3.134 \(\int \coth ^6(c+d x) (a+b \text {sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=69 \[ -\frac {\left (a^3+b^3\right ) \coth (c+d x)}{d}+a^3 x-\frac {(a+b)^3 \coth ^5(c+d x)}{5 d}-\frac {(a-2 b) (a+b)^2 \coth ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 69, 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} \[ -\frac {\left (a^3+b^3\right ) \coth (c+d x)}{d}+a^3 x-\frac {(a+b)^3 \coth ^5(c+d x)}{5 d}-\frac {(a-2 b) (a+b)^2 \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^6*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

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

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Mathematica [B]  time = 1.09, size = 303, normalized size = 4.39 \[ \frac {\text {csch}(c) \text {csch}^5(c+d x) \left (180 a^3 \sinh (2 c+d x)-140 a^3 \sinh (2 c+3 d x)-90 a^3 \sinh (4 c+3 d x)+46 a^3 \sinh (4 c+5 d x)+150 a^3 d x \cosh (2 c+d x)+75 a^3 d x \cosh (2 c+3 d x)-75 a^3 d x \cosh (4 c+3 d x)-15 a^3 d x \cosh (4 c+5 d x)+15 a^3 d x \cosh (6 c+5 d x)+280 a^3 \sinh (d x)-150 a^3 d x \cosh (d x)-90 a^2 b \sinh (4 c+3 d x)+18 a^2 b \sinh (4 c+5 d x)+180 a^2 b \sinh (d x)-180 a b^2 \sinh (2 c+d x)+60 a b^2 \sinh (2 c+3 d x)-12 a b^2 \sinh (4 c+5 d x)+60 a b^2 \sinh (d x)-80 b^3 \sinh (2 c+3 d x)+16 b^3 \sinh (4 c+5 d x)+160 b^3 \sinh (d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^6*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

(Csch[c]*Csch[c + d*x]^5*(-150*a^3*d*x*Cosh[d*x] + 150*a^3*d*x*Cosh[2*c + d*x] + 75*a^3*d*x*Cosh[2*c + 3*d*x]
- 75*a^3*d*x*Cosh[4*c + 3*d*x] - 15*a^3*d*x*Cosh[4*c + 5*d*x] + 15*a^3*d*x*Cosh[6*c + 5*d*x] + 280*a^3*Sinh[d*
x] + 180*a^2*b*Sinh[d*x] + 60*a*b^2*Sinh[d*x] + 160*b^3*Sinh[d*x] + 180*a^3*Sinh[2*c + d*x] - 180*a*b^2*Sinh[2
*c + d*x] - 140*a^3*Sinh[2*c + 3*d*x] + 60*a*b^2*Sinh[2*c + 3*d*x] - 80*b^3*Sinh[2*c + 3*d*x] - 90*a^3*Sinh[4*
c + 3*d*x] - 90*a^2*b*Sinh[4*c + 3*d*x] + 46*a^3*Sinh[4*c + 5*d*x] + 18*a^2*b*Sinh[4*c + 5*d*x] - 12*a*b^2*Sin
h[4*c + 5*d*x] + 16*b^3*Sinh[4*c + 5*d*x]))/(480*d)

