Integrand size = 10, antiderivative size = 50 \[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=-\frac {5}{2} \text {arcsinh}(\tanh (x))+2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {1+\tanh ^2(x)}}\right )-\frac {1}{2} \tanh (x) \sqrt {1+\tanh ^2(x)} \] Output:
-5/2*arcsinh(tanh(x))+2*2^(1/2)*arctanh(2^(1/2)*tanh(x)/(1+tanh(x)^2)^(1/2 ))-1/2*tanh(x)*(1+tanh(x)^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.48 \[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=-\frac {\left (-4 \sqrt {2} \text {arcsinh}\left (\sqrt {2} \sinh (x)\right ) \cosh ^3(x)+5 \text {arctanh}\left (\frac {\sinh (x)}{\sqrt {\cosh (2 x)}}\right ) \cosh ^3(x)+\cosh (x) \sqrt {\cosh (2 x)} \sinh (x)\right ) \left (1+\tanh ^2(x)\right )^{3/2}}{2 \cosh ^{\frac {3}{2}}(2 x)} \] Input:
Integrate[(1 + Tanh[x]^2)^(3/2),x]
Output:
-1/2*((-4*Sqrt[2]*ArcSinh[Sqrt[2]*Sinh[x]]*Cosh[x]^3 + 5*ArcTanh[Sinh[x]/S qrt[Cosh[2*x]]]*Cosh[x]^3 + Cosh[x]*Sqrt[Cosh[2*x]]*Sinh[x])*(1 + Tanh[x]^ 2)^(3/2))/Cosh[2*x]^(3/2)
Time = 0.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 4144, 318, 25, 398, 222, 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.
| \(\displaystyle \int \left (\tanh ^2(x)+1\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (1-\tan (i x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \int \frac {\left (\tanh ^2(x)+1\right )^{3/2}}{1-\tanh ^2(x)}d\tanh (x)\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle -\frac {1}{2} \int -\frac {5 \tanh ^2(x)+3}{\left (1-\tanh ^2(x)\right ) \sqrt {\tanh ^2(x)+1}}d\tanh (x)-\frac {1}{2} \sqrt {\tanh ^2(x)+1} \tanh (x)\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \int \frac {5 \tanh ^2(x)+3}{\left (1-\tanh ^2(x)\right ) \sqrt {\tanh ^2(x)+1}}d\tanh (x)-\frac {1}{2} \tanh (x) \sqrt {\tanh ^2(x)+1}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {\tanh ^2(x)+1}}d\tanh (x)-5 \int \frac {1}{\sqrt {\tanh ^2(x)+1}}d\tanh (x)\right )-\frac {1}{2} \tanh (x) \sqrt {\tanh ^2(x)+1}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{\left (1-\tanh ^2(x)\right ) \sqrt {\tanh ^2(x)+1}}d\tanh (x)-5 \text {arcsinh}(\tanh (x))\right )-\frac {1}{2} \tanh (x) \sqrt {\tanh ^2(x)+1}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{1-\frac {2 \tanh ^2(x)}{\tanh ^2(x)+1}}d\frac {\tanh (x)}{\sqrt {\tanh ^2(x)+1}}-5 \text {arcsinh}(\tanh (x))\right )-\frac {1}{2} \tanh (x) \sqrt {\tanh ^2(x)+1}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {\tanh ^2(x)+1}}\right )-5 \text {arcsinh}(\tanh (x))\right )-\frac {1}{2} \tanh (x) \sqrt {\tanh ^2(x)+1}\) |
Input:
Int[(1 + Tanh[x]^2)^(3/2),x]
Output:
(-5*ArcSinh[Tanh[x]] + 4*Sqrt[2]*ArcTanh[(Sqrt[2]*Tanh[x])/Sqrt[1 + Tanh[x ]^2]])/2 - (Tanh[x]*Sqrt[1 + Tanh[x]^2])/2
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] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 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[((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[(a + b* (ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Leaf count of result is larger than twice the leaf count of optimal. \(157\) vs. \(2(38)=76\).
Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 3.16
| method | result | size |
| derivativedivides | \(-\frac {\left (\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )\right )^{\frac {3}{2}}}{6}-\frac {\tanh \left (x \right ) \sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}{4}-\frac {5 \,\operatorname {arcsinh}\left (\tanh \left (x \right )\right )}{2}-\sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2+2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}\right )+\frac {\left (\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )\right )^{\frac {3}{2}}}{6}-\frac {\tanh \left (x \right ) \sqrt {\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )}}{4}+\sqrt {\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )}-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2-2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )}}\right )\) | \(158\) |
| default | \(-\frac {\left (\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )\right )^{\frac {3}{2}}}{6}-\frac {\tanh \left (x \right ) \sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}{4}-\frac {5 \,\operatorname {arcsinh}\left (\tanh \left (x \right )\right )}{2}-\sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2+2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (\tanh \left (x \right )-1\right )^{2}+2 \tanh \left (x \right )}}\right )+\frac {\left (\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )\right )^{\frac {3}{2}}}{6}-\frac {\tanh \left (x \right ) \sqrt {\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )}}{4}+\sqrt {\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )}-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2-2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {\left (\tanh \left (x \right )+1\right )^{2}-2 \tanh \left (x \right )}}\right )\) | \(158\) |
Input:
int((1+tanh(x)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/6*((tanh(x)-1)^2+2*tanh(x))^(3/2)-1/4*tanh(x)*((tanh(x)-1)^2+2*tanh(x)) ^(1/2)-5/2*arcsinh(tanh(x))-((tanh(x)-1)^2+2*tanh(x))^(1/2)+2^(1/2)*arctan h(1/4*(2+2*tanh(x))*2^(1/2)/((tanh(x)-1)^2+2*tanh(x))^(1/2))+1/6*((tanh(x) +1)^2-2*tanh(x))^(3/2)-1/4*tanh(x)*((tanh(x)+1)^2-2*tanh(x))^(1/2)+((tanh( x)+1)^2-2*tanh(x))^(1/2)-2^(1/2)*arctanh(1/4*(2-2*tanh(x))*2^(1/2)/((tanh( x)+1)^2-2*tanh(x))^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (38) = 76\).
Time = 0.10 (sec) , antiderivative size = 1027, normalized size of antiderivative = 20.54 \[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate((1+tanh(x)^2)^(3/2),x, algorithm="fricas")
Output:
1/4*(2*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^ 4 + 2*(3*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 2*sqrt(2)*cosh(x)^2 + 4* (sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(-2*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 - 3)*sinh(x)^6 - 3*cosh (x)^6 + 2*(28*cosh(x)^3 - 9*cosh(x))*sinh(x)^5 + 5*(14*cosh(x)^4 - 9*cosh( x)^2 + 1)*sinh(x)^4 + 5*cosh(x)^4 + 4*(14*cosh(x)^5 - 15*cosh(x)^3 + 5*cos h(x))*sinh(x)^3 + (28*cosh(x)^6 - 45*cosh(x)^4 + 30*cosh(x)^2 - 4)*sinh(x) ^2 - 4*cosh(x)^2 + 2*(4*cosh(x)^7 - 9*cosh(x)^5 + 10*cosh(x)^3 - 4*cosh(x) )*sinh(x) + (sqrt(2)*cosh(x)^6 + 6*sqrt(2)*cosh(x)*sinh(x)^5 + sqrt(2)*sin h(x)^6 + 3*(5*sqrt(2)*cosh(x)^2 - sqrt(2))*sinh(x)^4 - 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 - 3*sqrt(2)*cosh(x))*sinh(x)^3 + (15*sqrt(2)*cos h(x)^4 - 18*sqrt(2)*cosh(x)^2 + 4*sqrt(2))*sinh(x)^2 + 4*sqrt(2)*cosh(x)^2 + 2*(3*sqrt(2)*cosh(x)^5 - 6*sqrt(2)*cosh(x)^3 + 4*sqrt(2)*cosh(x))*sinh( x) - 4*sqrt(2))*sqrt((cosh(x)^2 + sinh(x)^2)/(cosh(x)^2 - 2*cosh(x)*sinh(x ) + sinh(x)^2)) + 4)/(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(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x) ^5 + sinh(x)^6)) + 2*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sq rt(2)*sinh(x)^4 + 2*(3*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 2*sqrt(2)* cosh(x)^2 + 4*(sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log (2*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sin...
\[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=\int \left (\tanh ^{2}{\left (x \right )} + 1\right )^{\frac {3}{2}}\, dx \] Input:
integrate((1+tanh(x)**2)**(3/2),x)
Output:
Integral((tanh(x)**2 + 1)**(3/2), x)
\[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=\int { {\left (\tanh \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((1+tanh(x)^2)^(3/2),x, algorithm="maxima")
Output:
integrate((tanh(x)^2 + 1)^(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (38) = 76\).
