Integrand size = 23, antiderivative size = 226 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=-\frac {1}{2} a \left (a^2+21 b^2\right ) x+\frac {b \left (3 a^2+b^2\right ) \cosh ^2(c+d x)}{2 d}-\frac {b \left (6 a^2+5 b^2\right ) \log (\cosh (c+d x))}{d}+\frac {a \left (a^2+3 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {9 a b^2 \tanh (c+d x)}{d}+\frac {b \left (3 a^2+4 b^2\right ) \tanh ^2(c+d x)}{2 d}+\frac {2 a b^2 \tanh ^3(c+d x)}{d}+\frac {3 b^3 \tanh ^4(c+d x)}{4 d}+\frac {3 a b^2 \tanh ^5(c+d x)}{5 d}+\frac {b^3 \tanh ^6(c+d x)}{3 d}+\frac {b^3 \tanh ^8(c+d x)}{8 d} \] Output:
-1/2*a*(a^2+21*b^2)*x+1/2*b*(3*a^2+b^2)*cosh(d*x+c)^2/d-b*(6*a^2+5*b^2)*ln (cosh(d*x+c))/d+1/2*a*(a^2+3*b^2)*cosh(d*x+c)*sinh(d*x+c)/d+9*a*b^2*tanh(d *x+c)/d+1/2*b*(3*a^2+4*b^2)*tanh(d*x+c)^2/d+2*a*b^2*tanh(d*x+c)^3/d+3/4*b^ 3*tanh(d*x+c)^4/d+3/5*a*b^2*tanh(d*x+c)^5/d+1/3*b^3*tanh(d*x+c)^6/d+1/8*b^ 3*tanh(d*x+c)^8/d
Time = 6.84 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.08 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=-\frac {a \left (a^2+21 b^2\right ) (c+d x)}{2 d}+\frac {b \left (3 a^2+b^2\right ) \cosh (2 (c+d x))}{4 d}+\frac {\left (-6 a^2 b-5 b^3\right ) \log (\cosh (c+d x))}{d}-\frac {b \left (3 a^2+10 b^2\right ) \text {sech}^2(c+d x)}{2 d}+\frac {5 b^3 \text {sech}^4(c+d x)}{2 d}-\frac {5 b^3 \text {sech}^6(c+d x)}{6 d}+\frac {b^3 \text {sech}^8(c+d x)}{8 d}+\frac {a \left (a^2+3 b^2\right ) \sinh (2 (c+d x))}{4 d}+\frac {58 a b^2 \tanh (c+d x)}{5 d}-\frac {16 a b^2 \text {sech}^2(c+d x) \tanh (c+d x)}{5 d}+\frac {3 a b^2 \text {sech}^4(c+d x) \tanh (c+d x)}{5 d} \] Input:
Integrate[Sinh[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^3,x]
Output:
-1/2*(a*(a^2 + 21*b^2)*(c + d*x))/d + (b*(3*a^2 + b^2)*Cosh[2*(c + d*x)])/ (4*d) + ((-6*a^2*b - 5*b^3)*Log[Cosh[c + d*x]])/d - (b*(3*a^2 + 10*b^2)*Se ch[c + d*x]^2)/(2*d) + (5*b^3*Sech[c + d*x]^4)/(2*d) - (5*b^3*Sech[c + d*x ]^6)/(6*d) + (b^3*Sech[c + d*x]^8)/(8*d) + (a*(a^2 + 3*b^2)*Sinh[2*(c + d* x)])/(4*d) + (58*a*b^2*Tanh[c + d*x])/(5*d) - (16*a*b^2*Sech[c + d*x]^2*Ta nh[c + d*x])/(5*d) + (3*a*b^2*Sech[c + d*x]^4*Tanh[c + d*x])/(5*d)
Time = 0.62 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.99, 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.
