### 3.114 $$\int \frac{\text{sech}^6(c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=75 $-\frac{(a+2 b) \tanh (c+d x)}{b^2 d}+\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2} d}+\frac{\tanh ^3(c+d x)}{3 b d}$

[Out]

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

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Rubi [A]  time = 0.0915841, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3675, 390, 205} $-\frac{(a+2 b) \tanh (c+d x)}{b^2 d}+\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2} d}+\frac{\tanh ^3(c+d x)}{3 b d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[c + d*x]^6/(a + b*Tanh[c + d*x]^2),x]

[Out]

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

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

Mathematica [A]  time = 0.34114, size = 71, normalized size = 0.95 $\frac{(a+b)^2 \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a} b^{5/2} d}-\frac{\tanh (c+d x) \left (3 a+b \text{sech}^2(c+d x)+5 b\right )}{3 b^2 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[c + d*x]^6/(a + b*Tanh[c + d*x]^2),x]

[Out]

((a + b)^2*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*b^(5/2)*d) - ((3*a + 5*b + b*Sech[c + d*x]^2)*Tan
h[c + d*x])/(3*b^2*d)

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Maple [B]  time = 0.078, size = 1077, normalized size = 14.4 \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)^6/(a+b*tanh(d*x+c)^2),x)

[Out]

-1/d*a^3/b^2/(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))-3/d*a^2/b/(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))-3*a/d/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*t
anh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d*a^2/b^2/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctan
h(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+2/d*a/b/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arcta
nh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d/((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/d*b/(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/d*a^3/b^2/(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))-3/d*a^2/b/(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)
)-3*a/d/(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/d*a^2/b^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))-2/d*a/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/d/((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/d*b/(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))-2/d/b^2/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^5*a-4/d/b/(tanh(1/2*d*x+
1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^5-4/d/b^2/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^3*a-16/3/d/b/(tanh
(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^3-2/d/b^2/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)*a-4/d/b/(
tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)

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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)^6/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.64971, size = 5189, normalized size = 69.19 \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)^6/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/6*(12*(a^2*b + a*b^2)*cosh(d*x + c)^4 + 48*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 12*(a^2*b + a*b^
2)*sinh(d*x + c)^4 + 12*a^2*b + 20*a*b^2 + 24*(a^2*b + 2*a*b^2)*cosh(d*x + c)^2 + 24*(a^2*b + 2*a*b^2 + 3*(a^2
*b + a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 3*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 6*(a^2 + 2*a*b + b^2)*
cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^6 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 +
3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(
d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 3*
(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x +
c)^2 + a^2 + 2*a*b + b^2 + 6*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 2*(a^2 + 2*a*b + 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)
) + 48*((a^2*b + a*b^2)*cosh(d*x + c)^3 + (a^2*b + 2*a*b^2)*cosh(d*x + c))*sinh(d*x + c))/(a*b^3*d*cosh(d*x +
c)^6 + 6*a*b^3*d*cosh(d*x + c)*sinh(d*x + c)^5 + a*b^3*d*sinh(d*x + c)^6 + 3*a*b^3*d*cosh(d*x + c)^4 + 3*a*b^3
*d*cosh(d*x + c)^2 + a*b^3*d + 3*(5*a*b^3*d*cosh(d*x + c)^2 + a*b^3*d)*sinh(d*x + c)^4 + 4*(5*a*b^3*d*cosh(d*x
+ c)^3 + 3*a*b^3*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a*b^3*d*cosh(d*x + c)^4 + 6*a*b^3*d*cosh(d*x + c)^2
+ a*b^3*d)*sinh(d*x + c)^2 + 6*(a*b^3*d*cosh(d*x + c)^5 + 2*a*b^3*d*cosh(d*x + c)^3 + a*b^3*d*cosh(d*x + c))*s
inh(d*x + c)), 1/3*(6*(a^2*b + a*b^2)*cosh(d*x + c)^4 + 24*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 6*(
a^2*b + a*b^2)*sinh(d*x + c)^4 + 6*a^2*b + 10*a*b^2 + 12*(a^2*b + 2*a*b^2)*cosh(d*x + c)^2 + 12*(a^2*b + 2*a*b
^2 + 3*(a^2*b + a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 6*(a^2 + 2*
a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^6 + 3*(a^2 + 2*a*b + b^2)*cosh(d*
x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 4*(5*(a^2 + 2*a*b +
b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x
+ c)^2 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)
*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 6*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 2*(a^2 + 2*a*b + 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)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(a*b)/(a*b)) + 24*((a^2*b + a*b^2)*
cosh(d*x + c)^3 + (a^2*b + 2*a*b^2)*cosh(d*x + c))*sinh(d*x + c))/(a*b^3*d*cosh(d*x + c)^6 + 6*a*b^3*d*cosh(d*
x + c)*sinh(d*x + c)^5 + a*b^3*d*sinh(d*x + c)^6 + 3*a*b^3*d*cosh(d*x + c)^4 + 3*a*b^3*d*cosh(d*x + c)^2 + a*b
^3*d + 3*(5*a*b^3*d*cosh(d*x + c)^2 + a*b^3*d)*sinh(d*x + c)^4 + 4*(5*a*b^3*d*cosh(d*x + c)^3 + 3*a*b^3*d*cosh
(d*x + c))*sinh(d*x + c)^3 + 3*(5*a*b^3*d*cosh(d*x + c)^4 + 6*a*b^3*d*cosh(d*x + c)^2 + a*b^3*d)*sinh(d*x + c)
^2 + 6*(a*b^3*d*cosh(d*x + c)^5 + 2*a*b^3*d*cosh(d*x + c)^3 + a*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}^{6}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**6/(a+b*tanh(d*x+c)**2),x)

[Out]

Integral(sech(c + d*x)**6/(a + b*tanh(c + d*x)**2), x)

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Giac [B]  time = 1.44931, size = 207, normalized size = 2.76 \begin{align*} \frac{\frac{3 \,{\left (a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )} + b^{2} 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} b^{2}} + \frac{2 \,{\left (3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a + 5 \, b\right )}}{b^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*(a^2*e^(2*c) + 2*a*b*e^(2*c) + b^2*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)*b^2) + 2*(3*a*e^(4*d*x + 4*c) + 3*b*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) + 12*
b*e^(2*d*x + 2*c) + 3*a + 5*b)/(b^2*(e^(2*d*x + 2*c) + 1)^3))/d