Integrand size = 23, antiderivative size = 133 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {1}{2} \left (a^2+7 b^2\right ) x+\frac {a b \cosh ^2(c+d x)}{d}-\frac {4 a b \log (\cosh (c+d x))}{d}+\frac {\left (a^2+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {3 b^2 \tanh (c+d x)}{d}+\frac {a b \tanh ^2(c+d x)}{d}+\frac {2 b^2 \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Output:
-1/2*(a^2+7*b^2)*x+a*b*cosh(d*x+c)^2/d-4*a*b*ln(cosh(d*x+c))/d+1/2*(a^2+b^ 2)*cosh(d*x+c)*sinh(d*x+c)/d+3*b^2*tanh(d*x+c)/d+a*b*tanh(d*x+c)^2/d+2/3*b ^2*tanh(d*x+c)^3/d+1/5*b^2*tanh(d*x+c)^5/d
Time = 1.40 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {-30 a^2 c-210 b^2 c-30 a^2 d x-210 b^2 d x+30 a b \cosh (2 (c+d x))-240 a b \log (\cosh (c+d x))+15 a^2 \sinh (2 (c+d x))+15 b^2 \sinh (2 (c+d x))+232 b^2 \tanh (c+d x)+12 b^2 \text {sech}^4(c+d x) \tanh (c+d x)-4 b \text {sech}^2(c+d x) (15 a+16 b \tanh (c+d x))}{60 d} \] Input:
Integrate[Sinh[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^2,x]
Output:
(-30*a^2*c - 210*b^2*c - 30*a^2*d*x - 210*b^2*d*x + 30*a*b*Cosh[2*(c + d*x )] - 240*a*b*Log[Cosh[c + d*x]] + 15*a^2*Sinh[2*(c + d*x)] + 15*b^2*Sinh[2 *(c + d*x)] + 232*b^2*Tanh[c + d*x] + 12*b^2*Sech[c + d*x]^4*Tanh[c + d*x] - 4*b*Sech[c + d*x]^2*(15*a + 16*b*Tanh[c + d*x]))/(60*d)
Time = 0.78 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4146, 2335, 25, 2333, 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.
| \(\displaystyle \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\sin (i c+i d x)^2 \left (a+i b \tan (i c+i d x)^3\right )^2dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \sin (i c+i d x)^2 \left (i b \tan (i c+i d x)^3+a\right )^2dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (b \tanh ^3(c+d x)+a\right )^2}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {\frac {1}{2} \int -\frac {\tanh (c+d x) \left (2 b^2 \tanh ^5(c+d x)+2 b^2 \tanh ^3(c+d x)+4 a b \tanh ^2(c+d x)+\left (a^2+3 b^2\right ) \tanh (c+d x)+4 a b\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {\tanh ^2(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tanh ^2(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int \frac {\tanh (c+d x) \left (2 b^2 \tanh ^5(c+d x)+2 b^2 \tanh ^3(c+d x)+4 a b \tanh ^2(c+d x)+\left (a^2+3 b^2\right ) \tanh (c+d x)+4 a b\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\frac {\tanh ^2(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int \left (-2 b^2 \tanh ^4(c+d x)-4 b^2 \tanh ^2(c+d x)-4 a b \tanh (c+d x)-a^2-7 b^2+\frac {a^2+8 b \tanh (c+d x) a+7 b^2}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} \left (-\left (a^2+7 b^2\right ) \text {arctanh}(\tanh (c+d x))+\left (a^2+7 b^2\right ) \tanh (c+d x)+2 a b \tanh ^2(c+d x)+4 a b \log \left (1-\tanh ^2(c+d x)\right )+\frac {2}{5} b^2 \tanh ^5(c+d x)+\frac {4}{3} b^2 \tanh ^3(c+d x)\right )+\frac {\tanh ^2(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
Input:
Int[Sinh[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^2,x]
Output:
((Tanh[c + d*x]^2*(2*a*b + (a^2 + b^2)*Tanh[c + d*x]))/(2*(1 - Tanh[c + d* x]^2)) + (-((a^2 + 7*b^2)*ArcTanh[Tanh[c + d*x]]) + 4*a*b*Log[1 - Tanh[c + d*x]^2] + (a^2 + 7*b^2)*Tanh[c + d*x] + 2*a*b*Tanh[c + d*x]^2 + (4*b^2*Ta nh[c + d*x]^3)/3 + (2*b^2*Tanh[c + d*x]^5)/5)/2)/d
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 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 = 4.52 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{4}}{2 \cosh \left (d x +c \right )^{2}}-2 \ln \left (\cosh \left (d x +c \right )\right )+\tanh \left (d x +c \right )^{2}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{7}}{2 \cosh \left (d x +c \right )^{5}}-\frac {7 d x}{2}-\frac {7 c}{2}+\frac {7 \tanh \left (d x +c \right )}{2}+\frac {7 \tanh \left (d x +c \right )^{3}}{6}+\frac {7 \tanh \left (d x +c \right )^{5}}{10}\right )}{d}\) | \(130\) |
