Integrand size = 12, antiderivative size = 48 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=-\frac {3 \arctan \left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d}+\frac {5 \sinh (c+d x)}{16 d (3+5 \cosh (c+d x))} \] Output:
-3/32*arctan(1/2*tanh(1/2*d*x+1/2*c))/d+5/16*sinh(d*x+c)/d/(3+5*cosh(d*x+c ))
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=\frac {3 \arctan \left (2 \coth \left (\frac {1}{2} (c+d x)\right )\right )+\frac {10 \sinh (c+d x)}{3+5 \cosh (c+d x)}}{32 d} \] Input:
Integrate[(3 + 5*Cosh[c + d*x])^(-2),x]
Output:
(3*ArcTan[2*Coth[(c + d*x)/2]] + (10*Sinh[c + d*x])/(3 + 5*Cosh[c + d*x])) /(32*d)
Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3143, 27, 3042, 3138, 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.
\(\displaystyle \int \frac {1}{(5 \cosh (c+d x)+3)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (3+5 \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {1}{16} \int -\frac {3}{5 \cosh (c+d x)+3}dx+\frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}-\frac {3}{16} \int \frac {1}{5 \cosh (c+d x)+3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}-\frac {3}{16} \int \frac {1}{5 \sin \left (i c+i d x+\frac {\pi }{2}\right )+3}dx\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}+\frac {3 i \int \frac {1}{2 \tanh ^2\left (\frac {1}{2} (c+d x)\right )+8}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}-\frac {3 \arctan \left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d}\) |
Input:
Int[(3 + 5*Cosh[c + d*x])^(-2),x]
Output:
(-3*ArcTan[Tanh[(c + d*x)/2]/2])/(32*d) + (5*Sinh[c + d*x])/(16*d*(3 + 5*C osh[c + d*x]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Time = 0.54 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )}-\frac {3 \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{32}}{d}\) | \(46\) |
default | \(\frac {\frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )}-\frac {3 \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{32}}{d}\) | \(46\) |
risch | \(-\frac {3 \,{\mathrm e}^{d x +c}+5}{8 d \left (5 \,{\mathrm e}^{2 d x +2 c}+6 \,{\mathrm e}^{d x +c}+5\right )}+\frac {3 i \ln \left ({\mathrm e}^{d x +c}+\frac {3}{5}-\frac {4 i}{5}\right )}{64 d}-\frac {3 i \ln \left ({\mathrm e}^{d x +c}+\frac {3}{5}+\frac {4 i}{5}\right )}{64 d}\) | \(74\) |
parallelrisch | \(\frac {\left (15 i \cosh \left (d x +c \right )+9 i\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 i\right )+\left (-15 i \cosh \left (d x +c \right )-9 i\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 i\right )+20 \sinh \left (d x +c \right )}{320 d \cosh \left (d x +c \right )+192 d}\) | \(78\) |
Input:
int(1/(3+5*cosh(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(5/16*tanh(1/2*d*x+1/2*c)/(tanh(1/2*d*x+1/2*c)^2+4)-3/32*arctan(1/2*ta nh(1/2*d*x+1/2*c)))
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (41) = 82\).
Time = 0.08 (sec) , antiderivative size = 147, normalized size of antiderivative = 3.06 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=-\frac {3 \, {\left (5 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right ) + 5 \, \sinh \left (d x + c\right )^{2} + 6 \, \cosh \left (d x + c\right ) + 5\right )} \arctan \left (\frac {5}{4} \, \cosh \left (d x + c\right ) + \frac {5}{4} \, \sinh \left (d x + c\right ) + \frac {3}{4}\right ) + 12 \, \cosh \left (d x + c\right ) + 12 \, \sinh \left (d x + c\right ) + 20}{32 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 5 \, d \sinh \left (d x + c\right )^{2} + 6 \, d \cosh \left (d x + c\right ) + 2 \, {\left (5 \, d \cosh \left (d x + c\right ) + 3 \, d\right )} \sinh \left (d x + c\right ) + 5 \, d\right )}} \] Input:
integrate(1/(3+5*cosh(d*x+c))^2,x, algorithm="fricas")
Output:
-1/32*(3*(5*cosh(d*x + c)^2 + 2*(5*cosh(d*x + c) + 3)*sinh(d*x + c) + 5*si nh(d*x + c)^2 + 6*cosh(d*x + c) + 5)*arctan(5/4*cosh(d*x + c) + 5/4*sinh(d *x + c) + 3/4) + 12*cosh(d*x + c) + 12*sinh(d*x + c) + 20)/(5*d*cosh(d*x + c)^2 + 5*d*sinh(d*x + c)^2 + 6*d*cosh(d*x + c) + 2*(5*d*cosh(d*x + c) + 3 *d)*sinh(d*x + c) + 5*d)
Result contains complex when optimal does not.
