Integrand size = 17, antiderivative size = 52 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\frac {a}{b (a+b) \sqrt {a+b \tanh ^2(x)}} \] Output:
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)+a/b/(a+b)/(a+b*tanh (x)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\frac {a}{b (a+b) \sqrt {a+b \tanh ^2(x)}} \] Input:
Integrate[Tanh[x]^3/(a + b*Tanh[x]^2)^(3/2),x]
Output:
ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]]/(a + b)^(3/2) + a/(b*(a + b)*Sq rt[a + b*Tanh[x]^2])
Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 26, 4153, 26, 354, 87, 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.
\(\displaystyle \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \tan (i x)^3}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\tan (i x)^3}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle i \int -\frac {i \tanh ^3(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\tanh ^3(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh ^2(x)}{a+b}+\frac {2 a}{b (a+b) \sqrt {a+b \tanh ^2(x)}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\tanh ^4(x)}{b}}d\sqrt {b \tanh ^2(x)+a}}{b (a+b)}+\frac {2 a}{b (a+b) \sqrt {a+b \tanh ^2(x)}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\frac {2 a}{b (a+b) \sqrt {a+b \tanh ^2(x)}}\right )\) |
Input:
Int[Tanh[x]^3/(a + b*Tanh[x]^2)^(3/2),x]
Output:
((2*ArcTanh[Sqrt[a + b*Tanh[x]^2]/Sqrt[a + b]])/(a + b)^(3/2) + (2*a)/(b*( a + b)*Sqrt[a + b*Tanh[x]^2]))/2
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && IntegerQ [(m - 1)/2]
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. \(286\) vs. \(2(44)=88\).
Time = 0.06 (sec) , antiderivative size = 287, normalized size of antiderivative = 5.52
method | result | size |
derivativedivides | \(\frac {1}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\) | \(287\) |
default | \(\frac {1}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\) | \(287\) |
Input:
int(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/b/(a+b*tanh(x)^2)^(1/2)-1/2/(a+b)/(b*(tanh(x)-1)^2+2*b*(tanh(x)-1)+a+b)^ (1/2)+b/(a+b)*(2*b*(tanh(x)-1)+2*b)/(4*b*(a+b)-4*b^2)/(b*(tanh(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*(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 )/(b*(tanh(x)+1)^2-2*b*(tanh(x)+1)+a+b)^(1/2)-b/(a+b)*(2*b*(tanh(x)+1)-2*b )/(4*b*(a+b)-4*b^2)/(b*(tanh(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*(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))
Leaf count of result is larger than twice the leaf count of optimal. 980 vs. \(2 (44) = 88\).
Time = 0.18 (sec) , antiderivative size = 2525, normalized size of antiderivative = 48.56 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x, algorithm="fricas")
Output:
[1/4*(((a*b + b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*sinh(x)^3 + (a*b + b^ 2)*sinh(x)^4 + 2*(a*b - b^2)*cosh(x)^2 + 2*(3*(a*b + b^2)*cosh(x)^2 + a*b - b^2)*sinh(x)^2 + a*b + b^2 + 4*((a*b + b^2)*cosh(x)^3 + (a*b - b^2)*cosh (x))*sinh(x))*sqrt(a + b)*log(((a^3 + a^2*b)*cosh(x)^8 + 8*(a^3 + a^2*b)*c osh(x)*sinh(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)*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)*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)*co sh(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*si nh(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)*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 ...
\[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(tanh(x)**3/(a+b*tanh(x)**2)**(3/2),x)
Output:
Integral(tanh(x)**3/(a + b*tanh(x)**2)**(3/2), x)
\[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")
Output:
integrate(tanh(x)^3/(b*tanh(x)^2 + a)^(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (44) = 88\).
Time = 0.26 (sec) , antiderivative size = 287, normalized size of antiderivative = 5.52 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\frac {{\left (a^{3} + a^{2} b\right )} e^{\left (2 \, x\right )}}{a^{3} b + 2 \, a^{2} b^{2} + a b^{3}} + \frac {a^{3} + a^{2} b}{a^{3} b + 2 \, a^{2} b^{2} + a b^{3}}}{\sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}} - \frac {\log \left ({\left | -{\left (\sqrt {a + b} e^{\left (2 \, x\right )} - \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b}\right )} {\left (a + b\right )} - \sqrt {a + b} {\left (a - b\right )} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} + \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} + \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} - \frac {\log \left ({\left | -\sqrt {a + b} e^{\left (2 \, x\right )} + \sqrt {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a + b} - \sqrt {a + b} \right |}\right )}{2 \, {\left (a + b\right )}^{\frac {3}{2}}} \] Input:
integrate(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x, algorithm="giac")
Output:
((a^3 + a^2*b)*e^(2*x)/(a^3*b + 2*a^2*b^2 + a*b^3) + (a^3 + a^2*b)/(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 = 3.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}}{\sqrt {a+b}}\right )}{{\left (a+b\right )}^{3/2}}+\frac {a}{\left (b^2+a\,b\right )\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}} \] Input:
int(tanh(x)^3/(a + b*tanh(x)^2)^(3/2),x)
Output:
atanh((a + b*tanh(x)^2)^(1/2)/(a + b)^(1/2))/(a + b)^(3/2) + a/((a*b + b^2 )*(a + b*tanh(x)^2)^(1/2))
\[ \int \frac {\tanh ^3(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\sqrt {\tanh \left (x \right )^{2} b +a}+\left (\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{4} b^{2}+2 \tanh \left (x \right )^{2} a b +a^{2}}d x \right ) \tanh \left (x \right )^{2} b^{2}+\left (\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{4} b^{2}+2 \tanh \left (x \right )^{2} a b +a^{2}}d x \right ) a b}{b \left (\tanh \left (x \right )^{2} b +a \right )} \] Input:
int(tanh(x)^3/(a+b*tanh(x)^2)^(3/2),x)
Output:
(sqrt(tanh(x)**2*b + a) + int((sqrt(tanh(x)**2*b + a)*tanh(x))/(tanh(x)**4 *b**2 + 2*tanh(x)**2*a*b + a**2),x)*tanh(x)**2*b**2 + int((sqrt(tanh(x)**2 *b + a)*tanh(x))/(tanh(x)**4*b**2 + 2*tanh(x)**2*a*b + a**2),x)*a*b)/(b*(t anh(x)**2*b + a))