Time = 0.13 (sec) , antiderivative size = 202, normalized size of antiderivative = 4.04 \[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=-\frac {1}{4} \, \sqrt {2} {\left (5 \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1}{\sqrt {2} + \sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} - 1}\right ) - \frac {4 \, {\left (3 \, {\left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{3} - {\left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} - \sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} - 1\right )}}{{\left ({\left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right )}^{2} - 2 \, \sqrt {e^{\left (4 \, x\right )} + 1} + 2 \, e^{\left (2 \, x\right )} - 1\right )}^{2}} + 4 \, \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) + 4 \, \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) - 4 \, \log \left (-\sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \] Input:
integrate((1+tanh(x)^2)^(3/2),x, algorithm="giac")
Output:
-1/4*sqrt(2)*(5*sqrt(2)*log((sqrt(2) - sqrt(e^(4*x) + 1) + e^(2*x) + 1)/(s qrt(2) + sqrt(e^(4*x) + 1) - e^(2*x) - 1)) - 4*(3*(sqrt(e^(4*x) + 1) - e^( 2*x))^3 - (sqrt(e^(4*x) + 1) - e^(2*x))^2 - sqrt(e^(4*x) + 1) + e^(2*x) - 1)/((sqrt(e^(4*x) + 1) - e^(2*x))^2 - 2*sqrt(e^(4*x) + 1) + 2*e^(2*x) - 1) ^2 + 4*log(sqrt(e^(4*x) + 1) - e^(2*x) + 1) + 4*log(sqrt(e^(4*x) + 1) - e^ (2*x)) - 4*log(-sqrt(e^(4*x) + 1) + e^(2*x) + 1))
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=\sqrt {2}\,\left (\ln \left (\mathrm {tanh}\left (x\right )+1\right )-\ln \left (\sqrt {2}\,\sqrt {{\mathrm {tanh}\left (x\right )}^2+1}-\mathrm {tanh}\left (x\right )+1\right )\right )-\frac {5\,\mathrm {asinh}\left (\mathrm {tanh}\left (x\right )\right )}{2}-\frac {\mathrm {tanh}\left (x\right )\,\sqrt {{\mathrm {tanh}\left (x\right )}^2+1}}{2}+\sqrt {2}\,\left (\ln \left (\mathrm {tanh}\left (x\right )+\sqrt {2}\,\sqrt {{\mathrm {tanh}\left (x\right )}^2+1}+1\right )-\ln \left (\mathrm {tanh}\left (x\right )-1\right )\right ) \] Input:
int((tanh(x)^2 + 1)^(3/2),x)
Output:
2^(1/2)*(log(tanh(x) + 1) - log(2^(1/2)*(tanh(x)^2 + 1)^(1/2) - tanh(x) + 1)) - (5*asinh(tanh(x)))/2 - (tanh(x)*(tanh(x)^2 + 1)^(1/2))/2 + 2^(1/2)*( log(tanh(x) + 2^(1/2)*(tanh(x)^2 + 1)^(1/2) + 1) - log(tanh(x) - 1))
\[ \int \left (1+\tanh ^2(x)\right )^{3/2} \, dx=\int \sqrt {\tanh \left (x \right )^{2}+1}d x +\int \sqrt {\tanh \left (x \right )^{2}+1}\, \tanh \left (x \right )^{2}d x \] Input:
int((1+tanh(x)^2)^(3/2),x)
Output:
int(sqrt(tanh(x)**2 + 1),x) + int(sqrt(tanh(x)**2 + 1)*tanh(x)**2,x)