Integrand size = 23, antiderivative size = 77 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\frac {a^2 \log (\cosh (c+d x))}{d}+\frac {a (a-2 b) \text {sech}^2(c+d x)}{2 d}+\frac {(2 a-b) b \text {sech}^4(c+d x)}{4 d}+\frac {b^2 \text {sech}^6(c+d x)}{6 d} \] Output:
a^2*ln(cosh(d*x+c))/d+1/2*a*(a-2*b)*sech(d*x+c)^2/d+1/4*(2*a-b)*b*sech(d*x +c)^4/d+1/6*b^2*sech(d*x+c)^6/d
Time = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.39 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\frac {\cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \left (12 a^2 \log (\cosh (c+d x))+6 a (a-2 b) \text {sech}^2(c+d x)+3 (2 a-b) b \text {sech}^4(c+d x)+2 b^2 \text {sech}^6(c+d x)\right )}{3 d (a+2 b+a \cosh (2 c+2 d x))^2} \] Input:
Integrate[(a + b*Sech[c + d*x]^2)^2*Tanh[c + d*x]^3,x]
Output:
(Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2*(12*a^2*Log[Cosh[c + d*x]] + 6* a*(a - 2*b)*Sech[c + d*x]^2 + 3*(2*a - b)*b*Sech[c + d*x]^4 + 2*b^2*Sech[c + d*x]^6))/(3*d*(a + 2*b + a*Cosh[2*c + 2*d*x])^2)
Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4626, 354, 85, 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.
| \(\displaystyle \int \tanh ^3(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int i \tan (i c+i d x)^3 \left (a+b \sec (i c+i d x)^2\right )^2dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \left (b \sec (i c+i d x)^2+a\right )^2 \tan (i c+i d x)^3dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \left (1-\cosh ^2(c+d x)\right ) \left (a \cosh ^2(c+d x)+b\right )^2 \text {sech}^7(c+d x)d\cosh (c+d x)}{d}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \left (1-\cosh ^2(c+d x)\right ) \left (a \cosh ^2(c+d x)+b\right )^2 \text {sech}^4(c+d x)d\cosh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle -\frac {\int \left (b^2 \text {sech}^4(c+d x)+(2 a-b) b \text {sech}^3(c+d x)+a (a-2 b) \text {sech}^2(c+d x)-a^2 \text {sech}(c+d x)\right )d\cosh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-a^2 \log \left (\cosh ^2(c+d x)\right )-\frac {1}{2} b (2 a-b) \text {sech}^2(c+d x)-a (a-2 b) \text {sech}(c+d x)-\frac {1}{3} b^2 \text {sech}^3(c+d x)}{2 d}\) |
Input:
Int[(a + b*Sech[c + d*x]^2)^2*Tanh[c + d*x]^3,x]
Output:
-1/2*(-(a^2*Log[Cosh[c + d*x]^2]) - a*(a - 2*b)*Sech[c + d*x] - ((2*a - b) *b*Sech[c + d*x]^2)/2 - (b^2*Sech[c + d*x]^3)/3)/d
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 20.75 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{2} \operatorname {sech}\left (d x +c \right )^{6}}{6}+\frac {a b \operatorname {sech}\left (d x +c \right )^{4}}{2}-\frac {\operatorname {sech}\left (d x +c \right )^{4} b^{2}}{4}+\frac {a^{2} \operatorname {sech}\left (d x +c \right )^{2}}{2}-\operatorname {sech}\left (d x +c \right )^{2} a b -a^{2} \ln \left (\operatorname {sech}\left (d x +c \right )\right )}{d}\) | \(81\) |
| default | \(\frac {\frac {b^{2} \operatorname {sech}\left (d x +c \right )^{6}}{6}+\frac {a b \operatorname {sech}\left (d x +c \right )^{4}}{2}-\frac {\operatorname {sech}\left (d x +c \right )^{4} b^{2}}{4}+\frac {a^{2} \operatorname {sech}\left (d x +c \right )^{2}}{2}-\operatorname {sech}\left (d x +c \right )^{2} a b -a^{2} \ln \left (\operatorname {sech}\left (d x +c \right )\right )}{d}\) | \(81\) |
