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

Optimal. Leaf size=63 \[ -\frac {a^2 \coth ^5(c+d x)}{5 d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {(a+b)^2 \coth (c+d x)}{d}+x (a+b)^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 63, 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} \[ -\frac {a^2 \coth ^5(c+d x)}{5 d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {(a+b)^2 \coth (c+d x)}{d}+x (a+b)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [C]  time = 0.09, size = 98, normalized size = 1.56 \[ -\frac {a^2 \coth ^5(c+d x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\tanh ^2(c+d x)\right )}{5 d}-\frac {2 a b \coth ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\tanh ^2(c+d x)\right )}{3 d}-\frac {b^2 \coth (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\tanh ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^2*Coth[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, Tanh[c + d*x]^2])/d - (2*a*b*Coth[c + d*x]^3*Hyperg
eometric2F1[-3/2, 1, -1/2, Tanh[c + d*x]^2])/(3*d) - (b^2*Coth[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[c
+ d*x]^2])/d

________________________________________________________________________________________

fricas [B]  time = 0.41, size = 473, normalized size = 7.51 \[ -\frac {{\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{2} + 16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 2 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{2} + 16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 3 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 46 \, a^{2} + 80 \, a b + 30 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*((23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^5 + 5*(23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)*sinh(d*x + c)^4
- (15*(a^2 + 2*a*b + b^2)*d*x + 23*a^2 + 40*a*b + 15*b^2)*sinh(d*x + c)^5 - 5*(5*a^2 + 16*a*b + 9*b^2)*cosh(d
*x + c)^3 + 5*(15*(a^2 + 2*a*b + b^2)*d*x - 2*(15*(a^2 + 2*a*b + b^2)*d*x + 23*a^2 + 40*a*b + 15*b^2)*cosh(d*x
+ c)^2 + 23*a^2 + 40*a*b + 15*b^2)*sinh(d*x + c)^3 + 5*(2*(23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^3 - 3*(5*a
^2 + 16*a*b + 9*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^2 + 4*a*b + 3*b^2)*cosh(d*x + c) - 5*((15*(a^2 +
2*a*b + b^2)*d*x + 23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^4 + 30*(a^2 + 2*a*b + b^2)*d*x - 3*(15*(a^2 + 2*a*
b + b^2)*d*x + 23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^2 + 46*a^2 + 80*a*b + 30*b^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)*sin
h(d*x + c))

________________________________________________________________________________________

giac [B]  time = 0.36, size = 218, normalized size = 3.46 \[ \frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 180 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 60 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 220 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 140 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 60 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(a^2 + 2*a*b + b^2)*(d*x + c) - 2*(45*a^2*e^(8*d*x + 8*c) + 60*a*b*e^(8*d*x + 8*c) + 15*b^2*e^(8*d*x
+ 8*c) - 90*a^2*e^(6*d*x + 6*c) - 180*a*b*e^(6*d*x + 6*c) - 60*b^2*e^(6*d*x + 6*c) + 140*a^2*e^(4*d*x + 4*c) +
220*a*b*e^(4*d*x + 4*c) + 90*b^2*e^(4*d*x + 4*c) - 70*a^2*e^(2*d*x + 2*c) - 140*a*b*e^(2*d*x + 2*c) - 60*b^2*
e^(2*d*x + 2*c) + 23*a^2 + 40*a*b + 15*b^2)/(e^(2*d*x + 2*c) - 1)^5)/d

________________________________________________________________________________________

maple [A]  time = 0.24, size = 87, normalized size = 1.38 \[ \frac {a^{2} \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\coth ^{5}\left (d x +c \right )\right )}{5}\right )+2 a b \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}\right )+b^{2} \left (d x +c -\coth \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 231, normalized size = 3.67 \[ \frac {1}{15} \, a^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {2}{3} \, a b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^2*(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))) + 2/3*a*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))) + b^2*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1)))

________________________________________________________________________________________

mupad [B]  time = 0.20, size = 529, normalized size = 8.40 \[ \frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}+\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+x\,{\left (a+b\right )}^2-\frac {\frac {2\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {2\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{15\,d}-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {2\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________