### 3.121 $$\int \frac{\text{sech}^4(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx$$

Optimal. Leaf size=77 $\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a+b \tanh ^2(c+d x)\right )}-\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2} d}$

[Out]

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

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Rubi [A]  time = 0.0788741, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3675, 385, 205} $\frac{(a+b) \tanh (c+d x)}{2 a b d \left (a+b \tanh ^2(c+d x)\right )}-\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2} d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

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

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.279209, size = 83, normalized size = 1.08 $\frac{(b-a) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )+\frac{\sqrt{a} \sqrt{b} (a+b) \sinh (2 (c+d x))}{(a+b) \cosh (2 (c+d x))+a-b}}{2 a^{3/2} b^{3/2} d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

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

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Maple [B]  time = 0.095, size = 746, normalized size = 9.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x)

[Out]

1/d/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)/b*tanh(1/2*d*x+1/2*c)^3+1/
d/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)/a*tanh(1/2*d*x+1/2*c)^3+1/d/
(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)/b*tanh(1/2*d*x+1/2*c)+1/d/(tan
h(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)/a*tanh(1/2*d*x+1/2*c)+1/2/d/b*a/(b
*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^
(1/2))-1/2/d/b/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)
^(1/2))+1/2/d/b*a/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b
))^(1/2)+a+2*b)*a)^(1/2))+1/2/d/b/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b
))^(1/2)+a+2*b)*a)^(1/2))+1/2/d/a/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+
b))^(1/2)-a-2*b)*a)^(1/2))-1/2/d/(b*(a+b))^(1/2)/a/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+
1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))*b-1/2/d/a/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*
x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/2/d/(b*(a+b))^(1/2)/a/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arct
an(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))*b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.22099, size = 3452, normalized size = 44.83 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[-1/4*(4*a^2*b + 4*a*b^2 + 4*(a^2*b - a*b^2)*cosh(d*x + c)^2 + 8*(a^2*b - a*b^2)*cosh(d*x + c)*sinh(d*x + c) +
4*(a^2*b - a*b^2)*sinh(d*x + c)^2 - ((a^2 - b^2)*cosh(d*x + c)^4 + 4*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c)^
3 + (a^2 - b^2)*sinh(d*x + c)^4 + 2*(a^2 - 2*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - b^2)*cosh(d*x + c)^2 + a
^2 - 2*a*b + b^2)*sinh(d*x + c)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(d*x + c)^3 + (a^2 - 2*a*b + b^2)*cosh(d*x
+ c))*sinh(d*x + c))*sqrt(-a*b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)
*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b
^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3
+ (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c
) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(-a*b))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x +
c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x +
c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)))/((a^3*b^2 + a^2*b^3)*d*co
sh(d*x + c)^4 + 4*(a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 + a^2*b^3)*d*sinh(d*x + c)^4
+ 2*(a^3*b^2 - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^2 + (a^3*b^2 - a^2*b^3)*d
)*sinh(d*x + c)^2 + (a^3*b^2 + a^2*b^3)*d + 4*((a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^3 + (a^3*b^2 - a^2*b^3)*d*c
osh(d*x + c))*sinh(d*x + c)), -1/2*(2*a^2*b + 2*a*b^2 + 2*(a^2*b - a*b^2)*cosh(d*x + c)^2 + 4*(a^2*b - a*b^2)*
cosh(d*x + c)*sinh(d*x + c) + 2*(a^2*b - a*b^2)*sinh(d*x + c)^2 + ((a^2 - b^2)*cosh(d*x + c)^4 + 4*(a^2 - b^2)
*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 - b^2)*sinh(d*x + c)^4 + 2*(a^2 - 2*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(a
^2 - b^2)*cosh(d*x + c)^2 + a^2 - 2*a*b + b^2)*sinh(d*x + c)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(d*x + c)^3 +
(a^2 - 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a*b)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*co
sh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(a*b)/(a*b)))/((a^3*b^2 + a^2*b^3)*d*cosh(d*x
+ c)^4 + 4*(a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3*b^2 + a^2*b^3)*d*sinh(d*x + c)^4 + 2*(a
^3*b^2 - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^2 + (a^3*b^2 - a^2*b^3)*d)*sinh
(d*x + c)^2 + (a^3*b^2 + a^2*b^3)*d + 4*((a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^3 + (a^3*b^2 - a^2*b^3)*d*cosh(d*
x + c))*sinh(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**4/(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Integral(sech(c + d*x)**4/(a + b*tanh(c + d*x)**2)**2, x)

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Giac [B]  time = 1.66815, size = 211, normalized size = 2.74 \begin{align*} -\frac{\frac{{\left (a e^{\left (2 \, c\right )} - b e^{\left (2 \, c\right )}\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right ) e^{\left (-2 \, c\right )}}{\sqrt{a b} a b} + \frac{2 \,{\left (a e^{\left (2 \, d x + 2 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} a b}}{2 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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