| \(\displaystyle \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\sin (i c+i d x)^2 \left (a+i b \tan (i c+i d x)^3\right )^3dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \sin (i c+i d x)^2 \left (i b \tan (i c+i d x)^3+a\right )^3dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (b \tanh ^3(c+d x)+a\right )^3}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle \frac {\frac {1}{2} \int -\frac {\tanh (c+d x) \left (2 b^3 \tanh ^8(c+d x)+2 b^3 \tanh ^6(c+d x)+6 a b^2 \tanh ^5(c+d x)+2 b^3 \tanh ^4(c+d x)+6 a b^2 \tanh ^3(c+d x)+2 b \left (3 a^2+b^2\right ) \tanh ^2(c+d x)+a \left (a^2+9 b^2\right ) \tanh (c+d x)+2 b \left (3 a^2+b^2\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {\tanh ^2(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tanh ^2(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int \frac {\tanh (c+d x) \left (2 b^3 \tanh ^8(c+d x)+2 b^3 \tanh ^6(c+d x)+6 a b^2 \tanh ^5(c+d x)+2 b^3 \tanh ^4(c+d x)+6 a b^2 \tanh ^3(c+d x)+2 b \left (3 a^2+b^2\right ) \tanh ^2(c+d x)+a \left (a^2+9 b^2\right ) \tanh (c+d x)+2 b \left (3 a^2+b^2\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\frac {\tanh ^2(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int \left (-2 b^3 \tanh ^7(c+d x)-4 b^3 \tanh ^5(c+d x)-6 a b^2 \tanh ^4(c+d x)-6 b^3 \tanh ^3(c+d x)-12 a b^2 \tanh ^2(c+d x)-2 b \left (3 a^2+4 b^2\right ) \tanh (c+d x)-a \left (a^2+21 b^2\right )+\frac {a^3+21 b^2 a+2 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} \left (-a \left (a^2+21 b^2\right ) \text {arctanh}(\tanh (c+d x))+b \left (3 a^2+4 b^2\right ) \tanh ^2(c+d x)+a \left (a^2+21 b^2\right ) \tanh (c+d x)+b \left (6 a^2+5 b^2\right ) \log \left (1-\tanh ^2(c+d x)\right )+\frac {6}{5} a b^2 \tanh ^5(c+d x)+4 a b^2 \tanh ^3(c+d x)+\frac {1}{4} b^3 \tanh ^8(c+d x)+\frac {2}{3} b^3 \tanh ^6(c+d x)+\frac {3}{2} b^3 \tanh ^4(c+d x)\right )+\frac {\tanh ^2(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
Input:
Int[Sinh[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^3,x]
Output:
((Tanh[c + d*x]^2*(b*(3*a^2 + b^2) + a*(a^2 + 3*b^2)*Tanh[c + d*x]))/(2*(1 - Tanh[c + d*x]^2)) + (-(a*(a^2 + 21*b^2)*ArcTanh[Tanh[c + d*x]]) + b*(6* a^2 + 5*b^2)*Log[1 - Tanh[c + d*x]^2] + a*(a^2 + 21*b^2)*Tanh[c + d*x] + b *(3*a^2 + 4*b^2)*Tanh[c + d*x]^2 + 4*a*b^2*Tanh[c + d*x]^3 + (3*b^3*Tanh[c + d*x]^4)/2 + (6*a*b^2*Tanh[c + d*x]^5)/5 + (2*b^3*Tanh[c + d*x]^6)/3 + ( b^3*Tanh[c + d*x]^8)/4)/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 = 18.96 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right )^{4}}{2 \cosh \left (d x +c \right )^{2}}-2 \ln \left (\cosh \left (d x +c \right )\right )+\tanh \left (d x +c \right )^{2}\right )+3 b^{2} a \left (\frac {\sinh \left (d x +c \right )^{7}}{2 \cosh \left (d x +c \right )^{5}}-\frac {7 d x}{2}-\frac {7 c}{2}+\frac {7 \tanh \left (d x +c \right )}{2}+\frac {7 \tanh \left (d x +c \right )^{3}}{6}+\frac {7 \tanh \left (d x +c \right )^{5}}{10}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{10}}{2 \cosh \left (d x +c \right )^{8}}-5 \ln \left (\cosh \left (d x +c \right )\right )+\frac {5 \tanh \left (d x +c \right )^{2}}{2}+\frac {5 \tanh \left (d x +c \right )^{4}}{4}+\frac {5 \tanh \left (d x +c \right )^{6}}{6}+\frac {5 \tanh \left (d x +c \right )^{8}}{8}\right )}{d}\) | \(206\) |
