\(\int \frac {\text {sech}^6(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [132]

Optimal result

Integrand size = 23, antiderivative size = 131 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{5/2} d}+\frac {(a+b) \text {sech}^2(c+d x) \tanh (c+d x)}{4 a b d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {3 \left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (c+d x)}{8 d \left (a+b \tanh ^2(c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3756, 424, 393, 211} \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {3 \left (\frac {1}{a^2}-\frac {1}{b^2}\right ) \tanh (c+d x)}{8 d \left (a+b \tanh ^2(c+d x)\right )}+\frac {\left (3 a^2-2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^{5/2} d}+\frac {(a+b) \tanh (c+d x) \text {sech}^2(c+d x)}{4 a b d \left (a+b \tanh ^2(c+d x)\right )^2} \]

[In]

Int[Sech[c + d*x]^6/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

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

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 393

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

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

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

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )-\frac {\sqrt {a} \sqrt {b} (a+b) \left (3 a^2-10 a b+3 b^2+3 \left (a^2-b^2\right ) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{(a-b+(a+b) \cosh (2 (c+d x)))^2}}{8 a^{5/2} b^{5/2} d} \]

[In]

Integrate[Sech[c + d*x]^6/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

((3*a^2 - 2*a*b + 3*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] - (Sqrt[a]*Sqrt[b]*(a + b)*(3*a^2 - 10*a*b +
3*b^2 + 3*(a^2 - b^2)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(a - b + (a + b)*Cosh[2*(c + d*x)])^2)/(8*a^(5/2)*
b^(5/2)*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(117)=234\).

Time = 0.19 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.87

\[\frac {-\frac {2 \left (\frac {\left (3 a^{2}-2 a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a \,b^{2}}+\frac {\left (9 a^{3}+14 a^{2} b -7 a \,b^{2}-12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2} b^{2}}+\frac {\left (9 a^{3}+14 a^{2} b -7 a \,b^{2}-12 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2} b^{2}}+\frac {\left (3 a^{2}-2 a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a \,b^{2}}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a \right )^{2}}-\frac {\left (3 a^{2}-2 a b +3 b^{2}\right ) \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a \,b^{2}}}{d}\]

[In]

int(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(-2*(1/8*(3*a^2-2*a*b-5*b^2)/a/b^2*tanh(1/2*d*x+1/2*c)^7+1/8*(9*a^3+14*a^2*b-7*a*b^2-12*b^3)/a^2/b^2*tanh(
1/2*d*x+1/2*c)^5+1/8*(9*a^3+14*a^2*b-7*a*b^2-12*b^3)/a^2/b^2*tanh(1/2*d*x+1/2*c)^3+1/8*(3*a^2-2*a*b-5*b^2)/a/b
^2*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)^2-1/4/
a*(3*a^2-2*a*b+3*b^2)/b^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)*arc
tan(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. 2464 vs. \(2 (117) = 234\).

Time = 0.34 (sec) , antiderivative size = 5233, normalized size of antiderivative = 39.95 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {\operatorname {sech}^{6}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(sech(d*x+c)**6/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral(sech(c + d*x)**6/(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. 332 vs. \(2 (117) = 234\).

Time = 0.49 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.53 \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {3 \, a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - 3 \, b^{3} + {\left (9 \, a^{3} - 13 \, a^{2} b - 13 \, a b^{2} + 9 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (3 \, a^{3} - 5 \, a^{2} b + 5 \, a b^{2} - 3 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (3 \, a^{3} + a^{2} b + a b^{2} + 3 \, b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4} + 4 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{4} b^{2} - 2 \, a^{3} b^{3} + 3 \, a^{2} b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} - \frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{2} d} \]

[In]

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{6}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]

[In]

integrate(sech(d*x+c)^6/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]

[In]

int(1/(cosh(c + d*x)^6*(a + b*tanh(c + d*x)^2)^3),x)

[Out]

int(1/(cosh(c + d*x)^6*(a + b*tanh(c + d*x)^2)^3), x)