Integrand size = 17, antiderivative size = 88 \[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {(a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{\sqrt {a+b}}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b} \] Output:
1/2*(a-2*b)*arctanh(b^(1/2)*tanh(x)/(a+b*tanh(x)^2)^(1/2))/b^(3/2)+arctanh ((a+b)^(1/2)*tanh(x)/(a+b*tanh(x)^2)^(1/2))/(a+b)^(1/2)-1/2*tanh(x)*(a+b*t anh(x)^2)^(1/2)/b
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 3.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.36 \[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\frac {\left (\sqrt {2} a (a+b) \sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}}}{\sqrt {2}}\right ),1\right )-2 \sqrt {2} a b \sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}} \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {\frac {(a-b+(a+b) \cosh (2 x)) \text {csch}^2(x)}{b}}}{\sqrt {2}}\right ),1\right )-(a+b) (a-b+(a+b) \cosh (2 x)) \text {sech}^2(x)\right ) \tanh (x)}{2 \sqrt {2} b (a+b) \sqrt {(a-b+(a+b) \cosh (2 x)) \text {sech}^2(x)}} \] Input:
Integrate[Tanh[x]^4/Sqrt[a + b*Tanh[x]^2],x]
Output:
((Sqrt[2]*a*(a + b)*Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]*Ellipt icF[ArcSin[Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]/Sqrt[2]], 1] - 2*Sqrt[2]*a*b*Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]*EllipticPi[b /(a + b), ArcSin[Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]/Sqrt[2]], 1] - (a + b)*(a - b + (a + b)*Cosh[2*x])*Sech[x]^2)*Tanh[x])/(2*Sqrt[2]*b *(a + b)*Sqrt[(a - b + (a + b)*Cosh[2*x])*Sech[x]^2])
Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4153, 381, 398, 224, 219, 291, 219}
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.
\(\displaystyle \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (i x)^4}{\sqrt {a-b \tan (i x)^2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \int \frac {\tanh ^4(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {a+b \tanh ^2(x)}}d\tanh (x)\) |
\(\Big \downarrow \) 381 |
\(\displaystyle \frac {\int \frac {a-(a-2 b) \tanh ^2(x)}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{2 b}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {(a-2 b) \int \frac {1}{\sqrt {b \tanh ^2(x)+a}}d\tanh (x)+2 b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)}{2 b}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {2 b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)+(a-2 b) \int \frac {1}{1-\frac {b \tanh ^2(x)}{b \tanh ^2(x)+a}}d\frac {\tanh (x)}{\sqrt {b \tanh ^2(x)+a}}}{2 b}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 b \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {b \tanh ^2(x)+a}}d\tanh (x)+\frac {(a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{\sqrt {b}}}{2 b}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {2 b \int \frac {1}{1-\frac {(a+b) \tanh ^2(x)}{b \tanh ^2(x)+a}}d\frac {\tanh (x)}{\sqrt {b \tanh ^2(x)+a}}+\frac {(a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{\sqrt {b}}}{2 b}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {(a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{\sqrt {b}}+\frac {2 b \text {arctanh}\left (\frac {\sqrt {a+b} \tanh (x)}{\sqrt {a+b \tanh ^2(x)}}\right )}{\sqrt {a+b}}}{2 b}-\frac {\tanh (x) \sqrt {a+b \tanh ^2(x)}}{2 b}\) |
Input:
Int[Tanh[x]^4/Sqrt[a + b*Tanh[x]^2],x]
Output:
(((a - 2*b)*ArcTanh[(Sqrt[b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]])/Sqrt[b] + (2 *b*ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]])/Sqrt[a + b])/(2*b ) - (Tanh[x]*Sqrt[a + b*Tanh[x]^2])/(2*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q }, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 , p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
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. \(177\) vs. \(2(70)=140\).
