Integrand size = 15, antiderivative size = 29 \[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}} \] Output:
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(1/2)
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}} \] Input:
Integrate[Tanh[x]/Sqrt[a + b*Tanh[x]^2],x]
Output:
ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]]/Sqrt[a + b]
Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 26, 4153, 26, 353, 73, 221}
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.
\(\displaystyle \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)}{\sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)}{\sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -i \int \frac {i \tanh (x)}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\tanh (x)}{\left (1-\tanh ^2(x)\right ) \sqrt {a+b \tanh ^2(x)}}d\tanh (x)\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}\) |
Input:
Int[Tanh[x]/Sqrt[a + b*Tanh[x]^2],x]
Output:
ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]]/Sqrt[a + b]
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/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(23)=46\).
Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.93
method | result | size |
derivativedivides | \(\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \sqrt {a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \sqrt {a +b}}\) | \(114\) |
default | \(\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \sqrt {a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \sqrt {a +b}}\) | \(114\) |
Input:
int(tanh(x)/(a+b*tanh(x)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2/(a+b)^(1/2)*ln((2*a+2*b+2*b*(tanh(x)-1)+2*(a+b)^(1/2)*(b*(tanh(x)-1)^2 +2*b*(tanh(x)-1)+a+b)^(1/2))/(tanh(x)-1))+1/2/(a+b)^(1/2)*ln((2*a+2*b-2*b* (tanh(x)+1)+2*(a+b)^(1/2)*(b*(tanh(x)+1)^2-2*b*(tanh(x)+1)+a+b)^(1/2))/(ta nh(x)+1))
Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (23) = 46\).
Time = 0.13 (sec) , antiderivative size = 1361, normalized size of antiderivative = 46.93 \[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)/(a+b*tanh(x)^2)^(1/2),x, algorithm="fricas")
Output:
[1/4*(sqrt(a + b)*log(((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)*cosh(x)*s inh(x)^7 + (a^3 + a^2*b)*sinh(x)^8 + 2*(2*a^3 + a^2*b)*cosh(x)^6 + 2*(2*a^ 3 + a^2*b + 14*(a^3 + a^2*b)*cosh(x)^2)*sinh(x)^6 + 4*(14*(a^3 + a^2*b)*co sh(x)^3 + 3*(2*a^3 + a^2*b)*cosh(x))*sinh(x)^5 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^4 + (70*(a^3 + a^2*b)*cosh(x)^4 + 6*a^3 + 4*a^2*b - a*b^2 + b^3 + 30*(2*a^3 + a^2*b)*cosh(x)^2)*sinh(x)^4 + 4*(14*(a^3 + a^2*b)*cosh( x)^5 + 10*(2*a^3 + a^2*b)*cosh(x)^3 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh (x))*sinh(x)^3 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^3 + 3*a^2*b - b^3) *cosh(x)^2 + 2*(14*(a^3 + a^2*b)*cosh(x)^6 + 15*(2*a^3 + a^2*b)*cosh(x)^4 + 2*a^3 + 3*a^2*b - b^3 + 3*(6*a^3 + 4*a^2*b - a*b^2 + b^3)*cosh(x)^2)*sin h(x)^2 + sqrt(2)*(a^2*cosh(x)^6 + 6*a^2*cosh(x)*sinh(x)^5 + a^2*sinh(x)^6 + 3*a^2*cosh(x)^4 + 3*(5*a^2*cosh(x)^2 + a^2)*sinh(x)^4 + 4*(5*a^2*cosh(x) ^3 + 3*a^2*cosh(x))*sinh(x)^3 + (3*a^2 + 2*a*b - b^2)*cosh(x)^2 + (15*a^2* cosh(x)^4 + 18*a^2*cosh(x)^2 + 3*a^2 + 2*a*b - b^2)*sinh(x)^2 + a^2 + 2*a* b + b^2 + 2*(3*a^2*cosh(x)^5 + 6*a^2*cosh(x)^3 + (3*a^2 + 2*a*b - b^2)*cos h(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a^3 + a^2*b)*co sh(x)^7 + 3*(2*a^3 + a^2*b)*cosh(x)^5 + (6*a^3 + 4*a^2*b - a*b^2 + b^3)*co sh(x)^3 + (2*a^3 + 3*a^2*b - b^3)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x) ^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(...
Time = 0.99 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.34 \[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=- \begin {cases} \frac {\operatorname {atan}{\left (\frac {\sqrt {a + b \tanh ^{2}{\left (x \right )}}}{\sqrt {- a - b}} \right )}}{\sqrt {- a - b}} & \text {for}\: b \neq 0 \\\begin {cases} \tilde {\infty } \tanh ^{2}{\left (x \right )} & \text {for}\: \sqrt {a} = 0 \\\frac {\log {\left (2 \sqrt {a} \tanh ^{2}{\left (x \right )} - 2 \sqrt {a} \right )}}{2 \sqrt {a}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \] Input:
integrate(tanh(x)/(a+b*tanh(x)**2)**(1/2),x)
Output:
-Piecewise((atan(sqrt(a + b*tanh(x)**2)/sqrt(-a - b))/sqrt(-a - b), Ne(b, 0)), (Piecewise((zoo*tanh(x)**2, Eq(sqrt(a), 0)), (log(2*sqrt(a)*tanh(x)** 2 - 2*sqrt(a))/(2*sqrt(a)), True)), True))
\[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int { \frac {\tanh \left (x\right )}{\sqrt {b \tanh \left (x\right )^{2} + a}} \,d x } \] Input:
integrate(tanh(x)/(a+b*tanh(x)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(tanh(x)/sqrt(b*tanh(x)^2 + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (23) = 46\).
Time = 0.21 (sec) , antiderivative size = 188, normalized size of antiderivative = 6.48 \[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=-\frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, \sqrt {a + b}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, \sqrt {a + b}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, \sqrt {a + b}} \] Input:
integrate(tanh(x)/(a+b*tanh(x)^2)^(1/2),x, algorithm="giac")
Output:
-1/2*log(abs(-(sqrt(a + b)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2 *x) - 2*b*e^(2*x) + a + b))*(a + b) - sqrt(a + b)*(a - b)))/sqrt(a + b) + 1/2*log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x ) - 2*b*e^(2*x) + a + b) + sqrt(a + b)))/sqrt(a + b) - 1/2*log(abs(-sqrt(a + b)*e^(2*x) + sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b) - sqrt(a + b)))/sqrt(a + b)
Time = 2.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}}{\sqrt {a+b}}\right )}{\sqrt {a+b}} \] Input:
int(tanh(x)/(a + b*tanh(x)^2)^(1/2),x)
Output:
atanh((a + b*tanh(x)^2)^(1/2)/(a + b)^(1/2))/(a + b)^(1/2)
\[ \int \frac {\tanh (x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{2} b +a}d x \] Input:
int(tanh(x)/(a+b*tanh(x)^2)^(1/2),x)
Output:
int((sqrt(tanh(x)**2*b + a)*tanh(x))/(tanh(x)**2*b + a),x)