Integrand size = 12, antiderivative size = 56 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\frac {5 x}{64}-\frac {5 \text {arctanh}\left (\frac {\sinh (c+d x)}{3+\cosh (c+d x)}\right )}{32 d}-\frac {3 \sinh (c+d x)}{16 d (5+3 \cosh (c+d x))} \] Output:
5/64*x-5/32*arctanh(sinh(d*x+c)/(3+cosh(d*x+c)))/d-3/16*sinh(d*x+c)/d/(5+3 *cosh(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(144\) vs. \(2(56)=112\).
Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.57 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\frac {-15 \cosh (c+d x) \left (\log \left (2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )\right )+25 \left (-\log \left (2 \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )\right )-12 \sinh (c+d x)}{64 d (5+3 \cosh (c+d x))} \] Input:
Integrate[(5 + 3*Cosh[c + d*x])^(-2),x]
Output:
(-15*Cosh[c + d*x]*(Log[2*Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2]] - Log[2*C osh[(c + d*x)/2] + Sinh[(c + d*x)/2]]) + 25*(-Log[2*Cosh[(c + d*x)/2] - Si nh[(c + d*x)/2]] + Log[2*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2]]) - 12*Sinh [c + d*x])/(64*d*(5 + 3*Cosh[c + d*x]))
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 3143, 27, 3042, 3136}
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.
\(\displaystyle \int \frac {1}{(3 \cosh (c+d x)+5)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (5+3 \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle -\frac {1}{16} \int -\frac {5}{3 \cosh (c+d x)+5}dx-\frac {3 \sinh (c+d x)}{16 d (3 \cosh (c+d x)+5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{16} \int \frac {1}{3 \cosh (c+d x)+5}dx-\frac {3 \sinh (c+d x)}{16 d (3 \cosh (c+d x)+5)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \sinh (c+d x)}{16 d (3 \cosh (c+d x)+5)}+\frac {5}{16} \int \frac {1}{3 \sin \left (i c+i d x+\frac {\pi }{2}\right )+5}dx\) |
\(\Big \downarrow \) 3136 |
\(\displaystyle \frac {5}{16} \left (\frac {x}{4}-\frac {\text {arctanh}\left (\frac {\sinh (c+d x)}{\cosh (c+d x)+3}\right )}{2 d}\right )-\frac {3 \sinh (c+d x)}{16 d (3 \cosh (c+d x)+5)}\) |
Input:
Int[(5 + 3*Cosh[c + d*x])^(-2),x]
Output:
(5*(x/4 - ArcTanh[Sinh[c + d*x]/(3 + Cosh[c + d*x])]/(2*d)))/16 - (3*Sinh[ c + d*x])/(16*d*(5 + 3*Cosh[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_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[ a^2 - b^2, 2]}, Simp[x/q, x] + Simp[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2, 0] && PosQ[a]
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 = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {\frac {3}{32 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {5 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{64}+\frac {3}{32 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {5 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64}}{d}\) | \(64\) |
default | \(\frac {\frac {3}{32 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {5 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{64}+\frac {3}{32 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {5 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64}}{d}\) | \(64\) |
risch | \(\frac {5 \,{\mathrm e}^{d x +c}+3}{8 d \left (3 \,{\mathrm e}^{2 d x +2 c}+10 \,{\mathrm e}^{d x +c}+3\right )}-\frac {5 \ln \left (3+{\mathrm e}^{d x +c}\right )}{64 d}+\frac {5 \ln \left ({\mathrm e}^{d x +c}+\frac {1}{3}\right )}{64 d}\) | \(68\) |
parallelrisch | \(\frac {-15 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right ) \cosh \left (d x +c \right )+15 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \cosh \left (d x +c \right )-25 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )+25 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )-12 \sinh \left (d x +c \right )}{64 d \left (5+3 \cosh \left (d x +c \right )\right )}\) | \(95\) |
Input:
int(1/(5+3*cosh(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(3/32/(tanh(1/2*d*x+1/2*c)-2)-5/64*ln(tanh(1/2*d*x+1/2*c)-2)+3/32/(tan h(1/2*d*x+1/2*c)+2)+5/64*ln(tanh(1/2*d*x+1/2*c)+2))
Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (50) = 100\).
Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 3.79 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\frac {5 \, {\left (3 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cosh \left (d x + c\right ) + 5\right )} \sinh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right )^{2} + 10 \, \cosh \left (d x + c\right ) + 3\right )} \log \left (3 \, \cosh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right ) + 1\right ) - 5 \, {\left (3 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cosh \left (d x + c\right ) + 5\right )} \sinh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right )^{2} + 10 \, \cosh \left (d x + c\right ) + 3\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 3\right ) + 40 \, \cosh \left (d x + c\right ) + 40 \, \sinh \left (d x + c\right ) + 24}{64 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} + 3 \, d \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right ) + 2 \, {\left (3 \, d \cosh \left (d x + c\right ) + 5 \, d\right )} \sinh \left (d x + c\right ) + 3 \, d\right )}} \] Input:
integrate(1/(5+3*cosh(d*x+c))^2,x, algorithm="fricas")
Output:
1/64*(5*(3*cosh(d*x + c)^2 + 2*(3*cosh(d*x + c) + 5)*sinh(d*x + c) + 3*sin h(d*x + c)^2 + 10*cosh(d*x + c) + 3)*log(3*cosh(d*x + c) + 3*sinh(d*x + c) + 1) - 5*(3*cosh(d*x + c)^2 + 2*(3*cosh(d*x + c) + 5)*sinh(d*x + c) + 3*s inh(d*x + c)^2 + 10*cosh(d*x + c) + 3)*log(cosh(d*x + c) + sinh(d*x + c) + 3) + 40*cosh(d*x + c) + 40*sinh(d*x + c) + 24)/(3*d*cosh(d*x + c)^2 + 3*d *sinh(d*x + c)^2 + 10*d*cosh(d*x + c) + 2*(3*d*cosh(d*x + c) + 5*d)*sinh(d *x + c) + 3*d)
Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (48) = 96\).
