\(\int \sinh ^4(c+d x) (a+b \tanh ^3(c+d x))^3 \, dx\) [65]

Optimal result

Integrand size = 23, antiderivative size = 286 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\frac {3}{8} a \left (a^2+63 b^2\right ) x-\frac {3 b \left (3 a^2+2 b^2\right ) \cosh ^2(c+d x)}{2 d}+\frac {b \left (3 a^2+b^2\right ) \cosh ^4(c+d x)}{4 d}+\frac {3 b \left (3 a^2+5 b^2\right ) \log (\cosh (c+d x))}{d}-\frac {a \left (5 a^2+51 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a \left (a^2+3 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac {18 a b^2 \tanh (c+d x)}{d}-\frac {b \left (3 a^2+10 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac {3 a b^2 \tanh ^3(c+d x)}{d}-\frac {3 b^3 \tanh ^4(c+d x)}{2 d}-\frac {3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac {b^3 \tanh ^6(c+d x)}{2 d}-\frac {b^3 \tanh ^8(c+d x)}{8 d} \] Output:

3/8*a*(a^2+63*b^2)*x-3/2*b*(3*a^2+2*b^2)*cosh(d*x+c)^2/d+1/4*b*(3*a^2+b^2) 
*cosh(d*x+c)^4/d+3*b*(3*a^2+5*b^2)*ln(cosh(d*x+c))/d-1/8*a*(5*a^2+51*b^2)* 
cosh(d*x+c)*sinh(d*x+c)/d+1/4*a*(a^2+3*b^2)*cosh(d*x+c)^3*sinh(d*x+c)/d-18 
*a*b^2*tanh(d*x+c)/d-1/2*b*(3*a^2+10*b^2)*tanh(d*x+c)^2/d-3*a*b^2*tanh(d*x 
+c)^3/d-3/2*b^3*tanh(d*x+c)^4/d-3/5*a*b^2*tanh(d*x+c)^5/d-1/2*b^3*tanh(d*x 
+c)^6/d-1/8*b^3*tanh(d*x+c)^8/d
 

Mathematica [A] (verified)

Time = 6.85 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.03 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\frac {3 a \left (a^2+63 b^2\right ) (c+d x)}{8 d}-\frac {b \left (15 a^2+11 b^2\right ) \cosh (2 (c+d x))}{8 d}+\frac {b \left (3 a^2+b^2\right ) \cosh (4 (c+d x))}{32 d}+\frac {3 \left (3 a^2 b+5 b^3\right ) \log (\cosh (c+d x))}{d}+\frac {b \left (3 a^2+20 b^2\right ) \text {sech}^2(c+d x)}{2 d}-\frac {15 b^3 \text {sech}^4(c+d x)}{4 d}+\frac {b^3 \text {sech}^6(c+d x)}{d}-\frac {b^3 \text {sech}^8(c+d x)}{8 d}-\frac {a \left (a^2+12 b^2\right ) \sinh (2 (c+d x))}{4 d}+\frac {a \left (a^2+3 b^2\right ) \sinh (4 (c+d x))}{32 d}-\frac {108 a b^2 \tanh (c+d x)}{5 d}+\frac {21 a b^2 \text {sech}^2(c+d x) \tanh (c+d x)}{5 d}-\frac {3 a b^2 \text {sech}^4(c+d x) \tanh (c+d x)}{5 d} \] Input:

Integrate[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^3)^3,x]
 

Output:

(3*a*(a^2 + 63*b^2)*(c + d*x))/(8*d) - (b*(15*a^2 + 11*b^2)*Cosh[2*(c + d* 
x)])/(8*d) + (b*(3*a^2 + b^2)*Cosh[4*(c + d*x)])/(32*d) + (3*(3*a^2*b + 5* 
b^3)*Log[Cosh[c + d*x]])/d + (b*(3*a^2 + 20*b^2)*Sech[c + d*x]^2)/(2*d) - 
(15*b^3*Sech[c + d*x]^4)/(4*d) + (b^3*Sech[c + d*x]^6)/d - (b^3*Sech[c + d 
*x]^8)/(8*d) - (a*(a^2 + 12*b^2)*Sinh[2*(c + d*x)])/(4*d) + (a*(a^2 + 3*b^ 
2)*Sinh[4*(c + d*x)])/(32*d) - (108*a*b^2*Tanh[c + d*x])/(5*d) + (21*a*b^2 
*Sech[c + d*x]^2*Tanh[c + d*x])/(5*d) - (3*a*b^2*Sech[c + d*x]^4*Tanh[c + 
d*x])/(5*d)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.99, 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.

