Integrand size = 23, antiderivative size = 46 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth (c+d x)}{d}+\frac {2 a b \tanh (c+d x)}{d}+\frac {b^2 \tanh ^3(c+d x)}{3 d} \] Output:
-a^2*coth(d*x+c)/d+2*a*b*tanh(d*x+c)/d+1/3*b^2*tanh(d*x+c)^3/d
Time = 0.76 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {-3 a^2 \coth (c+d x)+b \left (6 a+b-b \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{3 d} \] Input:
Integrate[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(-3*a^2*Coth[c + d*x] + b*(6*a + b - b*Sech[c + d*x]^2)*Tanh[c + d*x])/(3* d)
Time = 0.47 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 25, 4146, 244, 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.
\(\displaystyle \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\sin (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\sin (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 4146 |
\(\displaystyle \frac {\int \coth ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (a^2 \coth ^2(c+d x)+b^2 \tanh ^2(c+d x)+2 a b\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a^2 \coth (c+d x)+2 a b \tanh (c+d x)+\frac {1}{3} b^2 \tanh ^3(c+d x)}{d}\) |
Input:
Int[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(-(a^2*Coth[c + d*x]) + 2*a*b*Tanh[c + d*x] + (b^2*Tanh[c + d*x]^3)/3)/d
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 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 = 4.57 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48
method | result | size |
derivativedivides | \(\frac {-\coth \left (d x +c \right ) a^{2}+2 \tanh \left (d x +c \right ) a b +b^{2} \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(68\) |
default | \(\frac {-\coth \left (d x +c \right ) a^{2}+2 \tanh \left (d x +c \right ) a b +b^{2} \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(68\) |
risch | \(-\frac {2 \left (3 \,{\mathrm e}^{6 d x +6 c} a^{2}+6 \,{\mathrm e}^{6 d x +6 c} a b +3 \,{\mathrm e}^{6 d x +6 c} b^{2}+9 \,{\mathrm e}^{4 d x +4 c} a^{2}+6 \,{\mathrm e}^{4 d x +4 c} a b -3 \,{\mathrm e}^{4 d x +4 c} b^{2}+9 \,{\mathrm e}^{2 d x +2 c} a^{2}-6 \,{\mathrm e}^{2 d x +2 c} b a +b^{2} {\mathrm e}^{2 d x +2 c}+3 a^{2}-6 a b -b^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3} \left ({\mathrm e}^{2 d x +2 c}-1\right )}\) | \(169\) |
Input:
int(csch(d*x+c)^2*(a+tanh(d*x+c)^2*b)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-coth(d*x+c)*a^2+2*tanh(d*x+c)*a*b+b^2*(-1/2*sinh(d*x+c)/cosh(d*x+c)^ 3+1/2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (44) = 88\).
Time = 0.08 (sec) , antiderivative size = 264, normalized size of antiderivative = 5.74 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {4 \, {\left ({\left (3 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{3} + {\left (9 \, a^{2} - b^{2}\right )} \cosh \left (d x + c\right ) + 2 \, {\left (3 \, {\left (3 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 3 \, a b - b^{2}\right )} \sinh \left (d x + c\right )\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + d \sinh \left (d x + c\right )^{5} + d \cosh \left (d x + c\right )^{3} + {\left (10 \, d \cosh \left (d x + c\right )^{2} + 3 \, d\right )} \sinh \left (d x + c\right )^{3} + {\left (10 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 2 \, d \cosh \left (d x + c\right ) + {\left (5 \, d \cosh \left (d x + c\right )^{4} + 9 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
-4/3*((3*a^2 + b^2)*cosh(d*x + c)^3 + 3*(3*a^2 + b^2)*cosh(d*x + c)*sinh(d *x + c)^2 + 2*(3*a*b + b^2)*sinh(d*x + c)^3 + (9*a^2 - b^2)*cosh(d*x + c) + 2*(3*(3*a*b + b^2)*cosh(d*x + c)^2 + 3*a*b - b^2)*sinh(d*x + c))/(d*cosh (d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 + d*co sh(d*x + c)^3 + (10*d*cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^3 + (10*d*cosh( d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 - 2*d*cosh(d*x + c) + (5*d *cosh(d*x + c)^4 + 9*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))
\[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \operatorname {csch}^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(csch(d*x+c)**2*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Integral((a + b*tanh(c + d*x)**2)**2*csch(c + d*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (44) = 88\).
Time = 0.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.96 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {2}{3} \, b^{2} {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
2/3*b^2*(3*e^(-4*d*x - 4*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*a*b/(d*(e^(-2*d*x - 2*c) + 1)) + 2*a^2/(d*(e^(- 2*d*x - 2*c) - 1))
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.87 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (\frac {3 \, a^{2}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac {6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b + b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}\right )}}{3 \, d} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
-2/3*(3*a^2/(e^(2*d*x + 2*c) - 1) + (6*a*b*e^(4*d*x + 4*c) + 3*b^2*e^(4*d* x + 4*c) + 12*a*b*e^(2*d*x + 2*c) + 6*a*b + b^2)/(e^(2*d*x + 2*c) + 1)^3)/ d
Time = 2.34 (sec) , antiderivative size = 209, normalized size of antiderivative = 4.54 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\frac {2\,\left (2\,a\,b-b^2\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a\,b-b^2\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,\left (b^2+2\,a\,b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:
int((a + b*tanh(c + d*x)^2)^2/sinh(c + d*x)^2,x)
Output:
- ((2*(2*a*b - b^2))/(3*d) + (2*exp(2*c + 2*d*x)*(2*a*b + b^2))/(3*d))/(2* exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*(2*a*b + b^2))/(3*d) + (2*e xp(4*c + 4*d*x)*(2*a*b + b^2))/(3*d) + (4*exp(2*c + 2*d*x)*(2*a*b - b^2))/ (3*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - (2*a^2)/(d*(exp(2*c + 2*d*x) - 1)) - (2*(2*a*b + b^2))/(3*d*(exp(2*c + 2*d *x) + 1))
Time = 0.22 (sec) , antiderivative size = 176, normalized size of antiderivative = 3.83 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 e^{8 d x +8 c} a^{2}+6 e^{8 d x +8 c} a b +3 e^{8 d x +8 c} b^{2}-18 e^{4 d x +4 c} a^{2}-12 e^{4 d x +4 c} a b +6 e^{4 d x +4 c} b^{2}-24 e^{2 d x +2 c} a^{2}-8 e^{2 d x +2 c} b^{2}-9 a^{2}+6 a b -b^{2}}{3 d \left (e^{8 d x +8 c}+2 e^{6 d x +6 c}-2 e^{2 d x +2 c}-1\right )} \] Input:
int(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x)
Output:
(3*e**(8*c + 8*d*x)*a**2 + 6*e**(8*c + 8*d*x)*a*b + 3*e**(8*c + 8*d*x)*b** 2 - 18*e**(4*c + 4*d*x)*a**2 - 12*e**(4*c + 4*d*x)*a*b + 6*e**(4*c + 4*d*x )*b**2 - 24*e**(2*c + 2*d*x)*a**2 - 8*e**(2*c + 2*d*x)*b**2 - 9*a**2 + 6*a *b - b**2)/(3*d*(e**(8*c + 8*d*x) + 2*e**(6*c + 6*d*x) - 2*e**(2*c + 2*d*x ) - 1))