\(\int \frac {\text {sech}^5(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [131]

Optimal result

Integrand size = 23, antiderivative size = 104 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {a+b} d}+\frac {\sinh (c+d x)}{4 a d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {3 \sinh (c+d x)}{8 a^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3757, 205, 211} \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} d \sqrt {a+b}}+\frac {3 \sinh (c+d x)}{8 a^2 d \left ((a+b) \sinh ^2(c+d x)+a\right )}+\frac {\sinh (c+d x)}{4 a d \left ((a+b) \sinh ^2(c+d x)+a\right )^2} \]

[In]

Int[Sech[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(3*ArcTan[(Sqrt[a + b]*Sinh[c + d*x])/Sqrt[a]])/(8*a^(5/2)*Sqrt[a + b]*d) + Sinh[c + d*x]/(4*a*d*(a + (a + b)*
Sinh[c + d*x]^2)^2) + (3*Sinh[c + d*x])/(8*a^2*d*(a + (a + b)*Sinh[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 3757

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (a+(a+b) x^2\right )^3} \, dx,x,\sinh (c+d x)\right )}{d} \\ & = \frac {\sinh (c+d x)}{4 a d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {3 \text {Subst}\left (\int \frac {1}{\left (a+(a+b) x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{4 a d} \\ & = \frac {\sinh (c+d x)}{4 a d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {3 \sinh (c+d x)}{8 a^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\sinh (c+d x)\right )}{8 a^2 d} \\ & = \frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {a+b} d}+\frac {\sinh (c+d x)}{4 a d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}+\frac {3 \sinh (c+d x)}{8 a^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {3 \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+\frac {\sqrt {a} \sinh (c+d x) \left (5 a+3 (a+b) \sinh ^2(c+d x)\right )}{\left (a+(a+b) \sinh ^2(c+d x)\right )^2}}{8 a^{5/2} d} \]

[In]

Integrate[Sech[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(275\) vs. \(2(90)=180\).

Time = 0.22 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.65

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

[In]

int(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(2*(-5/8/a*tanh(1/2*d*x+1/2*c)^7+3/8*(a-4*b)/a^2*tanh(1/2*d*x+1/2*c)^5-3/8*(a-4*b)/a^2*tanh(1/2*d*x+1/2*c)
^3+5/8/a*tanh(1/2*d*x+1/2*c))/(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)^
2+3/4/a*(1/2*(((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+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. 2712 vs. \(2 (90) = 180\).

Time = 0.32 (sec) , antiderivative size = 5077, normalized size of antiderivative = 48.82 \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{5}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(sech(d*x+c)**5/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral(sech(c + d*x)**5/(a + b*tanh(c + d*x)**2)**3, x)

Maxima [F]

\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]

[In]

integrate(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/4*(3*(a*e^(7*c) + b*e^(7*c))*e^(7*d*x) + (11*a*e^(5*c) - 9*b*e^(5*c))*e^(5*d*x) - (11*a*e^(3*c) - 9*b*e^(3*c
))*e^(3*d*x) - 3*(a*e^c + b*e^c)*e^(d*x))/(a^4*d + 2*a^3*b*d + a^2*b^2*d + (a^4*d*e^(8*c) + 2*a^3*b*d*e^(8*c)
+ a^2*b^2*d*e^(8*c))*e^(8*d*x) + 4*(a^4*d*e^(6*c) - a^2*b^2*d*e^(6*c))*e^(6*d*x) + 2*(3*a^4*d*e^(4*c) - 2*a^3*
b*d*e^(4*c) + 3*a^2*b^2*d*e^(4*c))*e^(4*d*x) + 4*(a^4*d*e^(2*c) - a^2*b^2*d*e^(2*c))*e^(2*d*x)) + 32*integrate
(3/128*(e^(3*d*x + 3*c) + e^(d*x + c))/(a^3 + a^2*b + (a^3*e^(4*c) + a^2*b*e^(4*c))*e^(4*d*x) + 2*(a^3*e^(2*c)
- a^2*b*e^(2*c))*e^(2*d*x)), x)

Giac [F]

\[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{5}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]

[In]

integrate(sech(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^5\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]

[In]

int(1/(cosh(c + d*x)^5*(a + b*tanh(c + d*x)^2)^3),x)

[Out]

int(1/(cosh(c + d*x)^5*(a + b*tanh(c + d*x)^2)^3), x)