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

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 59, 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 = {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 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 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 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 ^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*}

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Mathematica [A]  time = 1.28, size = 82, normalized size = 1.39 \[ \frac {\tanh (c+d x) \left (-a^3 \coth ^4(c+d x)-3 a^2 (a+3 b) \coth ^2(c+d x)+3 (a+b)^3 \sqrt {\coth ^2(c+d x)} \tanh ^{-1}\left (\sqrt {\coth ^2(c+d x)}\right )-3 b^3\right )}{3 d} \]

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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fricas [B]  time = 0.44, size = 341, normalized size = 5.78 \[ -\frac {{\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 4 \, {\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} - 9 \, a^{2} b + 9 \, b^{3} + 4 \, {\left (a^{3} - 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} - 6 \, b^{3} + 3 \, {\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 4 \, {\left ({\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - {\left (4 \, a^{3} + 9 \, a^{2} b + 3 \, b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{12 \, {\left (d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]

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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giac [B]  time = 0.52, size = 135, normalized size = 2.29 \[ \frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} + \frac {6 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac {2 \, {\left (6 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} + 9 \, a^{2} b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

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]

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

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maple [A]  time = 0.25, size = 80, normalized size = 1.36 \[ \frac {a^{3} \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}\right )+3 a^{2} b \left (d x +c -\coth \left (d x +c \right )\right )+3 a \,b^{2} \left (d x +c \right )+b^{3} \left (d x +c -\tanh \left (d x +c \right )\right )}{d} \]

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 = 0.34, size = 147, normalized size = 2.49 \[ \frac {1}{3} \, a^{3} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + b^{3} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + 3 \, a b^{2} x \]

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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mupad [B]  time = 1.29, size = 219, normalized size = 3.71 \[ x\,{\left (a+b\right )}^3+\frac {\frac {2\,a^2\,b}{d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3+3\,b\,a^2\right )}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,\left (2\,a^3+3\,b\,a^2\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a^3+3\,b\,a^2\right )}{3\,d}-\frac {4\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}+\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,\left (2\,a^3+3\,b\,a^2\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \coth ^{4}{\left (c + d x \right )}\, dx \]

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]

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

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