3.88 \(\int (a+b \text{sech}(c+d x))^3 \, dx\)

Optimal. Leaf size=73 \[ \frac{b \left (6 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d}+a^3 x+\frac{5 a b^2 \tanh (c+d x)}{2 d}+\frac{b^2 \tanh (c+d x) (a+b \text{sech}(c+d x))}{2 d} \]

[Out]

a^3*x + (b*(6*a^2 + b^2)*ArcTan[Sinh[c + d*x]])/(2*d) + (5*a*b^2*Tanh[c + d*x])/(2*d) + (b^2*(a + b*Sech[c + d
*x])*Tanh[c + d*x])/(2*d)

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Rubi [A]  time = 0.0516835, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3782, 3770, 3767, 8} \[ \frac{b \left (6 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d}+a^3 x+\frac{5 a b^2 \tanh (c+d x)}{2 d}+\frac{b^2 \tanh (c+d x) (a+b \text{sech}(c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x])^3,x]

[Out]

a^3*x + (b*(6*a^2 + b^2)*ArcTan[Sinh[c + d*x]])/(2*d) + (5*a*b^2*Tanh[c + d*x])/(2*d) + (b^2*(a + b*Sech[c + d
*x])*Tanh[c + d*x])/(2*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \text{sech}(c+d x))^3 \, dx &=\frac{b^2 (a+b \text{sech}(c+d x)) \tanh (c+d x)}{2 d}+\frac{1}{2} \int \left (2 a^3+b \left (6 a^2+b^2\right ) \text{sech}(c+d x)+5 a b^2 \text{sech}^2(c+d x)\right ) \, dx\\ &=a^3 x+\frac{b^2 (a+b \text{sech}(c+d x)) \tanh (c+d x)}{2 d}+\frac{1}{2} \left (5 a b^2\right ) \int \text{sech}^2(c+d x) \, dx+\frac{1}{2} \left (b \left (6 a^2+b^2\right )\right ) \int \text{sech}(c+d x) \, dx\\ &=a^3 x+\frac{b \left (6 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{b^2 (a+b \text{sech}(c+d x)) \tanh (c+d x)}{2 d}+\frac{\left (5 i a b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{2 d}\\ &=a^3 x+\frac{b \left (6 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac{5 a b^2 \tanh (c+d x)}{2 d}+\frac{b^2 (a+b \text{sech}(c+d x)) \tanh (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.129793, size = 55, normalized size = 0.75 \[ \frac{b \left (6 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))+2 a^3 d x+b^2 \tanh (c+d x) (6 a+b \text{sech}(c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x])^3,x]

[Out]

(2*a^3*d*x + b*(6*a^2 + b^2)*ArcTan[Sinh[c + d*x]] + b^2*(6*a + b*Sech[c + d*x])*Tanh[c + d*x])/(2*d)

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Maple [A]  time = 0.023, size = 80, normalized size = 1.1 \begin{align*}{a}^{3}x+{\frac{{a}^{3}c}{d}}+6\,{\frac{{a}^{2}b\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}}+3\,{\frac{a{b}^{2}\tanh \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{3}\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(d*x+c))^3,x)

[Out]

a^3*x+1/d*a^3*c+6/d*a^2*b*arctan(exp(d*x+c))+3*a*b^2*tanh(d*x+c)/d+1/2*b^3*sech(d*x+c)*tanh(d*x+c)/d+1/d*b^3*a
rctan(exp(d*x+c))

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Maxima [A]  time = 1.8284, size = 154, normalized size = 2.11 \begin{align*} a^{3} x - b^{3}{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac{3 \, a^{2} b \arctan \left (\sinh \left (d x + c\right )\right )}{d} + \frac{6 \, a b^{2}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.36095, size = 1319, normalized size = 18.07 \begin{align*} \frac{a^{3} d x \cosh \left (d x + c\right )^{4} + a^{3} d x \sinh \left (d x + c\right )^{4} + b^{3} \cosh \left (d x + c\right )^{3} + a^{3} d x - b^{3} \cosh \left (d x + c\right ) +{\left (4 \, a^{3} d x \cosh \left (d x + c\right ) + b^{3}\right )} \sinh \left (d x + c\right )^{3} - 6 \, a b^{2} + 2 \,{\left (a^{3} d x - 3 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2} +{\left (6 \, a^{3} d x \cosh \left (d x + c\right )^{2} + 2 \, a^{3} d x + 3 \, b^{3} \cosh \left (d x + c\right ) - 6 \, a b^{2}\right )} \sinh \left (d x + c\right )^{2} +{\left ({\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 4 \,{\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (6 \, a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{4} + 6 \, a^{2} b + b^{3} + 2 \,{\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (6 \, a^{2} b + b^{3} + 3 \,{\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left ({\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{3} +{\left (6 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) +{\left (4 \, a^{3} d x \cosh \left (d x + c\right )^{3} + 3 \, b^{3} \cosh \left (d x + c\right )^{2} - b^{3} + 4 \,{\left (a^{3} d x - 3 \, a b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 2 \, d \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*d*x*cosh(d*x + c)^4 + a^3*d*x*sinh(d*x + c)^4 + b^3*cosh(d*x + c)^3 + a^3*d*x - b^3*cosh(d*x + c) + (4*a^
3*d*x*cosh(d*x + c) + b^3)*sinh(d*x + c)^3 - 6*a*b^2 + 2*(a^3*d*x - 3*a*b^2)*cosh(d*x + c)^2 + (6*a^3*d*x*cosh
(d*x + c)^2 + 2*a^3*d*x + 3*b^3*cosh(d*x + c) - 6*a*b^2)*sinh(d*x + c)^2 + ((6*a^2*b + b^3)*cosh(d*x + c)^4 +
4*(6*a^2*b + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (6*a^2*b + b^3)*sinh(d*x + c)^4 + 6*a^2*b + b^3 + 2*(6*a^2*b
+ b^3)*cosh(d*x + c)^2 + 2*(6*a^2*b + b^3 + 3*(6*a^2*b + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((6*a^2*b
+ b^3)*cosh(d*x + c)^3 + (6*a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + sinh(d*x + c)) +
(4*a^3*d*x*cosh(d*x + c)^3 + 3*b^3*cosh(d*x + c)^2 - b^3 + 4*(a^3*d*x - 3*a*b^2)*cosh(d*x + c))*sinh(d*x + c)
)/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*co
sh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c))**3,x)

[Out]

Integral((a + b*sech(c + d*x))**3, x)

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Giac [A]  time = 1.13269, size = 131, normalized size = 1.79 \begin{align*} \frac{{\left (d x + c\right )} a^{3}}{d} + \frac{{\left (6 \, a^{2} b + b^{3}\right )} \arctan \left (e^{\left (d x + c\right )}\right )}{d} + \frac{b^{3} e^{\left (3 \, d x + 3 \, c\right )} - 6 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{3} e^{\left (d x + c\right )} - 6 \, a b^{2}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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