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

Optimal. Leaf size=82 \[ \frac {1}{2} b^2 x (6 a-5 b)-\frac {(a-b)^3 \tanh ^3(c+d x)}{3 d}+\frac {(a-b)^2 (a+2 b) \tanh (c+d x)}{d}+\frac {b^3 \sinh (c+d x) \cosh (c+d x)}{2 d} \]

[Out]

1/2*(6*a-5*b)*b^2*x+1/2*b^3*cosh(d*x+c)*sinh(d*x+c)/d+(a-b)^2*(a+2*b)*tanh(d*x+c)/d-1/3*(a-b)^3*tanh(d*x+c)^3/
d

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Rubi [A]  time = 0.11, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3191, 390, 385, 206} \[ \frac {1}{2} b^2 x (6 a-5 b)-\frac {(a-b)^3 \tanh ^3(c+d x)}{3 d}+\frac {(a-b)^2 (a+2 b) \tanh (c+d x)}{d}+\frac {b^3 \sinh (c+d x) \cosh (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.76, size = 84, normalized size = 1.02 \[ \frac {6 b^2 (6 a-5 b) (c+d x)+2 (a-b)^2 \tanh (c+d x) \text {sech}^2(c+d x) ((2 a+7 b) \cosh (2 (c+d x))+4 a+5 b)+3 b^3 \sinh (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(6*a - 5*b)*b^2*(c + d*x) + 3*b^3*Sinh[2*(c + d*x)] + 2*(a - b)^2*(4*a + 5*b + (2*a + 7*b)*Cosh[2*(c + d*x)
])*Sech[c + d*x]^2*Tanh[c + d*x])/(12*d)

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fricas [B]  time = 0.74, size = 321, normalized size = 3.91 \[ \frac {3 \, b^{3} \sinh \left (d x + c\right )^{5} - 4 \, {\left (4 \, a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (4 \, a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, b^{3} \cosh \left (d x + c\right )^{2} + 16 \, a^{3} + 24 \, a^{2} b - 96 \, a b^{2} + 65 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (4 \, a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 16 \, a^{3} - 24 \, a^{2} b + 10 \, b^{3} + {\left (16 \, a^{3} + 24 \, a^{2} b - 96 \, a b^{2} + 65 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \]

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]

1/24*(3*b^3*sinh(d*x + c)^5 - 4*(4*a^3 + 6*a^2*b - 24*a*b^2 + 14*b^3 - 3*(6*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c)^
3 - 12*(4*a^3 + 6*a^2*b - 24*a*b^2 + 14*b^3 - 3*(6*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*b^3
*cosh(d*x + c)^2 + 16*a^3 + 24*a^2*b - 96*a*b^2 + 65*b^3)*sinh(d*x + c)^3 - 12*(4*a^3 + 6*a^2*b - 24*a*b^2 + 1
4*b^3 - 3*(6*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c) + 3*(5*b^3*cosh(d*x + c)^4 + 16*a^3 - 24*a^2*b + 10*b^3 + (16*a
^3 + 24*a^2*b - 96*a*b^2 + 65*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh
(d*x + c)^2 + 3*d*cosh(d*x + c))

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giac [B]  time = 0.23, size = 208, normalized size = 2.54 \[ \frac {3 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )} - 3 \, {\left (12 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - \frac {16 \, {\left (9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} + 3 \, a^{2} b - 12 \, a b^{2} + 7 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \]

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*b^3*e^(2*d*x + 2*c) + 12*(6*a*b^2 - 5*b^3)*(d*x + c) - 3*(12*a*b^2*e^(2*d*x + 2*c) - 10*b^3*e^(2*d*x +
2*c) + b^3)*e^(-2*d*x - 2*c) - 16*(9*a^2*b*e^(4*d*x + 4*c) - 18*a*b^2*e^(4*d*x + 4*c) + 9*b^3*e^(4*d*x + 4*c)
+ 6*a^3*e^(2*d*x + 2*c) - 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) + 2*a^3 + 3*a^2*b - 12*a*b^2 + 7*
b^3)/(e^(2*d*x + 2*c) + 1)^3)/d

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maple [A]  time = 0.12, size = 148, normalized size = 1.80 \[ \frac {a^{3} \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (d x +c -\tanh \left (d x +c \right )-\frac {\left (\tanh ^{3}\left (d x +c \right )\right )}{3}\right )+b^{3} \left (\frac {\sinh ^{5}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}\right )}{d} \]

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]

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

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maxima [B]  time = 0.38, size = 382, normalized size = 4.66 \[ a b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} - \frac {1}{24} \, b^{3} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} + \frac {4}{3} \, a^{3} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + 2 \, a^{2} b {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \]

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]

a*b^2*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*
c) + e^(-6*d*x - 6*c) + 1))) - 1/24*b^3*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*e^(-2*d*x - 2*c) + 201*e
^(-4*d*x - 4*c) + 147*e^(-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 3*e^(-6*d*x - 6*c) + e
^(-8*d*x - 8*c)))) + 4/3*a^3*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c
) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + 2*a^2*b*(3*e^(-4*d*x - 4*c
)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*
x - 4*c) + e^(-6*d*x - 6*c) + 1)))

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mupad [B]  time = 0.16, size = 273, normalized size = 3.33 \[ \frac {b^2\,x\,\left (6\,a-5\,b\right )}{2}-\frac {\frac {2\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3-3\,a^2\,b+b^3\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (2\,a^3-3\,a^2\,b+b^3\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {b^3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {2\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*x*(6*a - 5*b))/2 - ((2*(a^2*b - 2*a*b^2 + b^3))/d + (2*exp(4*c + 4*d*x)*(a^2*b - 2*a*b^2 + b^3))/d + (4*e
xp(2*c + 2*d*x)*(2*a^3 - 3*a^2*b + b^3))/(3*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) +
1) - ((2*(2*a^3 - 3*a^2*b + b^3))/(3*d) + (2*exp(2*c + 2*d*x)*(a^2*b - 2*a*b^2 + b^3))/d)/(2*exp(2*c + 2*d*x)
+ exp(4*c + 4*d*x) + 1) - (b^3*exp(- 2*c - 2*d*x))/(8*d) + (b^3*exp(2*c + 2*d*x))/(8*d) - (2*(a^2*b - 2*a*b^2
+ b^3))/(d*(exp(2*c + 2*d*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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