| default | \(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{4}}{2 \cosh \left (d x +c \right )^{2}}-2 \ln \left (\cosh \left (d x +c \right )\right )+\tanh \left (d x +c \right )^{2}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{7}}{2 \cosh \left (d x +c \right )^{5}}-\frac {7 d x}{2}-\frac {7 c}{2}+\frac {7 \tanh \left (d x +c \right )}{2}+\frac {7 \tanh \left (d x +c \right )^{3}}{6}+\frac {7 \tanh \left (d x +c \right )^{5}}{10}\right )}{d}\) | \(130\) |
| risch | \(-\frac {a^{2} x}{2}+4 a b x -\frac {7 b^{2} x}{2}+\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}+\frac {{\mathrm e}^{2 d x +2 c} a b}{4 d}+\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} a b}{4 d}-\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}+\frac {8 a b c}{d}-\frac {4 b \left (15 \,{\mathrm e}^{8 d x +8 c} a +45 \,{\mathrm e}^{8 d x +8 c} b +45 \,{\mathrm e}^{6 d x +6 c} a +120 \,{\mathrm e}^{6 d x +6 c} b +45 \,{\mathrm e}^{4 d x +4 c} a +170 b \,{\mathrm e}^{4 d x +4 c}+15 \,{\mathrm e}^{2 d x +2 c} a +100 \,{\mathrm e}^{2 d x +2 c} b +29 b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}-\frac {4 a b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(265\) |
Input:
int(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+2*a*b*(1/2*sinh(d*x+c )^4/cosh(d*x+c)^2-2*ln(cosh(d*x+c))+tanh(d*x+c)^2)+b^2*(1/2*sinh(d*x+c)^7/ cosh(d*x+c)^5-7/2*d*x-7/2*c+7/2*tanh(d*x+c)+7/6*tanh(d*x+c)^3+7/10*tanh(d* x+c)^5))
Leaf count of result is larger than twice the leaf count of optimal. 3649 vs. \(2 (125) = 250\).
Time = 0.13 (sec) , antiderivative size = 3649, normalized size of antiderivative = 27.44 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="fricas")
Output:
Too large to include
\[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right )^{2} \sinh ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(sinh(d*x+c)**2*(a+b*tanh(d*x+c)**3)**2,x)
Output:
Integral((a + b*tanh(c + d*x)**3)**2*sinh(c + d*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (125) = 250\).
Time = 0.13 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.26 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-\frac {1}{8} \, a^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{120} \, b^{2} {\left (\frac {420 \, {\left (d x + c\right )}}{d} + \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {1003 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3350 \, e^{\left (-4 \, d x - 4 \, c\right )} + 5590 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3915 \, e^{\left (-8 \, d x - 8 \, c\right )} + 1455 \, e^{\left (-10 \, d x - 10 \, c\right )} + 15}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 5 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )}\right )}}\right )} - \frac {1}{4} \, a b {\left (\frac {16 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} + \frac {16 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \] Input:
integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="maxima")
Output:
-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/120*b^2*(420*( d*x + c)/d + 15*e^(-2*d*x - 2*c)/d - (1003*e^(-2*d*x - 2*c) + 3350*e^(-4*d *x - 4*c) + 5590*e^(-6*d*x - 6*c) + 3915*e^(-8*d*x - 8*c) + 1455*e^(-10*d* x - 10*c) + 15)/(d*(e^(-2*d*x - 2*c) + 5*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 10*e^(-8*d*x - 8*c) + 5*e^(-10*d*x - 10*c) + e^(-12*d*x - 12*c)))) - 1/4*a*b*(16*(d*x + c)/d - e^(-2*d*x - 2*c)/d + 16*log(e^(-2*d*x - 2*c) + 1)/d - (2*e^(-2*d*x - 2*c) - 15*e^(-4*d*x - 4*c) + 1)/(d*(e^(-2*d*x - 2* c) + 2*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))))
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (125) = 250\).
