\(\int \frac {\text {sech}^7(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx\) [133]

Optimal result

Integrand size = 23, antiderivative size = 156 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\arctan (\sinh (c+d x))}{b^3 d}+\frac {\sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^3 d}+\frac {(a+b) \sinh (c+d x)}{4 a b d \left (a+(a+b) \sinh ^2(c+d x)\right )^2}-\frac {(4 a-3 b) (a+b) \sinh (c+d x)}{8 a^2 b^2 d \left (a+(a+b) \sinh ^2(c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3757, 425, 541, 536, 209, 211} \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {(4 a-3 b) (a+b) \sinh (c+d x)}{8 a^2 b^2 d \left ((a+b) \sinh ^2(c+d x)+a\right )}+\frac {\sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a+b} \sinh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^3 d}+\frac {(a+b) \sinh (c+d x)}{4 a b d \left ((a+b) \sinh ^2(c+d x)+a\right )^2}-\frac {\arctan (\sinh (c+d x))}{b^3 d} \]

[In]

Int[Sech[c + d*x]^7/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 425

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

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.69 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.03 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\frac {2 \sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )}{a^{5/2}}+\frac {2 \left (8 a^3+4 a^2 b-a b^2+3 b^3\right ) \arctan \left (\frac {\sqrt {a} \text {csch}(c+d x)}{\sqrt {a+b}}\right )}{a^{5/2} \sqrt {a+b}}+64 \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+\frac {i \sqrt {a+b} \left (8 a^2-4 a b+3 b^2\right ) \log (a-b+(a+b) \cosh (2 (c+d x)))}{a^{5/2}}-\frac {i \left (8 a^3+4 a^2 b-a b^2+3 b^3\right ) \log (a-b+(a+b) \cosh (2 (c+d x)))}{a^{5/2} \sqrt {a+b}}-\frac {32 b^2 (a+b) \sinh (c+d x)}{a (a-b+(a+b) \cosh (2 (c+d x)))^2}+\frac {8 b \left (4 a^2+a b-3 b^2\right ) \sinh (c+d x)}{a^2 (a-b+(a+b) \cosh (2 (c+d x)))}}{32 b^3 d} \]

[In]

Integrate[Sech[c + d*x]^7/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

-1/32*((2*Sqrt[a + b]*(8*a^2 - 4*a*b + 3*b^2)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[a + b]])/a^(5/2) + (2*(8*a^3
+ 4*a^2*b - a*b^2 + 3*b^3)*ArcTan[(Sqrt[a]*Csch[c + d*x])/Sqrt[a + b]])/(a^(5/2)*Sqrt[a + b]) + 64*ArcTan[Tan
h[(c + d*x)/2]] + (I*Sqrt[a + b]*(8*a^2 - 4*a*b + 3*b^2)*Log[a - b + (a + b)*Cosh[2*(c + d*x)]])/a^(5/2) - (I*
(8*a^3 + 4*a^2*b - a*b^2 + 3*b^3)*Log[a - b + (a + b)*Cosh[2*(c + d*x)]])/(a^(5/2)*Sqrt[a + b]) - (32*b^2*(a +
b)*Sinh[c + d*x])/(a*(a - b + (a + b)*Cosh[2*(c + d*x)])^2) + (8*b*(4*a^2 + a*b - 3*b^2)*Sinh[c + d*x])/(a^2*
(a - b + (a + b)*Cosh[2*(c + d*x)])))/(b^3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(388\) vs. \(2(142)=284\).

Time = 0.21 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.49

\[\frac {\frac {\frac {2 \left (\frac {b \left (4 a^{2}-a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a}+\frac {\left (4 a^{3}+23 a^{2} b +7 a \,b^{2}-12 b^{3}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2}}-\frac {\left (4 a^{3}+23 a^{2} b +7 a \,b^{2}-12 b^{3}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2}}-\frac {b \left (4 a^{2}-a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a \right )^{2}}+\frac {\left (8 a^{3}+4 a^{2} b -a \,b^{2}+3 b^{3}\right ) \left (\frac {\left (\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{4 a}}{b^{3}}-\frac {2 \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\]

[In]

int(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(2/b^3*((1/8*b*(4*a^2-a*b-5*b^2)/a*tanh(1/2*d*x+1/2*c)^7+1/8*(4*a^3+23*a^2*b+7*a*b^2-12*b^3)/a^2*b*tanh(1/
2*d*x+1/2*c)^5-1/8*(4*a^3+23*a^2*b+7*a*b^2-12*b^3)/a^2*b*tanh(1/2*d*x+1/2*c)^3-1/8*b*(4*a^2-a*b-5*b^2)/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+1/8/a*(8*a^3
+4*a^2*b-a*b^2+3*b^3)*(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))))-2/b^3*arctan(tan
h(1/2*d*x+1/2*c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4457 vs. \(2 (142) = 284\).

Time = 0.38 (sec) , antiderivative size = 8070, normalized size of antiderivative = 51.73 \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]

[In]

integrate(sech(d*x+c)**7/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{7}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]

[In]

integrate(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

-1/4*((4*a^3*e^(7*c) + 5*a^2*b*e^(7*c) - 2*a*b^2*e^(7*c) - 3*b^3*e^(7*c))*e^(7*d*x) + (4*a^3*e^(5*c) - 19*a^2*
b*e^(5*c) - 14*a*b^2*e^(5*c) + 9*b^3*e^(5*c))*e^(5*d*x) - (4*a^3*e^(3*c) - 19*a^2*b*e^(3*c) - 14*a*b^2*e^(3*c)
+ 9*b^3*e^(3*c))*e^(3*d*x) - (4*a^3*e^c + 5*a^2*b*e^c - 2*a*b^2*e^c - 3*b^3*e^c)*e^(d*x))/(a^4*b^2*d + 2*a^3*
b^3*d + a^2*b^4*d + (a^4*b^2*d*e^(8*c) + 2*a^3*b^3*d*e^(8*c) + a^2*b^4*d*e^(8*c))*e^(8*d*x) + 4*(a^4*b^2*d*e^(
6*c) - a^2*b^4*d*e^(6*c))*e^(6*d*x) + 2*(3*a^4*b^2*d*e^(4*c) - 2*a^3*b^3*d*e^(4*c) + 3*a^2*b^4*d*e^(4*c))*e^(4
*d*x) + 4*(a^4*b^2*d*e^(2*c) - a^2*b^4*d*e^(2*c))*e^(2*d*x)) - 2*arctan(e^(d*x + c))/(b^3*d) + 128*integrate(1
/512*((8*a^3*e^(3*c) + 4*a^2*b*e^(3*c) - a*b^2*e^(3*c) + 3*b^3*e^(3*c))*e^(3*d*x) + (8*a^3*e^c + 4*a^2*b*e^c -
a*b^2*e^c + 3*b^3*e^c)*e^(d*x))/(a^3*b^3 + a^2*b^4 + (a^3*b^3*e^(4*c) + a^2*b^4*e^(4*c))*e^(4*d*x) + 2*(a^3*b
^3*e^(2*c) - a^2*b^4*e^(2*c))*e^(2*d*x)), x)

Giac [F]

\[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int { \frac {\operatorname {sech}\left (d x + c\right )^{7}}{{\left (b \tanh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]

[In]

integrate(sech(d*x+c)^7/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^7(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^7\,{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]

[In]

int(1/(cosh(c + d*x)^7*(a + b*tanh(c + d*x)^2)^3),x)

[Out]

int(1/(cosh(c + d*x)^7*(a + b*tanh(c + d*x)^2)^3), x)