| parts | \(\frac {a^{2} \left (-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\tanh \left (d x +c \right )^{6}}{6}+\frac {\tanh \left (d x +c \right )^{4}}{4}\right )}{d}+\frac {a b \tanh \left (d x +c \right )^{4}}{2 d}\) | \(85\) |
| risch | \(-a^{2} x -\frac {2 a^{2} c}{d}+\frac {2 \,{\mathrm e}^{2 d x +2 c} \left (3 a^{2} {\mathrm e}^{8 d x +8 c}-6 a b \,{\mathrm e}^{8 d x +8 c}+12 a^{2} {\mathrm e}^{6 d x +6 c}-12 a b \,{\mathrm e}^{6 d x +6 c}-6 b^{2} {\mathrm e}^{6 d x +6 c}+18 a^{2} {\mathrm e}^{4 d x +4 c}-12 a b \,{\mathrm e}^{4 d x +4 c}+4 b^{2} {\mathrm e}^{4 d x +4 c}+12 a^{2} {\mathrm e}^{2 d x +2 c}-12 a b \,{\mathrm e}^{2 d x +2 c}-6 b^{2} {\mathrm e}^{2 d x +2 c}+3 a^{2}-6 a b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{6}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(223\) |
Input:
int((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/6*b^2*sech(d*x+c)^6+1/2*a*b*sech(d*x+c)^4-1/4*sech(d*x+c)^4*b^2+1/2 *a^2*sech(d*x+c)^2-sech(d*x+c)^2*a*b-a^2*ln(sech(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 2591 vs. \(2 (71) = 142\).
Time = 0.26 (sec) , antiderivative size = 2591, normalized size of antiderivative = 33.65 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^3,x, algorithm="fricas")
Output:
-1/3*(3*a^2*d*x*cosh(d*x + c)^12 + 36*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^ 11 + 3*a^2*d*x*sinh(d*x + c)^12 + 6*(3*a^2*d*x - a^2 + 2*a*b)*cosh(d*x + c )^10 + 6*(33*a^2*d*x*cosh(d*x + c)^2 + 3*a^2*d*x - a^2 + 2*a*b)*sinh(d*x + c)^10 + 60*(11*a^2*d*x*cosh(d*x + c)^3 + (3*a^2*d*x - a^2 + 2*a*b)*cosh(d *x + c))*sinh(d*x + c)^9 + 3*(15*a^2*d*x - 8*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c)^8 + 3*(495*a^2*d*x*cosh(d*x + c)^4 + 15*a^2*d*x + 90*(3*a^2*d*x - a^ 2 + 2*a*b)*cosh(d*x + c)^2 - 8*a^2 + 8*a*b + 4*b^2)*sinh(d*x + c)^8 + 24*( 99*a^2*d*x*cosh(d*x + c)^5 + 30*(3*a^2*d*x - a^2 + 2*a*b)*cosh(d*x + c)^3 + (15*a^2*d*x - 8*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c)^7 + 4* (15*a^2*d*x - 9*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^6 + 4*(693*a^2*d*x*cosh (d*x + c)^6 + 315*(3*a^2*d*x - a^2 + 2*a*b)*cosh(d*x + c)^4 + 15*a^2*d*x + 21*(15*a^2*d*x - 8*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c)^2 - 9*a^2 + 6*a*b - 2*b^2)*sinh(d*x + c)^6 + 24*(99*a^2*d*x*cosh(d*x + c)^7 + 63*(3*a^2*d*x - a^2 + 2*a*b)*cosh(d*x + c)^5 + 7*(15*a^2*d*x - 8*a^2 + 8*a*b + 4*b^2)*cos h(d*x + c)^3 + (15*a^2*d*x - 9*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c))*sinh(d* x + c)^5 + 3*(15*a^2*d*x - 8*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c)^4 + 3*(495 *a^2*d*x*cosh(d*x + c)^8 + 420*(3*a^2*d*x - a^2 + 2*a*b)*cosh(d*x + c)^6 + 70*(15*a^2*d*x - 8*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c)^4 + 15*a^2*d*x + 20 *(15*a^2*d*x - 9*a^2 + 6*a*b - 2*b^2)*cosh(d*x + c)^2 - 8*a^2 + 8*a*b + 4* b^2)*sinh(d*x + c)^4 + 3*a^2*d*x + 4*(165*a^2*d*x*cosh(d*x + c)^9 + 180...