\(\displaystyle \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (i c+i d x)^4 \left (a+i b \tan (i c+i d x)^3\right )^3dx\)

\(\Big \downarrow \) 4146

\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (b \tanh ^3(c+d x)+a\right )^3}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 2335

\(\displaystyle \frac {\frac {1}{4} \int -\frac {\tanh ^3(c+d x) \left (4 b^3 \tanh ^8(c+d x)+4 b^3 \tanh ^6(c+d x)+12 a b^2 \tanh ^5(c+d x)+4 b^3 \tanh ^4(c+d x)+12 a b^2 \tanh ^3(c+d x)+4 b \left (3 a^2+b^2\right ) \tanh ^2(c+d x)+a \left (a^2+15 b^2\right ) \tanh (c+d x)+4 b \left (3 a^2+b^2\right )\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {\tanh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\tanh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {\tanh ^3(c+d x) \left (4 b^3 \tanh ^8(c+d x)+4 b^3 \tanh ^6(c+d x)+12 a b^2 \tanh ^5(c+d x)+4 b^3 \tanh ^4(c+d x)+12 a b^2 \tanh ^3(c+d x)+4 b \left (3 a^2+b^2\right ) \tanh ^2(c+d x)+a \left (a^2+15 b^2\right ) \tanh (c+d x)+4 b \left (3 a^2+b^2\right )\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 2335

\(\displaystyle \frac {\frac {1}{4} \left (-\frac {1}{2} \int -\frac {\tanh ^2(c+d x) \left (8 b^3 \tanh ^7(c+d x)+16 b^3 \tanh ^5(c+d x)+24 a b^2 \tanh ^4(c+d x)+24 b^3 \tanh ^3(c+d x)+48 a b^2 \tanh ^2(c+d x)+72 b \left (a^2+b^2\right ) \tanh (c+d x)+3 a \left (a^2+39 b^2\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {\tanh ^3(c+d x) \left (4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)+a \left (a^2+39 b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\tanh ^2(c+d x) \left (8 b^3 \tanh ^7(c+d x)+16 b^3 \tanh ^5(c+d x)+24 a b^2 \tanh ^4(c+d x)+24 b^3 \tanh ^3(c+d x)+48 a b^2 \tanh ^2(c+d x)+72 b \left (a^2+b^2\right ) \tanh (c+d x)+3 a \left (a^2+39 b^2\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {\tanh ^3(c+d x) \left (4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)+a \left (a^2+39 b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \left (-8 b^3 \tanh ^7(c+d x)-24 b^3 \tanh ^5(c+d x)-24 a b^2 \tanh ^4(c+d x)-48 b^3 \tanh ^3(c+d x)-72 a b^2 \tanh ^2(c+d x)-24 b \left (3 a^2+5 b^2\right ) \tanh (c+d x)-3 a \left (a^2+63 b^2\right )+\frac {3 \left (a^3+63 b^2 a+8 b \left (3 a^2+5 b^2\right ) \tanh (c+d x)\right )}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)-\frac {\tanh ^3(c+d x) \left (4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)+a \left (a^2+39 b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a \left (a^2+63 b^2\right ) \text {arctanh}(\tanh (c+d x))-12 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)-3 a \left (a^2+63 b^2\right ) \tanh (c+d x)-12 b \left (3 a^2+5 b^2\right ) \log \left (1-\tanh ^2(c+d x)\right )-\frac {24}{5} a b^2 \tanh ^5(c+d x)-24 a b^2 \tanh ^3(c+d x)-b^3 \tanh ^8(c+d x)-4 b^3 \tanh ^6(c+d x)-12 b^3 \tanh ^4(c+d x)\right )-\frac {\tanh ^3(c+d x) \left (4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)+a \left (a^2+39 b^2\right )\right )}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\)

Input:

Int[Sinh[c + d*x]^4*(a + b*Tanh[c + d*x]^3)^3,x]
 

Output:

((Tanh[c + d*x]^4*(b*(3*a^2 + b^2) + a*(a^2 + 3*b^2)*Tanh[c + d*x]))/(4*(1 
 - Tanh[c + d*x]^2)^2) + (-1/2*(Tanh[c + d*x]^3*(a*(a^2 + 39*b^2) + 4*b*(6 
*a^2 + 5*b^2)*Tanh[c + d*x]))/(1 - Tanh[c + d*x]^2) + (3*a*(a^2 + 63*b^2)* 
ArcTanh[Tanh[c + d*x]] - 12*b*(3*a^2 + 5*b^2)*Log[1 - Tanh[c + d*x]^2] - 3 
*a*(a^2 + 63*b^2)*Tanh[c + d*x] - 12*b*(3*a^2 + 5*b^2)*Tanh[c + d*x]^2 - 2 
4*a*b^2*Tanh[c + d*x]^3 - 12*b^3*Tanh[c + d*x]^4 - (24*a*b^2*Tanh[c + d*x] 
^5)/5 - 4*b^3*Tanh[c + d*x]^6 - b^3*Tanh[c + d*x]^8)/2)/4)/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2333

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]
 

rule 2335

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4146

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]
 

Maple [A] (verified)

