Integrand size = 23, antiderivative size = 182 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {3}{8} \left (a^2+21 b^2\right ) x-\frac {3 a b \cosh ^2(c+d x)}{d}+\frac {a b \cosh ^4(c+d x)}{2 d}+\frac {6 a b \log (\cosh (c+d x))}{d}-\frac {\left (5 a^2+17 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {\left (a^2+b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac {6 b^2 \tanh (c+d x)}{d}-\frac {a b \tanh ^2(c+d x)}{d}-\frac {b^2 \tanh ^3(c+d x)}{d}-\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Output:
3/8*(a^2+21*b^2)*x-3*a*b*cosh(d*x+c)^2/d+1/2*a*b*cosh(d*x+c)^4/d+6*a*b*ln( cosh(d*x+c))/d-1/8*(5*a^2+17*b^2)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*(a^2+b^2)* cosh(d*x+c)^3*sinh(d*x+c)/d-6*b^2*tanh(d*x+c)/d-a*b*tanh(d*x+c)^2/d-b^2*ta nh(d*x+c)^3/d-1/5*b^2*tanh(d*x+c)^5/d
Time = 4.46 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.86 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {60 \left (a^2+21 b^2\right ) (c+d x)-200 a b \cosh (2 (c+d x))+10 a b \cosh (4 (c+d x))+960 a b \log (\cosh (c+d x))+160 a b \text {sech}^2(c+d x)-40 \left (a^2+4 b^2\right ) \sinh (2 (c+d x))+5 \left (a^2+b^2\right ) \sinh (4 (c+d x))-1152 b^2 \tanh (c+d x)+224 b^2 \text {sech}^2(c+d x) \tanh (c+d x)-32 b^2 \text {sech}^4(c+d x) \tanh (c+d x)}{160 d} \] Input:
Integrate[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^3)^2,x]
Output:
(60*(a^2 + 21*b^2)*(c + d*x) - 200*a*b*Cosh[2*(c + d*x)] + 10*a*b*Cosh[4*( c + d*x)] + 960*a*b*Log[Cosh[c + d*x]] + 160*a*b*Sech[c + d*x]^2 - 40*(a^2 + 4*b^2)*Sinh[2*(c + d*x)] + 5*(a^2 + b^2)*Sinh[4*(c + d*x)] - 1152*b^2*T anh[c + d*x] + 224*b^2*Sech[c + d*x]^2*Tanh[c + d*x] - 32*b^2*Sech[c + d*x ]^4*Tanh[c + d*x])/(160*d)
Time = 0.66 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4146, 2335, 25, 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.
| \(\displaystyle \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (i c+i d x)^4 \left (a+i b \tan (i c+i d x)^3\right )^2dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (b \tanh ^3(c+d x)+a\right )^2}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {\frac {1}{4} \int -\frac {\tanh ^3(c+d x) \left (4 b^2 \tanh ^5(c+d x)+4 b^2 \tanh ^3(c+d x)+8 a b \tanh ^2(c+d x)+\left (a^2+5 b^2\right ) \tanh (c+d x)+8 a b\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {\tanh ^4(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tanh ^4(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {\tanh ^3(c+d x) \left (4 b^2 \tanh ^5(c+d x)+4 b^2 \tanh ^3(c+d x)+8 a b \tanh ^2(c+d x)+\left (a^2+5 b^2\right ) \tanh (c+d x)+8 a b\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {\frac {1}{4} \left (-\frac {1}{2} \int -\frac {\tanh ^2(c+d x) \left (8 b^2 \tanh ^4(c+d x)+16 b^2 \tanh ^2(c+d x)+48 a b \tanh (c+d x)+3 \left (a^2+13 b^2\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {\tanh ^3(c+d x) \left (a^2+16 a b \tanh (c+d x)+13 b^2\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\tanh ^2(c+d x) \left (8 b^2 \tanh ^4(c+d x)+16 b^2 \tanh ^2(c+d x)+48 a b \tanh (c+d x)+3 \left (a^2+13 b^2\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {\tanh ^3(c+d x) \left (a^2+16 a b \tanh (c+d x)+13 b^2\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \left (-8 b^2 \tanh ^4(c+d x)-24 b^2 \tanh ^2(c+d x)-48 a b \tanh (c+d x)-3 \left (a^2+21 b^2\right )+\frac {3 \left (a^2+16 b \tanh (c+d x) a+21 b^2\right )}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)-\frac {\tanh ^3(c+d x) \left (a^2+16 a b \tanh (c+d x)+13 b^2\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 \left (a^2+21 b^2\right ) \text {arctanh}(\tanh (c+d x))-3 \left (a^2+21 b^2\right ) \tanh (c+d x)-24 a b \tanh ^2(c+d x)-24 a b \log \left (1-\tanh ^2(c+d x)\right )-\frac {8}{5} b^2 \tanh ^5(c+d x)-8 b^2 \tanh ^3(c+d x)\right )-\frac {\tanh ^3(c+d x) \left (a^2+16 a b \tanh (c+d x)+13 b^2\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (\left (a^2+b^2\right ) \tanh (c+d x)+2 a b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
