3.128 \(\int (a+b \text{sech}^2(c+d x))^3 \, dx\)

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

[Out]

a^3*x + (b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x])/d - (b^2*(3*a + 2*b)*Tanh[c + d*x]^3)/(3*d) + (b^3*Tanh[c + d*
x]^5)/(5*d)

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Rubi [A]  time = 0.0469913, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4128, 390, 206} \[ \frac{b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}+a^3 x-\frac{b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac{b^3 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^3,x]

[Out]

a^3*x + (b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x])/d - (b^2*(3*a + 2*b)*Tanh[c + d*x]^3)/(3*d) + (b^3*Tanh[c + d*
x]^5)/(5*d)

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a+b \text{sech}^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-b x^2\right )^3}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (b \left (3 a^2+3 a b+b^2\right )-b^2 (3 a+2 b) x^2+b^3 x^4+\frac{a^3}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac{b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac{b^3 \tanh ^5(c+d x)}{5 d}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=a^3 x+\frac{b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac{b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac{b^3 \tanh ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 0.91406, size = 268, normalized size = 3.67 \[ \frac{\text{sech}(c) \text{sech}^5(c+d x) \left (-360 a^2 b \sinh (2 c+d x)+360 a^2 b \sinh (2 c+3 d x)-90 a^2 b \sinh (4 c+3 d x)+90 a^2 b \sinh (4 c+5 d x)+540 a^2 b \sinh (d x)+150 a^3 d x \cosh (2 c+d x)+75 a^3 d x \cosh (2 c+3 d x)+75 a^3 d x \cosh (4 c+3 d x)+15 a^3 d x \cosh (4 c+5 d x)+15 a^3 d x \cosh (6 c+5 d x)+150 a^3 d x \cosh (d x)-180 a b^2 \sinh (2 c+d x)+300 a b^2 \sinh (2 c+3 d x)+60 a b^2 \sinh (4 c+5 d x)+420 a b^2 \sinh (d x)+80 b^3 \sinh (2 c+3 d x)+16 b^3 \sinh (4 c+5 d x)+160 b^3 \sinh (d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x]^2)^3,x]

[Out]

(Sech[c]*Sech[c + d*x]^5*(150*a^3*d*x*Cosh[d*x] + 150*a^3*d*x*Cosh[2*c + d*x] + 75*a^3*d*x*Cosh[2*c + 3*d*x] +
75*a^3*d*x*Cosh[4*c + 3*d*x] + 15*a^3*d*x*Cosh[4*c + 5*d*x] + 15*a^3*d*x*Cosh[6*c + 5*d*x] + 540*a^2*b*Sinh[d
*x] + 420*a*b^2*Sinh[d*x] + 160*b^3*Sinh[d*x] - 360*a^2*b*Sinh[2*c + d*x] - 180*a*b^2*Sinh[2*c + d*x] + 360*a^
2*b*Sinh[2*c + 3*d*x] + 300*a*b^2*Sinh[2*c + 3*d*x] + 80*b^3*Sinh[2*c + 3*d*x] - 90*a^2*b*Sinh[4*c + 3*d*x] +
90*a^2*b*Sinh[4*c + 5*d*x] + 60*a*b^2*Sinh[4*c + 5*d*x] + 16*b^3*Sinh[4*c + 5*d*x]))/(480*d)

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Maple [A]  time = 0.026, size = 83, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( dx+c \right ) +3\,{a}^{2}b\tanh \left ( dx+c \right ) +3\,a{b}^{2} \left ( 2/3+1/3\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) \tanh \left ( dx+c \right ) +{b}^{3} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) \tanh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(d*x+c)+3*a^2*b*tanh(d*x+c)+3*a*b^2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)+b^3*(8/15+1/5*sech(d*x+c)^4+4
/15*sech(d*x+c)^2)*tanh(d*x+c))

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Maxima [B]  time = 1.03359, size = 448, normalized size = 6.14 \begin{align*} a^{3} x + \frac{16}{15} \, b^{3}{\left (\frac{5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 4 \, a b^{2}{\left (\frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac{1}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac{6 \, a^{2} b}{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)^2)^3,x, algorithm="maxima")

[Out]

a^3*x + 16/15*b^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e
^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c)
+ 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*
x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 4*a*b^2*(3*e^(-2*d*x - 2*c)/
(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x
- 4*c) + e^(-6*d*x - 6*c) + 1))) + 6*a^2*b/(d*(e^(-2*d*x - 2*c) + 1))

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Fricas [B]  time = 2.49725, size = 1173, normalized size = 16.07 \begin{align*} \frac{{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} +{\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \,{\left (27 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3} + 2 \,{\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) + 5 \,{\left ({\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 18 \, a^{2} b + 24 \, a b^{2} + 16 \, b^{3} + 3 \,{\left (27 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)^5 + 5*(15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3
)*cosh(d*x + c)*sinh(d*x + c)^4 + (45*a^2*b + 30*a*b^2 + 8*b^3)*sinh(d*x + c)^5 + 5*(15*a^3*d*x - 45*a^2*b - 3
0*a*b^2 - 8*b^3)*cosh(d*x + c)^3 + 5*(27*a^2*b + 30*a*b^2 + 8*b^3 + 2*(45*a^2*b + 30*a*b^2 + 8*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^3 + 5*(2*(15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)^3 + 3*(15*a^3*d*x - 45*
a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(
d*x + c) + 5*((45*a^2*b + 30*a*b^2 + 8*b^3)*cosh(d*x + c)^4 + 18*a^2*b + 24*a*b^2 + 16*b^3 + 3*(27*a^2*b + 30*
a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*co
sh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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

[Out]

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

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Giac [B]  time = 1.16311, size = 246, normalized size = 3.37 \begin{align*} \frac{{\left (d x + c\right )} a^{3}}{d} - \frac{2 \,{\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 210 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 150 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 40 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)*a^3/d - 2/15*(45*a^2*b*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) + 90*a*b^2*e^(6*d*x + 6*c) + 270*
a^2*b*e^(4*d*x + 4*c) + 210*a*b^2*e^(4*d*x + 4*c) + 80*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 150*a
*b^2*e^(2*d*x + 2*c) + 40*b^3*e^(2*d*x + 2*c) + 45*a^2*b + 30*a*b^2 + 8*b^3)/(d*(e^(2*d*x + 2*c) + 1)^5)