Time = 0.66 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.68 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\begin {cases} a^{2} x - \frac {a^{2} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a^{2} \tanh ^{2}{\left (c + d x \right )}}{2 d} - \frac {a b \tanh ^{2}{\left (c + d x \right )} \operatorname {sech}^{2}{\left (c + d x \right )}}{2 d} - \frac {a b \operatorname {sech}^{2}{\left (c + d x \right )}}{2 d} - \frac {b^{2} \tanh ^{2}{\left (c + d x \right )} \operatorname {sech}^{4}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \operatorname {sech}^{4}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {sech}^{2}{\left (c \right )}\right )^{2} \tanh ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sech(d*x+c)**2)**2*tanh(d*x+c)**3,x)
Output:
Piecewise((a**2*x - a**2*log(tanh(c + d*x) + 1)/d - a**2*tanh(c + d*x)**2/ (2*d) - a*b*tanh(c + d*x)**2*sech(c + d*x)**2/(2*d) - a*b*sech(c + d*x)**2 /(2*d) - b**2*tanh(c + d*x)**2*sech(c + d*x)**4/(6*d) - b**2*sech(c + d*x) **4/(12*d), Ne(d, 0)), (x*(a + b*sech(c)**2)**2*tanh(c)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (71) = 142\).
Time = 0.13 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.32 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\frac {a b \tanh \left (d x + c\right )^{4}}{2 \, d} + a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} - \frac {4}{3} \, b^{2} {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}} - \frac {2 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}} + \frac {3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}}\right )} \] Input:
integrate((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^3,x, algorithm="maxima")
Output:
1/2*a*b*tanh(d*x + c)^4/d + a^2*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 2 *e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) - 4/3*b ^2*(3*e^(-4*d*x - 4*c)/(d*(6*e^(-2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e ^(-6*d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12*c) + 1)) - 2*e^(-6*d*x - 6*c)/(d*(6*e^(-2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e^(-6*d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12*c) + 1)) + 3*e^(-8*d*x - 8*c)/(d*(6*e^(-2*d*x - 2*c) + 15 *e^(-4*d*x - 4*c) + 20*e^(-6*d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d *x - 10*c) + e^(-12*d*x - 12*c) + 1)))
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (71) = 142\).
Time = 0.17 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.17 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=-\frac {60 \, {\left (d x + c\right )} a^{2} - 60 \, a^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac {147 \, a^{2} e^{\left (12 \, d x + 12 \, c\right )} + 762 \, a^{2} e^{\left (10 \, d x + 10 \, c\right )} + 240 \, a b e^{\left (10 \, d x + 10 \, c\right )} + 1725 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 480 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 240 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 2220 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 480 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 160 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 1725 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 480 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 240 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 762 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 240 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 147 \, a^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{6}}}{60 \, d} \] Input:
integrate((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^3,x, algorithm="giac")
Output:
-1/60*(60*(d*x + c)*a^2 - 60*a^2*log(e^(2*d*x + 2*c) + 1) + (147*a^2*e^(12 *d*x + 12*c) + 762*a^2*e^(10*d*x + 10*c) + 240*a*b*e^(10*d*x + 10*c) + 172 5*a^2*e^(8*d*x + 8*c) + 480*a*b*e^(8*d*x + 8*c) + 240*b^2*e^(8*d*x + 8*c) + 2220*a^2*e^(6*d*x + 6*c) + 480*a*b*e^(6*d*x + 6*c) - 160*b^2*e^(6*d*x + 6*c) + 1725*a^2*e^(4*d*x + 4*c) + 480*a*b*e^(4*d*x + 4*c) + 240*b^2*e^(4*d *x + 4*c) + 762*a^2*e^(2*d*x + 2*c) + 240*a*b*e^(2*d*x + 2*c) + 147*a^2)/( e^(2*d*x + 2*c) + 1)^6)/d
