Integrand size = 23, antiderivative size = 72 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {1}{8} \left (3 a^2+8 a b+8 b^2\right ) x+\frac {a (3 a+8 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d} \] Output:
1/8*(3*a^2+8*a*b+8*b^2)*x+1/8*a*(3*a+8*b)*cosh(d*x+c)*sinh(d*x+c)/d+1/4*a^ 2*cosh(d*x+c)^3*sinh(d*x+c)/d
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {4 \left (3 a^2+8 a b+8 b^2\right ) (c+d x)+8 a (a+2 b) \sinh (2 (c+d x))+a^2 \sinh (4 (c+d x))}{32 d} \] Input:
Integrate[Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2,x]
Output:
(4*(3*a^2 + 8*a*b + 8*b^2)*(c + d*x) + 8*a*(a + 2*b)*Sinh[2*(c + d*x)] + a ^2*Sinh[4*(c + d*x)])/(32*d)
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.46, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4634, 315, 25, 298, 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.
\(\displaystyle \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (i c+i d x)^2\right )^2}{\sec (i c+i d x)^4}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\left (-b \tanh ^2(c+d x)+a+b\right )^2}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\frac {a \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}-\frac {1}{4} \int -\frac {(a+b) (3 a+4 b)-b (a+4 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{4} \int \frac {(a+b) (3 a+4 b)-b (a+4 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {a \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^2+8 a b+8 b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {3 a (a+2 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^2+8 a b+8 b^2\right ) \text {arctanh}(\tanh (c+d x))+\frac {3 a (a+2 b) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {a \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
Input:
Int[Cosh[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2,x]
Output:
((a*Tanh[c + d*x]*(a + b - b*Tanh[c + d*x]^2))/(4*(1 - Tanh[c + d*x]^2)^2) + (((3*a^2 + 8*a*b + 8*b^2)*ArcTanh[Tanh[c + d*x]])/2 + (3*a*(a + 2*b)*Ta nh[c + d*x])/(2*(1 - Tanh[c + d*x]^2)))/4)/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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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 = 0.54 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(\frac {8 a \left (a +2 b \right ) \sinh \left (2 d x +2 c \right )+a^{2} \sinh \left (4 d x +4 c \right )+12 x \left (a^{2}+\frac {8}{3} a b +\frac {8}{3} b^{2}\right ) d}{32 d}\) | \(54\) |
derivativedivides | \(\frac {a^{2} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \left (d x +c \right )}{d}\) | \(79\) |
default | \(\frac {a^{2} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} \left (d x +c \right )}{d}\) | \(79\) |
risch | \(\frac {3 a^{2} x}{8}+a b x +b^{2} x +\frac {a^{2} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {a^{2} {\mathrm e}^{2 d x +2 c}}{8 d}+\frac {a \,{\mathrm e}^{2 d x +2 c} b}{4 d}-\frac {a^{2} {\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {a \,{\mathrm e}^{-2 d x -2 c} b}{4 d}-\frac {a^{2} {\mathrm e}^{-4 d x -4 c}}{64 d}\) | \(117\) |
Input:
int(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/32*(8*a*(a+2*b)*sinh(2*d*x+2*c)+a^2*sinh(4*d*x+4*c)+12*x*(a^2+8/3*a*b+8/ 3*b^2)*d)/d
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )} d x + {\left (a^{2} \cosh \left (d x + c\right )^{3} + 4 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{8 \, d} \] Input:
integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
Output:
1/8*(a^2*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a^2 + 8*a*b + 8*b^2)*d*x + (a^ 2*cosh(d*x + c)^3 + 4*(a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c))/d
Timed out. \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(cosh(d*x+c)**4*(a+b*sech(d*x+c)**2)**2,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.46 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {1}{64} \, a^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {1}{4} \, a b {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + b^{2} x \] Input:
integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/64*a^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) + 1/4*a*b*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) + b^2*x
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (66) = 132\).
Time = 0.14 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.10 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, {\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )} {\left (d x + c\right )} - {\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \] Input:
integrate(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/64*(a^2*e^(4*d*x + 4*c) + 8*a^2*e^(2*d*x + 2*c) + 16*a*b*e^(2*d*x + 2*c) + 8*(3*a^2 + 8*a*b + 8*b^2)*(d*x + c) - (18*a^2*e^(4*d*x + 4*c) + 48*a*b* e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) + 8*a^2*e^(2*d*x + 2*c) + 16*a*b* e^(2*d*x + 2*c) + a^2)*e^(-4*d*x - 4*c))/d
Time = 2.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {3\,a^2\,x}{8}+b^2\,x+a\,b\,x+\frac {a^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{32\,d}+\frac {a\,b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{2\,d} \] Input:
int(cosh(c + d*x)^4*(a + b/cosh(c + d*x)^2)^2,x)
Output:
(3*a^2*x)/8 + b^2*x + a*b*x + (a^2*sinh(2*c + 2*d*x))/(4*d) + (a^2*sinh(4* c + 4*d*x))/(32*d) + (a*b*sinh(2*c + 2*d*x))/(2*d)
Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.01 \[ \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {e^{8 d x +8 c} a^{2}+8 e^{6 d x +6 c} a^{2}+16 e^{6 d x +6 c} a b +24 e^{4 d x +4 c} a^{2} d x +64 e^{4 d x +4 c} a b d x +64 e^{4 d x +4 c} b^{2} d x -8 e^{2 d x +2 c} a^{2}-16 e^{2 d x +2 c} a b -a^{2}}{64 e^{4 d x +4 c} d} \] Input:
int(cosh(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x)
Output:
(e**(8*c + 8*d*x)*a**2 + 8*e**(6*c + 6*d*x)*a**2 + 16*e**(6*c + 6*d*x)*a*b + 24*e**(4*c + 4*d*x)*a**2*d*x + 64*e**(4*c + 4*d*x)*a*b*d*x + 64*e**(4*c + 4*d*x)*b**2*d*x - 8*e**(2*c + 2*d*x)*a**2 - 16*e**(2*c + 2*d*x)*a*b - a **2)/(64*e**(4*c + 4*d*x)*d)