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

Optimal. Leaf size=66 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} d}+\frac {\tanh (c+d x)}{2 a d \left (a+b \tanh ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3675, 199, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} d}+\frac {\tanh (c+d x)}{2 a d \left (a+b \tanh ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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]

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])

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 63, normalized size = 0.95 \[ \frac {\frac {\sqrt {a} \tanh (c+d x)}{a+b \tanh ^2(c+d x)}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {b}}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 1515, normalized size = 22.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^2/(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 + 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 + 2*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))*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)*co
sh(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^4*b + 2*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^4 + 4*(a^4*b + 2*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)*sinh(d*x + c
)^3 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*d*sinh(d*x + c)^4 + 2*(a^4*b - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^4*b +
2*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^2 + (a^4*b - a^2*b^3)*d)*sinh(d*x + c)^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*
d + 4*((a^4*b + 2*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^3 + (a^4*b - a^2*b^3)*d*cosh(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 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*s
inh(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 + 2*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))*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)))/((a^4*b + 2*a^3*b^2 + a^2*b^3)*d*cosh(d
*x + c)^4 + 4*(a^4*b + 2*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*d*
sinh(d*x + c)^4 + 2*(a^4*b - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^4*b + 2*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^2
+ (a^4*b - a^2*b^3)*d)*sinh(d*x + c)^2 + (a^4*b + 2*a^3*b^2 + a^2*b^3)*d + 4*((a^4*b + 2*a^3*b^2 + a^2*b^3)*d
*cosh(d*x + c)^3 + (a^4*b - a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [B]  time = 0.56, size = 138, normalized size = 2.09 \[ \frac {\frac {\arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {2 \, {\left (a e^{\left (2 \, d x + 2 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}}{{\left (a^{2} + a b\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.40, size = 498, normalized size = 7.55 \[ \frac {\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right ) a}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +4 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a \right ) a}-\frac {\arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{2 d \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}+\frac {\arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{2 d a \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}-\frac {\arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right ) b}{2 d \sqrt {b \left (a +b \right )}\, a \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}-\frac {\arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{2 d \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}-\frac {\arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{2 d a \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}-\frac {\arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right ) b}{2 d \sqrt {b \left (a +b \right )}\, a \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^2/(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)/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)/a*tanh(1/2*d*x+1/2*c)-1/2/d/
(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/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/(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/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)*arctan(
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 [B]  time = 0.52, size = 125, normalized size = 1.89 \[ \frac {{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a + b}{{\left (a^{3} + 2 \, a^{2} b + a b^{2} + 2 \, {\left (a^{3} - a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {\arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^2\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)^2*(a + b*tanh(c + d*x)^2)^2),x)

[Out]

int(1/(cosh(c + d*x)^2*(a + b*tanh(c + d*x)^2)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{2}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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