### 3.76 $$\int \sinh ^2(a+b x) \tanh ^3(a+b x) \, dx$$

Optimal. Leaf size=43 $\frac{\cosh ^2(a+b x)}{2 b}-\frac{\text{sech}^2(a+b x)}{2 b}-\frac{2 \log (\cosh (a+b x))}{b}$

[Out]

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

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Rubi [A]  time = 0.043562, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {2590, 266, 43} $\frac{\cosh ^2(a+b x)}{2 b}-\frac{\text{sech}^2(a+b x)}{2 b}-\frac{2 \log (\cosh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]^2*Tanh[a + b*x]^3,x]

[Out]

Cosh[a + b*x]^2/(2*b) - (2*Log[Cosh[a + b*x]])/b - Sech[a + b*x]^2/(2*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 ^2(a+b x) \tanh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^3} \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^2} \, dx,x,\cosh ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}-\frac{2}{x}\right ) \, dx,x,\cosh ^2(a+b x)\right )}{2 b}\\ &=\frac{\cosh ^2(a+b x)}{2 b}-\frac{2 \log (\cosh (a+b x))}{b}-\frac{\text{sech}^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.050776, size = 35, normalized size = 0.81 $-\frac{-\sinh ^2(a+b x)+\text{sech}^2(a+b x)+4 \log (\cosh (a+b x))}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[a + b*x]^2*Tanh[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.017, size = 47, normalized size = 1.1 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{2\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}-2\,{\frac{\ln \left ( \cosh \left ( bx+a \right ) \right ) }{b}}+{\frac{ \left ( \tanh \left ( bx+a \right ) \right ) ^{2}}{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.58946, size = 139, normalized size = 3.23 \begin{align*} -\frac{2 \,{\left (b x + a\right )}}{b} + \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, b} - \frac{2 \, \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac{2 \, e^{\left (-2 \, b x - 2 \, a\right )} - 15 \, e^{\left (-4 \, b x - 4 \, a\right )} + 1}{8 \, b{\left (e^{\left (-2 \, b x - 2 \, a\right )} + 2 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*x + a)/b + 1/8*e^(-2*b*x - 2*a)/b - 2*log(e^(-2*b*x - 2*a) + 1)/b + 1/8*(2*e^(-2*b*x - 2*a) - 15*e^(-4*b
*x - 4*a) + 1)/(b*(e^(-2*b*x - 2*a) + 2*e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a)))

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Fricas [B]  time = 1.89526, size = 2049, normalized size = 47.65 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 2*(8*b*x + 1)*cosh(b*x + a)^6 + 2*(
8*b*x + 14*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*(14*cosh(b*x + a)^3 + 3*(8*b*x + 1)*cosh(b*x + a))*sinh(b*
x + a)^5 + 2*(16*b*x - 7)*cosh(b*x + a)^4 + 2*(35*cosh(b*x + a)^4 + 15*(8*b*x + 1)*cosh(b*x + a)^2 + 16*b*x -
7)*sinh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 + 5*(8*b*x + 1)*cosh(b*x + a)^3 + (16*b*x - 7)*cosh(b*x + a))*sinh(b
*x + a)^3 + 2*(8*b*x + 1)*cosh(b*x + a)^2 + 2*(14*cosh(b*x + a)^6 + 15*(8*b*x + 1)*cosh(b*x + a)^4 + 6*(16*b*x
- 7)*cosh(b*x + a)^2 + 8*b*x + 1)*sinh(b*x + a)^2 - 16*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + s
inh(b*x + a)^6 + (15*cosh(b*x + a)^2 + 2)*sinh(b*x + a)^4 + 2*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 + 2*cosh(
b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 + 12*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + cosh(b*x + a)^2 +
2*(3*cosh(b*x + a)^5 + 4*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a))*log(2*cosh(b*x + a)/(cosh(b*x + a) -
sinh(b*x + a))) + 4*(2*cosh(b*x + a)^7 + 3*(8*b*x + 1)*cosh(b*x + a)^5 + 2*(16*b*x - 7)*cosh(b*x + a)^3 + (8*b
*x + 1)*cosh(b*x + a))*sinh(b*x + a) + 1)/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x
+ a)^6 + 2*b*cosh(b*x + a)^4 + (15*b*cosh(b*x + a)^2 + 2*b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 + 2*b*cos
h(b*x + a))*sinh(b*x + a)^3 + b*cosh(b*x + a)^2 + (15*b*cosh(b*x + a)^4 + 12*b*cosh(b*x + a)^2 + b)*sinh(b*x +
a)^2 + 2*(3*b*cosh(b*x + a)^5 + 4*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{2}{\left (a + b x \right )} \tanh ^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.29219, size = 130, normalized size = 3.02 \begin{align*} \frac{16 \, b x -{\left (8 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + \frac{8 \,{\left (3 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 3\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} + e^{\left (2 \, b x + 2 \, a\right )} - 16 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{8 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(16*b*x - (8*e^(2*b*x + 2*a) - 1)*e^(-2*b*x - 2*a) + 8*(3*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2*a) + 3)/(e^(2*b
*x + 2*a) + 1)^2 + e^(2*b*x + 2*a) - 16*log(e^(2*b*x + 2*a) + 1))/b