Input:
Int[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^3)^2,x]
Output:
((Tanh[c + d*x]^4*(2*a*b + (a^2 + b^2)*Tanh[c + d*x]))/(4*(1 - Tanh[c + d* x]^2)^2) + (-1/2*(Tanh[c + d*x]^3*(a^2 + 13*b^2 + 16*a*b*Tanh[c + d*x]))/( 1 - Tanh[c + d*x]^2) + (3*(a^2 + 21*b^2)*ArcTanh[Tanh[c + d*x]] - 24*a*b*L og[1 - Tanh[c + d*x]^2] - 3*(a^2 + 21*b^2)*Tanh[c + d*x] - 24*a*b*Tanh[c + d*x]^2 - 8*b^2*Tanh[c + d*x]^3 - (8*b^2*Tanh[c + d*x]^5)/5)/2)/4)/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 = 16.18 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{6}}{4 \cosh \left (d x +c \right )^{2}}-\frac {3 \sinh \left (d x +c \right )^{4}}{4 \cosh \left (d x +c \right )^{2}}+3 \ln \left (\cosh \left (d x +c \right )\right )-\frac {3 \tanh \left (d x +c \right )^{2}}{2}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{9}}{4 \cosh \left (d x +c \right )^{5}}-\frac {9 \sinh \left (d x +c \right )^{7}}{8 \cosh \left (d x +c \right )^{5}}+\frac {63 d x}{8}+\frac {63 c}{8}-\frac {63 \tanh \left (d x +c \right )}{8}-\frac {21 \tanh \left (d x +c \right )^{3}}{8}-\frac {63 \tanh \left (d x +c \right )^{5}}{40}\right )}{d}\) | \(180\) |
| default | \(\frac {a^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{6}}{4 \cosh \left (d x +c \right )^{2}}-\frac {3 \sinh \left (d x +c \right )^{4}}{4 \cosh \left (d x +c \right )^{2}}+3 \ln \left (\cosh \left (d x +c \right )\right )-\frac {3 \tanh \left (d x +c \right )^{2}}{2}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{9}}{4 \cosh \left (d x +c \right )^{5}}-\frac {9 \sinh \left (d x +c \right )^{7}}{8 \cosh \left (d x +c \right )^{5}}+\frac {63 d x}{8}+\frac {63 c}{8}-\frac {63 \tanh \left (d x +c \right )}{8}-\frac {21 \tanh \left (d x +c \right )^{3}}{8}-\frac {63 \tanh \left (d x +c \right )^{5}}{40}\right )}{d}\) | \(180\) |
| risch | \(\frac {3 a^{2} x}{8}-6 a b x +\frac {63 b^{2} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a^{2}}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} a b}{32 d}+\frac {{\mathrm e}^{4 d x +4 c} b^{2}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}-\frac {5 \,{\mathrm e}^{2 d x +2 c} a b}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{2 d}+\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}-\frac {5 \,{\mathrm e}^{-2 d x -2 c} a b}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{2 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{2}}{64 d}+\frac {{\mathrm e}^{-4 d x -4 c} a b}{32 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{2}}{64 d}-\frac {12 a b c}{d}+\frac {4 b \left (5 \,{\mathrm e}^{8 d x +8 c} a +25 \,{\mathrm e}^{8 d x +8 c} b +15 \,{\mathrm e}^{6 d x +6 c} a +75 \,{\mathrm e}^{6 d x +6 c} b +15 \,{\mathrm e}^{4 d x +4 c} a +105 b \,{\mathrm e}^{4 d x +4 c}+5 \,{\mathrm e}^{2 d x +2 c} a +65 \,{\mathrm e}^{2 d x +2 c} b +18 b \right )}{5 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}+\frac {6 a b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(365\) |
Input:
int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+2 *a*b*(1/4*sinh(d*x+c)^6/cosh(d*x+c)^2-3/4*sinh(d*x+c)^4/cosh(d*x+c)^2+3*ln (cosh(d*x+c))-3/2*tanh(d*x+c)^2)+b^2*(1/4*sinh(d*x+c)^9/cosh(d*x+c)^5-9/8* sinh(d*x+c)^7/cosh(d*x+c)^5+63/8*d*x+63/8*c-63/8*tanh(d*x+c)-21/8*tanh(d*x +c)^3-63/40*tanh(d*x+c)^5))
Leaf count of result is larger than twice the leaf count of optimal. 5034 vs. \(2 (172) = 344\).
