\(\int \frac {\text {sech}^2(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx\) [119]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 66 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} d}+\frac {\tanh (c+d x)}{2 a d \left (a+b \tanh ^2(c+d x)\right )} \]

[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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3756, 205, 211} \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b} d}+\frac {\tanh (c+d x)}{2 a d \left (a+b \tanh ^2(c+d x)\right )} \]

[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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

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

Rule 3756

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*} \text {integral}& = \frac {\text {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 {\text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{2 a d} \\ & = \frac {\arctan \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*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {b}}+\frac {\sqrt {a} \tanh (c+d x)}{a+b \tanh ^2(c+d x)}}{2 a^{3/2} d} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(209\) vs. \(2(54)=108\).

Time = 14.33 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.18

method result size
risch \(-\frac {{\mathrm e}^{2 d x +2 c} a -b \,{\mathrm e}^{2 d x +2 c}+a +b}{\left (a +b \right ) d a \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}-2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d a}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a \sqrt {-a b}-b \sqrt {-a b}+2 a b}{\left (a +b \right ) \sqrt {-a b}}\right )}{4 \sqrt {-a b}\, d a}\) \(210\)
derivativedivides \(\frac {-\frac {2 \left (-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a}-\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}+\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}}{d}\) \(232\)
default \(\frac {-\frac {2 \left (-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a}-\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}+\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}}{d}\) \(232\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (54) = 108\).

Time = 0.30 (sec) , antiderivative size = 1515, normalized size of antiderivative = 22.95 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]

[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))]

Sympy [F]

\[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {\operatorname {sech}^{2}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (54) = 108\).

Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.89 \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a + b}{{\left (a^{3} + 2 \, a^{2} b + a b^{2} + 2 \, {\left (a^{3} - a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac {\arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} a d} \]

[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)

Giac [F]

\[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{2}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^2\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]

[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)