Time = 0.24 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.23 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {15 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 15 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 480 \, a b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - 60 \, {\left (a^{2} - 8 \, a b + 7 \, b^{2}\right )} {\left (d x + c\right )} + 15 \, {\left (2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2} + 2 \, a b - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + \frac {8 \, {\left (137 \, a b e^{\left (10 \, d x + 10 \, c\right )} + 625 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 180 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 1190 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 480 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 1190 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 680 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 625 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 400 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 137 \, a b - 116 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="giac")
Output:
1/120*(15*a^2*e^(2*d*x + 2*c) + 30*a*b*e^(2*d*x + 2*c) + 15*b^2*e^(2*d*x + 2*c) - 480*a*b*log(e^(2*d*x + 2*c) + 1) - 60*(a^2 - 8*a*b + 7*b^2)*(d*x + c) + 15*(2*a^2*e^(2*d*x + 2*c) - 16*a*b*e^(2*d*x + 2*c) + 14*b^2*e^(2*d*x + 2*c) - a^2 + 2*a*b - b^2)*e^(-2*d*x - 2*c) + 8*(137*a*b*e^(10*d*x + 10* c) + 625*a*b*e^(8*d*x + 8*c) - 180*b^2*e^(8*d*x + 8*c) + 1190*a*b*e^(6*d*x + 6*c) - 480*b^2*e^(6*d*x + 6*c) + 1190*a*b*e^(4*d*x + 4*c) - 680*b^2*e^( 4*d*x + 4*c) + 625*a*b*e^(2*d*x + 2*c) - 400*b^2*e^(2*d*x + 2*c) + 137*a*b - 116*b^2)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 2.53 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.30 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-\frac {4\,\left (3\,b^2+a\,b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-x\,\left (\frac {a^2}{2}-4\,a\,b+\frac {7\,b^2}{2}\right )+\frac {4\,\left (4\,b^2+a\,b\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {64\,b^2}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a-b\right )}^2}{8\,d}+\frac {16\,b^2}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {32\,b^2}{5\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}-\frac {4\,a\,b\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d} \] Input:
int(sinh(c + d*x)^2*(a + b*tanh(c + d*x)^3)^2,x)
Output:
(exp(2*c + 2*d*x)*(a + b)^2)/(8*d) - (4*(a*b + 3*b^2))/(d*(exp(2*c + 2*d*x ) + 1)) - x*(a^2/2 - 4*a*b + (7*b^2)/2) + (4*(a*b + 4*b^2))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - (64*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)) - (exp(- 2*c - 2*d*x)*(a - b)^2) /(8*d) + (16*b^2)/(d*(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) - (32*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)) - (4*a*b*log(exp(2*c)*exp(2*d*x) + 1))/d
