Integrand size = 23, antiderivative size = 63 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {(a+b)^2 \tanh (c+d x)}{d}-\frac {b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Output:
(a+b)^2*x-(a+b)^2*tanh(d*x+c)/d-1/3*b*(2*a+b)*tanh(d*x+c)^3/d-1/5*b^2*tanh (d*x+c)^5/d
Leaf count is larger than twice the leaf count of optimal. \(137\) vs. \(2(63)=126\).
Time = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.17 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \text {arctanh}(\tanh (c+d x))}{d}+\frac {2 a b \text {arctanh}(\tanh (c+d x))}{d}+\frac {b^2 \text {arctanh}(\tanh (c+d x))}{d}-\frac {a^2 \tanh (c+d x)}{d}-\frac {2 a b \tanh (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{d}-\frac {2 a b \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d}-\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Input:
Integrate[Tanh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(a^2*ArcTanh[Tanh[c + d*x]])/d + (2*a*b*ArcTanh[Tanh[c + d*x]])/d + (b^2*A rcTanh[Tanh[c + d*x]])/d - (a^2*Tanh[c + d*x])/d - (2*a*b*Tanh[c + d*x])/d - (b^2*Tanh[c + d*x])/d - (2*a*b*Tanh[c + d*x]^3)/(3*d) - (b^2*Tanh[c + d *x]^3)/(3*d) - (b^2*Tanh[c + d*x]^5)/(5*d)
Time = 0.33 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4153, 25, 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.
\(\displaystyle \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\tan (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \tan (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -\frac {\int -\frac {\tanh ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left (-b^2 \tanh ^4(c+d x)-b (2 a+b) \tanh ^2(c+d x)-(a+b)^2+\frac {a^2+2 b a+b^2}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-(a+b)^2 \text {arctanh}(\tanh (c+d x))+\frac {1}{3} b (2 a+b) \tanh ^3(c+d x)+(a+b)^2 \tanh (c+d x)+\frac {1}{5} b^2 \tanh ^5(c+d x)}{d}\) |
Input:
Int[Tanh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-((-((a + b)^2*ArcTanh[Tanh[c + d*x]]) + (a + b)^2*Tanh[c + d*x] + (b*(2*a + b)*Tanh[c + d*x]^3)/3 + (b^2*Tanh[c + d*x]^5)/5)/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 = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.54
method | result | size |
parallelrisch | \(-\frac {3 b^{2} \tanh \left (d x +c \right )^{5}+10 a b \tanh \left (d x +c \right )^{3}+5 b^{2} \tanh \left (d x +c \right )^{3}-15 a^{2} d x -30 a b d x -15 b^{2} d x +15 a^{2} \tanh \left (d x +c \right )+30 a b \tanh \left (d x +c \right )+15 b^{2} \tanh \left (d x +c \right )}{15 d}\) | \(97\) |
derivativedivides | \(\frac {-2 a b \tanh \left (d x +c \right )-\frac {2 a b \tanh \left (d x +c \right )^{3}}{3}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )}{2}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {b^{2} \tanh \left (d x +c \right )^{5}}{5}-\frac {b^{2} \tanh \left (d x +c \right )^{3}}{3}-a^{2} \tanh \left (d x +c \right )-b^{2} \tanh \left (d x +c \right )}{d}\) | \(120\) |
default | \(\frac {-2 a b \tanh \left (d x +c \right )-\frac {2 a b \tanh \left (d x +c \right )^{3}}{3}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )}{2}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {b^{2} \tanh \left (d x +c \right )^{5}}{5}-\frac {b^{2} \tanh \left (d x +c \right )^{3}}{3}-a^{2} \tanh \left (d x +c \right )-b^{2} \tanh \left (d x +c \right )}{d}\) | \(120\) |
parts | \(\frac {a^{2} \left (-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\tanh \left (d x +c \right )^{5}}{5}-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 a b \left (-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2}\right )}{d}\) | \(146\) |
risch | \(a^{2} x +2 a b x +b^{2} x +\frac {2 a^{2} {\mathrm e}^{8 d x +8 c}+8 a b \,{\mathrm e}^{8 d x +8 c}+6 b^{2} {\mathrm e}^{8 d x +8 c}+8 a^{2} {\mathrm e}^{6 d x +6 c}+24 a b \,{\mathrm e}^{6 d x +6 c}+12 b^{2} {\mathrm e}^{6 d x +6 c}+12 a^{2} {\mathrm e}^{4 d x +4 c}+\frac {88 a b \,{\mathrm e}^{4 d x +4 c}}{3}+\frac {56 b^{2} {\mathrm e}^{4 d x +4 c}}{3}+8 a^{2} {\mathrm e}^{2 d x +2 c}+\frac {56 a b \,{\mathrm e}^{2 d x +2 c}}{3}+\frac {28 b^{2} {\mathrm e}^{2 d x +2 c}}{3}+2 a^{2}+\frac {16 a b}{3}+\frac {46 b^{2}}{15}}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(214\) |
Input:
int(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/15*(3*b^2*tanh(d*x+c)^5+10*a*b*tanh(d*x+c)^3+5*b^2*tanh(d*x+c)^3-15*a^2 *d*x-30*a*b*d*x-15*b^2*d*x+15*a^2*tanh(d*x+c)+30*a*b*tanh(d*x+c)+15*b^2*ta nh(d*x+c))/d
Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (59) = 118\).
