3.93 \(\int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx\)

Optimal. Leaf size=173 \[ \frac {x}{a^3}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \text {sech}(c+d x))}+\frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}} \]

[Out]

x/a^3-b*(6*a^4-5*a^2*b^2+2*b^4)*arctan((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2
)/d+1/2*b^2*tanh(d*x+c)/a/(a^2-b^2)/d/(a+b*sech(d*x+c))^2+1/2*b^2*(5*a^2-2*b^2)*tanh(d*x+c)/a^2/(a^2-b^2)^2/d/
(a+b*sech(d*x+c))

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Rubi [A]  time = 0.31, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3785, 4060, 3919, 3831, 2659, 208} \[ -\frac {b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \text {sech}(c+d x))}+\frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}+\frac {x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x])^(-3),x]

[Out]

x/a^3 - (b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(5/2)
*(a + b)^(5/2)*d) + (b^2*Tanh[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sech[c + d*x])^2) + (b^2*(5*a^2 - 2*b^2)*Tan
h[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Sech[c + d*x]))

Rule 208

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3785

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4060

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx &=\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}-\frac {\int \frac {-2 \left (a^2-b^2\right )+2 a b \text {sech}(c+d x)-b^2 \text {sech}^2(c+d x)}{(a+b \text {sech}(c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \text {sech}(c+d x))}+\frac {\int \frac {2 \left (a^2-b^2\right )^2-a b \left (4 a^2-b^2\right ) \text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {x}{a^3}+\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \text {sech}(c+d x))}-\frac {\left (b \left (6 a^4-5 a^2 b^2+2 b^4\right )\right ) \int \frac {\text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {x}{a^3}+\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \text {sech}(c+d x))}-\frac {\left (6 a^4-5 a^2 b^2+2 b^4\right ) \int \frac {1}{1+\frac {a \cosh (c+d x)}{b}} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {x}{a^3}+\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \text {sech}(c+d x))}+\frac {\left (i \left (6 a^4-5 a^2 b^2+2 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^3 \left (a^2-b^2\right )^2 d}\\ &=\frac {x}{a^3}-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \text {sech}(c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 205, normalized size = 1.18 \[ \frac {\text {sech}^3(c+d x) (a \cosh (c+d x)+b) \left (\frac {3 a b^2 \left (2 a^2-b^2\right ) \sinh (c+d x) (a \cosh (c+d x)+b)}{(a-b)^2 (a+b)^2}+\frac {2 b \left (6 a^4-5 a^2 b^2+2 b^4\right ) (a \cosh (c+d x)+b)^2 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^3 \sinh (c+d x)}{(b-a) (a+b)}+2 (c+d x) (a \cosh (c+d x)+b)^2\right )}{2 a^3 d (a+b \text {sech}(c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x])^(-3),x]

[Out]

