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

Optimal. Leaf size=40 \[ a^2 x+\frac{b (2 a+b) \tanh (c+d x)}{d}-\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0341698, antiderivative size = 40, 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} \[ a^2 x+\frac{b (2 a+b) \tanh (c+d x)}{d}-\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^2,x]

[Out]

a^2*x + (b*(2*a + b)*Tanh[c + d*x])/d - (b^2*Tanh[c + d*x]^3)/(3*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 )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-b x^2\right )^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (b (2 a+b)-b^2 x^2+\frac{a^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b (2 a+b) \tanh (c+d x)}{d}-\frac{b^2 \tanh ^3(c+d x)}{3 d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=a^2 x+\frac{b (2 a+b) \tanh (c+d x)}{d}-\frac{b^2 \tanh ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 0.420214, size = 106, normalized size = 2.65 \[ \frac{4 \text{sech}^3(c+d x) \left (a \cosh ^2(c+d x)+b\right )^2 \left (3 a^2 d x \cosh ^3(c+d x)+2 b (3 a+b) \text{sech}(c) \sinh (d x) \cosh ^2(c+d x)+b^2 \tanh (c) \cosh (c+d x)+b^2 \text{sech}(c) \sinh (d x)\right )}{3 d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 47, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( dx+c \right ) +2\,ab\tanh \left ( dx+c \right ) +{b}^{2} \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \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)^2,x)

[Out]

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

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Maxima [B]  time = 1.11098, size = 162, normalized size = 4.05 \begin{align*} a^{2} x + \frac{4}{3} \, 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{4 \, a 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.00682, size = 447, normalized size = 11.18 \begin{align*} \frac{{\left (3 \, a^{2} d x - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (3 \, a^{2} d x - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{3} + 3 \,{\left (3 \, a^{2} d x - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) + 6 \,{\left ({\left (3 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + a b + b^{2}\right )} \sinh \left (d x + c\right )}{3 \,{\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, 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)^2,x, algorithm="fricas")

[Out]

1/3*((3*a^2*d*x - 6*a*b - 2*b^2)*cosh(d*x + c)^3 + 3*(3*a^2*d*x - 6*a*b - 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^2
+ 2*(3*a*b + b^2)*sinh(d*x + c)^3 + 3*(3*a^2*d*x - 6*a*b - 2*b^2)*cosh(d*x + c) + 6*((3*a*b + b^2)*cosh(d*x +
c)^2 + a*b + b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + 3*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 )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1386, size = 107, normalized size = 2.68 \begin{align*} \frac{{\left (d x + c\right )} a^{2}}{d} - \frac{4 \,{\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b + b^{2}\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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