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fricas [B]  time = 0.41, size = 521, normalized size = 7.55 \[ -\frac {{\left (23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{3} d x + 23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{3} - 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (15 \, a^{3} d x + 23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3} - 2 \, {\left (15 \, a^{3} d x + 23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{3} - 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{3} + 9 \, a^{2} b + 12 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right ) - 5 \, {\left (30 \, a^{3} d x + {\left (15 \, a^{3} d x + 23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 46 \, a^{3} + 18 \, a^{2} b - 12 \, a b^{2} + 16 \, b^{3} - 3 \, {\left (15 \, a^{3} d x + 23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/15*((23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*cosh(d*x + c)^5 + 5*(23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*cosh(d*x
+ c)*sinh(d*x + c)^4 - (15*a^3*d*x + 23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*sinh(d*x + c)^5 - 5*(5*a^3 - 9*a^2*b
- 6*a*b^2 + 8*b^3)*cosh(d*x + c)^3 + 5*(15*a^3*d*x + 23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3 - 2*(15*a^3*d*x + 23*a
^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 5*(2*(23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*c
osh(d*x + c)^3 - 3*(5*a^3 - 9*a^2*b - 6*a*b^2 + 8*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^3 + 9*a^2*b +
12*a*b^2 + 8*b^3)*cosh(d*x + c) - 5*(30*a^3*d*x + (15*a^3*d*x + 23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*cosh(d*x +
c)^4 + 46*a^3 + 18*a^2*b - 12*a*b^2 + 16*b^3 - 3*(15*a^3*d*x + 23*a^3 + 9*a^2*b - 6*a*b^2 + 8*b^3)*cosh(d*x +
c)^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 = 0.43, size = 210, normalized size = 3.04 \[ \frac {15 \, a^{3} d x - \frac {2 \, {\left (45 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} + 45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 40 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{3} + 9 \, a^{2} b - 6 \, a b^{2} + 8 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/15*(15*a^3*d*x - 2*(45*a^3*e^(8*d*x + 8*c) + 45*a^2*b*e^(8*d*x + 8*c) - 90*a^3*e^(6*d*x + 6*c) + 90*a*b^2*e^
(6*d*x + 6*c) + 140*a^3*e^(4*d*x + 4*c) + 90*a^2*b*e^(4*d*x + 4*c) + 30*a*b^2*e^(4*d*x + 4*c) + 80*b^3*e^(4*d*
x + 4*c) - 70*a^3*e^(2*d*x + 2*c) + 30*a*b^2*e^(2*d*x + 2*c) - 40*b^3*e^(2*d*x + 2*c) + 23*a^3 + 9*a^2*b - 6*a
*b^2 + 8*b^3)/(e^(2*d*x + 2*c) - 1)^5)/d

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maple [B]  time = 0.42, size = 199, normalized size = 2.88 \[ \frac {a^{3} \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\coth ^{5}\left (d x +c \right )\right )}{5}\right )+3 a^{2} b \left (-\frac {\cosh ^{3}\left (d x +c \right )}{2 \sinh \left (d x +c \right )^{5}}+\frac {3 \cosh \left (d x +c \right )}{8 \sinh \left (d x +c \right )^{5}}+\frac {3 \left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{8}\right )+3 a \,b^{2} \left (-\frac {\cosh \left (d x +c \right )}{4 \sinh \left (d x +c \right )^{5}}-\frac {\left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{4}\right )+b^{3} \left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3-1/5*coth(d*x+c)^5)+3*a^2*b*(-1/2/sinh(d*x+c)^5*cosh(d*x+c)^3+3/8
/sinh(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))+3*a*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))+b^3*(-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 = 0.36, size = 826, normalized size = 11.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^6*(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/15*a^3*(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))) + 4/5*a*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))) - 16/15*b^3*(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)) - 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)) - 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))) + 6/5*a^2*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 = 1.60, size = 547, normalized size = 7.93 \[ a^3\,x-\frac {\frac {6\,\left (a^3+b\,a^2\right )}{5\,d}+\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a^3+b\,a^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^3+9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^3+b\,a^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^3+b\,a^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^3+9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (5\,a^3+9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{15\,d}+\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^3+b\,a^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {6\,\left (a^3+b\,a^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^6*(a + b/cosh(c + d*x)^2)^3,x)

[Out]

a^3*x - ((6*(a^2*b + a^3))/(5*d) + (24*exp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d) + (24*exp(6*c + 6*d*x)*(a*b^2 +
a^2*b))/(5*d) + (6*exp(8*c + 8*d*x)*(a^2*b + a^3))/(5*d) + (4*exp(4*c + 4*d*x)*(12*a*b^2 + 9*a^2*b + 5*a^3 +
8*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) - ((6*(a*b^2 + a^2*b))/(5*d) + (6*exp(2*c + 2*d*x)*(a^2*b + a^3))/(5*d))/(exp(4*c + 4*d*x) - 2
*exp(2*c + 2*d*x) + 1) - ((6*(a*b^2 + a^2*b))/(5*d) + (18*exp(4*c + 4*d*x)*(a*b^2 + a^2*b))/(5*d) + (6*exp(6*c
+ 6*d*x)*(a^2*b + a^3))/(5*d) + (2*exp(2*c + 2*d*x)*(12*a*b^2 + 9*a^2*b + 5*a^3 + 8*b^3))/(5*d))/(6*exp(4*c +
4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(12*a*b^2 + 9*a^2*b + 5*a^3 +
8*b^3))/(15*d) + (12*exp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d) + (6*exp(4*c + 4*d*x)*(a^2*b + a^3))/(5*d))/(3*ex
p(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (6*(a^2*b + a^3))/(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)**3,x)

[Out]

Timed out

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