Time = 2.35 (sec) , antiderivative size = 349, normalized size of antiderivative = 4.53 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\frac {4\,\left (2\,a\,b-9\,b^2\right )}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {32\,b^2}{3\,d\,\left (6\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,{\mathrm {e}}^{4\,c+4\,d\,x}+20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}+6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1\right )}-\frac {2\,\left (a^2-6\,a\,b+2\,b^2\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {2\,\left (2\,a\,b-a^2\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {8\,\left (6\,a\,b-7\,b^2\right )}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-a^2\,x+\frac {a^2\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d}+\frac {32\,b^2}{d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )} \] Input:
int(tanh(c + d*x)^3*(a + b/cosh(c + d*x)^2)^2,x)
Output:
(4*(2*a*b - 9*b^2))/(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)) - (32*b^2)/(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(12*c + 12*d*x) + 1)) - (2*(a^2 - 6*a*b + 2*b^2))/(d*(2* exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - (2*(2*a*b - a^2))/(d*(exp(2*c + 2*d*x) + 1)) - (8*(6*a*b - 7*b^2))/(3*d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - a^2*x + (a^2*log(exp(2*c)*exp(2*d*x) + 1))/d + (32*b^2)/(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))
Time = 0.23 (sec) , antiderivative size = 445, normalized size of antiderivative = 5.78 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \tanh ^3(c+d x) \, dx=\frac {12 e^{4 d x +4 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a^{2}-2 e^{4 d x +4 c} \mathrm {sech}\left (d x +c \right )^{4} \tanh \left (d x +c \right )^{2} b^{2}-e^{4 d x +4 c} \mathrm {sech}\left (d x +c \right )^{4} b^{2}-6 e^{4 d x +4 c} \mathrm {sech}\left (d x +c \right )^{2} \tanh \left (d x +c \right )^{2} a b -6 e^{4 d x +4 c} \mathrm {sech}\left (d x +c \right )^{2} a b -12 e^{4 d x +4 c} a^{2} d x -12 e^{4 d x +4 c} a^{2}+24 e^{2 d x +2 c} \mathrm {log}\left (e^{2 d x +2 c}+1\right ) a^{2}-4 e^{2 d x +2 c} \mathrm {sech}\left (d x +c \right )^{4} \tanh \left (d x +c \right )^{2} b^{2}-2 e^{2 d x +2 c} \mathrm {sech}\left (d x +c \right )^{4} b^{2}-12 e^{2 d x +2 c} \mathrm {sech}\left (d x +c \right )^{2} \tanh \left (d x +c \right )^{2} a b -12 e^{2 d x +2 c} \mathrm {sech}\left (d x +c \right )^{2} a b -24 e^{2 d x +2 c} a^{2} d x +12 \,\mathrm {log}\left (e^{2 d x +2 c}+1\right ) a^{2}-2 \mathrm {sech}\left (d x +c \right )^{4} \tanh \left (d x +c \right )^{2} b^{2}-\mathrm {sech}\left (d x +c \right )^{4} b^{2}-6 \mathrm {sech}\left (d x +c \right )^{2} \tanh \left (d x +c \right )^{2} a b -6 \mathrm {sech}\left (d x +c \right )^{2} a b -12 a^{2} d x -12 a^{2}}{12 d \left (e^{4 d x +4 c}+2 e^{2 d x +2 c}+1\right )} \] Input:
int((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^3,x)
Output:
(12*e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2 - 2*e**(4*c + 4*d*x)*s ech(c + d*x)**4*tanh(c + d*x)**2*b**2 - e**(4*c + 4*d*x)*sech(c + d*x)**4* b**2 - 6*e**(4*c + 4*d*x)*sech(c + d*x)**2*tanh(c + d*x)**2*a*b - 6*e**(4* c + 4*d*x)*sech(c + d*x)**2*a*b - 12*e**(4*c + 4*d*x)*a**2*d*x - 12*e**(4* c + 4*d*x)*a**2 + 24*e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x) + 1)*a**2 - 4*e **(2*c + 2*d*x)*sech(c + d*x)**4*tanh(c + d*x)**2*b**2 - 2*e**(2*c + 2*d*x )*sech(c + d*x)**4*b**2 - 12*e**(2*c + 2*d*x)*sech(c + d*x)**2*tanh(c + d* x)**2*a*b - 12*e**(2*c + 2*d*x)*sech(c + d*x)**2*a*b - 24*e**(2*c + 2*d*x) *a**2*d*x + 12*log(e**(2*c + 2*d*x) + 1)*a**2 - 2*sech(c + d*x)**4*tanh(c + d*x)**2*b**2 - sech(c + d*x)**4*b**2 - 6*sech(c + d*x)**2*tanh(c + d*x)* *2*a*b - 6*sech(c + d*x)**2*a*b - 12*a**2*d*x - 12*a**2)/(12*d*(e**(4*c + 4*d*x) + 2*e**(2*c + 2*d*x) + 1))