3.4.48 \(\int \frac {\text {sech}^4(c+d x)}{(a+b \sinh ^2(c+d x))^3} \, dx\) [348]

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

[Out]

1/8*b^2*(48*a^2-16*a*b+3*b^2)*arctanh((a-b)^(1/2)*tanh(d*x+c)/a^(1/2))/a^(5/2)/(a-b)^(9/2)/d+(a-4*b)*tanh(d*x+
c)/(a-b)^4/d-1/3*tanh(d*x+c)^3/(a-b)^3/d+1/4*b^4*tanh(d*x+c)/a/(a-b)^4/d/(a-(a-b)*tanh(d*x+c)^2)^2-1/8*(16*a-3
*b)*b^3*tanh(d*x+c)/a^2/(a-b)^4/d/(a-(a-b)*tanh(d*x+c)^2)

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Rubi [A]
time = 0.25, antiderivative size = 203, normalized size of antiderivative = 1.00, 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 = {3270, 398, 1171, 393, 214} \begin {gather*} -\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} \end {gather*}

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[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 398

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 3270

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]

Rubi steps

\begin {align*} \int \frac {\text {sech}^4(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 2.02, size = 169, normalized size = 0.83 \begin {gather*} \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 \left (-32 a^2+24 a b-3 b^2+b (-14 a+3 b) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{a^2 (2 a-b+b \cosh (2 (c+d x)))^2}+8 \left (2 a-11 b+(a-b) \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{(a-b)^4}}{24 d} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(444\) vs. \(2(187)=374\).
time = 2.08, size = 445, normalized size = 2.19 Too large to display

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,method=_RETURNVERBOSE)

[Out]

1/d*(-2*b^2/(a-b)^4*((1/8*b*(16*a-5*b)/a*tanh(1/2*d*x+1/2*c)^7-1/8*(16*a^2-61*a*b+12*b^2)/a^2*b*tanh(1/2*d*x+1
/2*c)^5-1/8*(16*a^2-61*a*b+12*b^2)/a^2*b*tanh(1/2*d*x+1/2*c)^3+1/8*b*(16*a-5*b)/a*tanh(1/2*d*x+1/2*c))/(a*tanh
(1/2*d*x+1/2*c)^4-2*a*tanh(1/2*d*x+1/2*c)^2+4*b*tanh(1/2*d*x+1/2*c)^2+a)^2+1/8/a*(48*a^2-16*a*b+3*b^2)*(-1/2*(
(-b*(a-b))^(1/2)-b)/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))+1/2*((-b*(a-b))^(1/2)+b)/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))))-2/(a-b)^4*((4*b-a)*tanh(1/2*d*x+1/2*
c)^5+(20/3*b-2/3*a)*tanh(1/2*d*x+1/2*c)^3+(4*b-a)*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^2+1)^3)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 9388 vs. \(2 (189) = 378\).
time = 0.56, size = 19032, normalized size = 93.75 \begin {gather*} \text {too large to display} \end {gather*}

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]

[1/48*(12*(48*a^4*b^3 - 64*a^3*b^4 + 19*a^2*b^5 - 3*a*b^6)*cosh(d*x + c)^12 + 144*(48*a^4*b^3 - 64*a^3*b^4 + 1
9*a^2*b^5 - 3*a*b^6)*cosh(d*x + c)*sinh(d*x + c)^11 + 12*(48*a^4*b^3 - 64*a^3*b^4 + 19*a^2*b^5 - 3*a*b^6)*sinh
(d*x + c)^12 + 72*(48*a^5*b^2 - 64*a^4*b^3 + 19*a^3*b^4 - 3*a^2*b^5)*cosh(d*x + c)^10 + 72*(48*a^5*b^2 - 64*a^
4*b^3 + 19*a^3*b^4 - 3*a^2*b^5 + 11*(48*a^4*b^3 - 64*a^3*b^4 + 19*a^2*b^5 - 3*a*b^6)*cosh(d*x + c)^2)*sinh(d*x
+ c)^10 + 240*(11*(48*a^4*b^3 - 64*a^3*b^4 + 19*a^2*b^5 - 3*a*b^6)*cosh(d*x + c)^3 + 3*(48*a^5*b^2 - 64*a^4*b
^3 + 19*a^3*b^4 - 3*a^2*b^5)*cosh(d*x + c))*sinh(d*x + c)^9 + 4*(768*a^6*b + 800*a^5*b^2 - 2560*a^4*b^3 + 1250
*a^3*b^4 - 285*a^2*b^5 + 27*a*b^6)*cosh(d*x + c)^8 + 4*(768*a^6*b + 800*a^5*b^2 - 2560*a^4*b^3 + 1250*a^3*b^4
- 285*a^2*b^5 + 27*a*b^6 + 1485*(48*a^4*b^3 - 64*a^3*b^4 + 19*a^2*b^5 - 3*a*b^6)*cosh(d*x + c)^4 + 810*(48*a^5
*b^2 - 64*a^4*b^3 + 19*a^3*b^4 - 3*a^2*b^5)*cosh(d*x + c)^2)*sinh(d*x + c)^8 + 32*(297*(48*a^4*b^3 - 64*a^3*b^
4 + 19*a^2 ...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 436 vs. \(2 (189) = 378\).
time = 0.94, size = 436, normalized size = 2.15 \begin {gather*} \frac {\frac {3 \, {\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 )}{{\left (a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} \sqrt {-a^{2} + a b}} + \frac {6 \, {\left (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}\right )}}{{\left (a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4}\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 {16 \, {\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 )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \end {gather*}

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]

1/24*(3*(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 - 4*a
^5*b + 6*a^4*b^2 - 4*a^3*b^3 + a^2*b^4)*sqrt(-a^2 + a*b)) + 6*(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 - 4*a^5*b + 6*a^4*b^2 - 4*a^3*b^3 + a^2*b^4)*(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) + 16*(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 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(e^(2*d*x + 2*c) + 1)^3))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x)^4*(a + b*sinh(c + d*x)^2)^3),x)

[Out]

int(1/(cosh(c + d*x)^4*(a + b*sinh(c + d*x)^2)^3), x)

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