Time = 0.84 (sec) , antiderivative size = 199, normalized size of antiderivative = 3.55 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\begin {cases} - \frac {5 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {20 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {5 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} - \frac {20 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {12 \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (3 \cosh {\left (c \right )} + 5\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(5+3*cosh(d*x+c))**2,x)
Output:
Piecewise((-5*log(tanh(c/2 + d*x/2) - 2)*tanh(c/2 + d*x/2)**2/(64*d*tanh(c /2 + d*x/2)**2 - 256*d) + 20*log(tanh(c/2 + d*x/2) - 2)/(64*d*tanh(c/2 + d *x/2)**2 - 256*d) + 5*log(tanh(c/2 + d*x/2) + 2)*tanh(c/2 + d*x/2)**2/(64* d*tanh(c/2 + d*x/2)**2 - 256*d) - 20*log(tanh(c/2 + d*x/2) + 2)/(64*d*tanh (c/2 + d*x/2)**2 - 256*d) + 12*tanh(c/2 + d*x/2)/(64*d*tanh(c/2 + d*x/2)** 2 - 256*d), Ne(d, 0)), (x/(3*cosh(c) + 5)**2, True))
Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=-\frac {5 \, \log \left (3 \, e^{\left (-d x - c\right )} + 1\right )}{64 \, d} + \frac {5 \, \log \left (e^{\left (-d x - c\right )} + 3\right )}{64 \, d} - \frac {5 \, e^{\left (-d x - c\right )} + 3}{8 \, d {\left (10 \, e^{\left (-d x - c\right )} + 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3\right )}} \] Input:
integrate(1/(5+3*cosh(d*x+c))^2,x, algorithm="maxima")
Output:
-5/64*log(3*e^(-d*x - c) + 1)/d + 5/64*log(e^(-d*x - c) + 3)/d - 1/8*(5*e^ (-d*x - c) + 3)/(d*(10*e^(-d*x - c) + 3*e^(-2*d*x - 2*c) + 3))
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\frac {\frac {8 \, {\left (5 \, e^{\left (d x + c\right )} + 3\right )}}{3 \, e^{\left (2 \, d x + 2 \, c\right )} + 10 \, e^{\left (d x + c\right )} + 3} + 5 \, \log \left (3 \, e^{\left (d x + c\right )} + 1\right ) - 5 \, \log \left (e^{\left (d x + c\right )} + 3\right )}{64 \, d} \] Input:
integrate(1/(5+3*cosh(d*x+c))^2,x, algorithm="giac")
Output:
1/64*(8*(5*e^(d*x + c) + 3)/(3*e^(2*d*x + 2*c) + 10*e^(d*x + c) + 3) + 5*l og(3*e^(d*x + c) + 1) - 5*log(e^(d*x + c) + 3))/d
Time = 2.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\frac {\frac {5\,{\mathrm {e}}^{c+d\,x}}{8\,d}+\frac {3}{8\,d}}{10\,{\mathrm {e}}^{c+d\,x}+3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3}-\frac {5\,\mathrm {atan}\left (\left (\frac {5}{4\,d}+\frac {3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4\,d}\right )\,\sqrt {-d^2}\right )}{32\,\sqrt {-d^2}} \] Input:
int(1/(3*cosh(c + d*x) + 5)^2,x)
Output:
((5*exp(c + d*x))/(8*d) + 3/(8*d))/(10*exp(c + d*x) + 3*exp(2*c + 2*d*x) + 3) - (5*atan((5/(4*d) + (3*exp(d*x)*exp(c))/(4*d))*(-d^2)^(1/2)))/(32*(-d ^2)^(1/2))
Time = 0.21 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(5+3 \cosh (c+d x))^2} \, dx=\frac {-15 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+3\right )+15 e^{2 d x +2 c} \mathrm {log}\left (3 e^{d x +c}+1\right )-12 e^{2 d x +2 c}-50 e^{d x +c} \mathrm {log}\left (e^{d x +c}+3\right )+50 e^{d x +c} \mathrm {log}\left (3 e^{d x +c}+1\right )-15 \,\mathrm {log}\left (e^{d x +c}+3\right )+15 \,\mathrm {log}\left (3 e^{d x +c}+1\right )+12}{64 d \left (3 e^{2 d x +2 c}+10 e^{d x +c}+3\right )} \] Input:
int(1/(5+3*cosh(d*x+c))^2,x)
Output:
( - 15*e**(2*c + 2*d*x)*log(e**(c + d*x) + 3) + 15*e**(2*c + 2*d*x)*log(3* e**(c + d*x) + 1) - 12*e**(2*c + 2*d*x) - 50*e**(c + d*x)*log(e**(c + d*x) + 3) + 50*e**(c + d*x)*log(3*e**(c + d*x) + 1) - 15*log(e**(c + d*x) + 3) + 15*log(3*e**(c + d*x) + 1) + 12)/(64*d*(3*e**(2*c + 2*d*x) + 10*e**(c + d*x) + 3))