3.22 \(\int \text {csch}^2(c+d x) (a+b \text {sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=70 \[ \frac {b^2 (a+b) \tanh ^3(c+d x)}{d}-\frac {3 b (a+b)^2 \tanh (c+d x)}{d}-\frac {(a+b)^3 \coth (c+d x)}{d}-\frac {b^3 \tanh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4132, 270} \[ \frac {b^2 (a+b) \tanh ^3(c+d x)}{d}-\frac {3 b (a+b)^2 \tanh (c+d x)}{d}-\frac {(a+b)^3 \coth (c+d x)}{d}-\frac {b^3 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

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

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Mathematica [B]  time = 2.79, size = 380, normalized size = 5.43 \[ -\frac {\text {csch}(c) \text {sech}(c) \coth (c+d x) \left (-25 a^3 \sinh (2 (c+d x))-20 a^3 \sinh (4 (c+d x))-5 a^3 \sinh (6 (c+d x))-25 a^3 \sinh (2 (c+2 d x))+25 a^3 \sinh (4 c+2 d x)+5 a^3 \sinh (6 c+4 d x)-5 a^3 \sinh (4 c+6 d x)+10 a \left (5 a^2+12 a b+8 b^2\right ) \sinh (2 c)-120 a^2 b \sinh (2 (c+2 d x))+30 a^2 b \sinh (4 c+2 d x)-30 a^2 b \sinh (4 c+6 d x)-10 \left (5 a^3+18 a^2 b+20 a b^2+8 b^3\right ) \sinh (2 d x)+50 a b^2 \sinh (2 (c+d x))+40 a b^2 \sinh (4 (c+d x))+10 a b^2 \sinh (6 (c+d x))-160 a b^2 \sinh (2 (c+2 d x))-40 a b^2 \sinh (4 c+6 d x)+30 b^3 \sinh (2 (c+d x))+24 b^3 \sinh (4 (c+d x))+6 b^3 \sinh (6 (c+d x))-64 b^3 \sinh (2 (c+2 d x))-16 b^3 \sinh (4 c+6 d x)\right ) \left (a+b \text {sech}^2(c+d x)\right )^3}{40 d (a \cosh (2 (c+d x))+a+2 b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/40*(Coth[c + d*x]*Csch[c]*Sech[c]*(a + b*Sech[c + d*x]^2)^3*(10*a*(5*a^2 + 12*a*b + 8*b^2)*Sinh[2*c] - 10*(
5*a^3 + 18*a^2*b + 20*a*b^2 + 8*b^3)*Sinh[2*d*x] - 25*a^3*Sinh[2*(c + d*x)] + 50*a*b^2*Sinh[2*(c + d*x)] + 30*
b^3*Sinh[2*(c + d*x)] - 20*a^3*Sinh[4*(c + d*x)] + 40*a*b^2*Sinh[4*(c + d*x)] + 24*b^3*Sinh[4*(c + d*x)] - 5*a
^3*Sinh[6*(c + d*x)] + 10*a*b^2*Sinh[6*(c + d*x)] + 6*b^3*Sinh[6*(c + d*x)] - 25*a^3*Sinh[2*(c + 2*d*x)] - 120
*a^2*b*Sinh[2*(c + 2*d*x)] - 160*a*b^2*Sinh[2*(c + 2*d*x)] - 64*b^3*Sinh[2*(c + 2*d*x)] + 25*a^3*Sinh[4*c + 2*
d*x] + 30*a^2*b*Sinh[4*c + 2*d*x] + 5*a^3*Sinh[6*c + 4*d*x] - 5*a^3*Sinh[4*c + 6*d*x] - 30*a^2*b*Sinh[4*c + 6*
d*x] - 40*a*b^2*Sinh[4*c + 6*d*x] - 16*b^3*Sinh[4*c + 6*d*x]))/(d*(a + 2*b + a*Cosh[2*(c + d*x)])^3)