| default | \(\frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right )^{4}}{2 \cosh \left (d x +c \right )^{2}}-2 \ln \left (\cosh \left (d x +c \right )\right )+\tanh \left (d x +c \right )^{2}\right )+3 b^{2} a \left (\frac {\sinh \left (d x +c \right )^{7}}{2 \cosh \left (d x +c \right )^{5}}-\frac {7 d x}{2}-\frac {7 c}{2}+\frac {7 \tanh \left (d x +c \right )}{2}+\frac {7 \tanh \left (d x +c \right )^{3}}{6}+\frac {7 \tanh \left (d x +c \right )^{5}}{10}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{10}}{2 \cosh \left (d x +c \right )^{8}}-5 \ln \left (\cosh \left (d x +c \right )\right )+\frac {5 \tanh \left (d x +c \right )^{2}}{2}+\frac {5 \tanh \left (d x +c \right )^{4}}{4}+\frac {5 \tanh \left (d x +c \right )^{6}}{6}+\frac {5 \tanh \left (d x +c \right )^{8}}{8}\right )}{d}\) | \(206\) |
| risch | \(-\frac {a^{3} x}{2}+6 a^{2} b x -\frac {21 a \,b^{2} x}{2}+5 b^{3} x +\frac {{\mathrm e}^{2 d x +2 c} a^{3}}{8 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} a^{2} b}{8 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} b^{2} a}{8 d}+\frac {{\mathrm e}^{2 d x +2 c} b^{3}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a^{3}}{8 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} a^{2} b}{8 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} b^{2} a}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{3}}{8 d}+\frac {12 b \,a^{2} c}{d}+\frac {10 b^{3} c}{d}-\frac {2 b \left (45 a^{2} {\mathrm e}^{14 d x +14 c}+270 a b \,{\mathrm e}^{14 d x +14 c}+150 b^{2} {\mathrm e}^{14 d x +14 c}+270 \,{\mathrm e}^{12 d x +12 c} a^{2}+1530 \,{\mathrm e}^{12 d x +12 c} a b +600 \,{\mathrm e}^{12 d x +12 c} b^{2}+675 \,{\mathrm e}^{10 d x +10 c} a^{2}+3990 \,{\mathrm e}^{10 d x +10 c} a b +1450 \,{\mathrm e}^{10 d x +10 c} b^{2}+900 \,{\mathrm e}^{8 d x +8 c} a^{2}+6090 \,{\mathrm e}^{8 d x +8 c} a b +1760 \,{\mathrm e}^{8 d x +8 c} b^{2}+675 \,{\mathrm e}^{6 d x +6 c} a^{2}+5754 \,{\mathrm e}^{6 d x +6 c} a b +1450 \,{\mathrm e}^{6 d x +6 c} b^{2}+270 \,{\mathrm e}^{4 d x +4 c} a^{2}+3342 \,{\mathrm e}^{4 d x +4 c} a b +600 \,{\mathrm e}^{4 d x +4 c} b^{2}+45 \,{\mathrm e}^{2 d x +2 c} a^{2}+1122 \,{\mathrm e}^{2 d x +2 c} b a +150 b^{2} {\mathrm e}^{2 d x +2 c}+174 a b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{8}}-\frac {6 b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{2}}{d}-\frac {5 b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(539\) |
Input:
int(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+3*a^2*b*(1/2*sinh(d*x +c)^4/cosh(d*x+c)^2-2*ln(cosh(d*x+c))+tanh(d*x+c)^2)+3*b^2*a*(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*t anh(d*x+c)^5)+b^3*(1/2*sinh(d*x+c)^10/cosh(d*x+c)^8-5*ln(cosh(d*x+c))+5/2* tanh(d*x+c)^2+5/4*tanh(d*x+c)^4+5/6*tanh(d*x+c)^6+5/8*tanh(d*x+c)^8))
Leaf count of result is larger than twice the leaf count of optimal. 9862 vs. \(2 (210) = 420\).
Time = 0.20 (sec) , antiderivative size = 9862, normalized size of antiderivative = 43.64 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**2*(a+b*tanh(d*x+c)**3)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 544 vs. \(2 (210) = 420\).
Time = 0.13 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.41 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^3,x, algorithm="maxima")
Output:
-1/8*a^3*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/40*a*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))) ) - 1/24*b^3*(120*(d*x + c)/d - 3*e^(-2*d*x - 2*c)/d + 120*log(e^(-2*d*x - 2*c) + 1)/d - (24*e^(-2*d*x - 2*c) - 396*e^(-4*d*x - 4*c) - 1752*e^(-6*d* x - 6*c) - 4430*e^(-8*d*x - 8*c) - 5464*e^(-10*d*x - 10*c) - 4556*e^(-12*d *x - 12*c) - 1896*e^(-14*d*x - 14*c) - 477*e^(-16*d*x - 16*c) + 3)/(d*(e^( -2*d*x - 2*c) + 8*e^(-4*d*x - 4*c) + 28*e^(-6*d*x - 6*c) + 56*e^(-8*d*x - 8*c) + 70*e^(-10*d*x - 10*c) + 56*e^(-12*d*x - 12*c) + 28*e^(-14*d*x - 14* c) + 8*e^(-16*d*x - 16*c) + e^(-18*d*x - 18*c)))) - 3/8*a^2*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. 577 vs. \(2 (210) = 420\).