Time = 1.32 (sec) , antiderivative size = 291, normalized size of antiderivative = 6.06 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=\begin {cases} - \frac {\log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )}}{25 d \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 30 d \cosh {\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 9 d} & \text {for}\: c = \log {\left (\left (-3 - 4 i\right ) e^{- d x} \right )} - \log {\left (5 \right )} \\\frac {x}{25 \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 30 \cosh {\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 9} & \text {for}\: c = \log {\left (\left (-3 + 4 i\right ) e^{- d x} \right )} - \log {\left (5 \right )} \\\frac {x}{\left (5 \cosh {\left (c \right )} + 3\right )^{2}} & \text {for}\: d = 0 \\- \frac {3 \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{32 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 128 d} + \frac {10 \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 128 d} - \frac {12 \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{32 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 128 d} & \text {otherwise} \end {cases} \] Input:
integrate(1/(3+5*cosh(d*x+c))**2,x)
Output:
Piecewise((-log(-3*exp(-d*x) - 4*I*exp(-d*x))/(25*d*cosh(d*x + log(-3*exp( -d*x) - 4*I*exp(-d*x)) - log(5))**2 + 30*d*cosh(d*x + log(-3*exp(-d*x) - 4 *I*exp(-d*x)) - log(5)) + 9*d), Eq(c, log((-3 - 4*I)*exp(-d*x)) - log(5))) , (x/(25*cosh(d*x + log(-3*exp(-d*x) + 4*I*exp(-d*x)) - log(5))**2 + 30*co sh(d*x + log(-3*exp(-d*x) + 4*I*exp(-d*x)) - log(5)) + 9), Eq(c, log((-3 + 4*I)*exp(-d*x)) - log(5))), (x/(5*cosh(c) + 3)**2, Eq(d, 0)), (-3*tanh(c/ 2 + d*x/2)**2*atan(tanh(c/2 + d*x/2)/2)/(32*d*tanh(c/2 + d*x/2)**2 + 128*d ) + 10*tanh(c/2 + d*x/2)/(32*d*tanh(c/2 + d*x/2)**2 + 128*d) - 12*atan(tan h(c/2 + d*x/2)/2)/(32*d*tanh(c/2 + d*x/2)**2 + 128*d), True))
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=\frac {3 \, \arctan \left (\frac {5}{4} \, e^{\left (-d x - c\right )} + \frac {3}{4}\right )}{32 \, d} + \frac {3 \, e^{\left (-d x - c\right )} + 5}{8 \, d {\left (6 \, e^{\left (-d x - c\right )} + 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5\right )}} \] Input:
integrate(1/(3+5*cosh(d*x+c))^2,x, algorithm="maxima")
Output:
3/32*arctan(5/4*e^(-d*x - c) + 3/4)/d + 1/8*(3*e^(-d*x - c) + 5)/(d*(6*e^( -d*x - c) + 5*e^(-2*d*x - 2*c) + 5))
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, e^{\left (d x + c\right )} + 5\right )}}{5 \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} + 5} + 3 \, \arctan \left (\frac {5}{4} \, e^{\left (d x + c\right )} + \frac {3}{4}\right )}{32 \, d} \] Input:
integrate(1/(3+5*cosh(d*x+c))^2,x, algorithm="giac")
Output:
-1/32*(4*(3*e^(d*x + c) + 5)/(5*e^(2*d*x + 2*c) + 6*e^(d*x + c) + 5) + 3*a rctan(5/4*e^(d*x + c) + 3/4))/d
Time = 2.00 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=-\frac {\frac {3\,{\mathrm {e}}^{c+d\,x}}{8\,d}+\frac {5}{8\,d}}{6\,{\mathrm {e}}^{c+d\,x}+5\,{\mathrm {e}}^{2\,c+2\,d\,x}+5}-\frac {3\,\mathrm {atan}\left (\left (\frac {3}{4\,d}+\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4\,d}\right )\,\sqrt {d^2}\right )}{32\,\sqrt {d^2}} \] Input:
int(1/(5*cosh(c + d*x) + 3)^2,x)
Output:
- ((3*exp(c + d*x))/(8*d) + 5/(8*d))/(6*exp(c + d*x) + 5*exp(2*c + 2*d*x) + 5) - (3*atan((3/(4*d) + (5*exp(d*x)*exp(c))/(4*d))*(d^2)^(1/2)))/(32*(d^ 2)^(1/2))
Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.15 \[ \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx=\frac {-15 e^{2 d x +2 c} \mathit {atan} \left (\frac {5 e^{d x +c}}{4}+\frac {3}{4}\right )-18 e^{d x +c} \mathit {atan} \left (\frac {5 e^{d x +c}}{4}+\frac {3}{4}\right )-15 \mathit {atan} \left (\frac {5 e^{d x +c}}{4}+\frac {3}{4}\right )+10 e^{2 d x +2 c}-10}{32 d \left (5 e^{2 d x +2 c}+6 e^{d x +c}+5\right )} \] Input:
int(1/(3+5*cosh(d*x+c))^2,x)
Output:
( - 15*e**(2*c + 2*d*x)*atan((5*e**(c + d*x) + 3)/4) - 18*e**(c + d*x)*ata n((5*e**(c + d*x) + 3)/4) - 15*atan((5*e**(c + d*x) + 3)/4) + 10*e**(2*c + 2*d*x) - 10)/(32*d*(5*e**(2*c + 2*d*x) + 6*e**(c + d*x) + 5))