Time = 51.99 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {a^{3} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right )^{6}}{4 \cosh \left (d x +c \right )^{2}}-\frac {3 \sinh \left (d x +c \right )^{4}}{4 \cosh \left (d x +c \right )^{2}}+3 \ln \left (\cosh \left (d x +c \right )\right )-\frac {3 \tanh \left (d x +c \right )^{2}}{2}\right )+3 b^{2} a \left (\frac {\sinh \left (d x +c \right )^{9}}{4 \cosh \left (d x +c \right )^{5}}-\frac {9 \sinh \left (d x +c \right )^{7}}{8 \cosh \left (d x +c \right )^{5}}+\frac {63 d x}{8}+\frac {63 c}{8}-\frac {63 \tanh \left (d x +c \right )}{8}-\frac {21 \tanh \left (d x +c \right )^{3}}{8}-\frac {63 \tanh \left (d x +c \right )^{5}}{40}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{12}}{4 \cosh \left (d x +c \right )^{8}}-\frac {3 \sinh \left (d x +c \right )^{10}}{2 \cosh \left (d x +c \right )^{8}}+15 \ln \left (\cosh \left (d x +c \right )\right )-\frac {15 \tanh \left (d x +c \right )^{2}}{2}-\frac {15 \tanh \left (d x +c \right )^{4}}{4}-\frac {5 \tanh \left (d x +c \right )^{6}}{2}-\frac {15 \tanh \left (d x +c \right )^{8}}{8}\right )}{d}\)	\(274\)
default	\(\frac {a^{3} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right )^{6}}{4 \cosh \left (d x +c \right )^{2}}-\frac {3 \sinh \left (d x +c \right )^{4}}{4 \cosh \left (d x +c \right )^{2}}+3 \ln \left (\cosh \left (d x +c \right )\right )-\frac {3 \tanh \left (d x +c \right )^{2}}{2}\right )+3 b^{2} a \left (\frac {\sinh \left (d x +c \right )^{9}}{4 \cosh \left (d x +c \right )^{5}}-\frac {9 \sinh \left (d x +c \right )^{7}}{8 \cosh \left (d x +c \right )^{5}}+\frac {63 d x}{8}+\frac {63 c}{8}-\frac {63 \tanh \left (d x +c \right )}{8}-\frac {21 \tanh \left (d x +c \right )^{3}}{8}-\frac {63 \tanh \left (d x +c \right )^{5}}{40}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{12}}{4 \cosh \left (d x +c \right )^{8}}-\frac {3 \sinh \left (d x +c \right )^{10}}{2 \cosh \left (d x +c \right )^{8}}+15 \ln \left (\cosh \left (d x +c \right )\right )-\frac {15 \tanh \left (d x +c \right )^{2}}{2}-\frac {15 \tanh \left (d x +c \right )^{4}}{4}-\frac {5 \tanh \left (d x +c \right )^{6}}{2}-\frac {15 \tanh \left (d x +c \right )^{8}}{8}\right )}{d}\)	\(274\)
risch	\(\frac {3 \,{\mathrm e}^{4 d x +4 c} a^{2} b}{64 d}+\frac {3 \,{\mathrm e}^{4 d x +4 c} b^{2} a}{64 d}-\frac {15 \,{\mathrm e}^{2 d x +2 c} a^{2} b}{16 d}-\frac {3 \,{\mathrm e}^{2 d x +2 c} b^{2} a}{2 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{3}}{64 d}+\frac {{\mathrm e}^{-4 d x -4 c} b^{3}}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} a^{3}}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} b^{3}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a^{3}}{8 d}-\frac {11 \,{\mathrm e}^{2 d x +2 c} b^{3}}{16 d}+\frac {{\mathrm e}^{-2 d x -2 c} a^{3}}{8 d}-\frac {11 \,{\mathrm e}^{-2 d x -2 c} b^{3}}{16 d}+\frac {3 a^{3} x}{8}-15 b^{3} x -\frac {15 \,{\mathrm e}^{-2 d x -2 c} a^{2} b}{16 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} b^{2} a}{2 d}+\frac {3 \,{\mathrm e}^{-4 d x -4 c} a^{2} b}{64 d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c} b^{2} a}{64 d}+\frac {189 a \,b^{2} x}{8}-9 a^{2} b x +\frac {9 b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{2}}{d}-\frac {30 b^{3} c}{d}+\frac {15 b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}-\frac {18 b \,a^{2} c}{d}+\frac {2 b \left (15 a^{2} {\mathrm e}^{14 d x +14 c}+150 a b \,{\mathrm e}^{14 d x +14 c}+100 b^{2} {\mathrm e}^{14 d x +14 c}+90 \,{\mathrm e}^{12 d x +12 c} a^{2}+900 \,{\mathrm e}^{12 d x +12 c} a b +450 \,{\mathrm e}^{12 d x +12 c} b^{2}+225 \,{\mathrm e}^{10 d x +10 c} a^{2}+2430 \,{\mathrm e}^{10 d x +10 c} a b +1060 \,{\mathrm e}^{10 d x +10 c} b^{2}+300 \,{\mathrm e}^{8 d x +8 c} a^{2}+3780 \,{\mathrm e}^{8 d x +8 c} a b +1340 \,{\mathrm e}^{8 d x +8 c} b^{2}+225 \,{\mathrm e}^{6 d x +6 c} a^{2}+3618 \,{\mathrm e}^{6 d x +6 c} a b +1060 \,{\mathrm e}^{6 d x +6 c} b^{2}+90 \,{\mathrm e}^{4 d x +4 c} a^{2}+2124 \,{\mathrm e}^{4 d x +4 c} a b +450 \,{\mathrm e}^{4 d x +4 c} b^{2}+15 \,{\mathrm e}^{2 d x +2 c} a^{2}+714 \,{\mathrm e}^{2 d x +2 c} b a +100 b^{2} {\mathrm e}^{2 d x +2 c}+108 a b \right )}{5 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{8}}\)	\(679\)

Input:

int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^3*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+3 
*a^2*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)+3*b^2*a*(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*ta 
nh(d*x+c)^3-63/40*tanh(d*x+c)^5)+b^3*(1/4*sinh(d*x+c)^12/cosh(d*x+c)^8-3/2 
*sinh(d*x+c)^10/cosh(d*x+c)^8+15*ln(cosh(d*x+c))-15/2*tanh(d*x+c)^2-15/4*t 
anh(d*x+c)^4-5/2*tanh(d*x+c)^6-15/8*tanh(d*x+c)^8))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12323 vs. \(2 (266) = 532\).

Time = 0.21 (sec) , antiderivative size = 12323, normalized size of antiderivative = 43.09 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\text {Too large to display} \] Input:

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:

integrate(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**3)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 647 vs. \(2 (266) = 532\).

