3.348 \(\int \frac{\text{sech}^4(c+d x)}{(a+b \sinh ^2(c+d x))^3} \, dx\)

Optimal. Leaf size=203 \[ -\frac{b^3 (16 a-3 b) \tanh (c+d x)}{8 a^2 d (a-b)^4 \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{b^2 \left (48 a^2-16 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} d (a-b)^{9/2}}+\frac{b^4 \tanh (c+d x)}{4 a d (a-b)^4 \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac{\tanh ^3(c+d x)}{3 d (a-b)^3}+\frac{(a-4 b) \tanh (c+d x)}{d (a-b)^4} \]

[Out]

(b^2*(48*a^2 - 16*a*b + 3*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(5/2)*(a - b)^(9/2)*d) + ((a
- 4*b)*Tanh[c + d*x])/((a - b)^4*d) - Tanh[c + d*x]^3/(3*(a - b)^3*d) + (b^4*Tanh[c + d*x])/(4*a*(a - b)^4*d*
(a - (a - b)*Tanh[c + d*x]^2)^2) - ((16*a - 3*b)*b^3*Tanh[c + d*x])/(8*a^2*(a - b)^4*d*(a - (a - b)*Tanh[c + d
*x]^2))

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Rubi [A]  time = 0.340881, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3191, 390, 1157, 385, 208} \[ -\frac{b^3 (16 a-3 b) \tanh (c+d x)}{8 a^2 d (a-b)^4 \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{b^2 \left (48 a^2-16 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} d (a-b)^{9/2}}+\frac{b^4 \tanh (c+d x)}{4 a d (a-b)^4 \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac{\tanh ^3(c+d x)}{3 d (a-b)^3}+\frac{(a-4 b) \tanh (c+d x)}{d (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(48*a^2 - 16*a*b + 3*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(5/2)*(a - b)^(9/2)*d) + ((a
- 4*b)*Tanh[c + d*x])/((a - b)^4*d) - Tanh[c + d*x]^3/(3*(a - b)^3*d) + (b^4*Tanh[c + d*x])/(4*a*(a - b)^4*d*
(a - (a - b)*Tanh[c + d*x]^2)^2) - ((16*a - 3*b)*b^3*Tanh[c + d*x])/(8*a^2*(a - b)^4*d*(a - (a - b)*Tanh[c + d
*x]^2))

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

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 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]

Rubi steps

\begin{align*} \int \frac{\text{sech}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{\left (a-(a-b) x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a-4 b}{(a-b)^4}-\frac{x^2}{(a-b)^3}+\frac{b^2 \left (6 a^2-4 a b+b^2\right )-4 (a-b) (3 a-b) b^2 x^2+6 (a-b)^2 b^2 x^4}{(a-b)^4 \left (a+(-a+b) x^2\right )^3}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a-4 b) \tanh (c+d x)}{(a-b)^4 d}-\frac{\tanh ^3(c+d x)}{3 (a-b)^3 d}+\frac{\operatorname{Subst}\left (\int \frac{b^2 \left (6 a^2-4 a b+b^2\right )-4 (a-b) (3 a-b) b^2 x^2+6 (a-b)^2 b^2 x^4}{\left (a+(-a+b) x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{(a-b)^4 d}\\ &=\frac{(a-4 b) \tanh (c+d x)}{(a-b)^4 d}-\frac{\tanh ^3(c+d x)}{3 (a-b)^3 d}+\frac{b^4 \tanh (c+d x)}{4 a (a-b)^4 d \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-b^2 \left (24 a^2-16 a b+3 b^2\right )+24 a (a-b) b^2 x^2}{\left (a+(-a+b) x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 a (a-b)^4 d}\\ &=\frac{(a-4 b) \tanh (c+d x)}{(a-b)^4 d}-\frac{\tanh ^3(c+d x)}{3 (a-b)^3 d}+\frac{b^4 \tanh (c+d x)}{4 a (a-b)^4 d \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac{(16 a-3 b) b^3 \tanh (c+d x)}{8 a^2 (a-b)^4 d \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac{\left (b^2 \left (48 a^2-16 a b+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\tanh (c+d x)\right )}{8 a^2 (a-b)^4 d}\\ &=\frac{b^2 \left (48 a^2-16 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{5/2} (a-b)^{9/2} d}+\frac{(a-4 b) \tanh (c+d x)}{(a-b)^4 d}-\frac{\tanh ^3(c+d x)}{3 (a-b)^3 d}+\frac{b^4 \tanh (c+d x)}{4 a (a-b)^4 d \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac{(16 a-3 b) b^3 \tanh (c+d x)}{8 a^2 (a-b)^4 d \left (a-(a-b) \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.19955, size = 169, normalized size = 0.83 \[ \frac{\frac{3 b^2 \left (48 a^2-16 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{5/2} (a-b)^{9/2}}+\frac{\frac{3 b^3 \sinh (2 (c+d x)) \left (-32 a^2+b (3 b-14 a) \cosh (2 (c+d x))+24 a b-3 b^2\right )}{a^2 (2 a+b \cosh (2 (c+d x))-b)^2}+8 \tanh (c+d x) \left ((a-b) \text{sech}^2(c+d x)+2 a-11 b\right )}{(a-b)^4}}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^2*(48*a^2 - 16*a*b + 3*b^2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(a^(5/2)*(a - b)^(9/2)) + ((3*
b^3*(-32*a^2 + 24*a*b - 3*b^2 + b*(-14*a + 3*b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(a^2*(2*a - b + b*Cosh[2
*(c + d*x)])^2) + 8*(2*a - 11*b + (a - b)*Sech[c + d*x]^2)*Tanh[c + d*x])/(a - b)^4)/(24*d)

