\(\int \frac {\text {sech}^4(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [130]

Optimal result

Integrand size = 23, antiderivative size = 115 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {(a-3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2} d}+\frac {(a+b) \tanh (c+d x)}{4 a b d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \tanh (c+d x)}{8 a^2 b d \left (a+b \tanh ^2(c+d x)\right )} \] Output:

-1/8*(a-3*b)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(5/2)/b^(3/2)/d+1/4*(a+ 
b)*tanh(d*x+c)/a/b/d/(a+b*tanh(d*x+c)^2)^2-1/8*(a-3*b)*tanh(d*x+c)/a^2/b/d 
/(a+b*tanh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {(-a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{b^{3/2}}+\frac {\sqrt {a} \left (a^2+6 a b-3 b^2+\left (a^2+4 a b+3 b^2\right ) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{b (a-b+(a+b) \cosh (2 (c+d x)))^2}}{8 a^{5/2} d} \] Input:

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

Output:

(((-a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/b^(3/2) + (Sqrt[a]*( 
a^2 + 6*a*b - 3*b^2 + (a^2 + 4*a*b + 3*b^2)*Cosh[2*(c + d*x)])*Sinh[2*(c + 
 d*x)])/(b*(a - b + (a + b)*Cosh[2*(c + d*x)])^2))/(8*a^(5/2)*d)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4158, 298, 215, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (i c+i d x)^4}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 4158

\(\displaystyle \frac {\int \frac {1-\tanh ^2(c+d x)}{\left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 a b \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \int \frac {1}{\left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{4 a b}}{d}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 a b \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \left (\frac {\int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{2 a}+\frac {\tanh (c+d x)}{2 a \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a b}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {(a+b) \tanh (c+d x)}{4 a b \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {(a-3 b) \left (\frac {\arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\tanh (c+d x)}{2 a \left (a+b \tanh ^2(c+d x)\right )}\right )}{4 a b}}{d}\)

Input:

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

Output:

(((a + b)*Tanh[c + d*x])/(4*a*b*(a + b*Tanh[c + d*x]^2)^2) - ((a - 3*b)*(A 
rcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]) + Tanh[c + d*x] 
/(2*a*(a + b*Tanh[c + d*x]^2))))/(4*a*b))/d
 

Defintions of rubi rules used

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 218

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

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4158

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]}, Sim 
p[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] && I 
ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] 
 || EqQ[n^2, 16])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(333\) vs. \(2(101)=202\).

Time = 0.69 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.90

\[\frac {-\frac {2 \left (-\frac {\left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a b}-\frac {\left (3 a^{2}+11 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2} b}-\frac {\left (3 a^{2}+11 a b +12 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2} b}-\frac {\left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a b}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}-\frac {\left (a -3 b \right ) \left (\frac {\left (-a -\sqrt {\left (a +b \right ) b}-b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (a -\sqrt {\left (a +b \right ) b}+b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a b}}{d}\]

Input:

int(sech(d*x+c)^4/(a+tanh(d*x+c)^2*b)^3,x)
 

Output:

1/d*(-2*(-1/8*(a+5*b)/a/b*tanh(1/2*d*x+1/2*c)^7-1/8*(3*a^2+11*a*b+12*b^2)/ 
a^2/b*tanh(1/2*d*x+1/2*c)^5-1/8*(3*a^2+11*a*b+12*b^2)/a^2/b*tanh(1/2*d*x+1 
/2*c)^3-1/8*(a+5*b)/a/b*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a+2*ta 
nh(1/2*d*x+1/2*c)^2*a+4*b*tanh(1/2*d*x+1/2*c)^2+a)^2-1/4/a*(a-3*b)/b*(1/2* 
(-a-((a+b)*b)^(1/2)-b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/ 
2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)+a+2*b)*a)^(1/2))-1/2*( 
a-((a+b)*b)^(1/2)+b)/a/((a+b)*b)^(1/2)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2) 
*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*((a+b)*b)^(1/2)-a-2*b)*a)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2677 vs. \(2 (101) = 202\).

Time = 0.17 (sec) , antiderivative size = 5659, normalized size of antiderivative = 49.21 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{4}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(sech(d*x+c)**4/(a+b*tanh(d*x+c)**2)**3,x)
 

Output:

Integral(sech(c + d*x)**4/(a + b*tanh(c + d*x)**2)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (101) = 202\).

