3.161 \(\int \coth (c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=72 \[ \frac {a^3 \log (\tanh (c+d x))}{d}-\frac {b^2 (3 a+b) \tanh ^2(c+d x)}{2 d}+\frac {(a+b)^3 \log (\cosh (c+d x))}{d}-\frac {b^3 \tanh ^4(c+d x)}{4 d} \]

[Out]

(a+b)^3*ln(cosh(d*x+c))/d+a^3*ln(tanh(d*x+c))/d-1/2*b^2*(3*a+b)*tanh(d*x+c)^2/d-1/4*b^3*tanh(d*x+c)^4/d

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Rubi [A]  time = 0.10, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3670, 446, 72} \[ \frac {a^3 \log (\tanh (c+d x))}{d}-\frac {b^2 (3 a+b) \tanh ^2(c+d x)}{2 d}+\frac {(a+b)^3 \log (\cosh (c+d x))}{d}-\frac {b^3 \tanh ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^3*Log[Cosh[c + d*x]])/d + (a^3*Log[Tanh[c + d*x]])/d - (b^2*(3*a + b)*Tanh[c + d*x]^2)/(2*d) - (b^3*T
anh[c + d*x]^4)/(4*d)

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.55, size = 67, normalized size = 0.93 \[ \frac {2 a^3 \log (\tanh (c+d x))-b^2 (3 a+b) \tanh ^2(c+d x)+2 (a+b)^3 \log (\cosh (c+d x))-\frac {1}{2} b^3 \tanh ^4(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b)^3*Log[Cosh[c + d*x]] + 2*a^3*Log[Tanh[c + d*x]] - b^2*(3*a + b)*Tanh[c + d*x]^2 - (b^3*Tanh[c + d*x
]^4)/2)/(2*d)

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fricas [B]  time = 0.47, size = 2381, normalized size = 33.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^8 + 8*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)*si
nh(d*x + c)^7 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*sinh(d*x + c)^8 - 2*(3*a*b^2 + 2*b^3 - 2*(a^3 + 3*a^2*b +
3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^6 + 2*(14*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^2 - 3*a*b^2 - 2*
b^3 + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*sinh(d*x + c)^6 + 4*(14*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(
d*x + c)^3 - 3*(3*a*b^2 + 2*b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^5 - 2*(6
*a*b^2 + 2*b^3 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 + 2*(35*(a^3 + 3*a^2*b + 3*a*b^2 + b^3
)*d*x*cosh(d*x + c)^4 - 6*a*b^2 - 2*b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 15*(3*a*b^2 + 2*b^3 - 2*(a^3
+ 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*c
osh(d*x + c)^5 - 5*(3*a*b^2 + 2*b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^3 - (6*a*b^2 + 2*b^
3 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*
x - 2*(3*a*b^2 + 2*b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2 + 2*(14*(a^3 + 3*a^2*b + 3*a*b
^2 + b^3)*d*x*cosh(d*x + c)^6 - 15*(3*a*b^2 + 2*b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 -
3*a*b^2 - 2*b^3 + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 6*(6*a*b^2 + 2*b^3 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b
^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - ((3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^8 + 8*(3*a^2*b + 3*a*b^2
+ b^3)*cosh(d*x + c)*sinh(d*x + c)^7 + (3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^8 + 4*(3*a^2*b + 3*a*b^2 + b^3)
*cosh(d*x + c)^6 + 4*(3*a^2*b + 3*a*b^2 + b^3 + 7*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 +
8*(7*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + 3*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^5 +
6*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 2*(35*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 9*a^2*b + 9*a
*b^2 + 3*b^3 + 30*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(3*a^2*b + 3*a*b^2 + b^3)*
cosh(d*x + c)^5 + 10*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + 3*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c))*si
nh(d*x + c)^3 + 3*a^2*b + 3*a*b^2 + b^3 + 4*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2 + 4*(7*(3*a^2*b + 3*a*b^
2 + b^3)*cosh(d*x + c)^6 + 15*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 3*a^2*b + 3*a*b^2 + b^3 + 9*(3*a^2*b
+ 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 8*((3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^7 + 3*(3*a^2*b
+ 3*a*b^2 + b^3)*cosh(d*x + c)^5 + 3*(3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (3*a^2*b + 3*a*b^2 + b^3)*co
sh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - (a^3*cosh(d*x + c)^8 + 8*a^
3*cosh(d*x + c)*sinh(d*x + c)^7 + a^3*sinh(d*x + c)^8 + 4*a^3*cosh(d*x + c)^6 + 6*a^3*cosh(d*x + c)^4 + 4*(7*a
^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^6 + 8*(7*a^3*cosh(d*x + c)^3 + 3*a^3*cosh(d*x + c))*sinh(d*x + c)^5 +
4*a^3*cosh(d*x + c)^2 + 2*(35*a^3*cosh(d*x + c)^4 + 30*a^3*cosh(d*x + c)^2 + 3*a^3)*sinh(d*x + c)^4 + 8*(7*a^3
*cosh(d*x + c)^5 + 10*a^3*cosh(d*x + c)^3 + 3*a^3*cosh(d*x + c))*sinh(d*x + c)^3 + a^3 + 4*(7*a^3*cosh(d*x + c
)^6 + 15*a^3*cosh(d*x + c)^4 + 9*a^3*cosh(d*x + c)^2 + a^3)*sinh(d*x + c)^2 + 8*(a^3*cosh(d*x + c)^7 + 3*a^3*c
osh(d*x + c)^5 + 3*a^3*cosh(d*x + c)^3 + a^3*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c)
- sinh(d*x + c))) + 4*(2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^7 - 3*(3*a*b^2 + 2*b^3 - 2*(a^3 + 3
*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^5 - 2*(6*a*b^2 + 2*b^3 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cos
h(d*x + c)^3 - (3*a*b^2 + 2*b^3 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh
(d*x + c)^8 + 8*d*cosh(d*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 + 4*d*cosh(d*x + c)^6 + 4*(7*d*cosh(d*x +
c)^2 + d)*sinh(d*x + c)^6 + 8*(7*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4
+ 2*(35*d*cosh(d*x + c)^4 + 30*d*cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d
*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^3 + 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 + 15*d*cosh(d*x
+ c)^4 + 9*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 8*(d*cosh(d*x + c)^7 + 3*d*cosh(d*x + c)^5 + 3*d*cosh(d*x
+ c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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giac [B]  time = 0.33, size = 267, normalized size = 3.71 \[ \frac {2 \, a^{3} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right ) + 2 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right ) - \frac {9 \, a^{2} b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 9 \, a b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 3 \, b^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 36 \, a^{2} b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 12 \, a b^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 4 \, b^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 36 \, a^{2} b - 12 \, a b^{2} - 4 \, b^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right )}^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a^3*log(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) - 2) + 2*(3*a^2*b + 3*a*b^2 + b^3)*log(e^(2*d*x + 2*c) + e^(
-2*d*x - 2*c) + 2) - (9*a^2*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 9*a*b^2*(e^(2*d*x + 2*c) + e^(-2*d*x -
2*c))^2 + 3*b^3*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 36*a^2*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 12*a*
b^2*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 4*b^3*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 36*a^2*b - 12*a*b^2 -
4*b^3)/(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2)^2)/d