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fricas [B]  time = 0.47, size = 622, normalized size = 8.89 \[ -\frac {4 \, {\left ({\left (5 \, a^{3} + 15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (5 \, a^{3} + 15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + {\left (25 \, a^{3} + 75 \, a^{2} b + 80 \, a b^{2} + 32 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - {\left (45 \, a^{2} b + 80 \, a b^{2} + 32 \, b^{3} + 10 \, {\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + {\left (10 \, {\left (5 \, a^{3} + 15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (25 \, a^{3} + 75 \, a^{2} b + 80 \, a b^{2} + 32 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{3} + 15 \, a^{2} b + 14 \, a b^{2} + 4 \, b^{3}\right )} \cosh \left (d x + c\right ) - {\left (5 \, {\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 30 \, a^{2} b + 60 \, a b^{2} + 40 \, b^{3} + 3 \, {\left (45 \, a^{2} b + 80 \, a b^{2} + 32 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )\right )}}{5 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 3 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (7 \, d \cosh \left (d x + c\right )^{5} + 10 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*((5*a^3 + 15*a^2*b + 20*a*b^2 + 8*b^3)*cosh(d*x + c)^5 + 5*(5*a^3 + 15*a^2*b + 20*a*b^2 + 8*b^3)*cosh(d*x
+ c)*sinh(d*x + c)^4 - (15*a^2*b + 20*a*b^2 + 8*b^3)*sinh(d*x + c)^5 + (25*a^3 + 75*a^2*b + 80*a*b^2 + 32*b^3
)*cosh(d*x + c)^3 - (45*a^2*b + 80*a*b^2 + 32*b^3 + 10*(15*a^2*b + 20*a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x
+ c)^3 + (10*(5*a^3 + 15*a^2*b + 20*a*b^2 + 8*b^3)*cosh(d*x + c)^3 + 3*(25*a^3 + 75*a^2*b + 80*a*b^2 + 32*b^3
)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^3 + 15*a^2*b + 14*a*b^2 + 4*b^3)*cosh(d*x + c) - (5*(15*a^2*b + 20*
a*b^2 + 8*b^3)*cosh(d*x + c)^4 + 30*a^2*b + 60*a*b^2 + 40*b^3 + 3*(45*a^2*b + 80*a*b^2 + 32*b^3)*cosh(d*x + c)
^2)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 3*d*cosh(d*x +
c)^5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x +
c)^4 + d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + 3*(7*d*cosh(d
*x + c)^5 + 10*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 - 5*d*cosh(d*x + c) + (7*d*cosh(d*x + c)^6
+ 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c))

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giac [B]  time = 0.18, size = 249, normalized size = 3.56 \[ -\frac {2 \, {\left (\frac {5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} - \frac {15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 30 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 160 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 110 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 50 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b + 25 \, a b^{2} + 11 \, b^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}\right )}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)/(e^(2*d*x + 2*c) - 1) - (15*a^2*b*e^(8*d*x + 8*c) + 15*a*b^2*e^(8*d*x
+ 8*c) + 5*b^3*e^(8*d*x + 8*c) + 60*a^2*b*e^(6*d*x + 6*c) + 90*a*b^2*e^(6*d*x + 6*c) + 30*b^3*e^(6*d*x + 6*c)
+ 90*a^2*b*e^(4*d*x + 4*c) + 160*a*b^2*e^(4*d*x + 4*c) + 80*b^3*e^(4*d*x + 4*c) + 60*a^2*b*e^(2*d*x + 2*c) + 1
10*a*b^2*e^(2*d*x + 2*c) + 50*b^3*e^(2*d*x + 2*c) + 15*a^2*b + 25*a*b^2 + 11*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [B]  time = 0.59, size = 148, normalized size = 2.11 \[ \frac {-a^{3} \coth \left (d x +c \right )+3 a^{2} b \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}-2 \tanh \left (d x +c \right )\right )+3 a \,b^{2} \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}-4 \left (\frac {2}{3}+\frac {\mathrm {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )\right )+b^{3} \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{5}}-6 \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x)

[Out]

1/d*(-a^3*coth(d*x+c)+3*a^2*b*(-1/sinh(d*x+c)/cosh(d*x+c)-2*tanh(d*x+c))+3*a*b^2*(-1/sinh(d*x+c)/cosh(d*x+c)^3
-4*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c))+b^3*(-1/sinh(d*x+c)/cosh(d*x+c)^5-6*(8/15+1/5*sech(d*x+c)^4+4/15*sech(
d*x+c)^2)*tanh(d*x+c)))

