Integrand size = 15, antiderivative size = 49 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {1}{(a+b) \sqrt {a+b \tanh ^2(x)}} \]
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tanh ^2(x)}{a+b}\right )}{(a+b) \sqrt {a+b \tanh ^2(x)}} \]
-(Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tanh[x]^2)/(a + b)]/((a + b)*Sqrt [a + b*Tanh[x]^2]))
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 26, 4153, 26, 353, 61, 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.
\(\displaystyle \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -i \int \frac {i \tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\tanh (x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)}{a+b}-\frac {2}{(a+b) \sqrt {a+b \tanh ^2(x)}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b (a+b)}-\frac {2}{(a+b) \sqrt {a+b \tanh ^2(x)}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {2}{(a+b) \sqrt {a+b \tanh ^2(x)}}\right )\) |
((2*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]])/(a + b)^(3/2) - 2/((a + b) *Sqrt[a + b*Tanh[x]^2]))/2
3.3.42.3.1 Defintions of rubi rules used
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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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. \(272\) vs. \(2(41)=82\).
Time = 0.07 (sec) , antiderivative size = 273, normalized size of antiderivative = 5.57
method | result | size |
derivativedivides | \(-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {b \left (2 b \left (1+\tanh \left (x \right )\right )-2 b \right )}{\left (a +b \right ) \left (4 \left (a +b \right ) b -4 b^{2}\right ) \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 \left (a +b \right ) b -4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\) | \(273\) |
default | \(-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}-\frac {b \left (2 b \left (1+\tanh \left (x \right )\right )-2 b \right )}{\left (a +b \right ) \left (4 \left (a +b \right ) b -4 b^{2}\right ) \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tanh \left (x \right )\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tanh \left (x \right )\right )^{2}-2 b \left (1+\tanh \left (x \right )\right )+a +b}}{1+\tanh \left (x \right )}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 \left (a +b \right ) b -4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\) | \(273\) |
-1/2/(a+b)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)-b/(a+b)*(2*b*(1+tan h(x))-2*b)/(4*(a+b)*b-4*b^2)/(b*(1+tanh(x))^2-2*b*(1+tanh(x))+a+b)^(1/2)+1 /2/(a+b)^(3/2)*ln((2*a+2*b-2*b*(1+tanh(x))+2*(a+b)^(1/2)*(b*(1+tanh(x))^2- 2*b*(1+tanh(x))+a+b)^(1/2))/(1+tanh(x)))-1/2/(a+b)/(b*(tanh(x)-1)^2+2*b*(t anh(x)-1)+a+b)^(1/2)+b/(a+b)*(2*b*(tanh(x)-1)+2*b)/(4*(a+b)*b-4*b^2)/(b*(t anh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2)+1/2/(a+b)^(3/2)*ln((2*a+2*b+2*b*(ta nh(x)-1)+2*(a+b)^(1/2)*(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^(1/2))/(tanh( x)-1))
Leaf count of result is larger than twice the leaf count of optimal. 856 vs. \(2 (41) = 82\).
Time = 0.36 (sec) , antiderivative size = 2277, normalized size of antiderivative = 46.47 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(((a + b)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 + 2*(a - b)*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 + a - b)*sinh(x)^2 + 4*((a + b)*cosh(x)^3 + (a - b)*cosh(x))*sinh(x) + a + b)*sqrt(a + b)*log(((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)*cosh(x)*sinh(x)^7 + (a^3 + a^2*b)*sin h(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)*cosh(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)*co sh(x)^2)*sinh(x)^4 + 4*(14*(a^3 + a^2*b)*cosh(x)^5 + 10*(2*a^3 + a^2*b)*co sh(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)*sinh(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)*cosh(x))*sinh(x))*sqrt(a + b)*s qrt(((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)*cosh(x)^7 + 3*(2*a^3 + a^2*...
Time = 10.71 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.88 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=- \begin {cases} \frac {2 \left (\frac {b}{2 \left (a + b\right ) \sqrt {a + b \tanh ^{2}{\left (x \right )}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {a + b \tanh ^{2}{\left (x \right )}}}{\sqrt {- a - b}} \right )}}{2 \sqrt {- a - b} \left (a + b\right )}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \tilde {\infty } \tanh ^{2}{\left (x \right )} & \text {for}\: a^{\frac {3}{2}} = 0 \\\frac {\log {\left (2 a^{\frac {3}{2}} \tanh ^{2}{\left (x \right )} - 2 a^{\frac {3}{2}} \right )}}{2 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]
-Piecewise((2*(b/(2*(a + b)*sqrt(a + b*tanh(x)**2)) + b*atan(sqrt(a + b*ta nh(x)**2)/sqrt(-a - b))/(2*sqrt(-a - b)*(a + b)))/b, Ne(b, 0)), (Piecewise ((zoo*tanh(x)**2, Eq(a**(3/2), 0)), (log(2*a**(3/2)*tanh(x)**2 - 2*a**(3/2 ))/(2*a**(3/2)), True)), True))
\[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (41) = 82\).
Time = 0.45 (sec) , antiderivative size = 292, normalized size of antiderivative = 5.96 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\frac {{\left (a^{2} b + a b^{2}\right )} e^{\left (2 \, x\right )}}{a^{3} b + 2 \, a^{2} b^{2} + a b^{3}} + \frac {a^{2} b + a b^{2}}{a^{3} b + 2 \, a^{2} b^{2} + a b^{3}}}{\sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} \]
-((a^2*b + a*b^2)*e^(2*x)/(a^3*b + 2*a^2*b^2 + a*b^3) + (a^2*b + a*b^2)/(a ^3*b + 2*a^2*b^2 + a*b^3))/sqrt(a*e^(4*x) + b*e^(4*x) + 2*a*e^(2*x) - 2*b* e^(2*x) + 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))*(a + b) - sqrt(a + b)*(a - b )))/(a + b)^(3/2) + 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)))/(a + b)^(3/2) - 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)))/(a + b)^(3/2)
Time = 2.51 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {\tanh (x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}}{\sqrt {a+b}}\right )}{{\left (a+b\right )}^{3/2}}-\frac {1}{\left (a+b\right )\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}} \]