3.209 \(\int \coth (5 x) \sinh (x) \, dx\)

Optimal. Leaf size=82 \[ \sinh (x)-\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{5+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sinh (x)\right ) \]

[Out]

sinh(x)-1/10*arctan(1/5*sinh(x)*(50+10*5^(1/2))^(1/2))*(10-2*5^(1/2))^(1/2)-1/10*arctan(2*sinh(x)*2^(1/2)/(5+5
^(1/2))^(1/2))*(10+2*5^(1/2))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1676, 1166, 203} \[ \sinh (x)-\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{5+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sinh (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[(5 + Sqrt[5])/2]*ArcTan[2*Sqrt[2/(5 + Sqrt[5])]*Sinh[x]])/5 - (Sqrt[(5 - Sqrt[5])/2]*ArcTan[Sqrt[(2*(5
+ Sqrt[5]))/5]*Sinh[x]])/5 + Sinh[x]

Rule 203

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

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[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 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

Rubi steps

\begin {align*} \int \coth (5 x) \sinh (x) \, dx &=\operatorname {Subst}\left (\int \frac {1+12 x^2+16 x^4}{5+20 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=\operatorname {Subst}\left (\int \left (1-\frac {4 \left (1+2 x^2\right )}{5+20 x^2+16 x^4}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-4 \operatorname {Subst}\left (\int \frac {1+2 x^2}{5+20 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac {1}{5} \left (4 \left (5-\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{10-2 \sqrt {5}+16 x^2} \, dx,x,\sinh (x)\right )-\frac {1}{5} \left (4 \left (5+\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{10+2 \sqrt {5}+16 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{5+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sinh (x)\right )+\sinh (x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 76, normalized size = 0.93 \[ \frac {1}{10} \left (10 \sinh (x)-\sqrt {10-2 \sqrt {5}} \tan ^{-1}\left (\sqrt {2+\frac {2}{\sqrt {5}}} \sinh (x)\right )-\sqrt {2 \left (5+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{5+\sqrt {5}}} \sinh (x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 230, normalized size = 2.80 \[ -\frac {1}{10} \, {\left (2 \, \sqrt {2} \sqrt {\sqrt {5} + 5} \arctan \left (\frac {1}{40} \, {\left (\sqrt {2 \, {\left (\sqrt {5} + 1\right )} e^{\left (2 \, x\right )} + 4 \, e^{\left (4 \, x\right )} + 4} {\left (\sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 5} - 2 \, {\left ({\left (\sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} e^{\left (2 \, x\right )} - \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 5}\right )} e^{\left (-x\right )}\right ) e^{x} - 2 \, \sqrt {2} \sqrt {-\sqrt {5} + 5} \arctan \left (\frac {1}{40} \, {\left (\sqrt {-2 \, {\left (\sqrt {5} - 1\right )} e^{\left (2 \, x\right )} + 4 \, e^{\left (4 \, x\right )} + 4} {\left (\sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 5} - 2 \, {\left ({\left (\sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} e^{\left (2 \, x\right )} - \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 5}\right )} e^{\left (-x\right )}\right ) e^{x} - 5 \, e^{\left (2 \, x\right )} + 5\right )} e^{\left (-x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(2*sqrt(2)*sqrt(sqrt(5) + 5)*arctan(1/40*(sqrt(2*(sqrt(5) + 1)*e^(2*x) + 4*e^(4*x) + 4)*(sqrt(5)*sqrt(2)
- 5*sqrt(2))*sqrt(sqrt(5) + 5) - 2*((sqrt(5)*sqrt(2) - 5*sqrt(2))*e^(2*x) - sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt
(sqrt(5) + 5))*e^(-x))*e^x - 2*sqrt(2)*sqrt(-sqrt(5) + 5)*arctan(1/40*(sqrt(-2*(sqrt(5) - 1)*e^(2*x) + 4*e^(4*
x) + 4)*(sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt(-sqrt(5) + 5) - 2*((sqrt(5)*sqrt(2) + 5*sqrt(2))*e^(2*x) - sqrt(5)*
sqrt(2) - 5*sqrt(2))*sqrt(-sqrt(5) + 5))*e^(-x))*e^x - 5*e^(2*x) + 5)*e^(-x)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 75, normalized size = 0.91 \[ -\frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (-\frac {e^{\left (-x\right )} - e^{x}}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}}\right ) - \frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (-\frac {e^{\left (-x\right )} - e^{x}}{\sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.40, size = 246, normalized size = 3.00 \[ -\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\frac {\arctan \left (\frac {\tanh \left (\frac {x}{2}\right )}{\sqrt {5+2 \sqrt {5}}}\right )}{2 \sqrt {5+2 \sqrt {5}}}-\frac {\sqrt {5}\, \arctan \left (\frac {\tanh \left (\frac {x}{2}\right )}{\sqrt {5+2 \sqrt {5}}}\right )}{10 \sqrt {5+2 \sqrt {5}}}+\frac {\sqrt {5}\, \arctan \left (\frac {\tanh \left (\frac {x}{2}\right )}{\sqrt {5-2 \sqrt {5}}}\right )}{10 \sqrt {5-2 \sqrt {5}}}-\frac {\arctan \left (\frac {\tanh \left (\frac {x}{2}\right )}{\sqrt {5-2 \sqrt {5}}}\right )}{2 \sqrt {5-2 \sqrt {5}}}+\frac {\sqrt {5}\, \arctan \left (\frac {5 \tanh \left (\frac {x}{2}\right )}{\sqrt {25-10 \sqrt {5}}}\right )}{2 \sqrt {25-10 \sqrt {5}}}-\frac {3 \arctan \left (\frac {5 \tanh \left (\frac {x}{2}\right )}{\sqrt {25-10 \sqrt {5}}}\right )}{2 \sqrt {25-10 \sqrt {5}}}-\frac {\sqrt {5}\, \arctan \left (\frac {5 \tanh \left (\frac {x}{2}\right )}{\sqrt {25+10 \sqrt {5}}}\right )}{2 \sqrt {25+10 \sqrt {5}}}-\frac {3 \arctan \left (\frac {5 \tanh \left (\frac {x}{2}\right )}{\sqrt {25+10 \sqrt {5}}}\right )}{2 \sqrt {25+10 \sqrt {5}}}-\frac {1}{\tanh \left (\frac {x}{2}\right )-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(tanh(1/2*x)+1)-1/2/(5+2*5^(1/2))^(1/2)*arctan(tanh(1/2*x)/(5+2*5^(1/2))^(1/2))-1/10*5^(1/2)/(5+2*5^(1/2))^
(1/2)*arctan(tanh(1/2*x)/(5+2*5^(1/2))^(1/2))+1/10*5^(1/2)/(5-2*5^(1/2))^(1/2)*arctan(tanh(1/2*x)/(5-2*5^(1/2)
)^(1/2))-1/2/(5-2*5^(1/2))^(1/2)*arctan(tanh(1/2*x)/(5-2*5^(1/2))^(1/2))+1/2*5^(1/2)/(25-10*5^(1/2))^(1/2)*arc
tan(5*tanh(1/2*x)/(25-10*5^(1/2))^(1/2))-3/2/(25-10*5^(1/2))^(1/2)*arctan(5*tanh(1/2*x)/(25-10*5^(1/2))^(1/2))
-1/2*5^(1/2)/(25+10*5^(1/2))^(1/2)*arctan(5*tanh(1/2*x)/(25+10*5^(1/2))^(1/2))-3/2/(25+10*5^(1/2))^(1/2)*arcta
n(5*tanh(1/2*x)/(25+10*5^(1/2))^(1/2))-1/(tanh(1/2*x)-1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )} - \frac {1}{2} \, \int \frac {e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x}}{e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1}\,{d x} - \frac {1}{2} \, \int \frac {e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} + e^{x}}{e^{\left (4 \, x\right )} - e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - e^{x} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^(2*x) - 1)*e^(-x) - 1/2*integrate((e^(3*x) + e^(2*x) + e^x)/(e^(4*x) + e^(3*x) + e^(2*x) + e^x + 1), x)
- 1/2*integrate((e^(3*x) - e^(2*x) + e^x)/(e^(4*x) - e^(3*x) + e^(2*x) - e^x + 1), x)

