### 3.80 $$\int \sinh ^4(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=40 $\frac{\cosh ^4(a+b x)}{4 b}-\frac{\cosh ^2(a+b x)}{b}+\frac{\log (\cosh (a+b x))}{b}$

[Out]

-(Cosh[a + b*x]^2/b) + Cosh[a + b*x]^4/(4*b) + Log[Cosh[a + b*x]]/b

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Rubi [A]  time = 0.0321603, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2590, 266, 43} $\frac{\cosh ^4(a+b x)}{4 b}-\frac{\cosh ^2(a+b x)}{b}+\frac{\log (\cosh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]^4*Tanh[a + b*x],x]

[Out]

-(Cosh[a + b*x]^2/b) + Cosh[a + b*x]^4/(4*b) + Log[Cosh[a + b*x]]/b

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sinh ^4(a+b x) \tanh (a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x} \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x} \, dx,x,\cosh ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2+\frac{1}{x}+x\right ) \, dx,x,\cosh ^2(a+b x)\right )}{2 b}\\ &=-\frac{\cosh ^2(a+b x)}{b}+\frac{\cosh ^4(a+b x)}{4 b}+\frac{\log (\cosh (a+b x))}{b}\\ \end{align*}

Mathematica [A]  time = 0.0302297, size = 34, normalized size = 0.85 $\frac{\frac{1}{4} \cosh ^4(a+b x)-\cosh ^2(a+b x)+\log (\cosh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[a + b*x]^4*Tanh[a + b*x],x]

[Out]

(-Cosh[a + b*x]^2 + Cosh[a + b*x]^4/4 + Log[Cosh[a + b*x]])/b

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Maple [A]  time = 0.017, size = 39, normalized size = 1. \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{4\,b}}-{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+{\frac{\ln \left ( \cosh \left ( bx+a \right ) \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^4*tanh(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.5845, size = 109, normalized size = 2.72 \begin{align*} -\frac{{\left (12 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{64 \, b} + \frac{b x + a}{b} - \frac{12 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )}}{64 \, b} + \frac{\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^4*tanh(b*x+a),x, algorithm="maxima")

[Out]

-1/64*(12*e^(-2*b*x - 2*a) - 1)*e^(4*b*x + 4*a)/b + (b*x + a)/b - 1/64*(12*e^(-2*b*x - 2*a) - e^(-4*b*x - 4*a)
)/b + log(e^(-2*b*x - 2*a) + 1)/b

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Fricas [B]  time = 1.81901, size = 1270, normalized size = 31.75 \begin{align*} \frac{\cosh \left (b x + a\right )^{8} + 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + \sinh \left (b x + a\right )^{8} + 4 \,{\left (7 \, \cosh \left (b x + a\right )^{2} - 3\right )} \sinh \left (b x + a\right )^{6} - 64 \, b x \cosh \left (b x + a\right )^{4} - 12 \, \cosh \left (b x + a\right )^{6} + 8 \,{\left (7 \, \cosh \left (b x + a\right )^{3} - 9 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + 2 \,{\left (35 \, \cosh \left (b x + a\right )^{4} - 32 \, b x - 90 \, \cosh \left (b x + a\right )^{2}\right )} \sinh \left (b x + a\right )^{4} + 8 \,{\left (7 \, \cosh \left (b x + a\right )^{5} - 32 \, b x \cosh \left (b x + a\right ) - 30 \, \cosh \left (b x + a\right )^{3}\right )} \sinh \left (b x + a\right )^{3} + 4 \,{\left (7 \, \cosh \left (b x + a\right )^{6} - 96 \, b x \cosh \left (b x + a\right )^{2} - 45 \, \cosh \left (b x + a\right )^{4} - 3\right )} \sinh \left (b x + a\right )^{2} - 12 \, \cosh \left (b x + a\right )^{2} + 64 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4}\right )} \log \left (\frac{2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 8 \,{\left (\cosh \left (b x + a\right )^{7} - 32 \, b x \cosh \left (b x + a\right )^{3} - 9 \, \cosh \left (b x + a\right )^{5} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{64 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^4*tanh(b*x+a),x, algorithm="fricas")

[Out]

1/64*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 - 3)*sinh(b*x
+ a)^6 - 64*b*x*cosh(b*x + a)^4 - 12*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 - 9*cosh(b*x + a))*sinh(b*x + a)^
5 + 2*(35*cosh(b*x + a)^4 - 32*b*x - 90*cosh(b*x + a)^2)*sinh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 32*b*x*cosh(
b*x + a) - 30*cosh(b*x + a)^3)*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 96*b*x*cosh(b*x + a)^2 - 45*cosh(b*x +
a)^4 - 3)*sinh(b*x + a)^2 - 12*cosh(b*x + a)^2 + 64*(cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*co
sh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4)*log(2*cosh(b*x + a)/(cosh(b
*x + a) - sinh(b*x + a))) + 8*(cosh(b*x + a)^7 - 32*b*x*cosh(b*x + a)^3 - 9*cosh(b*x + a)^5 - 3*cosh(b*x + a))
*sinh(b*x + a) + 1)/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*cosh(b*x + a)^2*sinh(b*x + a)
^2 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{4}{\left (a + b x \right )} \tanh{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)**4*tanh(b*x+a),x)

[Out]

Integral(sinh(a + b*x)**4*tanh(a + b*x), x)

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Giac [B]  time = 1.19127, size = 116, normalized size = 2.9 \begin{align*} -\frac{64 \, b x -{\left (48 \, e^{\left (4 \, b x + 4 \, a\right )} - 12 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )} -{\left (e^{\left (4 \, b x + 16 \, a\right )} - 12 \, e^{\left (2 \, b x + 14 \, a\right )}\right )} e^{\left (-12 \, a\right )} - 64 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{64 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^4*tanh(b*x+a),x, algorithm="giac")

[Out]

-1/64*(64*b*x - (48*e^(4*b*x + 4*a) - 12*e^(2*b*x + 2*a) + 1)*e^(-4*b*x - 4*a) - (e^(4*b*x + 16*a) - 12*e^(2*b
*x + 14*a))*e^(-12*a) - 64*log(e^(2*b*x + 2*a) + 1))/b