Integrand size = 18, antiderivative size = 44 \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=-\frac {2 \text {arctanh}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2} d} \]
Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=\frac {2 \arctan \left (\frac {b-2 a \tanh (c+d x)}{\sqrt {-4 a^2-b^2}}\right )}{\sqrt {-4 a^2-b^2} d} \]
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3145, 3042, 3139, 1083, 217}
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.
\(\displaystyle \int \frac {1}{a+b \sinh (c+d x) \cosh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a-i b \sin (i c+i d x) \cos (i c+i d x)}dx\) |
\(\Big \downarrow \) 3145 |
\(\displaystyle \int \frac {1}{a+\frac {1}{2} b \sinh (2 c+2 d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{a-\frac {1}{2} i b \sin (2 i c+2 i d x)}dx\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {i \int \frac {1}{-a \tanh ^2(c+d x)+b \tanh (c+d x)+a}d(i \tanh (c+d x))}{d}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {2 i \int \frac {1}{-4 a^2-b^2+\tanh ^2(c+d x)}d(2 i a \tanh (c+d x)-i b)}{d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 \text {arctanh}\left (\frac {\tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{d \sqrt {4 a^2+b^2}}\) |
3.9.58.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(132\) vs. \(2(40)=80\).
Time = 1.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 3.02
method | result | size |
risch | \(\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \sqrt {4 a^{2}+b^{2}}-4 a^{2}-b^{2}}{\sqrt {4 a^{2}+b^{2}}\, b}\right )}{\sqrt {4 a^{2}+b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \sqrt {4 a^{2}+b^{2}}+4 a^{2}+b^{2}}{\sqrt {4 a^{2}+b^{2}}\, b}\right )}{\sqrt {4 a^{2}+b^{2}}\, d}\) | \(133\) |
derivativedivides | \(\frac {2 a \left (\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{2 \sqrt {4 a^{2}+b^{2}}\, a}+\frac {\left (-4 a^{2}-b^{2}\right ) \ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a}\right )}{d}\) | \(151\) |
default | \(\frac {2 a \left (\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{2 \sqrt {4 a^{2}+b^{2}}\, a}+\frac {\left (-4 a^{2}-b^{2}\right ) \ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a}\right )}{d}\) | \(151\) |
1/(4*a^2+b^2)^(1/2)/d*ln(exp(2*d*x+2*c)+(2*a*(4*a^2+b^2)^(1/2)-4*a^2-b^2)/ (4*a^2+b^2)^(1/2)/b)-1/(4*a^2+b^2)^(1/2)/d*ln(exp(2*d*x+2*c)+(2*a*(4*a^2+b ^2)^(1/2)+4*a^2+b^2)/(4*a^2+b^2)^(1/2)/b)
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (43) = 86\).
Time = 0.26 (sec) , antiderivative size = 299, normalized size of antiderivative = 6.80 \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=\frac {\log \left (\frac {b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + 8 \, a^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 2 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2 \, {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + 2 \, a\right )} \sqrt {4 \, a^{2} + b^{2}}}{b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cosh \left (d x + c\right )^{2} + 2 \, a\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b \cosh \left (d x + c\right )^{3} + 2 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - b}\right )}{\sqrt {4 \, a^{2} + b^{2}} d} \]
log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh( d*x + c)^4 + 4*a*b*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b)*sin h(d*x + c)^2 + 8*a^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + 2*a*b*cosh(d*x + c)) *sinh(d*x + c) - 2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2*a)*sqrt(4*a^2 + b^2))/(b*cosh(d*x + c)^4 + 4*b*cosh( d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 4*a*cosh(d*x + c)^2 + 2*(3* b*cosh(d*x + c)^2 + 2*a)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + 2*a*cosh (d*x + c))*sinh(d*x + c) - b))/(sqrt(4*a^2 + b^2)*d)
Timed out. \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.66 \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=\frac {\log \left (\frac {b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a - \sqrt {4 \, a^{2} + b^{2}}}{b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a + \sqrt {4 \, a^{2} + b^{2}}}\right )}{\sqrt {4 \, a^{2} + b^{2}} d} \]
log((b*e^(-2*d*x - 2*c) - 2*a - sqrt(4*a^2 + b^2))/(b*e^(-2*d*x - 2*c) - 2 *a + sqrt(4*a^2 + b^2)))/(sqrt(4*a^2 + b^2)*d)
Time = 0.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.80 \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=\frac {\log \left (\frac {{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a - 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}\right )}{\sqrt {4 \, a^{2} + b^{2}} d} \]
log(abs(2*b*e^(2*d*x + 2*c) + 4*a - 2*sqrt(4*a^2 + b^2))/abs(2*b*e^(2*d*x + 2*c) + 4*a + 2*sqrt(4*a^2 + b^2)))/(sqrt(4*a^2 + b^2)*d)
Time = 2.95 (sec) , antiderivative size = 343, normalized size of antiderivative = 7.80 \[ \int \frac {1}{a+b \cosh (c+d x) \sinh (c+d x)} \, dx=\frac {2\,\mathrm {atan}\left (\left (\frac {b^4\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}}{16}+\frac {a^2\,b^2\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}}{4}\right )\,\left (\frac {32\,a\,\left (8\,a^2+b^2\right )}{b^4\,d\,{\left (4\,a^2+b^2\right )}^2}-{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {64\,a\,\left (16\,d\,a^3+4\,d\,a\,b^2\right )}{b^5\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\,\left (4\,a^2+b^2\right )\,\sqrt {-d^2\,\left (4\,a^2+b^2\right )}}+\frac {16\,\left (8\,a^2+b^2\right )\,\left (8\,a^2\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}+b^2\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\right )}{b^5\,d\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\,{\left (4\,a^2+b^2\right )}^2}\right )+\frac {64\,a\,\left (4\,d\,a^2\,b+d\,b^3\right )}{b^5\,\sqrt {-4\,a^2\,d^2-b^2\,d^2}\,\left (4\,a^2+b^2\right )\,\sqrt {-d^2\,\left (4\,a^2+b^2\right )}}\right )\right )}{\sqrt {-4\,a^2\,d^2-b^2\,d^2}} \]
(2*atan(((b^4*(- 4*a^2*d^2 - b^2*d^2)^(1/2))/16 + (a^2*b^2*(- 4*a^2*d^2 - b^2*d^2)^(1/2))/4)*((32*a*(8*a^2 + b^2))/(b^4*d*(4*a^2 + b^2)^2) - exp(2*c )*exp(2*d*x)*((64*a*(16*a^3*d + 4*a*b^2*d))/(b^5*(- 4*a^2*d^2 - b^2*d^2)^( 1/2)*(4*a^2 + b^2)*(-d^2*(4*a^2 + b^2))^(1/2)) + (16*(8*a^2 + b^2)*(8*a^2* (- 4*a^2*d^2 - b^2*d^2)^(1/2) + b^2*(- 4*a^2*d^2 - b^2*d^2)^(1/2)))/(b^5*d *(- 4*a^2*d^2 - b^2*d^2)^(1/2)*(4*a^2 + b^2)^2)) + (64*a*(b^3*d + 4*a^2*b* d))/(b^5*(- 4*a^2*d^2 - b^2*d^2)^(1/2)*(4*a^2 + b^2)*(-d^2*(4*a^2 + b^2))^ (1/2)))))/(- 4*a^2*d^2 - b^2*d^2)^(1/2)