### 3.205 $$\int \sinh (x) \tanh (n x) \, dx$$

Optimal. Leaf size=81 $-e^{-x} \, _2F_1\left (1,-\frac{1}{2 n};1-\frac{1}{2 n};-e^{2 n x}\right )-e^x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-e^{2 n x}\right )+\frac{e^{-x}}{2}+\frac{e^x}{2}$

[Out]

1/(2*E^x) + E^x/2 - Hypergeometric2F1[1, -1/(2*n), 1 - 1/(2*n), -E^(2*n*x)]/E^x - E^x*Hypergeometric2F1[1, 1/(
2*n), (2 + n^(-1))/2, -E^(2*n*x)]

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Rubi [A]  time = 0.0696508, antiderivative size = 81, 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 = {5601, 2194, 2251} $-e^{-x} \, _2F_1\left (1,-\frac{1}{2 n};1-\frac{1}{2 n};-e^{2 n x}\right )-e^x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-e^{2 n x}\right )+\frac{e^{-x}}{2}+\frac{e^x}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[x]*Tanh[n*x],x]

[Out]

1/(2*E^x) + E^x/2 - Hypergeometric2F1[1, -1/(2*n), 1 - 1/(2*n), -E^(2*n*x)]/E^x - E^x*Hypergeometric2F1[1, 1/(
2*n), (2 + n^(-1))/2, -E^(2*n*x)]

Rule 5601

Int[Sinh[(a_.) + (b_.)*(x_)]*Tanh[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[-(1/(E^(a + b*x)*2)) + E^(a + b*x)/2 +
1/(E^(a + b*x)*(1 + E^(2*(c + d*x)))) - E^(a + b*x)/(1 + E^(2*(c + d*x))), x] /; FreeQ[{a, b, c, d}, x] && Ne
Q[b^2 - d^2, 0]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int \sinh (x) \tanh (n x) \, dx &=\int \left (-\frac{e^{-x}}{2}+\frac{e^x}{2}+\frac{e^{-x}}{1+e^{2 n x}}-\frac{e^x}{1+e^{2 n x}}\right ) \, dx\\ &=-\left (\frac{1}{2} \int e^{-x} \, dx\right )+\frac{\int e^x \, dx}{2}+\int \frac{e^{-x}}{1+e^{2 n x}} \, dx-\int \frac{e^x}{1+e^{2 n x}} \, dx\\ &=\frac{e^{-x}}{2}+\frac{e^x}{2}-e^{-x} \, _2F_1\left (1,-\frac{1}{2 n};1-\frac{1}{2 n};-e^{2 n x}\right )-e^x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-e^{2 n x}\right )\\ \end{align*}

Mathematica [B]  time = 0.170531, size = 164, normalized size = 2.02 $\frac{1}{2} e^{-2 x} \left (-\frac{e^{2 n x+x} \, _2F_1\left (1,1-\frac{1}{2 n};2-\frac{1}{2 n};-e^{2 n x}\right )}{2 n-1}+\frac{e^{(2 n+3) x} \, _2F_1\left (1,1+\frac{1}{2 n};2+\frac{1}{2 n};-e^{2 n x}\right )}{2 n+1}-e^x \left (\, _2F_1\left (1,-\frac{1}{2 n};1-\frac{1}{2 n};-e^{2 n x}\right )+e^{2 x} \, _2F_1\left (1,\frac{1}{2 n};1+\frac{1}{2 n};-e^{2 n x}\right )\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[x]*Tanh[n*x],x]

[Out]

(-((E^(x + 2*n*x)*Hypergeometric2F1[1, 1 - 1/(2*n), 2 - 1/(2*n), -E^(2*n*x)])/(-1 + 2*n)) + (E^((3 + 2*n)*x)*H
ypergeometric2F1[1, 1 + 1/(2*n), 2 + 1/(2*n), -E^(2*n*x)])/(1 + 2*n) - E^x*(Hypergeometric2F1[1, -1/(2*n), 1 -
1/(2*n), -E^(2*n*x)] + E^(2*x)*Hypergeometric2F1[1, 1/(2*n), 1 + 1/(2*n), -E^(2*n*x)]))/(2*E^(2*x))

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int \sinh \left ( x \right ) \tanh \left ( nx \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)*tanh(n*x),x)

[Out]

int(sinh(x)*tanh(n*x),x)

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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{2 \,{\left (e^{\left (2 \, x\right )} - 1\right )}}{e^{\left (2 \, n x + x\right )} + e^{x}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(n*x),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sinh \left (x\right ) \tanh \left (n x\right ), x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(n*x),x, algorithm="fricas")

[Out]

integral(sinh(x)*tanh(n*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(n*x),x)

[Out]

Integral(sinh(x)*tanh(n*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh \left (x\right ) \tanh \left (n x\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(n*x),x, algorithm="giac")

[Out]

integrate(sinh(x)*tanh(n*x), x)