3.122 \(\int \coth ^6(c+d x) (a+b \text {sech}^2(c+d x))^2 \, dx\)

Optimal. Leaf size=64 \[ -\frac {\left (a^2-b^2\right ) \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth (c+d x)}{d}+a^2 x-\frac {(a+b)^2 \coth ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4141, 1802, 207} \[ -\frac {\left (a^2-b^2\right ) \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth (c+d x)}{d}+a^2 x-\frac {(a+b)^2 \coth ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^6*(a + b*Sech[c + d*x]^2)^2,x]

[Out]

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

Rule 207

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

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

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

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Mathematica [B]  time = 1.10, size = 256, normalized size = 4.00 \[ \frac {\text {csch}(c) \text {csch}^5(c+d x) \left (180 a^2 \sinh (2 c+d x)-140 a^2 \sinh (2 c+3 d x)-90 a^2 \sinh (4 c+3 d x)+46 a^2 \sinh (4 c+5 d x)+150 a^2 d x \cosh (2 c+d x)+75 a^2 d x \cosh (2 c+3 d x)-75 a^2 d x \cosh (4 c+3 d x)-15 a^2 d x \cosh (4 c+5 d x)+15 a^2 d x \cosh (6 c+5 d x)+280 a^2 \sinh (d x)-150 a^2 d x \cosh (d x)-60 a b \sinh (4 c+3 d x)+12 a b \sinh (4 c+5 d x)+120 a b \sinh (d x)-60 b^2 \sinh (2 c+d x)+20 b^2 \sinh (2 c+3 d x)-4 b^2 \sinh (4 c+5 d x)+20 b^2 \sinh (d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^6*(a + b*Sech[c + d*x]^2)^2,x]

[Out]

(Csch[c]*Csch[c + d*x]^5*(-150*a^2*d*x*Cosh[d*x] + 150*a^2*d*x*Cosh[2*c + d*x] + 75*a^2*d*x*Cosh[2*c + 3*d*x]
- 75*a^2*d*x*Cosh[4*c + 3*d*x] - 15*a^2*d*x*Cosh[4*c + 5*d*x] + 15*a^2*d*x*Cosh[6*c + 5*d*x] + 280*a^2*Sinh[d*
x] + 120*a*b*Sinh[d*x] + 20*b^2*Sinh[d*x] + 180*a^2*Sinh[2*c + d*x] - 60*b^2*Sinh[2*c + d*x] - 140*a^2*Sinh[2*
c + 3*d*x] + 20*b^2*Sinh[2*c + 3*d*x] - 90*a^2*Sinh[4*c + 3*d*x] - 60*a*b*Sinh[4*c + 3*d*x] + 46*a^2*Sinh[4*c
+ 5*d*x] + 12*a*b*Sinh[4*c + 5*d*x] - 4*b^2*Sinh[4*c + 5*d*x]))/(480*d)

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fricas [B]  time = 0.42, size = 425, normalized size = 6.64 \[ -\frac {{\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{2} - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (15 \, a^{2} d x - 2 \, {\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{2} - 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{2} + 6 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 30 \, a^{2} d x - 3 \, {\left (15 \, a^{2} d x + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 46 \, a^{2} + 12 \, a b - 4 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-1/15*((23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^5 + 5*(23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 - (
15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*sinh(d*x + c)^5 - 5*(5*a^2 - 6*a*b - 2*b^2)*cosh(d*x + c)^3 + 5*(15*a^2*d
*x - 2*(15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^2 + 23*a^2 + 6*a*b - 2*b^2)*sinh(d*x + c)^3 + 5*(2*
(23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^3 - 3*(5*a^2 - 6*a*b - 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^
2 + 6*a*b + 4*b^2)*cosh(d*x + c) - 5*((15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^4 + 30*a^2*d*x - 3*(
15*a^2*d*x + 23*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^2 + 46*a^2 + 12*a*b - 4*b^2)*sinh(d*x + c))/(d*sinh(d*x + c
)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x
+ c))

