### 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]

-(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]

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Rubi [A]  time = 0.187413, antiderivative size = 82, normalized size of antiderivative = 1., 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 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

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 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])

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 = 0.223715, 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

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Maple [B]  time = 0.155, size = 246, normalized size = 3. \begin{align*}{\frac{\sqrt{5}}{10\,\sqrt{5-2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5-2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{1}{2\,\sqrt{5-2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5-2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{\sqrt{5}}{10\,\sqrt{5+2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5+2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }-{\frac{1}{2\,\sqrt{5+2\,\sqrt{5}}}\arctan \left ({\frac{1}{\sqrt{5+2\,\sqrt{5}}}\tanh \left ({\frac{x}{2}} \right ) } \right ) }+{\frac{\sqrt{5}}{2\,\sqrt{25-10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25-10\,\sqrt{5}}}} \right ) }-{\frac{3}{2\,\sqrt{25-10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25-10\,\sqrt{5}}}} \right ) }-{\frac{\sqrt{5}}{2\,\sqrt{25+10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25+10\,\sqrt{5}}}} \right ) }-{\frac{3}{2\,\sqrt{25+10\,\sqrt{5}}}\arctan \left ( 5\,{\frac{\tanh \left ( x/2 \right ) }{\sqrt{25+10\,\sqrt{5}}}} \right ) }- \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}- \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)

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Fricas [B]  time = 2.38223, size = 699, normalized size = 8.52 \begin{align*} -\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 )} \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (x \right )} \coth{\left (5 x \right )}\, dx \end{align*}

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)

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Giac [A]  time = 1.25231, size = 101, normalized size = 1.23 \begin{align*} -\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} \end{align*}

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