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Maple [B]  time = 0.108, size = 1892, normalized size = 9.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/d*b^3/(a-b)^4/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a
-b))^(1/2)-a+2*b)*a)^(1/2))-6/d*b^3/(a-b)^4/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*ta
nh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+5/4/d*b^4/(a-b)^4/(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/a*tanh(1/2*d*x+1/2*c)^7-61/4/d*b^4/(a-b)^4/(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/a*tanh(1/2*d*x+1/2*c)^5+3/d*b^5/(a-b)^4/(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/a^2*tanh(1/2*d*x+1/2*c)^5-61/4/d*b^4/
(a-b)^4/(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/a*tanh(1/2*d*x+1/2*c
)^3+3/d*b^5/(a-b)^4/(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/a^2*tanh
(1/2*d*x+1/2*c)^3+5/4/d*b^4/(a-b)^4/(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/a*tanh(1/2*d*x+1/2*c)-2/d*b^3/(a-b)^4/a/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/
2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+3/8/d*b^4/(a-b)^4/a^2/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(
a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+2/d*b^3/(a-b)^4/a/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(
1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))-3/8/d*b^4/(a-b)^4/a^2/((2*(-b*(a-b))^(
1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))+2/d*b^4/(a-b)^4/a/(-b*
(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^
(1/2))-3/8/d*b^5/(a-b)^4/a^2/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c
)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))+2/d*b^4/(a-b)^4/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2
)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))-3/8/d*b^5/(a-b)^4/a^2/(-b*(a-b))^(1/2)/(
(2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+2/d/(a
-b)^4/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^5*a-8/d/(a-b)^4/(tanh(1/2*d*x+1/2*c)^2+1)^3*b*tanh(1/2*d
*x+1/2*c)^5+4/3/d/(a-b)^4/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)^3*a-40/3/d/(a-b)^4/(tanh(1/2*d*x+1/2
*c)^2+1)^3*b*tanh(1/2*d*x+1/2*c)^3+2/d/(a-b)^4/(tanh(1/2*d*x+1/2*c)^2+1)^3*tanh(1/2*d*x+1/2*c)*a-8/d/(a-b)^4/(
tanh(1/2*d*x+1/2*c)^2+1)^3*b*tanh(1/2*d*x+1/2*c)-4/d*b^3/(a-b)^4/(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*tanh(1/2*d*x+1/2*c)+4/d*b^3/(a-b)^4/(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*tanh(1/2*d*x+1/2*c)^3+4/d*b^3/(a-b)^4/(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*tanh(1/2*d*x+1/2*c)^5-6/d*b^2/(a-b)^4/((2*(-b*(a-b))^(1
/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))-4/d*b^3/(a-b)^4/(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*tanh(1/2*d*x+1/2*c)^7+6/d*b^2/(a-b)
^4/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.49803, size = 608, normalized size = 3. \begin{align*} \frac{{\left (48 \, a^{2} b^{2} - 16 \, a b^{3} + 3 \, b^{4}\right )} \arctan \left (\frac{b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt{-a^{2} + a b}}\right )}{8 \,{\left (a^{6} d - 4 \, a^{5} b d + 6 \, a^{4} b^{2} d - 4 \, a^{3} b^{3} d + a^{2} b^{4} d\right )} \sqrt{-a^{2} + a b}} + \frac{24 \, a^{2} b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 16 \, a b^{4} e^{\left (6 \, d x + 6 \, c\right )} + 3 \, b^{5} e^{\left (6 \, d x + 6 \, c\right )} + 112 \, a^{3} b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 136 \, a^{2} b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 66 \, a b^{4} e^{\left (4 \, d x + 4 \, c\right )} - 9 \, b^{5} e^{\left (4 \, d x + 4 \, c\right )} + 88 \, a^{2} b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 64 \, a b^{4} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, b^{5} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, a b^{4} - 3 \, b^{5}}{4 \,{\left (a^{6} d - 4 \, a^{5} b d + 6 \, a^{4} b^{2} d - 4 \, a^{3} b^{3} d + a^{2} b^{4} d\right )}{\left (b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )}^{2}} + \frac{2 \,{\left (9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 24 \, b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + 11 \, b\right )}}{3 \,{\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(48*a^2*b^2 - 16*a*b^3 + 3*b^4)*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/((a^6*d - 4*a^5
*b*d + 6*a^4*b^2*d - 4*a^3*b^3*d + a^2*b^4*d)*sqrt(-a^2 + a*b)) + 1/4*(24*a^2*b^3*e^(6*d*x + 6*c) - 16*a*b^4*e
^(6*d*x + 6*c) + 3*b^5*e^(6*d*x + 6*c) + 112*a^3*b^2*e^(4*d*x + 4*c) - 136*a^2*b^3*e^(4*d*x + 4*c) + 66*a*b^4*
e^(4*d*x + 4*c) - 9*b^5*e^(4*d*x + 4*c) + 88*a^2*b^3*e^(2*d*x + 2*c) - 64*a*b^4*e^(2*d*x + 2*c) + 9*b^5*e^(2*d
*x + 2*c) + 14*a*b^4 - 3*b^5)/((a^6*d - 4*a^5*b*d + 6*a^4*b^2*d - 4*a^3*b^3*d + a^2*b^4*d)*(b*e^(4*d*x + 4*c)
+ 4*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + b)^2) + 2/3*(9*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) + 24*b*e^
(2*d*x + 2*c) - 2*a + 11*b)/((a^4*d - 4*a^3*b*d + 6*a^2*b^2*d - 4*a*b^3*d + b^4*d)*(e^(2*d*x + 2*c) + 1)^3)