Integrand size = 23, antiderivative size = 65 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=-\frac {1}{2} a (a-4 b) x+\frac {a^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {2 a b \tanh (c+d x)}{d}+\frac {b^2 \tanh ^3(c+d x)}{3 d} \] Output:
-1/2*a*(a-4*b)*x+1/2*a^2*cosh(d*x+c)*sinh(d*x+c)/d-2*a*b*tanh(d*x+c)/d+1/3 *b^2*tanh(d*x+c)^3/d
Time = 1.36 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.94 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=\frac {\left (b+a \cosh ^2(c+d x)\right )^2 \text {sech}^3(c+d x) \left (-4 b^2 \text {sech}(c) \sinh (d x)-4 (6 a-b) b \cosh ^2(c+d x) \text {sech}(c) \sinh (d x)+3 a \cosh ^3(c+d x) (-2 (a-4 b) d x+a \sinh (2 (c+d x)))-4 b^2 \cosh (c+d x) \tanh (c)\right )}{3 d (a+2 b+a \cosh (2 (c+d x)))^2} \] Input:
Integrate[(a + b*Sech[c + d*x]^2)^2*Sinh[c + d*x]^2,x]
Output:
((b + a*Cosh[c + d*x]^2)^2*Sech[c + d*x]^3*(-4*b^2*Sech[c]*Sinh[d*x] - 4*( 6*a - b)*b*Cosh[c + d*x]^2*Sech[c]*Sinh[d*x] + 3*a*Cosh[c + d*x]^3*(-2*(a - 4*b)*d*x + a*Sinh[2*(c + d*x)]) - 4*b^2*Cosh[c + d*x]*Tanh[c]))/(3*d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4620, 366, 363, 262, 219}
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.
| \(\displaystyle \int \sinh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (i c+i d x)^2 \left (-\left (a+b \sec (i c+i d x)^2\right )^2\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \left (b \sec (i c+i d x)^2+a\right )^2 \sin (i c+i d x)^2dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )^2}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle \frac {\frac {a^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int \frac {\tanh ^2(c+d x) \left (3 a^2-2 (a+b)^2+2 b^2 \tanh ^2(c+d x)\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} b^2 \tanh ^3(c+d x)-a (a-4 b) \int \frac {\tanh ^2(c+d x)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)\right )+\frac {a^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} b^2 \tanh ^3(c+d x)-a (a-4 b) \left (\int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\tanh (c+d x)\right )\right )+\frac {a^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}+\frac {1}{2} \left (\frac {2}{3} b^2 \tanh ^3(c+d x)-a (a-4 b) (\text {arctanh}(\tanh (c+d x))-\tanh (c+d x))\right )}{d}\) |
Input:
Int[(a + b*Sech[c + d*x]^2)^2*Sinh[c + d*x]^2,x]
Output:
((a^2*Tanh[c + d*x]^3)/(2*(1 - Tanh[c + d*x]^2)) + (-(a*(a - 4*b)*(ArcTanh [Tanh[c + d*x]] - Tanh[c + d*x])) + (2*b^2*Tanh[c + d*x]^3)/3)/2)/d
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 23.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (d x +c -\tanh \left (d x +c \right )\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(90\) |
| default | \(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (d x +c -\tanh \left (d x +c \right )\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d}\) | \(90\) |
| parts | \(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d}+\frac {b^{2} \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )}{d}+\frac {2 a b \left (d x +c -\tanh \left (d x +c \right )\right )}{d}\) | \(95\) |
| risch | \(-\frac {a^{2} x}{2}+2 a b x +\frac {a^{2} {\mathrm e}^{2 d x +2 c}}{8 d}-\frac {a^{2} {\mathrm e}^{-2 d x -2 c}}{8 d}+\frac {2 b \left (6 \,{\mathrm e}^{4 d x +4 c} a -3 \,{\mathrm e}^{4 d x +4 c} b +12 a \,{\mathrm e}^{2 d x +2 c}+6 a -b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}\) | \(109\) |
Input:
int((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+2*a*b*(d*x+c-tanh(d*x +c))+b^2*(-1/2*sinh(d*x+c)/cosh(d*x+c)^3+1/2*(2/3+1/3*sech(d*x+c)^2)*tanh( d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (59) = 118\).
Time = 0.18 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.88 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=\frac {3 \, a^{2} \sinh \left (d x + c\right )^{5} - 4 \, {\left (3 \, {\left (a^{2} - 4 \, a b\right )} d x - 12 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (3 \, {\left (a^{2} - 4 \, a b\right )} d x - 12 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, a^{2} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} - 48 \, a b + 8 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (3 \, {\left (a^{2} - 4 \, a b\right )} d x - 12 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, a^{2} \cosh \left (d x + c\right )^{4} + {\left (9 \, a^{2} - 48 \, a b + 8 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} - 16 \, a b - 8 \, b^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \] Input:
integrate((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^2,x, algorithm="fricas")
Output:
1/24*(3*a^2*sinh(d*x + c)^5 - 4*(3*(a^2 - 4*a*b)*d*x - 12*a*b + 2*b^2)*cos h(d*x + c)^3 - 12*(3*(a^2 - 4*a*b)*d*x - 12*a*b + 2*b^2)*cosh(d*x + c)*sin h(d*x + c)^2 + (30*a^2*cosh(d*x + c)^2 + 9*a^2 - 48*a*b + 8*b^2)*sinh(d*x + c)^3 - 12*(3*(a^2 - 4*a*b)*d*x - 12*a*b + 2*b^2)*cosh(d*x + c) + 3*(5*a^ 2*cosh(d*x + c)^4 + (9*a^2 - 48*a*b + 8*b^2)*cosh(d*x + c)^2 + 2*a^2 - 16* a*b - 8*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh(d* x + c)^2 + 3*d*cosh(d*x + c))
\[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2} \sinh ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*sech(d*x+c)**2)**2*sinh(d*x+c)**2,x)
Output:
Integral((a + b*sech(c + d*x)**2)**2*sinh(c + d*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (59) = 118\).
