Integrand size = 13, antiderivative size = 24 \[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \cosh (x)}}{\sqrt {a}}\right )}{\sqrt {a}} \] Output:
-2*arctanh((a+b*cosh(x))^(1/2)/a^(1/2))/a^(1/2)
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \cosh (x)}}{\sqrt {a}}\right )}{\sqrt {a}} \] Input:
Integrate[Tanh[x]/Sqrt[a + b*Cosh[x]],x]
Output:
(-2*ArcTanh[Sqrt[a + b*Cosh[x]]/Sqrt[a]])/Sqrt[a]
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 26, 3200, 73, 220}
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.
\(\displaystyle \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\tan \left (-\frac {\pi }{2}+i x\right ) \sqrt {a-b \sin \left (-\frac {\pi }{2}+i x\right )}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\sqrt {a-b \sin \left (i x-\frac {\pi }{2}\right )} \tan \left (i x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle \int \frac {\text {sech}(x)}{b \sqrt {a+b \cosh (x)}}d(b \cosh (x))\) |
\(\Big \downarrow \) 73 |
\(\displaystyle 2 \int \frac {1}{b^2 \cosh ^2(x)-a}d\sqrt {a+b \cosh (x)}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \cosh (x)}}{\sqrt {a}}\right )}{\sqrt {a}}\) |
Input:
Int[Tanh[x]/Sqrt[a + b*Cosh[x]],x]
Output:
(-2*ArcTanh[Sqrt[a + b*Cosh[x]]/Sqrt[a]])/Sqrt[a]
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(103\) vs. \(2(18)=36\).
Time = 0.43 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.33
method | result | size |
default | \(-\frac {\ln \left (\frac {4 \cosh \left (\frac {x}{2}\right ) b \sqrt {2}+4 \sqrt {a}\, \sqrt {2 b \sinh \left (\frac {x}{2}\right )^{2}+a +b}+4 a -4 b}{2 \cosh \left (\frac {x}{2}\right )-\sqrt {2}}\right )+\ln \left (-\frac {4 \left (\cosh \left (\frac {x}{2}\right ) b \sqrt {2}-\sqrt {a}\, \sqrt {2 b \sinh \left (\frac {x}{2}\right )^{2}+a +b}-a +b \right )}{2 \cosh \left (\frac {x}{2}\right )+\sqrt {2}}\right )}{\sqrt {a}}\) | \(104\) |
Input:
int(tanh(x)/(a+b*cosh(x))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/a^(1/2)*(ln(4/(2*cosh(1/2*x)-2^(1/2))*(cosh(1/2*x)*b*2^(1/2)+a^(1/2)*(2 *b*sinh(1/2*x)^2+a+b)^(1/2)+a-b))+ln(-4/(2*cosh(1/2*x)+2^(1/2))*(cosh(1/2* x)*b*2^(1/2)-a^(1/2)*(2*b*sinh(1/2*x)^2+a+b)^(1/2)-a+b)))
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (18) = 36\).