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giac [B]  time = 0.37, size = 167, normalized size = 2.61 \[ \frac {15 \, a^{2} d x - \frac {2 \, {\left (45 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 30 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 30 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/15*(15*a^2*d*x - 2*(45*a^2*e^(8*d*x + 8*c) + 30*a*b*e^(8*d*x + 8*c) - 90*a^2*e^(6*d*x + 6*c) + 30*b^2*e^(6*d
*x + 6*c) + 140*a^2*e^(4*d*x + 4*c) + 60*a*b*e^(4*d*x + 4*c) + 10*b^2*e^(4*d*x + 4*c) - 70*a^2*e^(2*d*x + 2*c)
+ 10*b^2*e^(2*d*x + 2*c) + 23*a^2 + 6*a*b - 2*b^2)/(e^(2*d*x + 2*c) - 1)^5)/d

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maple [B]  time = 0.44, size = 163, normalized size = 2.55 \[ \frac {a^{2} \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\coth ^{5}\left (d x +c \right )\right )}{5}\right )+2 a b \left (-\frac {\cosh ^{3}\left (d x +c \right )}{2 \sinh \left (d x +c \right )^{5}}+\frac {3 \cosh \left (d x +c \right )}{8 \sinh \left (d x +c \right )^{5}}+\frac {3 \left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{8}\right )+b^{2} \left (-\frac {\cosh \left (d x +c \right )}{4 \sinh \left (d x +c \right )^{5}}-\frac {\left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^2,x)

[Out]

1/d*(a^2*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3-1/5*coth(d*x+c)^5)+2*a*b*(-1/2/sinh(d*x+c)^5*cosh(d*x+c)^3+3/8/s
inh(d*x+c)^5*cosh(d*x+c)+3/8*(-8/15-1/5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c))+b^2*(-1/4/sinh(d*x+c)^5
*cosh(d*x+c)-1/4*(-8/15-1/5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c)))

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maxima [B]  time = 0.54, size = 613, normalized size = 9.58 \[ \frac {1}{15} \, a^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {4}{15} \, b^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} + \frac {5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} + \frac {15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {4}{5} \, a b {\left (\frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} + \frac {5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/15*a^2*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) - 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) - 45*e^(-8*d*x -
8*c) - 23)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d
*x - 10*c) - 1))) + 4/15*b^2*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x -
6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) + 5*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d
*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) + 15*e^(-6*d*x - 6*c)/(d*(5*e^
(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 1
/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c)
- 1))) + 4/5*a*b*(10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e
^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) + 5*e^(-8*d*x - 8*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) +
10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) + 1/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x
- 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)))

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mupad [B]  time = 1.44, size = 511, normalized size = 7.98 \[ a^2\,x-\frac {\frac {2\,\left (5\,a^2+6\,a\,b+4\,b^2\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^2+6\,a\,b+4\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2+2\,b\,a\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2+2\,b\,a\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^2+6\,a\,b+4\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {2\,\left (3\,a^2+2\,b\,a\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^6*(a + b/cosh(c + d*x)^2)^2,x)

[Out]

a^2*x - ((2*(6*a*b + 5*a^2 + 4*b^2))/(15*d) + (4*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d) + (2*exp(4*c + 4*d*x)*(
2*a*b + 3*a^2))/(5*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*b + b^2))/(
5*d) + (2*exp(2*c + 2*d*x)*(2*a*b + 3*a^2))/(5*d))/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1) - ((2*(2*a*b +
b^2))/(5*d) + (6*exp(4*c + 4*d*x)*(2*a*b + b^2))/(5*d) + (2*exp(6*c + 6*d*x)*(2*a*b + 3*a^2))/(5*d) + (2*exp(2
*c + 2*d*x)*(6*a*b + 5*a^2 + 4*b^2))/(5*d))/(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + ex
p(8*c + 8*d*x) + 1) - ((2*(2*a*b + 3*a^2))/(5*d) + (8*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d) + (8*exp(6*c + 6*d
*x)*(2*a*b + b^2))/(5*d) + (2*exp(8*c + 8*d*x)*(2*a*b + 3*a^2))/(5*d) + (4*exp(4*c + 4*d*x)*(6*a*b + 5*a^2 + 4
*b^2))/(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*(2*a*b + 3*a^2))/(5*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(coth(d*x+c)**6*(a+b*sech(d*x+c)**2)**2,x)

[Out]

Timed out

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