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maxima [B]  time = 0.34, size = 358, normalized size = 5.11 \[ -\frac {32}{5} \, b^{3} {\left (\frac {4 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} - 4 \, e^{\left (-10 \, d x - 10 \, c\right )} - e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}} + \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} - 4 \, e^{\left (-10 \, d x - 10 \, c\right )} - e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}} + \frac {1}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} - 4 \, e^{\left (-10 \, d x - 10 \, c\right )} - e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}}\right )} - 16 \, a b^{2} {\left (\frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}} + \frac {1}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + \frac {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} + \frac {12 \, a^{2} b}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/5*b^3*(4*e^(-2*d*x - 2*c)/(d*(4*e^(-2*d*x - 2*c) + 5*e^(-4*d*x - 4*c) - 5*e^(-8*d*x - 8*c) - 4*e^(-10*d*x
- 10*c) - e^(-12*d*x - 12*c) + 1)) + 5*e^(-4*d*x - 4*c)/(d*(4*e^(-2*d*x - 2*c) + 5*e^(-4*d*x - 4*c) - 5*e^(-8*
d*x - 8*c) - 4*e^(-10*d*x - 10*c) - e^(-12*d*x - 12*c) + 1)) + 1/(d*(4*e^(-2*d*x - 2*c) + 5*e^(-4*d*x - 4*c) -
5*e^(-8*d*x - 8*c) - 4*e^(-10*d*x - 10*c) - e^(-12*d*x - 12*c) + 1))) - 16*a*b^2*(2*e^(-2*d*x - 2*c)/(d*(2*e^
(-2*d*x - 2*c) - 2*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) + 1)) + 1/(d*(2*e^(-2*d*x - 2*c) - 2*e^(-6*d*x - 6*c) -
e^(-8*d*x - 8*c) + 1))) + 2*a^3/(d*(e^(-2*d*x - 2*c) - 1)) + 12*a^2*b/(d*(e^(-4*d*x - 4*c) - 1))

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mupad [B]  time = 1.48, size = 644, normalized size = 9.20 \[ \frac {\frac {2\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+7\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {12\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+7\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b+7\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}+\frac {2\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d) + (2*exp(2*c + 2*d*x)*(3*a*b^2 + 3*a^2*b + b^3))/(5*d))/(2*exp(2*c + 2*
d*x) + exp(4*c + 4*d*x) + 1) + ((2*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d) + (2*exp(6*c + 6*d*x)*(3*a*b^2 + 3*a^2*b
+ b^3))/(5*d) + (6*exp(2*c + 2*d*x)*(7*a*b^2 + 3*a^2*b + 5*b^3))/(5*d) + (6*exp(4*c + 4*d*x)*(6*a*b^2 + 3*a^2
*b + 2*b^3))/(5*d))/(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) + ((
2*(3*a*b^2 + 3*a^2*b + b^3))/(5*d) + (2*exp(8*c + 8*d*x)*(3*a*b^2 + 3*a^2*b + b^3))/(5*d) + (8*exp(2*c + 2*d*x
)*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d) + (12*exp(4*c + 4*d*x)*(7*a*b^2 + 3*a^2*b + 5*b^3))/(5*d) + (8*exp(6*c +
6*d*x)*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) + 5
*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) + ((2*(7*a*b^2 + 3*a^2*b + 5*b^3))/(5*d) + (2*exp(4*c + 4*d*x)*(3*
a*b^2 + 3*a^2*b + b^3))/(5*d) + (4*exp(2*c + 2*d*x)*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d))/(3*exp(2*c + 2*d*x) +
3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - (2*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(d*(exp(2*c + 2*d*x) - 1)) +
(2*(3*a*b^2 + 3*a^2*b + b^3))/(5*d*(exp(2*c + 2*d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3} \operatorname {csch}^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**2*(a+b*sech(d*x+c)**2)**3,x)

[Out]

Integral((a + b*sech(c + d*x)**2)**3*csch(c + d*x)**2, x)

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