Time = 0.12 (sec) , antiderivative size = 356, normalized size of antiderivative = 14.83 \[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=\left [\frac {\log \left (\frac {b^{2} \cosh \left (x\right )^{4} + b^{2} \sinh \left (x\right )^{4} + 16 \, a b \cosh \left (x\right )^{3} + 4 \, {\left (b^{2} \cosh \left (x\right ) + 4 \, a b\right )} \sinh \left (x\right )^{3} + 16 \, a b \cosh \left (x\right ) + 2 \, {\left (16 \, a^{2} + b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (x\right )^{2} + 24 \, a b \cosh \left (x\right ) + 16 \, a^{2} + b^{2}\right )} \sinh \left (x\right )^{2} - 8 \, {\left (b \cosh \left (x\right )^{3} + b \sinh \left (x\right )^{3} + 4 \, a \cosh \left (x\right )^{2} + {\left (3 \, b \cosh \left (x\right ) + 4 \, a\right )} \sinh \left (x\right )^{2} + b \cosh \left (x\right ) + {\left (3 \, b \cosh \left (x\right )^{2} + 8 \, a \cosh \left (x\right ) + b\right )} \sinh \left (x\right )\right )} \sqrt {b \cosh \left (x\right ) + a} \sqrt {a} + b^{2} + 4 \, {\left (b^{2} \cosh \left (x\right )^{3} + 12 \, a b \cosh \left (x\right )^{2} + 4 \, a b + {\left (16 \, a^{2} + b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {{\left (b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 4 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + 2 \, a\right )} \sinh \left (x\right ) + b\right )} \sqrt {b \cosh \left (x\right ) + a} \sqrt {-a}}{2 \, {\left (a b \cosh \left (x\right )^{2} + a b \sinh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) + a b + 2 \, {\left (a b \cosh \left (x\right ) + a^{2}\right )} \sinh \left (x\right )\right )}}\right )}{a}\right ] \] Input:
integrate(tanh(x)/(a+b*cosh(x))^(1/2),x, algorithm="fricas")
Output:
[1/2*log((b^2*cosh(x)^4 + b^2*sinh(x)^4 + 16*a*b*cosh(x)^3 + 4*(b^2*cosh(x ) + 4*a*b)*sinh(x)^3 + 16*a*b*cosh(x) + 2*(16*a^2 + b^2)*cosh(x)^2 + 2*(3* b^2*cosh(x)^2 + 24*a*b*cosh(x) + 16*a^2 + b^2)*sinh(x)^2 - 8*(b*cosh(x)^3 + b*sinh(x)^3 + 4*a*cosh(x)^2 + (3*b*cosh(x) + 4*a)*sinh(x)^2 + b*cosh(x) + (3*b*cosh(x)^2 + 8*a*cosh(x) + b)*sinh(x))*sqrt(b*cosh(x) + a)*sqrt(a) + b^2 + 4*(b^2*cosh(x)^3 + 12*a*b*cosh(x)^2 + 4*a*b + (16*a^2 + b^2)*cosh(x ))*sinh(x))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1))/sqrt( a), sqrt(-a)*arctan(1/2*(b*cosh(x)^2 + b*sinh(x)^2 + 4*a*cosh(x) + 2*(b*co sh(x) + 2*a)*sinh(x) + b)*sqrt(b*cosh(x) + a)*sqrt(-a)/(a*b*cosh(x)^2 + a* b*sinh(x)^2 + 2*a^2*cosh(x) + a*b + 2*(a*b*cosh(x) + a^2)*sinh(x)))/a]
\[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=\int \frac {\tanh {\left (x \right )}}{\sqrt {a + b \cosh {\left (x \right )}}}\, dx \] Input:
integrate(tanh(x)/(a+b*cosh(x))**(1/2),x)
Output:
Integral(tanh(x)/sqrt(a + b*cosh(x)), x)
\[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=\int { \frac {\tanh \left (x\right )}{\sqrt {b \cosh \left (x\right ) + a}} \,d x } \] Input:
integrate(tanh(x)/(a+b*cosh(x))^(1/2),x, algorithm="maxima")
Output:
integrate(tanh(x)/sqrt(b*cosh(x) + a), x)
\[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=\int { \frac {\tanh \left (x\right )}{\sqrt {b \cosh \left (x\right ) + a}} \,d x } \] Input:
integrate(tanh(x)/(a+b*cosh(x))^(1/2),x, algorithm="giac")
Output:
integrate(tanh(x)/sqrt(b*cosh(x) + a), x)
Timed out. \[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=\int \frac {\mathrm {tanh}\left (x\right )}{\sqrt {a+b\,\mathrm {cosh}\left (x\right )}} \,d x \] Input:
int(tanh(x)/(a + b*cosh(x))^(1/2),x)
Output:
int(tanh(x)/(a + b*cosh(x))^(1/2), x)
\[ \int \frac {\tanh (x)}{\sqrt {a+b \cosh (x)}} \, dx=\int \frac {\sqrt {\cosh \left (x \right ) b +a}\, \tanh \left (x \right )}{\cosh \left (x \right ) b +a}d x \] Input:
int(tanh(x)/(a+b*cosh(x))^(1/2),x)
Output:
int((sqrt(cosh(x)*b + a)*tanh(x))/(cosh(x)*b + a),x)