3.70 \(\int \sinh (a+b x) \tanh (a+b x) \, dx\)

Optimal. Leaf size=23 \[ \frac {\sinh (a+b x)}{b}-\frac {\tan ^{-1}(\sinh (a+b x))}{b} \]

[Out]

-arctan(sinh(b*x+a))/b+sinh(b*x+a)/b

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2592, 321, 203} \[ \frac {\sinh (a+b x)}{b}-\frac {\tan ^{-1}(\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]*Tanh[a + b*x],x]

[Out]

-(ArcTan[Sinh[a + b*x]]/b) + Sinh[a + b*x]/b

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rubi steps

\begin {align*} \int \sinh (a+b x) \tanh (a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\sinh (a+b x)\right )}{b}\\ &=\frac {\sinh (a+b x)}{b}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (a+b x)\right )}{b}\\ &=-\frac {\tan ^{-1}(\sinh (a+b x))}{b}+\frac {\sinh (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ \frac {\sinh (a+b x)}{b}-\frac {\tan ^{-1}(\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]*Tanh[a + b*x],x]

[Out]

-(ArcTan[Sinh[a + b*x]]/b) + Sinh[a + b*x]/b

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fricas [B]  time = 0.55, size = 86, normalized size = 3.74 \[ -\frac {4 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - \cosh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - \sinh \left (b x + a\right )^{2} + 1}{2 \, {\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a),x, algorithm="fricas")

[Out]

-1/2*(4*(cosh(b*x + a) + sinh(b*x + a))*arctan(cosh(b*x + a) + sinh(b*x + a)) - cosh(b*x + a)^2 - 2*cosh(b*x +
a)*sinh(b*x + a) - sinh(b*x + a)^2 + 1)/(b*cosh(b*x + a) + b*sinh(b*x + a))

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giac [A]  time = 0.13, size = 32, normalized size = 1.39 \[ -\frac {4 \, \arctan \left (e^{\left (b x + a\right )}\right ) - e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a),x, algorithm="giac")

[Out]

-1/2*(4*arctan(e^(b*x + a)) - e^(b*x + a) + e^(-b*x - a))/b

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maple [A]  time = 0.14, size = 24, normalized size = 1.04 \[ \frac {\sinh \left (b x +a \right )}{b}-\frac {2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)*tanh(b*x+a),x)

[Out]

sinh(b*x+a)/b-2*arctan(exp(b*x+a))/b

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maxima [A]  time = 0.40, size = 41, normalized size = 1.78 \[ \frac {2 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{2 \, b} - \frac {e^{\left (-b x - a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a),x, algorithm="maxima")

[Out]

2*arctan(e^(-b*x - a))/b + 1/2*e^(b*x + a)/b - 1/2*e^(-b*x - a)/b

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mupad [B]  time = 1.47, size = 49, normalized size = 2.13 \[ \frac {{\mathrm {e}}^{a+b\,x}}{2\,b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {{\mathrm {e}}^{-a-b\,x}}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)*tanh(a + b*x),x)

[Out]

exp(a + b*x)/(2*b) - (2*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) - exp(- a - b*x)/(2*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\left (a + b x \right )} \tanh {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a),x)

[Out]

Integral(sinh(a + b*x)*tanh(a + b*x), x)

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