Time = 0.30 (sec) , antiderivative size = 360, normalized size of antiderivative = 3.13 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {a^{3} + 5 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3} + {\left (3 \, a^{3} + 13 \, a^{2} b + a b^{2} - 9 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (3 \, a^{3} + 7 \, a^{2} b - 3 \, a b^{2} + 9 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{3} - a^{2} b - 5 \, a b^{2} - 3 \, b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4} + 4 \, {\left (a^{5} b + a^{4} b^{2} - a^{3} b^{3} - a^{2} b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} b + a^{4} b^{2} + a^{3} b^{3} + 3 \, a^{2} b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} b + a^{4} b^{2} - a^{3} b^{3} - a^{2} b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {{\left (a - 3 \, b\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b d} \] Input:

integrate(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

1/4*(a^3 + 5*a^2*b + 7*a*b^2 + 3*b^3 + (3*a^3 + 13*a^2*b + a*b^2 - 9*b^3)* 
e^(-2*d*x - 2*c) + (3*a^3 + 7*a^2*b - 3*a*b^2 + 9*b^3)*e^(-4*d*x - 4*c) + 
(a^3 - a^2*b - 5*a*b^2 - 3*b^3)*e^(-6*d*x - 6*c))/((a^5*b + 3*a^4*b^2 + 3* 
a^3*b^3 + a^2*b^4 + 4*(a^5*b + a^4*b^2 - a^3*b^3 - a^2*b^4)*e^(-2*d*x - 2* 
c) + 2*(3*a^5*b + a^4*b^2 + a^3*b^3 + 3*a^2*b^4)*e^(-4*d*x - 4*c) + 4*(a^5 
*b + a^4*b^2 - a^3*b^3 - a^2*b^4)*e^(-6*d*x - 6*c) + (a^5*b + 3*a^4*b^2 + 
3*a^3*b^3 + a^2*b^4)*e^(-8*d*x - 8*c))*d) + 1/8*(a - 3*b)*arctan(1/2*((a + 
 b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a^2*b*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (101) = 202\).

Time = 0.61 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.77 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (a - 3 \, b\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} a^{2} b} + \frac {2 \, {\left (a^{3} e^{\left (6 \, d x + 6 \, c\right )} - a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 5 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 3 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 3 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 7 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 13 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 9 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 5 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3}\right )}}{{\left (a^{3} b + a^{2} b^{2}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2}}}{8 \, d} \] Input:

integrate(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

-1/8*((a - 3*b)*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*b) + 2*(a^3*e^(6*d*x + 6*c) - a^2*b*e^(6*d*x + 
6*c) - 5*a*b^2*e^(6*d*x + 6*c) - 3*b^3*e^(6*d*x + 6*c) + 3*a^3*e^(4*d*x + 
4*c) + 7*a^2*b*e^(4*d*x + 4*c) - 3*a*b^2*e^(4*d*x + 4*c) + 9*b^3*e^(4*d*x 
+ 4*c) + 3*a^3*e^(2*d*x + 2*c) + 13*a^2*b*e^(2*d*x + 2*c) + a*b^2*e^(2*d*x 
 + 2*c) - 9*b^3*e^(2*d*x + 2*c) + a^3 + 5*a^2*b + 7*a*b^2 + 3*b^3)/((a^3*b 
 + a^2*b^2)*(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)^2))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \] Input:

int(1/(cosh(c + d*x)^4*(a + b*tanh(c + d*x)^2)^3),x)
 

Output:

int(1/(cosh(c + d*x)^4*(a + b*tanh(c + d*x)^2)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 2949, normalized size of antiderivative = 25.64 \[ \int \frac {\text {sech}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sech(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x)
 

Output:

( - 2*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sq 
rt(b))/sqrt(a))*a**5 + 2*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d* 
x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**4*b + 12*e**(8*c + 8*d*x)*sqrt(b)*sq 
rt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b**2 + 4*e** 
(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sq 
rt(a))*a**2*b**3 - 10*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)* 
sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**4 - 6*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a 
)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**5 - 8*e**(6*c + 6* 
d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a* 
*5 + 24*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - 
sqrt(b))/sqrt(a))*a**4*b + 16*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c 
 + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b**2 - 48*e**(6*c + 6*d*x)*sq 
rt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**3 
 - 8*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqr 
t(b))/sqrt(a))*a*b**4 + 24*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + 
d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**5 - 12*e**(4*c + 4*d*x)*sqrt(b)*sq 
rt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**5 + 44*e**(4*c 
 + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a 
))*a**4*b - 24*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a 
+ b) - sqrt(b))/sqrt(a))*a**3*b**2 - 8*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)...