Integrand size = 17, antiderivative size = 72 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}-\frac {\sqrt {a+b \tanh ^2(x)}}{b^2} \] Output:
arctanh((a+b*tanh(x)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)-a^2/b^2/(a+b)/(a+b* tanh(x)^2)^(1/2)-(a+b*tanh(x)^2)^(1/2)/b^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {-b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tanh ^2(x)}{a+b}\right )-(a+b) \left (2 a-b+b \tanh ^2(x)\right )}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}} \] Input:
Integrate[Tanh[x]^5/(a + b*Tanh[x]^2)^(3/2),x]
Output:
(-(b^2*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tanh[x]^2)/(a + b)]) - (a + b)*(2*a - b + b*Tanh[x]^2))/(b^2*(a + b)*Sqrt[a + b*Tanh[x]^2])
Time = 0.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 26, 4153, 26, 354, 98, 2009}
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.
\(\displaystyle \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)^5}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)^5}{\left (a-b \tan (i x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle -i \int \frac {i \tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\tanh ^5(x)}{\left (1-\tanh ^2(x)\right ) \left (a+b \tanh ^2(x)\right )^{3/2}}d\tanh (x)\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\tanh ^4(x)}{\left (1-\tanh ^2(x)\right ) \left (b \tanh ^2(x)+a\right )^{3/2}}d\tanh ^2(x)\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a^2}{b (a+b) \left (b \tanh ^2(x)+a\right )^{3/2}}-\frac {1}{b \sqrt {b \tanh ^2(x)+a}}-\frac {1}{(a+b) \left (\tanh ^2(x)-1\right ) \sqrt {b \tanh ^2(x)+a}}\right )d\tanh ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 a^2}{b^2 (a+b) \sqrt {a+b \tanh ^2(x)}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tanh ^2(x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {2 \sqrt {a+b \tanh ^2(x)}}{b^2}\right )\) |
Input:
Int[Tanh[x]^5/(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^2)/(b ^2*(a + b)*Sqrt[a + b*Tanh[x]^2]) - (2*Sqrt[a + b*Tanh[x]^2])/b^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[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
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. \(321\) vs. \(2(62)=124\).
Time = 0.08 (sec) , antiderivative size = 322, normalized size of antiderivative = 4.47
method | result | size |
derivativedivides | \(\frac {1}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {\tanh \left (x \right )^{2}}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {2 a}{b^{2} \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\) | \(322\) |
default | \(\frac {1}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {\tanh \left (x \right )^{2}}{b \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {2 a}{b^{2} \sqrt {a +b \tanh \left (x \right )^{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {b \left (2 b \left (\tanh \left (x \right )-1\right )+2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}-\frac {1}{2 \left (a +b \right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}-\frac {b \left (2 b \left (\tanh \left (x \right )+1\right )-2 b \right )}{\left (a +b \right ) \left (4 b \left (a +b \right )-4 b^{2}\right ) \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \left (a +b \right )^{\frac {3}{2}}}\) | \(322\) |
Input:
int(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/b/(a+b*tanh(x)^2)^(1/2)-tanh(x)^2/b/(a+b*tanh(x)^2)^(1/2)-2*a/b^2/(a+b*t anh(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. 1713 vs. \(2 (62) = 124\).
Time = 0.33 (sec) , antiderivative size = 3991, normalized size of antiderivative = 55.43 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\int \frac {\tanh ^{5}{\left (x \right )}}{\left (a + b \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(tanh(x)**5/(a+b*tanh(x)**2)**(3/2),x)
Output:
Integral(tanh(x)**5/(a + b*tanh(x)**2)**(3/2), x)
\[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x, algorithm="maxima")
Output:
integrate(tanh(x)^5/(b*tanh(x)^2 + a)^(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 460 vs. \(2 (62) = 124\).
Time = 0.39 (sec) , antiderivative size = 460, normalized size of antiderivative = 6.39 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=-\frac {\frac {{\left (a^{4} b + a^{3} b^{2}\right )} e^{\left (2 \, x\right )}}{a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}} + \frac {a^{4} b + a^{3} b^{2}}{a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}}}{\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}}} - \frac {4 \, {\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 )}}{{\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 )}^{2} + 2 \, {\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 )} \sqrt {a + b} + a - 3 \, b\right )} b} \] Input:
integrate(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x, algorithm="giac")
Output:
-((a^4*b + a^3*b^2)*e^(2*x)/(a^3*b^3 + 2*a^2*b^4 + a*b^5) + (a^4*b + a^3*b ^2)/(a^3*b^3 + 2*a^2*b^4 + a*b^5))/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) - 4*(sqr t(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)*e^(2*x) - sqrt(a*e^(4*x) + b*e^(4*x ) + 2*a*e^(2*x) - 2*b*e^(2*x) + a + b))^2 + 2*(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 - 3*b)*b)
Time = 3.84 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}\,\left (2\,a+2\,b\right )}{2\,{\left (a+b\right )}^{3/2}}\right )}{{\left (a+b\right )}^{3/2}}-\frac {\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}}{b^2}-\frac {a^2}{b^2\,\left (a+b\right )\,\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}} \] Input:
int(tanh(x)^5/(a + b*tanh(x)^2)^(3/2),x)
Output:
atanh(((a + b*tanh(x)^2)^(1/2)*(2*a + 2*b))/(2*(a + b)^(3/2)))/(a + b)^(3/ 2) - (a + b*tanh(x)^2)^(1/2)/b^2 - a^2/(b^2*(a + b)*(a + b*tanh(x)^2)^(1/2 ))
\[ \int \frac {\tanh ^5(x)}{\left (a+b \tanh ^2(x)\right )^{3/2}} \, dx=\frac {-\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )^{2} b -2 \sqrt {\tanh \left (x \right )^{2} b +a}\, a +\sqrt {\tanh \left (x \right )^{2} b +a}\, b +\left (\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{4} b^{2}+2 \tanh \left (x \right )^{2} a b +a^{2}}d x \right ) \tanh \left (x \right )^{2} b^{3}+\left (\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )}{\tanh \left (x \right )^{4} b^{2}+2 \tanh \left (x \right )^{2} a b +a^{2}}d x \right ) a \,b^{2}}{b^{2} \left (\tanh \left (x \right )^{2} b +a \right )} \] Input:
int(tanh(x)^5/(a+b*tanh(x)^2)^(3/2),x)
Output:
( - sqrt(tanh(x)**2*b + a)*tanh(x)**2*b - 2*sqrt(tanh(x)**2*b + a)*a + sqr t(tanh(x)**2*b + a)*b + 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**3 + 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**2)/(b** 2*(tanh(x)**2*b + a))