Integrand size = 21, antiderivative size = 44 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {(a-b) \coth (c+d x)}{d}-\frac {a \coth ^3(c+d x)}{3 d}-\frac {b \tanh (c+d x)}{d} \] Output:
(a-b)*coth(d*x+c)/d-1/3*a*coth(d*x+c)^3/d-b*tanh(d*x+c)/d
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.39 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {2 a \coth (c+d x)}{3 d}-\frac {b \coth (c+d x)}{d}-\frac {a \coth (c+d x) \text {csch}^2(c+d x)}{3 d}-\frac {b \tanh (c+d x)}{d} \] Input:
Integrate[Csch[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]
Output:
(2*a*Coth[c + d*x])/(3*d) - (b*Coth[c + d*x])/d - (a*Coth[c + d*x]*Csch[c + d*x]^2)/(3*d) - (b*Tanh[c + d*x])/d
Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4146, 355, 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.
\(\displaystyle \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-b \tan (i c+i d x)^2}{\sin (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \coth ^4(c+d x) \left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {\int \left (a \coth ^4(c+d x)+(b-a) \coth ^2(c+d x)-b\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a-b) \coth (c+d x)-\frac {1}{3} a \coth ^3(c+d x)-b \tanh (c+d x)}{d}\) |
Input:
Int[Csch[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]
Output:
((a - b)*Coth[c + d*x] - (a*Coth[c + d*x]^3)/3 - b*Tanh[c + d*x])/d
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 6.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {a \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}-2 \tanh \left (d x +c \right )\right )}{d}\) | \(55\) |
default | \(\frac {a \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}-2 \tanh \left (d x +c \right )\right )}{d}\) | \(55\) |
risch | \(-\frac {4 \left (3 \,{\mathrm e}^{4 d x +4 c} a +3 b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -6 \,{\mathrm e}^{2 d x +2 c} b -a +3 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(87\) |
Input:
int(csch(d*x+c)^4*(a+tanh(d*x+c)^2*b),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+b*(-1/sinh(d*x+c)/cosh(d*x+c)-2 *tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (42) = 84\).
Time = 0.09 (sec) , antiderivative size = 244, normalized size of antiderivative = 5.55 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {8 \, {\left ({\left (a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + 3 \, b\right )} \sinh \left (d x + c\right )^{2} + a - 3 \, b\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} - 2 \, d \cosh \left (d x + c\right )^{4} + {\left (15 \, d \cosh \left (d x + c\right )^{2} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} - 2 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )^{2} + {\left (15 \, d \cosh \left (d x + c\right )^{4} - 12 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} - 4 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2 \, d\right )}} \] Input:
integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="fricas")
Output:
-8/3*((a + 3*b)*cosh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c) + (a + 3 *b)*sinh(d*x + c)^2 + a - 3*b)/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh (d*x + c)^5 + d*sinh(d*x + c)^6 - 2*d*cosh(d*x + c)^4 + (15*d*cosh(d*x + c )^2 - 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - 2*d*cosh(d*x + c))*s inh(d*x + c)^3 - d*cosh(d*x + c)^2 + (15*d*cosh(d*x + c)^4 - 12*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 - 4*d*cosh(d*x + c)^ 3 + d*cosh(d*x + c))*sinh(d*x + c) + 2*d)
\[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname {csch}^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(csch(d*x+c)**4*(a+b*tanh(d*x+c)**2),x)
Output:
Integral((a + b*tanh(c + d*x)**2)*csch(c + d*x)**4, x)
Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (42) = 84\).
Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.57 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {4}{3} \, a {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {4 \, b}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \] Input:
integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="maxima")
Output:
4/3*a*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^ (-6*d*x - 6*c) - 1)) - 1/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^( -6*d*x - 6*c) - 1))) + 4*b/(d*(e^(-4*d*x - 4*c) - 1))
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.82 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (\frac {3 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac {3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + 3 \, b}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}\right )}}{3 \, d} \] Input:
integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="giac")
Output:
2/3*(3*b/(e^(2*d*x + 2*c) + 1) - (3*b*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c ) - 6*b*e^(2*d*x + 2*c) - 2*a + 3*b)/(e^(2*d*x + 2*c) - 1)^3)/d
Time = 2.49 (sec) , antiderivative size = 173, normalized size of antiderivative = 3.93 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {\frac {2\,\left (2\,a-b\right )}{3\,d}+\frac {2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {2\,b}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {\frac {2\,b}{3\,d}+\frac {2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1} \] Input:
int((a + b*tanh(c + d*x)^2)/sinh(c + d*x)^4,x)
Output:
(2*b)/(d*(exp(2*c + 2*d*x) + 1)) - ((2*(2*a - b))/(3*d) + (2*b*exp(2*c + 2 *d*x))/(3*d))/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1) - (2*b)/(3*d*(ex p(2*c + 2*d*x) - 1)) - ((2*b)/(3*d) + (2*b*exp(4*c + 4*d*x))/(3*d) + (4*ex p(2*c + 2*d*x)*(2*a - b))/(3*d))/(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1)
Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.27 \[ \int \text {csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {-4 e^{4 d x +4 c} a -4 e^{4 d x +4 c} b -\frac {8 e^{2 d x +2 c} a}{3}+8 e^{2 d x +2 c} b +\frac {4 a}{3}-4 b}{d \left (e^{8 d x +8 c}-2 e^{6 d x +6 c}+2 e^{2 d x +2 c}-1\right )} \] Input:
int(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x)
Output:
(4*( - 3*e**(4*c + 4*d*x)*a - 3*e**(4*c + 4*d*x)*b - 2*e**(2*c + 2*d*x)*a + 6*e**(2*c + 2*d*x)*b + a - 3*b))/(3*d*(e**(8*c + 8*d*x) - 2*e**(6*c + 6* d*x) + 2*e**(2*c + 2*d*x) - 1))