### 3.164 $$\int \coth ^4(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=59 $-\frac{a^2 (a+3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}+x (a+b)^3-\frac{b^3 \tanh (c+d x)}{d}$

[Out]

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

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Rubi [A]  time = 0.0812287, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3670, 461, 207} $-\frac{a^2 (a+3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}+x (a+b)^3-\frac{b^3 \tanh (c+d x)}{d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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]))

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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])

Rubi steps

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

Mathematica [A]  time = 1.18732, size = 82, normalized size = 1.39 $\frac{\tanh (c+d x) \left (-3 a^2 (a+3 b) \coth ^2(c+d x)-a^3 \coth ^4(c+d x)+3 (a+b)^3 \sqrt{\coth ^2(c+d x)} \tanh ^{-1}\left (\sqrt{\coth ^2(c+d x)}\right )-3 b^3\right )}{3 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-3*b^3 - 3*a^2*(a + 3*b)*Coth[c + d*x]^2 - a^3*Coth[c + d*x]^4 + 3*(a + b)^3*ArcTanh[Sqrt[Coth[c + d*x]^2]]*
Sqrt[Coth[c + d*x]^2])*Tanh[c + d*x])/(3*d)

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Maple [A]  time = 0.05, size = 80, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( dx+c-{\rm coth} \left (dx+c\right )-{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}}{3}} \right ) +3\,{a}^{2}b \left ( dx+c-{\rm coth} \left (dx+c\right ) \right ) +3\,a{b}^{2} \left ( dx+c \right ) +{b}^{3} \left ( dx+c-\tanh \left ( dx+c \right ) \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^4*(a+b*tanh(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)+3*a^2*b*(d*x+c-coth(d*x+c))+3*a*b^2*(d*x+c)+b^3*(d*x+c-tanh(d*x
+c)))

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Maxima [B]  time = 1.07444, size = 198, normalized size = 3.36 \begin{align*} \frac{1}{3} \, a^{3}{\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 )} + b^{3}{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + 3 \, a^{2} b{\left (x + \frac{c}{d} + \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + 3 \, a b^{2} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.00627, size = 811, normalized size = 13.75 \begin{align*} -\frac{{\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 4 \,{\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3} + 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} - 9 \, a^{2} b + 9 \, b^{3} + 4 \,{\left (a^{3} - 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (2 \, a^{3} - 6 \, b^{3} + 3 \,{\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 4 \,{\left ({\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3} + 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} -{\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3} + 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{12 \,{\left (d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/12*((4*a^3 + 9*a^2*b + 3*b^3)*cosh(d*x + c)^4 - 4*(4*a^3 + 9*a^2*b + 3*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 + (4*a^3 + 9*a^2*b + 3*b^3)*sinh(d*x + c)^4 - 9*a^2*b + 9*b^3 + 4*(a^3
- 3*b^3)*cosh(d*x + c)^2 + 2*(2*a^3 - 6*b^3 + 3*(4*a^3 + 9*a^2*b + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 4
*((4*a^3 + 9*a^2*b + 3*b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^3 - (4*a^3 + 9*a^2*b + 3*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))/(d*cosh(d*x + c)*sinh(d*x + c)^3 + (d*
cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**4*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.4275, size = 186, normalized size = 3.15 \begin{align*} \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (d x + c\right )}}{d} + \frac{2 \, b^{3}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} - \frac{2 \,{\left (6 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} + 9 \, a^{2} b\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

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