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.02
method | result | size |
derivativedivides | \(-\frac {\ln \left (\sqrt {b}\, \tanh \left (x \right )+\sqrt {a +b \tanh \left (x \right )^{2}}\right )}{\sqrt {b}}-\frac {\tanh \left (x \right ) \sqrt {a +b \tanh \left (x \right )^{2}}}{2 b}+\frac {a \ln \left (\sqrt {b}\, \tanh \left (x \right )+\sqrt {a +b \tanh \left (x \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \sqrt {a +b}}\) | \(178\) |
default | \(-\frac {\ln \left (\sqrt {b}\, \tanh \left (x \right )+\sqrt {a +b \tanh \left (x \right )^{2}}\right )}{\sqrt {b}}-\frac {\tanh \left (x \right ) \sqrt {a +b \tanh \left (x \right )^{2}}}{2 b}+\frac {a \ln \left (\sqrt {b}\, \tanh \left (x \right )+\sqrt {a +b \tanh \left (x \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\ln \left (\frac {2 a +2 b +2 b \left (\tanh \left (x \right )-1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )-1\right )^{2}+2 b \left (\tanh \left (x \right )-1\right )+a +b}}{\tanh \left (x \right )-1}\right )}{2 \sqrt {a +b}}-\frac {\ln \left (\frac {2 a +2 b -2 b \left (\tanh \left (x \right )+1\right )+2 \sqrt {a +b}\, \sqrt {b \left (\tanh \left (x \right )+1\right )^{2}-2 b \left (\tanh \left (x \right )+1\right )+a +b}}{\tanh \left (x \right )+1}\right )}{2 \sqrt {a +b}}\) | \(178\) |
Input:
int(tanh(x)^4/(a+b*tanh(x)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-ln(b^(1/2)*tanh(x)+(a+b*tanh(x)^2)^(1/2))/b^(1/2)-1/2*tanh(x)*(a+b*tanh(x )^2)^(1/2)/b+1/2*a/b^(3/2)*ln(b^(1/2)*tanh(x)+(a+b*tanh(x)^2)^(1/2))+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))/(tanh(x) +1))
Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (70) = 140\).
Time = 0.32 (sec) , antiderivative size = 5494, normalized size of antiderivative = 62.43 \[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)^4/(a+b*tanh(x)^2)^(1/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\tanh ^{4}{\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )}}}\, dx \] Input:
integrate(tanh(x)**4/(a+b*tanh(x)**2)**(1/2),x)
Output:
Integral(tanh(x)**4/sqrt(a + b*tanh(x)**2), x)
\[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{4}}{\sqrt {b \tanh \left (x\right )^{2} + a}} \,d x } \] Input:
integrate(tanh(x)^4/(a+b*tanh(x)^2)^(1/2),x, algorithm="maxima")
Output:
integrate(tanh(x)^4/sqrt(b*tanh(x)^2 + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (70) = 140\).
Time = 0.41 (sec) , antiderivative size = 559, normalized size of antiderivative = 6.35 \[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx =\text {Too large to display} \] Input:
integrate(tanh(x)^4/(a+b*tanh(x)^2)^(1/2),x, algorithm="giac")
Output:
(a - 2*b)*arctan(-1/2*(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))/sqrt(-b))/(sqrt(-b)*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)))/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) - 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))^3*(a + 2*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*(3*a - 2*b)*sqrt(a + b) + (3*a^2 - 3*a*b - 2*b^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)) + (a^2 - a*b + 2*b^2)*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)^2*b)
Timed out. \[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^4}{\sqrt {b\,{\mathrm {tanh}\left (x\right )}^2+a}} \,d x \] Input:
int(tanh(x)^4/(a + b*tanh(x)^2)^(1/2),x)
Output:
int(tanh(x)^4/(a + b*tanh(x)^2)^(1/2), x)
\[ \int \frac {\tanh ^4(x)}{\sqrt {a+b \tanh ^2(x)}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )^{2} b +a}\, \tanh \left (x \right )^{4}}{\tanh \left (x \right )^{2} b +a}d x \] Input:
int(tanh(x)^4/(a+b*tanh(x)^2)^(1/2),x)
Output:
int((sqrt(tanh(x)**2*b + a)*tanh(x)**4)/(tanh(x)**2*b + a),x)