________________________________________________________________________________________

mupad [B]  time = 3.33, size = 141, normalized size = 1.72 \[ \frac {{\mathrm {e}}^x}{2}-\frac {{\mathrm {e}}^{-x}}{2}+\ln \left (40\,{\mathrm {e}}^x\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}-4\,{\mathrm {e}}^{2\,x}+4\right )\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}+\ln \left (40\,{\mathrm {e}}^x\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}-4\,{\mathrm {e}}^{2\,x}+4\right )\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}-\ln \left (4\,{\mathrm {e}}^{2\,x}+40\,{\mathrm {e}}^x\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}-4\right )\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}-\ln \left (4\,{\mathrm {e}}^{2\,x}+40\,{\mathrm {e}}^x\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}-4\right )\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/2 - exp(-x)/2 + log(40*exp(x)*(- 5^(1/2)/200 - 1/40)^(1/2) - 4*exp(2*x) + 4)*(- 5^(1/2)/200 - 1/40)^(1/
2) + log(40*exp(x)*(5^(1/2)/200 - 1/40)^(1/2) - 4*exp(2*x) + 4)*(5^(1/2)/200 - 1/40)^(1/2) - log(4*exp(2*x) +
40*exp(x)*(- 5^(1/2)/200 - 1/40)^(1/2) - 4)*(- 5^(1/2)/200 - 1/40)^(1/2) - log(4*exp(2*x) + 40*exp(x)*(5^(1/2)
/200 - 1/40)^(1/2) - 4)*(5^(1/2)/200 - 1/40)^(1/2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \coth {\left (5 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(x)*coth(5*x), x)

________________________________________________________________________________________