Time = 0.16 (sec) , antiderivative size = 5034, normalized size of antiderivative = 27.66 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**3)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (172) = 344\).
Time = 0.14 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.08 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {1}{64} \, a^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{320} \, b^{2} {\left (\frac {2520 \, {\left (d x + c\right )}}{d} + \frac {5 \, {\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}\right )}}{d} - \frac {135 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5358 \, e^{\left (-4 \, d x - 4 \, c\right )} + 18190 \, e^{\left (-6 \, d x - 6 \, c\right )} + 28455 \, e^{\left (-8 \, d x - 8 \, c\right )} + 19995 \, e^{\left (-10 \, d x - 10 \, c\right )} + 6560 \, e^{\left (-12 \, d x - 12 \, c\right )} - 5}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 5 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 10 \, e^{\left (-10 \, d x - 10 \, c\right )} + 5 \, e^{\left (-12 \, d x - 12 \, c\right )} + e^{\left (-14 \, d x - 14 \, c\right )}\right )}}\right )} + \frac {1}{32} \, a b {\left (\frac {192 \, {\left (d x + c\right )}}{d} - \frac {20 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} + \frac {192 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 39 \, e^{\left (-4 \, d x - 4 \, c\right )} - 108 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} \] Input:
integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^2,x, algorithm="maxima")
Output:
1/64*a^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2 *c)/d - e^(-4*d*x - 4*c)/d) + 1/320*b^2*(2520*(d*x + c)/d + 5*(32*e^(-2*d* x - 2*c) - e^(-4*d*x - 4*c))/d - (135*e^(-2*d*x - 2*c) + 5358*e^(-4*d*x - 4*c) + 18190*e^(-6*d*x - 6*c) + 28455*e^(-8*d*x - 8*c) + 19995*e^(-10*d*x - 10*c) + 6560*e^(-12*d*x - 12*c) - 5)/(d*(e^(-4*d*x - 4*c) + 5*e^(-6*d*x - 6*c) + 10*e^(-8*d*x - 8*c) + 10*e^(-10*d*x - 10*c) + 5*e^(-12*d*x - 12*c ) + e^(-14*d*x - 14*c)))) + 1/32*a*b*(192*(d*x + c)/d - (20*e^(-2*d*x - 2* c) - e^(-4*d*x - 4*c))/d + 192*log(e^(-2*d*x - 2*c) + 1)/d - (18*e^(-2*d*x - 2*c) + 39*e^(-4*d*x - 4*c) - 108*e^(-6*d*x - 6*c) - 1)/(d*(e^(-4*d*x - 4*c) + 2*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c))))
Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (172) = 344\).