Time = 0.15 (sec) , antiderivative size = 647, normalized size of antiderivative = 2.26 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^3,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/64*a^3*(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) + 3/320*a*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)))) + 1/64*b^3*(960*(d*x + c)/d - (44*e^(-2*d*x - 
2*c) - e^(-4*d*x - 4*c))/d + 960*log(e^(-2*d*x - 2*c) + 1)/d - (36*e^(-2*d 
*x - 2*c) + 324*e^(-4*d*x - 4*c) - 1384*e^(-6*d*x - 6*c) - 9126*e^(-8*d*x 
- 8*c) - 24112*e^(-10*d*x - 10*c) - 31868*e^(-12*d*x - 12*c) - 25912*e^(-1 
4*d*x - 14*c) - 11169*e^(-16*d*x - 16*c) - 2516*e^(-18*d*x - 18*c) - 1)/(d 
*(e^(-4*d*x - 4*c) + 8*e^(-6*d*x - 6*c) + 28*e^(-8*d*x - 8*c) + 56*e^(-10* 
d*x - 10*c) + 70*e^(-12*d*x - 12*c) + 56*e^(-14*d*x - 14*c) + 28*e^(-16*d* 
x - 16*c) + 8*e^(-18*d*x - 18*c) + e^(-20*d*x - 20*c)))) + 3/64*a^2*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))))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 694 vs. \(2 (266) = 532\).

Time = 0.64 (sec) , antiderivative size = 694, normalized size of antiderivative = 2.43 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^3,x, algorithm="giac")
 

Output:

1/2240*(35*a^3*e^(4*d*x + 4*c) + 105*a^2*b*e^(4*d*x + 4*c) + 105*a*b^2*e^( 
4*d*x + 4*c) + 35*b^3*e^(4*d*x + 4*c) - 280*a^3*e^(2*d*x + 2*c) - 2100*a^2 
*b*e^(2*d*x + 2*c) - 3360*a*b^2*e^(2*d*x + 2*c) - 1540*b^3*e^(2*d*x + 2*c) 
 + 840*(a^3 - 24*a^2*b + 63*a*b^2 - 40*b^3)*(d*x + c) - 35*(18*a^3*e^(4*d* 
x + 4*c) - 432*a^2*b*e^(4*d*x + 4*c) + 1134*a*b^2*e^(4*d*x + 4*c) - 720*b^ 
3*e^(4*d*x + 4*c) - 8*a^3*e^(2*d*x + 2*c) + 60*a^2*b*e^(2*d*x + 2*c) - 96* 
a*b^2*e^(2*d*x + 2*c) + 44*b^3*e^(2*d*x + 2*c) + a^3 - 3*a^2*b + 3*a*b^2 - 
 b^3)*e^(-4*d*x - 4*c) + 6720*(3*a^2*b + 5*b^3)*log(e^(2*d*x + 2*c) + 1) - 
 8*(6849*a^2*b*e^(16*d*x + 16*c) + 11415*b^3*e^(16*d*x + 16*c) + 53112*a^2 
*b*e^(14*d*x + 14*c) - 16800*a*b^2*e^(14*d*x + 14*c) + 80120*b^3*e^(14*d*x 
 + 14*c) + 181692*a^2*b*e^(12*d*x + 12*c) - 100800*a*b^2*e^(12*d*x + 12*c) 
 + 269220*b^3*e^(12*d*x + 12*c) + 358344*a^2*b*e^(10*d*x + 10*c) - 272160* 
a*b^2*e^(10*d*x + 10*c) + 520520*b^3*e^(10*d*x + 10*c) + 445830*a^2*b*e^(8 
*d*x + 8*c) - 423360*a*b^2*e^(8*d*x + 8*c) + 648970*b^3*e^(8*d*x + 8*c) + 
358344*a^2*b*e^(6*d*x + 6*c) - 405216*a*b^2*e^(6*d*x + 6*c) + 520520*b^3*e 
^(6*d*x + 6*c) + 181692*a^2*b*e^(4*d*x + 4*c) - 237888*a*b^2*e^(4*d*x + 4* 
c) + 269220*b^3*e^(4*d*x + 4*c) + 53112*a^2*b*e^(2*d*x + 2*c) - 79968*a*b^ 
2*e^(2*d*x + 2*c) + 80120*b^3*e^(2*d*x + 2*c) + 6849*a^2*b - 12096*a*b^2 + 
 11415*b^3)/(e^(2*d*x + 2*c) + 1)^8)/d
 

