Integrand size = 14, antiderivative size = 73 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=a^3 x+\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {b^3 \tanh ^5(c+d x)}{5 d} \]
a^3*x+b*(3*a^2+3*a*b+b^2)*tanh(d*x+c)/d-1/3*b^2*(3*a+2*b)*tanh(d*x+c)^3/d+ 1/5*b^3*tanh(d*x+c)^5/d
Time = 1.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=a^3 x+\frac {3 a^2 b \tanh (c+d x)}{d}+\frac {3 a b^2 \tanh (c+d x)}{d}+\frac {b^3 \tanh (c+d x)}{d}-\frac {a b^2 \tanh ^3(c+d x)}{d}-\frac {2 b^3 \tanh ^3(c+d x)}{3 d}+\frac {b^3 \tanh ^5(c+d x)}{5 d} \]
a^3*x + (3*a^2*b*Tanh[c + d*x])/d + (3*a*b^2*Tanh[c + d*x])/d + (b^3*Tanh[ c + d*x])/d - (a*b^2*Tanh[c + d*x]^3)/d - (2*b^3*Tanh[c + d*x]^3)/(3*d) + (b^3*Tanh[c + d*x]^5)/(5*d)
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4616, 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.
| \(\displaystyle \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \sec (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 4616 |
| \(\displaystyle \frac {\int \frac {\left (-b \tanh ^2(c+d x)+a+b\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (b^3 \tanh ^4(c+d x)-b^2 (3 a+2 b) \tanh ^2(c+d x)+b \left (3 a^2+3 b a+b^2\right )+\frac {a^3}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \text {arctanh}(\tanh (c+d x))+b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)-\frac {1}{3} b^2 (3 a+2 b) \tanh ^3(c+d x)+\frac {1}{5} b^3 \tanh ^5(c+d x)}{d}\) |
(a^3*ArcTanh[Tanh[c + d*x]] + b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x] - (b^2 *(3*a + 2*b)*Tanh[c + d*x]^3)/3 + (b^3*Tanh[c + d*x]^5)/5)/d
3.2.28.3.1 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[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Time = 1.51 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (d x +c \right )+3 a^{2} b \tanh \left (d x +c \right )+3 a \,b^{2} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b^{3} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d}\) | \(83\) |
| default | \(\frac {a^{3} \left (d x +c \right )+3 a^{2} b \tanh \left (d x +c \right )+3 a \,b^{2} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b^{3} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d}\) | \(83\) |
| parts | \(a^{3} x +\frac {b^{3} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d}+\frac {3 a^{2} b \tanh \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}\) | \(84\) |
| risch | \(a^{3} x -\frac {2 b \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+180 a^{2} {\mathrm e}^{6 d x +6 c}+90 a b \,{\mathrm e}^{6 d x +6 c}+270 a^{2} {\mathrm e}^{4 d x +4 c}+210 a b \,{\mathrm e}^{4 d x +4 c}+80 \,{\mathrm e}^{4 d x +4 c} b^{2}+180 a^{2} {\mathrm e}^{2 d x +2 c}+150 a b \,{\mathrm e}^{2 d x +2 c}+40 \,{\mathrm e}^{2 d x +2 c} b^{2}+45 a^{2}+30 a b +8 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(164\) |
| parallelrisch | \(\frac {15 x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} a^{3} d +30 \left (3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+75 x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{3} d +360 \left (a +\frac {b}{3}\right )^{2} b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+150 x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{3} d +4 \left (135 a^{2} b +75 a \,b^{2}+29 b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+150 x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{3} d +360 \left (a +\frac {b}{3}\right )^{2} b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+75 x \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3} d +30 \left (3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+15 a^{3} d x}{15 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{5}}\) | \(248\) |
1/d*(a^3*(d*x+c)+3*a^2*b*tanh(d*x+c)+3*a*b^2*(2/3+1/3*sech(d*x+c)^2)*tanh( d*x+c)+b^3*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (69) = 138\).