Time = 0.29 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.05 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {5 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 5 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 200 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 160 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 1920 \, a b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + 120 \, {\left (a^{2} - 16 \, a b + 21 \, b^{2}\right )} {\left (d x + c\right )} - 5 \, {\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 288 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 378 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 40 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 32 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {32 \, {\left (137 \, a b e^{\left (10 \, d x + 10 \, c\right )} + 645 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 200 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 1250 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 600 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 1250 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 840 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 645 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 520 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 137 \, a b - 144 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{320 \, d} \] Input:
integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^2,x, algorithm="giac")
Output:
1/320*(5*a^2*e^(4*d*x + 4*c) + 10*a*b*e^(4*d*x + 4*c) + 5*b^2*e^(4*d*x + 4 *c) - 40*a^2*e^(2*d*x + 2*c) - 200*a*b*e^(2*d*x + 2*c) - 160*b^2*e^(2*d*x + 2*c) + 1920*a*b*log(e^(2*d*x + 2*c) + 1) + 120*(a^2 - 16*a*b + 21*b^2)*( d*x + c) - 5*(18*a^2*e^(4*d*x + 4*c) - 288*a*b*e^(4*d*x + 4*c) + 378*b^2*e ^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) + 40*a*b*e^(2*d*x + 2*c) - 32*b^2*e ^(2*d*x + 2*c) + a^2 - 2*a*b + b^2)*e^(-4*d*x - 4*c) - 32*(137*a*b*e^(10*d *x + 10*c) + 645*a*b*e^(8*d*x + 8*c) - 200*b^2*e^(8*d*x + 8*c) + 1250*a*b* e^(6*d*x + 6*c) - 600*b^2*e^(6*d*x + 6*c) + 1250*a*b*e^(4*d*x + 4*c) - 840 *b^2*e^(4*d*x + 4*c) + 645*a*b*e^(2*d*x + 2*c) - 520*b^2*e^(2*d*x + 2*c) + 137*a*b - 144*b^2)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 0.49 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.97 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=x\,\left (\frac {3\,a^2}{8}-6\,a\,b+\frac {63\,b^2}{8}\right )+\frac {4\,\left (5\,b^2+a\,b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a^2-5\,a\,b+4\,b^2\right )}{8\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2+5\,a\,b+4\,b^2\right )}{8\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^2}{64\,d}-\frac {4\,\left (5\,b^2+a\,b\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {24\,b^2}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a-b\right )}^2}{64\,d}-\frac {16\,b^2}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {32\,b^2}{5\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}+\frac {6\,a\,b\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d} \] Input:
int(sinh(c + d*x)^4*(a + b*tanh(c + d*x)^3)^2,x)
Output:
x*((3*a^2)/8 - 6*a*b + (63*b^2)/8) + (4*(a*b + 5*b^2))/(d*(exp(2*c + 2*d*x ) + 1)) + (exp(- 2*c - 2*d*x)*(a^2 - 5*a*b + 4*b^2))/(8*d) - (exp(2*c + 2* d*x)*(5*a*b + a^2 + 4*b^2))/(8*d) + (exp(4*c + 4*d*x)*(a + b)^2)/(64*d) - (4*(a*b + 5*b^2))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) + (24*b^ 2)/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - (exp(- 4*c - 4*d*x)*(a - b)^2)/(64*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)) + (6*a*b*log(exp(2*c)*exp(2*d *x) + 1))/d