Time = 0.47 (sec) , antiderivative size = 577, normalized size of antiderivative = 2.55 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^3,x, algorithm="giac")
Output:
1/840*(105*a^3*e^(2*d*x + 2*c) + 315*a^2*b*e^(2*d*x + 2*c) + 315*a*b^2*e^( 2*d*x + 2*c) + 105*b^3*e^(2*d*x + 2*c) - 420*(a^3 - 12*a^2*b + 21*a*b^2 - 10*b^3)*(d*x + c) + 105*(2*a^3*e^(2*d*x + 2*c) - 24*a^2*b*e^(2*d*x + 2*c) + 42*a*b^2*e^(2*d*x + 2*c) - 20*b^3*e^(2*d*x + 2*c) - a^3 + 3*a^2*b - 3*a* b^2 + b^3)*e^(-2*d*x - 2*c) - 840*(6*a^2*b + 5*b^3)*log(e^(2*d*x + 2*c) + 1) + (13698*a^2*b*e^(16*d*x + 16*c) + 11415*b^3*e^(16*d*x + 16*c) + 104544 *a^2*b*e^(14*d*x + 14*c) - 30240*a*b^2*e^(14*d*x + 14*c) + 74520*b^3*e^(14 *d*x + 14*c) + 353304*a^2*b*e^(12*d*x + 12*c) - 171360*a*b^2*e^(12*d*x + 1 2*c) + 252420*b^3*e^(12*d*x + 12*c) + 691488*a^2*b*e^(10*d*x + 10*c) - 446 880*a*b^2*e^(10*d*x + 10*c) + 476840*b^3*e^(10*d*x + 10*c) + 858060*a^2*b* e^(8*d*x + 8*c) - 682080*a*b^2*e^(8*d*x + 8*c) + 601930*b^3*e^(8*d*x + 8*c ) + 691488*a^2*b*e^(6*d*x + 6*c) - 644448*a*b^2*e^(6*d*x + 6*c) + 476840*b ^3*e^(6*d*x + 6*c) + 353304*a^2*b*e^(4*d*x + 4*c) - 374304*a*b^2*e^(4*d*x + 4*c) + 252420*b^3*e^(4*d*x + 4*c) + 104544*a^2*b*e^(2*d*x + 2*c) - 12566 4*a*b^2*e^(2*d*x + 2*c) + 74520*b^3*e^(2*d*x + 2*c) + 13698*a^2*b - 19488* a*b^2 + 11415*b^3)/(e^(2*d*x + 2*c) + 1)^8)/d
Time = 0.60 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.73 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\frac {8\,\left (29\,b^3+6\,a\,b^2\right )}{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 {736\,b^3}{3\,d\,\left (6\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,{\mathrm {e}}^{4\,c+4\,d\,x}+20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}+6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1\right )}-\frac {\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )\,\left (6\,a^2\,b+5\,b^3\right )}{d}-\frac {2\,\left (3\,a^2\,b+18\,a\,b^2+10\,b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {96\,\left (15\,b^3+a\,b^2\right )}{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 {128\,b^3}{d\,\left (7\,{\mathrm {e}}^{2\,c+2\,d\,x}+21\,{\mathrm {e}}^{4\,c+4\,d\,x}+35\,{\mathrm {e}}^{6\,c+6\,d\,x}+35\,{\mathrm {e}}^{8\,c+8\,d\,x}+21\,{\mathrm {e}}^{10\,c+10\,d\,x}+7\,{\mathrm {e}}^{12\,c+12\,d\,x}+{\mathrm {e}}^{14\,c+14\,d\,x}+1\right )}-\frac {x\,{\left (a-b\right )}^2\,\left (a-10\,b\right )}{2}+\frac {6\,\left (a^2\,b+8\,a\,b^2+10\,b^3\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {32\,b^3}{d\,\left (8\,{\mathrm {e}}^{2\,c+2\,d\,x}+28\,{\mathrm {e}}^{4\,c+4\,d\,x}+56\,{\mathrm {e}}^{6\,c+6\,d\,x}+70\,{\mathrm {e}}^{8\,c+8\,d\,x}+56\,{\mathrm {e}}^{10\,c+10\,d\,x}+28\,{\mathrm {e}}^{12\,c+12\,d\,x}+8\,{\mathrm {e}}^{14\,c+14\,d\,x}+{\mathrm {e}}^{16\,c+16\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a-b\right )}^3}{8\,d}-\frac {16\,\left (25\,b^3+12\,a\,b^2\right )}{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 )} \] Input:
int(sinh(c + d*x)^2*(a + b*tanh(c + d*x)^3)^3,x)
Output:
(8*(6*a*b^2 + 29*b^3))/(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)) + (736*b^3)/(3*d*(6*exp(2*c + 2*d*x ) + 15*exp(4*c + 4*d*x) + 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8*d*x) + 6*ex p(10*c + 10*d*x) + exp(12*c + 12*d*x) + 1)) - (log(exp(2*c)*exp(2*d*x) + 1 )*(6*a^2*b + 5*b^3))/d - (2*(18*a*b^2 + 3*a^2*b + 10*b^3))/(d*(exp(2*c + 2 *d*x) + 1)) - (96*(a*b^2 + 15*b^3))/(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)) - (128*b^3)/(d*(7*exp(2*c + 2*d*x) + 21*exp(4*c + 4*d*x) + 35*exp(6*c + 6*d*x) + 35*exp(8*c + 8*d*x) + 21*exp(10*c + 10*d*x) + 7*exp(12*c + 12* d*x) + exp(14*c + 14*d*x) + 1)) - (x*(a - b)^2*(a - 10*b))/2 + (6*(8*a*b^2 + a^2*b + 10*b^3))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) + (32* b^3)/(d*(8*exp(2*c + 2*d*x) + 28*exp(4*c + 4*d*x) + 56*exp(6*c + 6*d*x) + 70*exp(8*c + 8*d*x) + 56*exp(10*c + 10*d*x) + 28*exp(12*c + 12*d*x) + 8*ex p(14*c + 14*d*x) + exp(16*c + 16*d*x) + 1)) + (exp(2*c + 2*d*x)*(a + b)^3) /(8*d) - (exp(- 2*c - 2*d*x)*(a - b)^3)/(8*d) - (16*(12*a*b^2 + 25*b^3))/( 3*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1))
Time = 0.27 (sec) , antiderivative size = 1808, normalized size of antiderivative = 8.00 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:
int(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^3)^3,x)
Output:
(120*e**(20*c + 20*d*x)*a**3 + 360*e**(20*c + 20*d*x)*a**2*b + 360*e**(20* c + 20*d*x)*a*b**2 + 120*e**(20*c + 20*d*x)*b**3 - 5760*e**(18*c + 18*d*x) *log(e**(2*c + 2*d*x) + 1)*a**2*b - 4800*e**(18*c + 18*d*x)*log(e**(2*c + 2*d*x) + 1)*b**3 - 480*e**(18*c + 18*d*x)*a**3*d*x + 555*e**(18*c + 18*d*x )*a**3 + 5760*e**(18*c + 18*d*x)*a**2*b*d*x + 2295*e**(18*c + 18*d*x)*a**2 *b - 10080*e**(18*c + 18*d*x)*a*b**2*d*x + 5985*e**(18*c + 18*d*x)*a*b**2 + 4800*e**(18*c + 18*d*x)*b**3*d*x + 2925*e**(18*c + 18*d*x)*b**3 - 46080* e**(16*c + 16*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b - 38400*e**(16*c + 16* d*x)*log(e**(2*c + 2*d*x) + 1)*b**3 - 3840*e**(16*c + 16*d*x)*a**3*d*x + 4 6080*e**(16*c + 16*d*x)*a**2*b*d*x - 80640*e**(16*c + 16*d*x)*a*b**2*d*x + 38400*e**(16*c + 16*d*x)*b**3*d*x - 161280*e**(14*c + 14*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b - 134400*e**(14*c + 14*d*x)*log(e**(2*c + 2*d*x) + 1 )*b**3 - 13440*e**(14*c + 14*d*x)*a**3*d*x - 5580*e**(14*c + 14*d*x)*a**3 + 161280*e**(14*c + 14*d*x)*a**2*b*d*x - 27900*e**(14*c + 14*d*x)*a**2*b - 282240*e**(14*c + 14*d*x)*a*b**2*d*x - 91620*e**(14*c + 14*d*x)*a*b**2 + 134400*e**(14*c + 14*d*x)*b**3*d*x - 14100*e**(14*c + 14*d*x)*b**3 - 32256 0*e**(12*c + 12*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b - 268800*e**(12*c + 12*d*x)*log(e**(2*c + 2*d*x) + 1)*b**3 - 26880*e**(12*c + 12*d*x)*a**3*d*x - 17640*e**(12*c + 12*d*x)*a**3 + 322560*e**(12*c + 12*d*x)*a**2*b*d*x - 83880*e**(12*c + 12*d*x)*a**2*b - 564480*e**(12*c + 12*d*x)*a*b**2*d*x ...