Integrand size = 23, antiderivative size = 91 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {3}{8} (a+b) \left (a^2-2 a b+5 b^2\right ) x+\frac {3 (a-3 b) (a+b)^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b)^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac {b^3 \tanh (c+d x)}{d} \]
3/8*(a+b)*(a^2-2*a*b+5*b^2)*x+3/8*(a-3*b)*(a+b)^2*cosh(d*x+c)*sinh(d*x+c)/ d+1/4*(a+b)^3*cosh(d*x+c)^3*sinh(d*x+c)/d-b^3*tanh(d*x+c)/d
Time = 6.86 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {12 \left (a^3-a^2 b+3 a b^2+5 b^3\right ) (c+d x)+8 (a-2 b) (a+b)^2 \sinh (2 (c+d x))+(a+b)^3 \sinh (4 (c+d x))-32 b^3 \tanh (c+d x)}{32 d} \]
(12*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*(c + d*x) + 8*(a - 2*b)*(a + b)^2*Sinh [2*(c + d*x)] + (a + b)^3*Sinh[4*(c + d*x)] - 32*b^3*Tanh[c + d*x])/(32*d)
Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 300, 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.
\(\displaystyle \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\sec (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \frac {\left (b \tanh ^2(c+d x)+a\right )^3}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (\frac {3 b^2 (a+b) \tanh ^4(c+d x)+3 b \left (a^2-b^2\right ) \tanh ^2(c+d x)+a^3+b^3}{\left (1-\tanh ^2(c+d x)\right )^3}-b^3\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3}{8} (a+b) \left (a^2-2 a b+5 b^2\right ) \text {arctanh}(\tanh (c+d x))+\frac {3 (a-3 b) (a+b)^2 \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )}+\frac {(a+b)^3 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}+b^3 (-\tanh (c+d x))}{d}\) |
((3*(a + b)*(a^2 - 2*a*b + 5*b^2)*ArcTanh[Tanh[c + d*x]])/8 - b^3*Tanh[c + d*x] + ((a + b)^3*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + (3*(a - 3* b)*(a + b)^2*Tanh[c + d*x])/(8*(1 - Tanh[c + d*x]^2)))/d
3.1.97.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(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[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Leaf count of result is larger than twice the leaf count of optimal. \(183\) vs. \(2(85)=170\).
Time = 55.73 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.02
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+3 a \,b^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(184\) |
default | \(\frac {a^{3} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+3 a \,b^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{4 \cosh \left (d x +c \right )}-\frac {5 \sinh \left (d x +c \right )^{3}}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(184\) |
risch | \(\frac {3 a^{3} x}{8}-\frac {3 b \,a^{2} x}{8}+\frac {9 a \,b^{2} x}{8}+\frac {15 b^{3} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a^{3}}{64 d}+\frac {3 \,{\mathrm e}^{4 d x +4 c} a^{2} b}{64 d}+\frac {3 \,{\mathrm e}^{4 d x +4 c} a \,b^{2}}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} b^{3}}{64 d}+\frac {{\mathrm e}^{2 d x +2 c} a^{3}}{8 d}-\frac {3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{3}}{4 d}-\frac {{\mathrm e}^{-2 d x -2 c} a^{3}}{8 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{3}}{4 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{3}}{64 d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c} a^{2} b}{64 d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c} a \,b^{2}}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{3}}{64 d}+\frac {2 b^{3}}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(293\) |
1/d*(a^3*((1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*sinh(d*x+c)+3/8*d*x+3/8*c)+3 *a^2*b*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sinh(d*x+c)-1/8*d*x- 1/8*c)+3*a*b^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)+b^3*(1/4*sinh(d*x+c)^5/cosh(d*x+c)-5/8*sinh(d*x+c)^3/cosh(d*x+c)+15/8 *d*x+15/8*c-15/8*tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (85) = 170\).
Time = 0.26 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.49 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{5} + {\left (9 \, a^{3} + 3 \, a^{2} b - 21 \, a b^{2} - 15 \, b^{3} + 10 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 8 \, {\left (8 \, b^{3} + 3 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + {\left (5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 8 \, a^{3} - 24 \, a b^{2} - 80 \, b^{3} + 9 \, {\left (3 \, a^{3} + a^{2} b - 7 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \]
1/64*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^5 + (9*a^3 + 3*a^2*b - 21*a*b^2 - 15*b^3 + 10*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*s inh(d*x + c)^3 + 8*(8*b^3 + 3*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*d*x)*cosh(d* x + c) + (5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 8*a^3 - 24*a *b^2 - 80*b^3 + 9*(3*a^3 + a^2*b - 7*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh( d*x + c))/(d*cosh(d*x + c))
\[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \cosh ^{4}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (85) = 170\).
Time = 0.20 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.93 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{64} \, a^{3} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {3}{64} \, a b^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{64} \, b^{3} {\left (\frac {120 \, {\left (d x + c\right )}}{d} + \frac {16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} - \frac {3}{64} \, a^{2} b {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \]
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) + 3/64*a*b^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) + 1/64*b^3*(12 0*(d*x + c)/d + (16*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d - (15*e^(-2*d*x - 2*c) + 144*e^(-4*d*x - 4*c) - 1)/(d*(e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c )))) - 3/64*a^2*b*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d)
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (85) = 170\).
Time = 0.54 (sec) , antiderivative size = 282, normalized size of antiderivative = 3.10 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 24 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (a^{3} - a^{2} b + 3 \, a b^{2} + 5 \, b^{3}\right )} {\left (d x + c\right )} + \frac {128 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - {\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 54 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 24 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]
1/64*(a^3*e^(4*d*x + 4*c) + 3*a^2*b*e^(4*d*x + 4*c) + 3*a*b^2*e^(4*d*x + 4 *c) + b^3*e^(4*d*x + 4*c) + 8*a^3*e^(2*d*x + 2*c) - 24*a*b^2*e^(2*d*x + 2* c) - 16*b^3*e^(2*d*x + 2*c) + 24*(a^3 - a^2*b + 3*a*b^2 + 5*b^3)*(d*x + c) + 128*b^3/(e^(2*d*x + 2*c) + 1) - (18*a^3*e^(4*d*x + 4*c) - 18*a^2*b*e^(4 *d*x + 4*c) + 54*a*b^2*e^(4*d*x + 4*c) + 90*b^3*e^(4*d*x + 4*c) + 8*a^3*e^ (2*d*x + 2*c) - 24*a*b^2*e^(2*d*x + 2*c) - 16*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))/d
Time = 2.04 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.46 \[ \int \cosh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,\left (\frac {3\,a^3}{8}-\frac {3\,a^2\,b}{8}+\frac {9\,a\,b^2}{8}+\frac {15\,b^3}{8}\right )+\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{8\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{8\,d} \]