Time = 0.24 (sec) , antiderivative size = 894, normalized size of antiderivative = 4.91 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {10 e^{18 d x +18 c} a b -150 e^{16 d x +16 c} a b +2540 e^{8 d x +8 c} a b -88 e^{14 d x +14 c} a^{2}-1752 e^{14 d x +14 c} b^{2}+440 e^{10 d x +10 c} a^{2}+8400 e^{10 d x +10 c} b^{2}+1920 e^{14 d x +14 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b +1200 e^{10 d x +10 c} a^{2} d x +25200 e^{10 d x +10 c} b^{2} d x +1200 e^{8 d x +8 c} a^{2} d x +25200 e^{8 d x +8 c} b^{2} d x +600 e^{6 d x +6 c} a^{2} d x +12600 e^{6 d x +6 c} b^{2} d x +120 e^{4 d x +4 c} a^{2} d x +2520 e^{4 d x +4 c} b^{2} d x +120 e^{14 d x +14 c} a^{2} d x +2520 e^{14 d x +14 c} b^{2} d x +9600 e^{12 d x +12 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b +600 e^{12 d x +12 c} a^{2} d x +12600 e^{12 d x +12 c} b^{2} d x +19200 e^{10 d x +10 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b +19200 e^{8 d x +8 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b +9600 e^{6 d x +6 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b +1920 e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a b -5 a^{2}-19200 e^{10 d x +10 c} a b d x -19200 e^{8 d x +8 c} a b d x -9600 e^{6 d x +6 c} a b d x -1920 e^{4 d x +4 c} a b d x +5 e^{18 d x +18 c} a^{2}+5 e^{18 d x +18 c} b^{2}-15 e^{16 d x +16 c} a^{2}-135 e^{16 d x +16 c} b^{2}+135 e^{2 d x +2 c} b^{2}-1920 e^{14 d x +14 c} a b d x -9600 e^{12 d x +12 c} a b d x -5 b^{2}-736 e^{14 d x +14 c} a b +2540 e^{10 d x +10 c} a b -736 e^{4 d x +4 c} a b -150 e^{2 d x +2 c} a b +10 a b +800 e^{8 d x +8 c} a^{2}+17640 e^{8 d x +8 c} b^{2}+620 e^{6 d x +6 c} a^{2}+13020 e^{6 d x +6 c} b^{2}+212 e^{4 d x +4 c} a^{2}+4356 e^{4 d x +4 c} b^{2}+15 e^{2 d x +2 c} a^{2}}{320 e^{4 d x +4 c} d \left (e^{10 d x +10 c}+5 e^{8 d x +8 c}+10 e^{6 d x +6 c}+10 e^{4 d x +4 c}+5 e^{2 d x +2 c}+1\right )} \] Input:
int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^2,x)
Output:
(5*e**(18*c + 18*d*x)*a**2 + 10*e**(18*c + 18*d*x)*a*b + 5*e**(18*c + 18*d *x)*b**2 - 15*e**(16*c + 16*d*x)*a**2 - 150*e**(16*c + 16*d*x)*a*b - 135*e **(16*c + 16*d*x)*b**2 + 1920*e**(14*c + 14*d*x)*log(e**(2*c + 2*d*x) + 1) *a*b + 120*e**(14*c + 14*d*x)*a**2*d*x - 88*e**(14*c + 14*d*x)*a**2 - 1920 *e**(14*c + 14*d*x)*a*b*d*x - 736*e**(14*c + 14*d*x)*a*b + 2520*e**(14*c + 14*d*x)*b**2*d*x - 1752*e**(14*c + 14*d*x)*b**2 + 9600*e**(12*c + 12*d*x) *log(e**(2*c + 2*d*x) + 1)*a*b + 600*e**(12*c + 12*d*x)*a**2*d*x - 9600*e* *(12*c + 12*d*x)*a*b*d*x + 12600*e**(12*c + 12*d*x)*b**2*d*x + 19200*e**(1 0*c + 10*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b + 1200*e**(10*c + 10*d*x)*a**2 *d*x + 440*e**(10*c + 10*d*x)*a**2 - 19200*e**(10*c + 10*d*x)*a*b*d*x + 25 40*e**(10*c + 10*d*x)*a*b + 25200*e**(10*c + 10*d*x)*b**2*d*x + 8400*e**(1 0*c + 10*d*x)*b**2 + 19200*e**(8*c + 8*d*x)*log(e**(2*c + 2*d*x) + 1)*a*b + 1200*e**(8*c + 8*d*x)*a**2*d*x + 800*e**(8*c + 8*d*x)*a**2 - 19200*e**(8 *c + 8*d*x)*a*b*d*x + 2540*e**(8*c + 8*d*x)*a*b + 25200*e**(8*c + 8*d*x)*b **2*d*x + 17640*e**(8*c + 8*d*x)*b**2 + 9600*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 + 620*e**(6*c + 6*d*x)*a **2 - 9600*e**(6*c + 6*d*x)*a*b*d*x + 12600*e**(6*c + 6*d*x)*b**2*d*x + 13 020*e**(6*c + 6*d*x)*b**2 + 1920*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1 )*a*b + 120*e**(4*c + 4*d*x)*a**2*d*x + 212*e**(4*c + 4*d*x)*a**2 - 1920*e **(4*c + 4*d*x)*a*b*d*x - 736*e**(4*c + 4*d*x)*a*b + 2520*e**(4*c + 4*d...