Time = 0.23 (sec) , antiderivative size = 806, normalized size of antiderivative = 6.06 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {-690 e^{8 d x +8 c} a b +15 e^{14 d x +14 c} a^{2}+15 e^{14 d x +14 c} b^{2}+48 e^{12 d x +12 c} a^{2}+336 e^{12 d x +12 c} b^{2}-60 e^{2 d x +2 c} a^{2} d x -300 e^{10 d x +10 c} a^{2} d x -2100 e^{10 d x +10 c} b^{2} d x -600 e^{8 d x +8 c} a^{2} d x -4200 e^{8 d x +8 c} b^{2} d x -600 e^{6 d x +6 c} a^{2} d x -4200 e^{6 d x +6 c} b^{2} d x -300 e^{4 d x +4 c} a^{2} d x -2100 e^{4 d x +4 c} b^{2} d x -480 e^{12 d x +12 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b -60 e^{12 d x +12 c} a^{2} d x -420 e^{12 d x +12 c} b^{2} d x -2400 e^{10 d x +10 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b -4800 e^{8 d x +8 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b -4800 e^{6 d x +6 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b -2400 e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b -15 a^{2}+480 e^{2 d x +2 c} a b d x +2400 e^{10 d x +10 c} a b d x +4800 e^{8 d x +8 c} a b d x +4800 e^{6 d x +6 c} a b d x +2400 e^{4 d x +4 c} a b d x -742 e^{2 d x +2 c} b^{2}-480 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b +480 e^{12 d x +12 c} a b d x -420 e^{2 d x +2 c} b^{2} d x -15 b^{2}+30 e^{14 d x +14 c} a b +180 e^{12 d x +12 c} a b -690 e^{6 d x +6 c} a b +180 e^{2 d x +2 c} a b +30 a b -195 e^{8 d x +8 c} a^{2}-1155 e^{8 d x +8 c} b^{2}-345 e^{6 d x +6 c} a^{2}-2905 e^{6 d x +6 c} b^{2}-270 e^{4 d x +4 c} a^{2}-2030 e^{4 d x +4 c} b^{2}-102 e^{2 d x +2 c} a^{2}}{120 e^{2 d x +2 c} d \left (e^{10 d x +10 c}+5 e^{8 d x +8 c}+10 e^{6 d x +6 c}+10 e^{4 d x +4 c}+5 e^{2 d x +2 c}+1\right )} \] Input:
int(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x)
Output:
(15*e**(14*c + 14*d*x)*a**2 + 30*e**(14*c + 14*d*x)*a*b + 15*e**(14*c + 14 *d*x)*b**2 - 480*e**(12*c + 12*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b - 60*e** (12*c + 12*d*x)*a**2*d*x + 48*e**(12*c + 12*d*x)*a**2 + 480*e**(12*c + 12* d*x)*a*b*d*x + 180*e**(12*c + 12*d*x)*a*b - 420*e**(12*c + 12*d*x)*b**2*d* x + 336*e**(12*c + 12*d*x)*b**2 - 2400*e**(10*c + 10*d*x)*log(e**(2*c + 2* d*x) + 1)*a*b - 300*e**(10*c + 10*d*x)*a**2*d*x + 2400*e**(10*c + 10*d*x)* a*b*d*x - 2100*e**(10*c + 10*d*x)*b**2*d*x - 4800*e**(8*c + 8*d*x)*log(e** (2*c + 2*d*x) + 1)*a*b - 600*e**(8*c + 8*d*x)*a**2*d*x - 195*e**(8*c + 8*d *x)*a**2 + 4800*e**(8*c + 8*d*x)*a*b*d*x - 690*e**(8*c + 8*d*x)*a*b - 4200 *e**(8*c + 8*d*x)*b**2*d*x - 1155*e**(8*c + 8*d*x)*b**2 - 4800*e**(6*c + 6 *d*x)*log(e**(2*c + 2*d*x) + 1)*a*b - 600*e**(6*c + 6*d*x)*a**2*d*x - 345* e**(6*c + 6*d*x)*a**2 + 4800*e**(6*c + 6*d*x)*a*b*d*x - 690*e**(6*c + 6*d* x)*a*b - 4200*e**(6*c + 6*d*x)*b**2*d*x - 2905*e**(6*c + 6*d*x)*b**2 - 240 0*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b - 300*e**(4*c + 4*d*x)*a* *2*d*x - 270*e**(4*c + 4*d*x)*a**2 + 2400*e**(4*c + 4*d*x)*a*b*d*x - 2100* e**(4*c + 4*d*x)*b**2*d*x - 2030*e**(4*c + 4*d*x)*b**2 - 480*e**(2*c + 2*d *x)*log(e**(2*c + 2*d*x) + 1)*a*b - 60*e**(2*c + 2*d*x)*a**2*d*x - 102*e** (2*c + 2*d*x)*a**2 + 480*e**(2*c + 2*d*x)*a*b*d*x + 180*e**(2*c + 2*d*x)*a *b - 420*e**(2*c + 2*d*x)*b**2*d*x - 742*e**(2*c + 2*d*x)*b**2 - 15*a**2 + 30*a*b - 15*b**2)/(120*e**(2*c + 2*d*x)*d*(e**(10*c + 10*d*x) + 5*e**(...