Mupad [B] (verification not implemented)

Time = 2.95 (sec) , antiderivative size = 682, normalized size of antiderivative = 2.38 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

int(sinh(c + d*x)^4*(a + b*tanh(c + d*x)^3)^3,x)
 

Output:

x*((189*a*b^2)/8 - 9*a^2*b + (3*a^3)/8 - 15*b^3) - (4*(12*a*b^2 + 71*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)) - (256*b^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*exp(10*c + 10*d*x) + exp(1 
2*c + 12*d*x) + 1)) + (log(exp(2*c)*exp(2*d*x) + 1)*(9*a^2*b + 15*b^3))/d 
+ (2*(30*a*b^2 + 3*a^2*b + 20*b^3))/(d*(exp(2*c + 2*d*x) + 1)) + (32*(3*a* 
b^2 + 50*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)) - (2*(30*a*b^2 + 3*a^2*b + 50*b^3))/(d*(2*exp(2*c + 2*d*x) + ex 
p(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) + 2 
8*exp(12*c + 12*d*x) + 8*exp(14*c + 14*d*x) + exp(16*c + 16*d*x) + 1)) + ( 
exp(4*c + 4*d*x)*(a + b)^3)/(64*d) - (exp(- 4*c - 4*d*x)*(a - b)^3)/(64*d) 
 + (8*(9*a*b^2 + 23*b^3))/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + ex 
p(6*c + 6*d*x) + 1)) - (exp(2*c + 2*d*x)*(a + b)^2*(2*a + 11*b))/(16*d) + 
(exp(- 2*c - 2*d*x)*(a - b)^2*(2*a - 11*b))/(16*d)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1902, normalized size of antiderivative = 6.65 \[ \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^3)^3,x)
 

Output:

(10*e**(24*c + 24*d*x)*a**3 + 30*e**(24*c + 24*d*x)*a**2*b + 30*e**(24*c + 
 24*d*x)*a*b**2 + 10*e**(24*c + 24*d*x)*b**3 - 360*e**(22*c + 22*d*x)*a**2 
*b - 720*e**(22*c + 22*d*x)*a*b**2 - 360*e**(22*c + 22*d*x)*b**3 + 5760*e* 
*(20*c + 20*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b + 9600*e**(20*c + 20*d*x 
)*log(e**(2*c + 2*d*x) + 1)*b**3 + 240*e**(20*c + 20*d*x)*a**3*d*x - 160*e 
**(20*c + 20*d*x)*a**3 - 5760*e**(20*c + 20*d*x)*a**2*b*d*x - 2475*e**(20* 
c + 20*d*x)*a**2*b + 15120*e**(20*c + 20*d*x)*a*b**2*d*x - 8610*e**(20*c + 
 20*d*x)*a*b**2 - 9600*e**(20*c + 20*d*x)*b**3*d*x - 4915*e**(20*c + 20*d* 
x)*b**3 + 46080*e**(18*c + 18*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b + 7680 
0*e**(18*c + 18*d*x)*log(e**(2*c + 2*d*x) + 1)*b**3 + 1920*e**(18*c + 18*d 
*x)*a**3*d*x - 46080*e**(18*c + 18*d*x)*a**2*b*d*x + 120960*e**(18*c + 18* 
d*x)*a*b**2*d*x - 76800*e**(18*c + 18*d*x)*b**3*d*x + 161280*e**(16*c + 16 
*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b + 268800*e**(16*c + 16*d*x)*log(e** 
(2*c + 2*d*x) + 1)*b**3 + 6720*e**(16*c + 16*d*x)*a**3*d*x + 2450*e**(16*c 
 + 16*d*x)*a**3 - 161280*e**(16*c + 16*d*x)*a**2*b*d*x + 28350*e**(16*c + 
16*d*x)*a**2*b + 423360*e**(16*c + 16*d*x)*a*b**2*d*x + 136830*e**(16*c + 
16*d*x)*a*b**2 - 268800*e**(16*c + 16*d*x)*b**3*d*x + 40850*e**(16*c + 16* 
d*x)*b**3 + 322560*e**(14*c + 14*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2*b + 5 
37600*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 + 8320*e**(14*c + 14*d*x)*a**3 - 322560*e**(14*c + 14...