Time = 0.25 (sec) , antiderivative size = 470, normalized size of antiderivative = 6.44 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {{\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (27 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3} + 2 \, {\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, a^{3} d x - 45 \, a^{2} b - 30 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) + 5 \, {\left ({\left (45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 18 \, a^{2} b + 24 \, a b^{2} + 16 \, b^{3} + 3 \, {\left (27 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]
1/15*((15*a^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)^5 + 5*(15*a ^3*d*x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)*sinh(d*x + c)^4 + (45* a^2*b + 30*a*b^2 + 8*b^3)*sinh(d*x + c)^5 + 5*(15*a^3*d*x - 45*a^2*b - 30* a*b^2 - 8*b^3)*cosh(d*x + c)^3 + 5*(27*a^2*b + 30*a*b^2 + 8*b^3 + 2*(45*a^ 2*b + 30*a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 5*(2*(15*a^3*d* x - 45*a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c)^3 + 3*(15*a^3*d*x - 45*a^2* b - 30*a*b^2 - 8*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(15*a^3*d*x - 45 *a^2*b - 30*a*b^2 - 8*b^3)*cosh(d*x + c) + 5*((45*a^2*b + 30*a*b^2 + 8*b^3 )*cosh(d*x + c)^4 + 18*a^2*b + 24*a*b^2 + 16*b^3 + 3*(27*a^2*b + 30*a*b^2 + 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d *cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))
\[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (69) = 138\).
Time = 0.19 (sec) , antiderivative size = 332, normalized size of antiderivative = 4.55 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=a^{3} x + \frac {16}{15} \, b^{3} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 4 \, a b^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {6 \, a^{2} b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
a^3*x + 16/15*b^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d* x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d *(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8 *d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 4*a*b^2*(3*e^(-2*d*x - 2*c)/(d*( 3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3 *e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + 6*a^2*b /(d*(e^(-2*d*x - 2*c) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (69) = 138\).
Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.49 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {15 \, {\left (d x + c\right )} a^{3} - \frac {2 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 210 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 150 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 40 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 30 \, a b^{2} + 8 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]
1/15*(15*(d*x + c)*a^3 - 2*(45*a^2*b*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) + 90*a*b^2*e^(6*d*x + 6*c) + 270*a^2*b*e^(4*d*x + 4*c) + 210*a*b^2* e^(4*d*x + 4*c) + 80*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 150 *a*b^2*e^(2*d*x + 2*c) + 40*b^3*e^(2*d*x + 2*c) + 45*a^2*b + 30*a*b^2 + 8* b^3)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 2.12 (sec) , antiderivative size = 502, normalized size of antiderivative = 6.88 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=a^3\,x-\frac {\frac {2\,\left (9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{15\,d}+\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {6\,a^2\,b}{5\,d}+\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (9\,a^2\,b+12\,a\,b^2+8\,b^3\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {6\,a^2\,b}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
a^3*x - ((2*(12*a*b^2 + 9*a^2*b + 8*b^3))/(15*d) + (12*exp(2*c + 2*d*x)*(a *b^2 + a^2*b))/(5*d) + (6*a^2*b*exp(4*c + 4*d*x))/(5*d))/(3*exp(2*c + 2*d* x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - ((6*a^2*b)/(5*d) + (24*e xp(2*c + 2*d*x)*(a*b^2 + a^2*b))/(5*d) + (24*exp(6*c + 6*d*x)*(a*b^2 + a^2 *b))/(5*d) + (4*exp(4*c + 4*d*x)*(12*a*b^2 + 9*a^2*b + 8*b^3))/(5*d) + (6* a^2*b*exp(8*c + 8*d*x))/(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^2 + a^2*b))/(5*d) + (18*exp(4*c + 4*d*x)*(a*b^2 + a^2*b))/(5*d) + (2* exp(2*c + 2*d*x)*(12*a*b^2 + 9*a^2*b + 8*b^3))/(5*d) + (6*a^2*b*exp(6*c + 6*d*x))/(5*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) - ((6*(a*b^2 + a^2*b))/(5*d) + (6*a^2*b*exp(2*c + 2*d*x))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - (6*a^2*b)/ (5*d*(exp(2*c + 2*d*x) + 1))