Integrand size = 23, antiderivative size = 84 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3}{8} a \left (a^2+4 a b+8 b^2\right ) x+\frac {3 a^2 (a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {b^3 \tanh (c+d x)}{d} \]
3/8*a*(a^2+4*a*b+8*b^2)*x+3/8*a^2*(a+4*b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*a^ 3*cosh(d*x+c)^3*sinh(d*x+c)/d+b^3*tanh(d*x+c)/d
Time = 1.69 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {12 a \left (a^2+4 a b+8 b^2\right ) (c+d x)+8 a^2 (a+3 b) \sinh (2 (c+d x))+a^3 \sinh (4 (c+d x))+32 b^3 \tanh (c+d x)}{32 d} \]
(12*a*(a^2 + 4*a*b + 8*b^2)*(c + d*x) + 8*a^2*(a + 3*b)*Sinh[2*(c + d*x)] + a^3*Sinh[4*(c + d*x)] + 32*b^3*Tanh[c + d*x])/(32*d)
Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 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.
\(\displaystyle \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (i c+i d x)^2\right )^3}{\sec (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\left (-b \tanh ^2(c+d x)+a+b\right )^3}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (b^3+\frac {3 a b^2 \tanh ^4(c+d x)-3 a b (a+2 b) \tanh ^2(c+d x)+a \left (a^2+3 b a+3 b^2\right )}{\left (1-\tanh ^2(c+d x)\right )^3}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^3 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}+\frac {3}{8} a \left (a^2+4 a b+8 b^2\right ) \text {arctanh}(\tanh (c+d x))+\frac {3 a^2 (a+4 b) \tanh (c+d x)}{8 \left (1-\tanh ^2(c+d x)\right )}+b^3 \tanh (c+d x)}{d}\) |
((3*a*(a^2 + 4*a*b + 8*b^2)*ArcTanh[Tanh[c + d*x]])/8 + b^3*Tanh[c + d*x] + (a^3*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + (3*a^2*(a + 4*b)*Tanh[ c + d*x])/(8*(1 - Tanh[c + d*x]^2)))/d
3.1.65.3.1 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_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Time = 1.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a \,b^{2} \left (d x +c \right )+b^{3} \tanh \left (d x +c \right )}{d}\) | \(93\) |
default | \(\frac {a^{3} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a \,b^{2} \left (d x +c \right )+b^{3} \tanh \left (d x +c \right )}{d}\) | \(93\) |
parallelrisch | \(\frac {\left (9 a^{3}+24 a^{2} b \right ) \sinh \left (3 d x +3 c \right )+a^{3} \sinh \left (5 d x +5 c \right )+24 a d x \left (a^{2}+4 a b +8 b^{2}\right ) \cosh \left (d x +c \right )+8 \sinh \left (d x +c \right ) \left (a^{3}+3 a^{2} b +8 b^{3}\right )}{64 d \cosh \left (d x +c \right )}\) | \(97\) |
risch | \(\frac {3 a^{3} x}{8}+\frac {3 b \,a^{2} x}{2}+3 a \,b^{2} x +\frac {a^{3} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {a^{3} {\mathrm e}^{2 d x +2 c}}{8 d}+\frac {3 a^{2} {\mathrm e}^{2 d x +2 c} b}{8 d}-\frac {a^{3} {\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {3 a^{2} {\mathrm e}^{-2 d x -2 c} b}{8 d}-\frac {a^{3} {\mathrm e}^{-4 d x -4 c}}{64 d}-\frac {2 b^{3}}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )}\) | \(147\) |
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*(1/2*cosh(d*x+c)*sinh(d*x+c)+1/2*d*x+1/2*c)+3*a*b^2*(d*x+c)+b^3*tan h(d*x+c))
Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.82 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {a^{3} \sinh \left (d x + c\right )^{5} + {\left (10 \, a^{3} \cosh \left (d x + c\right )^{2} + 9 \, a^{3} + 24 \, a^{2} b\right )} \sinh \left (d x + c\right )^{3} - 8 \, {\left (8 \, b^{3} - 3 \, {\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right ) + {\left (5 \, a^{3} \cosh \left (d x + c\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 64 \, b^{3} + 9 \, {\left (3 \, a^{3} + 8 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \]
1/64*(a^3*sinh(d*x + c)^5 + (10*a^3*cosh(d*x + c)^2 + 9*a^3 + 24*a^2*b)*si nh(d*x + c)^3 - 8*(8*b^3 - 3*(a^3 + 4*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c) + (5*a^3*cosh(d*x + c)^4 + 8*a^3 + 24*a^2*b + 64*b^3 + 9*(3*a^3 + 8*a^2*b) *cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c))
Timed out. \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.55 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {1}{64} \, a^{3} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {3}{8} \, a^{2} b {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a b^{2} x + \frac {2 \, b^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
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) + 3/8*a^2*b*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d *x - 2*c)/d) + 3*a*b^2*x + 2*b^3/(d*(e^(-2*d*x - 2*c) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (78) = 156\).
Time = 0.31 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.11 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {128 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - {\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 72 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]
1/64*(a^3*e^(4*d*x + 4*c) + 8*a^3*e^(2*d*x + 2*c) + 24*a^2*b*e^(2*d*x + 2* c) + 24*(a^3 + 4*a^2*b + 8*a*b^2)*(d*x + c) - 128*b^3/(e^(2*d*x + 2*c) + 1 ) - (18*a^3*e^(4*d*x + 4*c) + 72*a^2*b*e^(4*d*x + 4*c) + 144*a*b^2*e^(4*d* x + 4*c) + 8*a^3*e^(2*d*x + 2*c) + 24*a^2*b*e^(2*d*x + 2*c) + a^3)*e^(-4*d *x - 4*c))/d
Time = 2.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.39 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3\,a\,x\,\left (a^2+4\,a\,b+8\,b^2\right )}{8}-\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {a^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {a^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+3\,b\right )}{8\,d}+\frac {a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+3\,b\right )}{8\,d} \]