Time = 0.09 (sec) , antiderivative size = 483, normalized size of antiderivative = 7.67 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {{\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (2 \, {\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 16 \, a b + 5 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (9 \, a^{2} + 16 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a^{2} + 8 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \] Input:
integrate(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
1/15*((15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c )^5 + 5*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 - (15*a^2 + 40*a*b + 23*b^2)*sinh(d*x + c)^5 + 5*(15*( a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^3 - 5*(2* (15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^2 + 9*a^2 + 16*a*b + 5*b^2)*sinh( d*x + c)^3 + 5*(2*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)* cosh(d*x + c)^3 + 3*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2 )*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c) - 5*((15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^4 + 3*(9*a^2 + 16*a*b + 5*b^2)*cosh(d*x + c)^2 + 6*a^2 + 8*a*b + 10* b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh( d*x + c)^2 + 10*d*cosh(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (53) = 106\).
Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.86 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\begin {cases} a^{2} x - \frac {a^{2} \tanh {\left (c + d x \right )}}{d} + 2 a b x - \frac {2 a b \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a b \tanh {\left (c + d x \right )}}{d} + b^{2} x - \frac {b^{2} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{2} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{2} \tanh ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tanh(d*x+c)**2*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Piecewise((a**2*x - a**2*tanh(c + d*x)/d + 2*a*b*x - 2*a*b*tanh(c + d*x)** 3/(3*d) - 2*a*b*tanh(c + d*x)/d + b**2*x - b**2*tanh(c + d*x)**5/(5*d) - b **2*tanh(c + d*x)**3/(3*d) - b**2*tanh(c + d*x)/d, Ne(d, 0)), (x*(a + b*ta nh(c)**2)**2*tanh(c)**2, True))
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (59) = 118\).
Time = 0.05 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.67 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{15} \, b^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {2}{3} \, a b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + a^{2} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} \] Input:
integrate(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/15*b^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) + 1 0*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))) + a^2*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1)))
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (59) = 118\).
Time = 0.18 (sec) , antiderivative size = 218, normalized size of antiderivative = 3.46 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 180 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 220 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 140 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 70 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \] Input:
integrate(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/15*(15*(a^2 + 2*a*b + b^2)*(d*x + c) + 2*(15*a^2*e^(8*d*x + 8*c) + 60*a* b*e^(8*d*x + 8*c) + 45*b^2*e^(8*d*x + 8*c) + 60*a^2*e^(6*d*x + 6*c) + 180* a*b*e^(6*d*x + 6*c) + 90*b^2*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) + 22 0*a*b*e^(4*d*x + 4*c) + 140*b^2*e^(4*d*x + 4*c) + 60*a^2*e^(2*d*x + 2*c) + 140*a*b*e^(2*d*x + 2*c) + 70*b^2*e^(2*d*x + 2*c) + 15*a^2 + 40*a*b + 23*b ^2)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 2.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=x\,\left (a^2+2\,a\,b+b^2\right )-\frac {\mathrm {tanh}\left (c+d\,x\right )\,{\left (a+b\right )}^2}{d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^3\,\left (b^2+2\,a\,b\right )}{3\,d}-\frac {b^2\,{\mathrm {tanh}\left (c+d\,x\right )}^5}{5\,d} \] Input:
int(tanh(c + d*x)^2*(a + b*tanh(c + d*x)^2)^2,x)
Output:
x*(2*a*b + a^2 + b^2) - (tanh(c + d*x)*(a + b)^2)/d - (tanh(c + d*x)^3*(2* a*b + b^2))/(3*d) - (b^2*tanh(c + d*x)^5)/(5*d)
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.52 \[ \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {-3 \tanh \left (d x +c \right )^{5} b^{2}-10 \tanh \left (d x +c \right )^{3} a b -5 \tanh \left (d x +c \right )^{3} b^{2}-15 \tanh \left (d x +c \right ) a^{2}-30 \tanh \left (d x +c \right ) a b -15 \tanh \left (d x +c \right ) b^{2}+15 a^{2} d x +30 a b d x +15 b^{2} d x}{15 d} \] Input:
int(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x)
Output:
( - 3*tanh(c + d*x)**5*b**2 - 10*tanh(c + d*x)**3*a*b - 5*tanh(c + d*x)**3 *b**2 - 15*tanh(c + d*x)*a**2 - 30*tanh(c + d*x)*a*b - 15*tanh(c + d*x)*b* *2 + 15*a**2*d*x + 30*a*b*d*x + 15*b**2*d*x)/(15*d)