Time = 0.04 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.46 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=-\frac {1}{8} \, a^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 2 \, a b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac {2}{3} \, b^{2} {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \] Input:
integrate((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^2,x, algorithm="maxima")
Output:
-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 2*a*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + 2/3*b^2*(3*e^(-4*d*x - 4*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)))
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (59) = 118\).
Time = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.22 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=\frac {3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 12 \, {\left (a^{2} - 4 \, a b\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} - a^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + \frac {16 \, {\left (6 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b - b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \] Input:
integrate((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^2,x, algorithm="giac")
Output:
1/24*(3*a^2*e^(2*d*x + 2*c) - 12*(a^2 - 4*a*b)*(d*x + c) + 3*(2*a^2*e^(2*d *x + 2*c) - 8*a*b*e^(2*d*x + 2*c) - a^2)*e^(-2*d*x - 2*c) + 16*(6*a*b*e^(4 *d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) + 12*a*b*e^(2*d*x + 2*c) + 6*a*b - b^2 )/(e^(2*d*x + 2*c) + 1)^3)/d
Time = 0.15 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.63 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a\,b-b^2\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+x\,\left (2\,a\,b-\frac {a^2}{2}\right )+\frac {\frac {2\,\left (2\,a\,b-b^2\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a\,b-b^2\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {2\,\left (2\,a\,b-b^2\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d} \] Input:
int(sinh(c + d*x)^2*(a + b/cosh(c + d*x)^2)^2,x)
Output:
((2*(2*a*b + b^2))/(3*d) + (2*exp(2*c + 2*d*x)*(2*a*b - b^2))/(3*d))/(2*ex p(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) + x*(2*a*b - a^2/2) + ((2*(2*a*b - b^2))/(3*d) + (4*exp(2*c + 2*d*x)*(2*a*b + b^2))/(3*d) + (2*exp(4*c + 4*d* x)*(2*a*b - 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) + (2*(2*a*b - b^2))/(3*d*(exp(2*c + 2*d*x) + 1)) - (a^2*ex p(- 2*c - 2*d*x))/(8*d) + (a^2*exp(2*c + 2*d*x))/(8*d)
Time = 0.22 (sec) , antiderivative size = 340, normalized size of antiderivative = 5.23 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^2 \sinh ^2(c+d x) \, dx=\frac {3 e^{10 d x +10 c} a^{2}-12 e^{8 d x +8 c} a^{2} d x +9 e^{8 d x +8 c} a^{2}+48 e^{8 d x +8 c} a b d x -96 e^{8 d x +8 c} a b +16 e^{8 d x +8 c} b^{2}-36 e^{6 d x +6 c} a^{2} d x +6 e^{6 d x +6 c} a^{2}+144 e^{6 d x +6 c} a b d x -192 e^{6 d x +6 c} a b -36 e^{4 d x +4 c} a^{2} d x -6 e^{4 d x +4 c} a^{2}+144 e^{4 d x +4 c} a b d x -96 e^{4 d x +4 c} a b +48 e^{4 d x +4 c} b^{2}-12 e^{2 d x +2 c} a^{2} d x -9 e^{2 d x +2 c} a^{2}+48 e^{2 d x +2 c} a b d x -3 a^{2}}{24 e^{2 d x +2 c} d \left (e^{6 d x +6 c}+3 e^{4 d x +4 c}+3 e^{2 d x +2 c}+1\right )} \] Input:
int((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^2,x)
Output:
(3*e**(10*c + 10*d*x)*a**2 - 12*e**(8*c + 8*d*x)*a**2*d*x + 9*e**(8*c + 8* d*x)*a**2 + 48*e**(8*c + 8*d*x)*a*b*d*x - 96*e**(8*c + 8*d*x)*a*b + 16*e** (8*c + 8*d*x)*b**2 - 36*e**(6*c + 6*d*x)*a**2*d*x + 6*e**(6*c + 6*d*x)*a** 2 + 144*e**(6*c + 6*d*x)*a*b*d*x - 192*e**(6*c + 6*d*x)*a*b - 36*e**(4*c + 4*d*x)*a**2*d*x - 6*e**(4*c + 4*d*x)*a**2 + 144*e**(4*c + 4*d*x)*a*b*d*x - 96*e**(4*c + 4*d*x)*a*b + 48*e**(4*c + 4*d*x)*b**2 - 12*e**(2*c + 2*d*x) *a**2*d*x - 9*e**(2*c + 2*d*x)*a**2 + 48*e**(2*c + 2*d*x)*a*b*d*x - 3*a**2 )/(24*e**(2*c + 2*d*x)*d*(e**(6*c + 6*d*x) + 3*e**(4*c + 4*d*x) + 3*e**(2* c + 2*d*x) + 1))