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

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fricas [B]  time = 0.50, size = 4125, normalized size = 23.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/2*(12*a^6*b^2 - 18*a^4*b^4 + 6*a^2*b^6 - 2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cosh(d*x + c)^4 - 2
*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*sinh(d*x + c)^4 + 2*(7*a^5*b^3 - 11*a^3*b^5 + 4*a*b^7 - 4*(a^7*b
- 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*cosh(d*x + c)^3 + 2*(7*a^5*b^3 - 11*a^3*b^5 + 4*a*b^7 - 4*(a^8 - 3*a^6*b
^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cosh(d*x + c) - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*sinh(d*x + c)^3 -
2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x + 2*(6*a^6*b^2 + 3*a^4*b^4 - 15*a^2*b^6 + 6*b^8 - 2*(a^8 - a^6*
b^2 - 3*a^4*b^4 + 5*a^2*b^6 - 2*b^8)*d*x)*cosh(d*x + c)^2 + 2*(6*a^6*b^2 + 3*a^4*b^4 - 15*a^2*b^6 + 6*b^8 - 6*
(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cosh(d*x + c)^2 - 2*(a^8 - a^6*b^2 - 3*a^4*b^4 + 5*a^2*b^6 - 2*b^8
)*d*x + 3*(7*a^5*b^3 - 11*a^3*b^5 + 4*a*b^7 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*cosh(d*x + c))*si
nh(d*x + c)^2 + (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5 + (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cosh(d*x + c)^4 + (6*a^6*
b - 5*a^4*b^3 + 2*a^2*b^5)*sinh(d*x + c)^4 + 4*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c)^3 + 4*(6*a^5*b^
2 - 5*a^3*b^4 + 2*a*b^6 + (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(6*a^6*b + 7*a^
4*b^3 - 8*a^2*b^5 + 4*b^7)*cosh(d*x + c)^2 + 2*(6*a^6*b + 7*a^4*b^3 - 8*a^2*b^5 + 4*b^7 + 3*(6*a^6*b - 5*a^4*b
^3 + 2*a^2*b^5)*cosh(d*x + c)^2 + 6*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c))*sinh(d*x + c)^2 + 4*(6*a^
5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c) + 4*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6 + (6*a^6*b - 5*a^4*b^3 + 2*a^2
*b^5)*cosh(d*x + c)^3 + 3*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c)^2 + (6*a^6*b + 7*a^4*b^3 - 8*a^2*b^5
+ 4*b^7)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-a^2 + b^2)*log((a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*
b*cosh(d*x + c) - a^2 + 2*b^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(-a^2 + b^2)*(a*cosh(d*x + c
) + a*sinh(d*x + c) + b))/(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + 2*b*cosh(d*x + c) + 2*(a*cosh(d*x + c) + b)
*sinh(d*x + c) + a)) + 2*(17*a^5*b^3 - 25*a^3*b^5 + 8*a*b^7 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*c
osh(d*x + c) + 2*(17*a^5*b^3 - 25*a^3*b^5 + 8*a*b^7 - 4*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cosh(d*x +
c)^3 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x + 3*(7*a^5*b^3 - 11*a^3*b^5 + 4*a*b^7 - 4*(a^7*b - 3*a^5
*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*cosh(d*x + c)^2 + 2*(6*a^6*b^2 + 3*a^4*b^4 - 15*a^2*b^6 + 6*b^8 - 2*(a^8 - a^6*
b^2 - 3*a^4*b^4 + 5*a^2*b^6 - 2*b^8)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b
^6)*d*cosh(d*x + c)^4 + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*sinh(d*x + c)^4 + 4*(a^10*b - 3*a^8*b^3 + 3
*a^6*b^5 - a^4*b^7)*d*cosh(d*x + c)^3 + 2*(a^11 - a^9*b^2 - 3*a^7*b^4 + 5*a^5*b^6 - 2*a^3*b^8)*d*cosh(d*x + c)
^2 + 4*((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cosh(d*x + c) + (a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*
d)*sinh(d*x + c)^3 + 4*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cosh(d*x + c) + 2*(3*(a^11 - 3*a^9*b^2 + 3
*a^7*b^4 - a^5*b^6)*d*cosh(d*x + c)^2 + 6*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cosh(d*x + c) + (a^11 -
a^9*b^2 - 3*a^7*b^4 + 5*a^5*b^6 - 2*a^3*b^8)*d)*sinh(d*x + c)^2 + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d
+ 4*((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cosh(d*x + c)^3 + 3*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)
*d*cosh(d*x + c)^2 + (a^11 - a^9*b^2 - 3*a^7*b^4 + 5*a^5*b^6 - 2*a^3*b^8)*d*cosh(d*x + c) + (a^10*b - 3*a^8*b^
3 + 3*a^6*b^5 - a^4*b^7)*d)*sinh(d*x + c)), -(6*a^6*b^2 - 9*a^4*b^4 + 3*a^2*b^6 - (a^8 - 3*a^6*b^2 + 3*a^4*b^4
- a^2*b^6)*d*x*cosh(d*x + c)^4 - (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*sinh(d*x + c)^4 + (7*a^5*b^3 - 1
1*a^3*b^5 + 4*a*b^7 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*cosh(d*x + c)^3 + (7*a^5*b^3 - 11*a^3*b^5
+ 4*a*b^7 - 4*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cosh(d*x + c) - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 -
a*b^7)*d*x)*sinh(d*x + c)^3 - (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x + (6*a^6*b^2 + 3*a^4*b^4 - 15*a^2*b^
6 + 6*b^8 - 2*(a^8 - a^6*b^2 - 3*a^4*b^4 + 5*a^2*b^6 - 2*b^8)*d*x)*cosh(d*x + c)^2 + (6*a^6*b^2 + 3*a^4*b^4 -
15*a^2*b^6 + 6*b^8 - 6*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cosh(d*x + c)^2 - 2*(a^8 - a^6*b^2 - 3*a^4*
b^4 + 5*a^2*b^6 - 2*b^8)*d*x + 3*(7*a^5*b^3 - 11*a^3*b^5 + 4*a*b^7 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)
*d*x)*cosh(d*x + c))*sinh(d*x + c)^2 - (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5 + (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*co
sh(d*x + c)^4 + (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*sinh(d*x + c)^4 + 4*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d
*x + c)^3 + 4*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6 + (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cosh(d*x + c))*sinh(d*x + c
)^3 + 2*(6*a^6*b + 7*a^4*b^3 - 8*a^2*b^5 + 4*b^7)*cosh(d*x + c)^2 + 2*(6*a^6*b + 7*a^4*b^3 - 8*a^2*b^5 + 4*b^7
+ 3*(6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cosh(d*x + c)^2 + 6*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c))*si
nh(d*x + c)^2 + 4*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c) + 4*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6 + (6*a^
6*b - 5*a^4*b^3 + 2*a^2*b^5)*cosh(d*x + c)^3 + 3*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*cosh(d*x + c)^2 + (6*a^6*b
+ 7*a^4*b^3 - 8*a^2*b^5 + 4*b^7)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cosh(d*x + c) + a*si
nh(d*x + c) + b)/sqrt(a^2 - b^2)) + (17*a^5*b^3 - 25*a^3*b^5 + 8*a*b^7 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*
b^7)*d*x)*cosh(d*x + c) + (17*a^5*b^3 - 25*a^3*b^5 + 8*a*b^7 - 4*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*c
osh(d*x + c)^3 - 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x + 3*(7*a^5*b^3 - 11*a^3*b^5 + 4*a*b^7 - 4*(a^7*
b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x)*cosh(d*x + c)^2 + 2*(6*a^6*b^2 + 3*a^4*b^4 - 15*a^2*b^6 + 6*b^8 - 2*(a
^8 - a^6*b^2 - 3*a^4*b^4 + 5*a^2*b^6 - 2*b^8)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^
4 - a^5*b^6)*d*cosh(d*x + c)^4 + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*sinh(d*x + c)^4 + 4*(a^10*b - 3*a^
8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cosh(d*x + c)^3 + 2*(a^11 - a^9*b^2 - 3*a^7*b^4 + 5*a^5*b^6 - 2*a^3*b^8)*d*cosh
(d*x + c)^2 + 4*((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cosh(d*x + c) + (a^10*b - 3*a^8*b^3 + 3*a^6*b^5 -
a^4*b^7)*d)*sinh(d*x + c)^3 + 4*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cosh(d*x + c) + 2*(3*(a^11 - 3*a^
9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cosh(d*x + c)^2 + 6*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*cosh(d*x + c)
+ (a^11 - a^9*b^2 - 3*a^7*b^4 + 5*a^5*b^6 - 2*a^3*b^8)*d)*sinh(d*x + c)^2 + (a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^
5*b^6)*d + 4*((a^11 - 3*a^9*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cosh(d*x + c)^3 + 3*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 -
a^4*b^7)*d*cosh(d*x + c)^2 + (a^11 - a^9*b^2 - 3*a^7*b^4 + 5*a^5*b^6 - 2*a^3*b^8)*d*cosh(d*x + c) + (a^10*b -
3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d)*sinh(d*x + c))]

