3.166 \(\int \coth ^6(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=74 \[ -\frac {a^3 \coth ^5(c+d x)}{5 d}-\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}+x (a+b)^3 \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 74, 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} \[ -\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}-\frac {a^3 \coth ^5(c+d x)}{5 d}+x (a+b)^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 1.78, size = 100, normalized size = 1.35 \[ \frac {(a+b)^3 \tanh ^{-1}\left (\sqrt {\tanh ^2(c+d x)}\right ) \tanh (c+d x)}{d \sqrt {\tanh ^2(c+d x)}}-\frac {a \coth (c+d x) \left (15 \left (a^2+3 a b+3 b^2\right )+3 a^2 \coth ^4(c+d x)+5 a (a+3 b) \coth ^2(c+d x)\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(a*Coth[c + d*x]*(15*(a^2 + 3*a*b + 3*b^2) + 5*a*(a + 3*b)*Coth[c + d*x]^2 + 3*a^2*Coth[c + d*x]^4))/d +
((a + b)^3*ArcTanh[Sqrt[Tanh[c + d*x]^2]]*Tanh[c + d*x])/(d*Sqrt[Tanh[c + d*x]^2])

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fricas [B]  time = 0.44, size = 557, normalized size = 7.53 \[ -\frac {{\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{3} + 24 \, a^{2} b + 27 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 2 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{3} + 24 \, a^{2} b + 27 \, a b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{4} + 46 \, a^{3} + 120 \, a^{2} b + 90 \, a b^{2} + 30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 3 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\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*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/15*((23*a^3 + 60*a^2*b + 45*a*b^2)*cosh(d*x + c)^5 + 5*(23*a^3 + 60*a^2*b + 45*a*b^2)*cosh(d*x + c)*sinh(d*
x + c)^4 - (23*a^3 + 60*a^2*b + 45*a*b^2 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*sinh(d*x + c)^5 - 5*(5*a^3
+ 24*a^2*b + 27*a*b^2)*cosh(d*x + c)^3 + 5*(23*a^3 + 60*a^2*b + 45*a*b^2 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*
d*x - 2*(23*a^3 + 60*a^2*b + 45*a*b^2 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)
^3 + 5*(2*(23*a^3 + 60*a^2*b + 45*a*b^2)*cosh(d*x + c)^3 - 3*(5*a^3 + 24*a^2*b + 27*a*b^2)*cosh(d*x + c))*sinh
(d*x + c)^2 + 10*(5*a^3 + 6*a^2*b + 9*a*b^2)*cosh(d*x + c) - 5*((23*a^3 + 60*a^2*b + 45*a*b^2 + 15*(a^3 + 3*a^
2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 + 46*a^3 + 120*a^2*b + 90*a*b^2 + 30*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)
*d*x - 3*(23*a^3 + 60*a^2*b + 45*a*b^2 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*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 = 0.56, size = 241, normalized size = 3.26 \[ \frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} - 270 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 180 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 210 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 180 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c) - 2*(45*a^3*e^(8*d*x + 8*c) + 90*a^2*b*e^(8*d*x + 8*c) + 45
*a*b^2*e^(8*d*x + 8*c) - 90*a^3*e^(6*d*x + 6*c) - 270*a^2*b*e^(6*d*x + 6*c) - 180*a*b^2*e^(6*d*x + 6*c) + 140*
a^3*e^(4*d*x + 4*c) + 330*a^2*b*e^(4*d*x + 4*c) + 270*a*b^2*e^(4*d*x + 4*c) - 70*a^3*e^(2*d*x + 2*c) - 210*a^2
*b*e^(2*d*x + 2*c) - 180*a*b^2*e^(2*d*x + 2*c) + 23*a^3 + 60*a^2*b + 45*a*b^2)/(e^(2*d*x + 2*c) - 1)^5)/d

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maple [A]  time = 0.28, size = 100, normalized size = 1.35 \[ \frac {a^{3} \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\coth ^{5}\left (d x +c \right )\right )}{5}\right )+3 a^{2} b \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}\right )+3 a \,b^{2} \left (d x +c -\coth \left (d x +c \right )\right )+b^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^6*(a+b*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)^3-1/5*coth(d*x+c)^5)+3*a^2*b*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3)
+3*a*b^2*(d*x+c-coth(d*x+c))+b^3*(d*x+c))

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maxima [B]  time = 0.35, size = 239, normalized size = 3.23 \[ \frac {1}{15} \, a^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + a^{2} b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 3 \, a b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + b^{3} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.30, size = 568, normalized size = 7.68 \[ x\,{\left (a+b\right )}^3-\frac {\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {2\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{15\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}-\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}+\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \coth ^{6}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**6*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*coth(c + d*x)**6, x)

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