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maple [A]  time = 0.29, size = 111, normalized size = 1.54 \[ \frac {a^{3} \ln \left (\sinh \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \ln \left (\cosh \left (d x +c \right )\right )}{d}+\frac {3 a \,b^{2} \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {3 \left (\tanh ^{2}\left (d x +c \right )\right ) a \,b^{2}}{2 d}+\frac {b^{3} \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {b^{3} \left (\tanh ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b^{3} \left (\tanh ^{4}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^3*ln(sinh(d*x+c))+3/d*a^2*b*ln(cosh(d*x+c))+3/d*a*b^2*ln(cosh(d*x+c))-3/2/d*tanh(d*x+c)^2*a*b^2+1/d*b^3*
ln(cosh(d*x+c))-1/2*b^3*tanh(d*x+c)^2/d-1/4*b^3*tanh(d*x+c)^4/d

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maxima [B]  time = 0.44, size = 214, normalized size = 2.97 \[ b^{3} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + 3 \, a b^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {3 \, a^{2} b \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}{d} + \frac {a^{3} \log \left (\sinh \left (d x + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 4*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/(d*(4*
e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) + 3*a*b^2*(x + c/d + log(
e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) + 3*a^2*b*log(e^
(d*x + c) + e^(-d*x - c))/d + a^3*log(sinh(d*x + c))/d

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mupad [B]  time = 0.40, size = 380, normalized size = 5.28 \[ \frac {\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )\,\left (a^3\,d+d\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )\right )}{2\,d^2}-x\,{\left (a+b\right )}^3+\frac {2\,\left (2\,b^3+3\,a\,b^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {8\,b^3}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (b^3\,\sqrt {-d^2}-a^3\,\sqrt {-d^2}+3\,a\,b^2\,\sqrt {-d^2}+3\,a^2\,b\,\sqrt {-d^2}\right )}{d\,\sqrt {a^6-6\,a^5\,b+3\,a^4\,b^2+16\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}}\right )\,\sqrt {a^6-6\,a^5\,b+3\,a^4\,b^2+16\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+b^6}}{\sqrt {-d^2}}-\frac {2\,\left (4\,b^3+3\,a\,b^2\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {4\,b^3}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(exp(4*c + 4*d*x) - 1)*(a^3*d + d*(3*a*b^2 + 3*a^2*b + b^3)))/(2*d^2) - x*(a + b)^3 + (2*(3*a*b^2 + 2*b^3)
)/(d*(exp(2*c + 2*d*x) + 1)) + (8*b^3)/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) +
(atan((exp(2*c)*exp(2*d*x)*(b^3*(-d^2)^(1/2) - a^3*(-d^2)^(1/2) + 3*a*b^2*(-d^2)^(1/2) + 3*a^2*b*(-d^2)^(1/2))
)/(d*(6*a*b^5 - 6*a^5*b + a^6 + b^6 + 15*a^2*b^4 + 16*a^3*b^3 + 3*a^4*b^2)^(1/2)))*(6*a*b^5 - 6*a^5*b + a^6 +
b^6 + 15*a^2*b^4 + 16*a^3*b^3 + 3*a^4*b^2)^(1/2))/(-d^2)^(1/2) - (2*(3*a*b^2 + 4*b^3))/(d*(2*exp(2*c + 2*d*x)
+ exp(4*c + 4*d*x) + 1)) - (4*b^3)/(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*coth(c + d*x), x)

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