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giac [A]  time = 0.14, size = 261, normalized size = 1.51 \[ -\frac {\frac {{\left (6 \, a^{4} b - 5 \, a^{2} b^{3} + 2 \, b^{5}\right )} \arctan \left (\frac {a e^{\left (d x + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {7 \, a^{3} b^{3} e^{\left (3 \, d x + 3 \, c\right )} - 4 \, a b^{5} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a^{4} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a^{2} b^{4} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b^{6} e^{\left (2 \, d x + 2 \, c\right )} + 17 \, a^{3} b^{3} e^{\left (d x + c\right )} - 8 \, a b^{5} e^{\left (d x + c\right )} + 6 \, a^{4} b^{2} - 3 \, a^{2} b^{4}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (d x + c\right )} + a\right )}^{2}} - \frac {d x + c}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))^3,x, algorithm="giac")

[Out]

-((6*a^4*b - 5*a^2*b^3 + 2*b^5)*arctan((a*e^(d*x + c) + b)/sqrt(a^2 - b^2))/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(
a^2 - b^2)) + (7*a^3*b^3*e^(3*d*x + 3*c) - 4*a*b^5*e^(3*d*x + 3*c) + 6*a^4*b^2*e^(2*d*x + 2*c) + 9*a^2*b^4*e^(
2*d*x + 2*c) - 6*b^6*e^(2*d*x + 2*c) + 17*a^3*b^3*e^(d*x + c) - 8*a*b^5*e^(d*x + c) + 6*a^4*b^2 - 3*a^2*b^4)/(
(a^7 - 2*a^5*b^2 + a^3*b^4)*(a*e^(2*d*x + 2*c) + 2*b*e^(d*x + c) + a)^2) - (d*x + c)/a^3)/d

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maple [B]  time = 0.27, size = 660, normalized size = 3.82 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}+\frac {6 b^{2} \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b^{3} \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b^{4} \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {6 b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {b^{3} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 b^{4} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {6 a b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {5 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {2 b^{5} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d \,a^{3} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sech(d*x+c))^3,x)

[Out]

-1/d/a^3*ln(tanh(1/2*d*x+1/2*c)-1)+6/d*b^2/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+
2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+1/d/a*b^3/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+
2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3-2/d/a^2*b^4/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^
2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+6/d*b^2/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-
2*a*b+b^2)*tanh(1/2*d*x+1/2*c)-1/d/a*b^3/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*
a*b+b^2)*tanh(1/2*d*x+1/2*c)-2/d/a^2*b^4/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*
a*b+b^2)*tanh(1/2*d*x+1/2*c)-6/d*a*b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*d*x+1/2*c)/
((a+b)*(a-b))^(1/2))+5/d/a*b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*d*x+1/2*c)/((a+b)
*(a-b))^(1/2))-2/d/a^3*b^5/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*d*x+1/2*c)/((a+b)*(a-
b))^(1/2))+1/d/a^3*ln(tanh(1/2*d*x+1/2*c)+1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cosh(c + d*x))^3,x)

[Out]

int(1/(a + b/cosh(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sech(d*x+c))**3,x)

[Out]

Integral((a + b*sech(c + d*x))**(-3), x)

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