{}\mathrm {i}-\ln \left (\frac {9\,\sqrt {5}}{25}-\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}-\frac {4}{5}+{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,5{}\mathrm {i}-\frac {\sqrt {5}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}+\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,1{}\mathrm {i}-\frac {\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}\right )\,1{}\mathrm {i}\right ) \] Input:

int(sinh(x)/sinh(5*x),x)
 

Output:

2*atan((exp(2*x)/5 + (9*5^(1/2))/25 + (6*5^(1/2)*exp(2*x))/25 + 4/5)/(5*ex 
p(2*x)*(5^(1/2)/200 + 1/40)^(1/2) + (9*5^(1/2)*(5^(1/2)/200 + 1/40)^(1/2)) 
/5 + (5^(1/2)/200 + 1/40)^(1/2) + (9*5^(1/2)*exp(2*x)*(5^(1/2)/200 + 1/40) 
^(1/2))/5))*(5^(1/2)/200 + 1/40)^(1/2) + (1/40 - 5^(1/2)/200)^(1/2)*(log(( 
5^(1/2)*(1/40 - 5^(1/2)/200)^(1/2)*9i)/5 - exp(2*x)*(1/40 - 5^(1/2)/200)^( 
1/2)*5i - exp(2*x)/5 + (9*5^(1/2))/25 - (1/40 - 5^(1/2)/200)^(1/2)*1i + (6 
*5^(1/2)*exp(2*x))/25 + (5^(1/2)*exp(2*x)*(1/40 - 5^(1/2)/200)^(1/2)*9i)/5 
 - 4/5)*1i - log(exp(2*x)*(1/40 - 5^(1/2)/200)^(1/2)*5i - exp(2*x)/5 - (5^ 
(1/2)*(1/40 - 5^(1/2)/200)^(1/2)*9i)/5 + (9*5^(1/2))/25 + (1/40 - 5^(1/2)/ 
200)^(1/2)*1i + (6*5^(1/2)*exp(2*x))/25 - (5^(1/2)*exp(2*x)*(1/40 - 5^(1/2 
)/200)^(1/2)*9i)/5 - 4/5)*1i)
 

Reduce [F]

\[ \int \text {csch}(5 x) \sinh (x) \, dx=\int \mathrm {csch}\left (5 x \right ) \sinh \left (x \right )d x \] Input:

int(csch(5*x)*sinh(x),x)
 

Output:

int(csch(5*x)*sinh(x),x)