Integrand size = 23, antiderivative size = 43 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {a (a+2 b) \coth (c+d x)}{d}-\frac {a^2 \coth ^3(c+d x)}{3 d} \] Output:
(a+b)^2*x-a*(a+2*b)*coth(d*x+c)/d-1/3*a^2*coth(d*x+c)^3/d
Time = 0.42 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.51 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\coth (c+d x) \left (a \left (3 a+6 b+a \coth ^2(c+d x)\right )-3 (a+b)^2 \text {arctanh}\left (\sqrt {\tanh ^2(c+d x)}\right ) \sqrt {\tanh ^2(c+d x)}\right )}{3 d} \] Input:
Integrate[Coth[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-1/3*(Coth[c + d*x]*(a*(3*a + 6*b + a*Coth[c + d*x]^2) - 3*(a + b)^2*ArcTa nh[Sqrt[Tanh[c + d*x]^2]]*Sqrt[Tanh[c + d*x]^2]))/d
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4153, 364, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left (a^2 \coth ^4(c+d x)+a (a+2 b) \coth ^2(c+d x)-\frac {(a+b)^2}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} a^2 \coth ^3(c+d x)+(a+b)^2 \text {arctanh}(\tanh (c+d x))-a (a+2 b) \coth (c+d x)}{d}\) |
Input:
Int[Coth[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
((a + b)^2*ArcTanh[Tanh[c + d*x]] - a*(a + 2*b)*Coth[c + d*x] - (a^2*Coth[ c + d*x]^3)/3)/d
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^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] && Ratio nalQ[n]))
Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {-\coth \left (d x +c \right )^{3} a^{2}-3 a \left (a +2 b \right ) \coth \left (d x +c \right )+3 d x \left (a +b \right )^{2}}{3 d}\) | \(43\) |
derivativedivides | \(-\frac {\frac {a^{2}}{3 \tanh \left (d x +c \right )^{3}}+\frac {a \left (a +2 b \right )}{\tanh \left (d x +c \right )}+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\) | \(84\) |
default | \(-\frac {\frac {a^{2}}{3 \tanh \left (d x +c \right )^{3}}+\frac {a \left (a +2 b \right )}{\tanh \left (d x +c \right )}+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\) | \(84\) |
risch | \(a^{2} x +2 a b x +b^{2} x -\frac {4 a \left (3 a \,{\mathrm e}^{4 d x +4 c}+3 b \,{\mathrm e}^{4 d x +4 c}-3 a \,{\mathrm e}^{2 d x +2 c}-6 b \,{\mathrm e}^{2 d x +2 c}+2 a +3 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\) | \(91\) |
Input:
int(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/3*(-coth(d*x+c)^3*a^2-3*a*(a+2*b)*coth(d*x+c)+3*d*x*(a+b)^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (41) = 82\).
Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 4.58 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} - {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 4 \, a^{2} + 6 \, a b\right )} \sinh \left (d x + c\right )^{3} - 6 \, a b \cosh \left (d x + c\right ) + 3 \, {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 4 \, a^{2} + 6 \, a b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a^{2} + 6 \, a b\right )} \sinh \left (d x + c\right )}{3 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \] Input:
integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
-1/3*(2*(2*a^2 + 3*a*b)*cosh(d*x + c)^3 + 6*(2*a^2 + 3*a*b)*cosh(d*x + c)* sinh(d*x + c)^2 - (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*sinh(d*x + c )^3 - 6*a*b*cosh(d*x + c) + 3*(3*(a^2 + 2*a*b + b^2)*d*x - (3*(a^2 + 2*a*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^2 + 4*a^2 + 6*a*b)*sinh(d*x + c ))/(d*sinh(d*x + c)^3 + 3*(d*cosh(d*x + c)^2 - d)*sinh(d*x + c))
\[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(coth(d*x+c)**4*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Integral((a + b*tanh(c + d*x)**2)**2*coth(c + d*x)**4, x)
Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (41) = 82\).
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.65 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{3} \, a^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 2 \, a b {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + b^{2} x \] Input:
integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/3*a^2*(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))) + 2*a*b *(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1))) + b^2*x
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (41) = 82\).
Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.40 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, {\left (3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} + 3 \, a b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \] Input:
integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/3*(3*(a^2 + 2*a*b + b^2)*(d*x + c) - 4*(3*a^2*e^(4*d*x + 4*c) + 3*a*b*e^ (4*d*x + 4*c) - 3*a^2*e^(2*d*x + 2*c) - 6*a*b*e^(2*d*x + 2*c) + 2*a^2 + 3* a*b)/(e^(2*d*x + 2*c) - 1)^3)/d
Time = 0.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 4.07 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=x\,{\left (a+b\right )}^2-\frac {\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2+b\,a\right )}{3\,d}-\frac {4\,a\,b}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {4\,\left (a^2+b\,a\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2+b\,a\right )}{3\,d}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {4\,\left (a^2+b\,a\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \] Input:
int(coth(c + d*x)^4*(a + b*tanh(c + d*x)^2)^2,x)
Output:
x*(a + b)^2 - ((4*exp(2*c + 2*d*x)*(a*b + a^2))/(3*d) - (4*a*b)/(3*d))/(ex p(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1) - ((4*(a*b + a^2))/(3*d) + (4*exp (4*c + 4*d*x)*(a*b + a^2))/(3*d) - (8*a*b*exp(2*c + 2*d*x))/(3*d))/(3*exp( 2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (4*(a*b + a^2) )/(3*d*(exp(2*c + 2*d*x) - 1))
Time = 0.30 (sec) , antiderivative size = 266, normalized size of antiderivative = 6.19 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 e^{6 d x +6 c} a^{2} d x -4 e^{6 d x +6 c} a^{2}+6 e^{6 d x +6 c} a b d x -4 e^{6 d x +6 c} a b +3 e^{6 d x +6 c} b^{2} d x -9 e^{4 d x +4 c} a^{2} d x -18 e^{4 d x +4 c} a b d x -9 e^{4 d x +4 c} b^{2} d x +9 e^{2 d x +2 c} a^{2} d x +18 e^{2 d x +2 c} a b d x +12 e^{2 d x +2 c} a b +9 e^{2 d x +2 c} b^{2} d x -3 a^{2} d x -4 a^{2}-6 a b d x -8 a b -3 b^{2} d x}{3 d \left (e^{6 d x +6 c}-3 e^{4 d x +4 c}+3 e^{2 d x +2 c}-1\right )} \] Input:
int(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x)
Output:
(3*e**(6*c + 6*d*x)*a**2*d*x - 4*e**(6*c + 6*d*x)*a**2 + 6*e**(6*c + 6*d*x )*a*b*d*x - 4*e**(6*c + 6*d*x)*a*b + 3*e**(6*c + 6*d*x)*b**2*d*x - 9*e**(4 *c + 4*d*x)*a**2*d*x - 18*e**(4*c + 4*d*x)*a*b*d*x - 9*e**(4*c + 4*d*x)*b* *2*d*x + 9*e**(2*c + 2*d*x)*a**2*d*x + 18*e**(2*c + 2*d*x)*a*b*d*x + 12*e* *(2*c + 2*d*x)*a*b + 9*e**(2*c + 2*d*x)*b**2*d*x - 3*a**2*d*x - 4*a**2 - 6 *a*b*d*x - 8*a*b - 3*b**2*d*x)/(3*d*(e**(6*c + 6*d*x) - 3*e**(4*c + 4*d*x) + 3*e**(2*c + 2*d*x) - 1))