3.386 \(\int x \sinh (a+b x) \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac {\sinh (a+b x)}{b^2}-\frac {\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac {x \cosh (a+b x)}{b}+\frac {x \text {sech}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5449, 3296, 2637, 5418, 3770} \[ -\frac {\sinh (a+b x)}{b^2}-\frac {\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac {x \cosh (a+b x)}{b}+\frac {x \text {sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

-(ArcTan[Sinh[a + b*x]]/b^2) + (x*Cosh[a + b*x])/b + (x*Sech[a + b*x])/b - Sinh[a + b*x]/b^2

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 5449

Int[((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int
[(c + d*x)^m*Sinh[a + b*x]^n*Tanh[a + b*x]^(p - 2), x] - Int[(c + d*x)^m*Sinh[a + b*x]^(n - 2)*Tanh[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x \sinh (a+b x) \tanh ^2(a+b x) \, dx &=\int x \sinh (a+b x) \, dx-\int x \text {sech}(a+b x) \tanh (a+b x) \, dx\\ &=\frac {x \cosh (a+b x)}{b}+\frac {x \text {sech}(a+b x)}{b}-\frac {\int \cosh (a+b x) \, dx}{b}-\frac {\int \text {sech}(a+b x) \, dx}{b}\\ &=-\frac {\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac {x \cosh (a+b x)}{b}+\frac {x \text {sech}(a+b x)}{b}-\frac {\sinh (a+b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 50, normalized size = 1.09 \[ -\frac {\sinh (a+b x)}{b^2}-\frac {2 \tan ^{-1}\left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )}{b^2}+\frac {x \cosh (a+b x)}{b}+\frac {x \text {sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

(-2*ArcTan[Tanh[(a + b*x)/2]])/b^2 + (x*Cosh[a + b*x])/b + (x*Sech[a + b*x])/b - Sinh[a + b*x]/b^2

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fricas [B]  time = 0.53, size = 283, normalized size = 6.15 \[ \frac {{\left (b x - 1\right )} \cosh \left (b x + a\right )^{4} + 4 \, {\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + {\left (b x - 1\right )} \sinh \left (b x + a\right )^{4} + 6 \, b x \cosh \left (b x + a\right )^{2} + 6 \, {\left ({\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + b x\right )} \sinh \left (b x + a\right )^{2} + b x - 4 \, {\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} + {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + \cosh \left (b x + a\right )\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 4 \, {\left ({\left (b x - 1\right )} \cosh \left (b x + a\right )^{3} + 3 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{2 \, {\left (b^{2} \cosh \left (b x + a\right )^{3} + 3 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b^{2} \sinh \left (b x + a\right )^{3} + b^{2} \cosh \left (b x + a\right ) + {\left (3 \, b^{2} \cosh \left (b x + a\right )^{2} + b^{2}\right )} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*((b*x - 1)*cosh(b*x + a)^4 + 4*(b*x - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x - 1)*sinh(b*x + a)^4 + 6*b*x
*cosh(b*x + a)^2 + 6*((b*x - 1)*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^2 + b*x - 4*(cosh(b*x + a)^3 + 3*cosh(b*x
+ a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + cosh(b*x + a))*arctan(cosh(b
*x + a) + sinh(b*x + a)) + 4*((b*x - 1)*cosh(b*x + a)^3 + 3*b*x*cosh(b*x + a))*sinh(b*x + a) + 1)/(b^2*cosh(b*
x + a)^3 + 3*b^2*cosh(b*x + a)*sinh(b*x + a)^2 + b^2*sinh(b*x + a)^3 + b^2*cosh(b*x + a) + (3*b^2*cosh(b*x + a
)^2 + b^2)*sinh(b*x + a))

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giac [B]  time = 0.19, size = 102, normalized size = 2.22 \[ \frac {b x e^{\left (4 \, b x + 4 \, a\right )} + 6 \, b x e^{\left (2 \, b x + 2 \, a\right )} + b x - 4 \, \arctan \left (e^{\left (b x + a\right )}\right ) e^{\left (3 \, b x + 3 \, a\right )} - 4 \, \arctan \left (e^{\left (b x + a\right )}\right ) e^{\left (b x + a\right )} - e^{\left (4 \, b x + 4 \, a\right )} + 1}{2 \, {\left (b^{2} e^{\left (3 \, b x + 3 \, a\right )} + b^{2} e^{\left (b x + a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

1/2*(b*x*e^(4*b*x + 4*a) + 6*b*x*e^(2*b*x + 2*a) + b*x - 4*arctan(e^(b*x + a))*e^(3*b*x + 3*a) - 4*arctan(e^(b
*x + a))*e^(b*x + a) - e^(4*b*x + 4*a) + 1)/(b^2*e^(3*b*x + 3*a) + b^2*e^(b*x + a))

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maple [C]  time = 0.40, size = 94, normalized size = 2.04 \[ \frac {\left (b x -1\right ) {\mathrm e}^{b x +a}}{2 b^{2}}+\frac {\left (b x +1\right ) {\mathrm e}^{-b x -a}}{2 b^{2}}+\frac {2 x \,{\mathrm e}^{b x +a}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}+\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{b^{2}}-\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sech(b*x+a)^2*sinh(b*x+a)^3,x)

[Out]

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

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maxima [A]  time = 0.80, size = 81, normalized size = 1.76 \[ \frac {6 \, b x e^{\left (b x + 2 \, a\right )} + {\left (b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (3 \, b x\right )} + {\left (b x + 1\right )} e^{\left (-b x\right )}}{2 \, {\left (b^{2} e^{\left (2 \, b x + 3 \, a\right )} + b^{2} e^{a}\right )}} - \frac {2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/2*(6*b*x*e^(b*x + 2*a) + (b*x*e^(4*a) - e^(4*a))*e^(3*b*x) + (b*x + 1)*e^(-b*x))/(b^2*e^(2*b*x + 3*a) + b^2*
e^a) - 2*arctan(e^(b*x + a))/b^2

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mupad [B]  time = 0.10, size = 90, normalized size = 1.96 \[ {\mathrm {e}}^{-a-b\,x}\,\left (\frac {x}{2\,b}+\frac {1}{2\,b^2}\right )-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^4}}{b^2}\right )}{\sqrt {b^4}}+{\mathrm {e}}^{a+b\,x}\,\left (\frac {x}{2\,b}-\frac {1}{2\,b^2}\right )+\frac {2\,x\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sinh(a + b*x)^3)/cosh(a + b*x)^2,x)

[Out]

exp(- a - b*x)*(x/(2*b) + 1/(2*b^2)) - (2*atan((exp(b*x)*exp(a)*(b^4)^(1/2))/b^2))/(b^4)^(1/2) + exp(a + b*x)*
(x/(2*b) - 1/(2*b^2)) + (2*x*exp(a + b*x))/(b*(exp(2*a + 2*b*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sech(b*x+a)**2*sinh(b*x+a)**3,x)

[Out]

Integral(x*sinh(a + b*x)**3*sech(a + b*x)**2, x)

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