Integrand size = 10, antiderivative size = 69 \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=-a \coth (x) \sqrt {a \tanh ^4(x)}+a x \coth ^2(x) \sqrt {a \tanh ^4(x)}-\frac {1}{3} a \tanh (x) \sqrt {a \tanh ^4(x)}-\frac {1}{5} a \tanh ^3(x) \sqrt {a \tanh ^4(x)} \] Output:
-a*coth(x)*(a*tanh(x)^4)^(1/2)+a*x*coth(x)^2*(a*tanh(x)^4)^(1/2)-1/3*a*tan h(x)*(a*tanh(x)^4)^(1/2)-1/5*a*tanh(x)^3*(a*tanh(x)^4)^(1/2)
Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57 \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=\frac {1}{15} \coth (x) \left (-3-5 \coth ^2(x)-15 \coth ^4(x)+15 \text {arctanh}(\tanh (x)) \coth ^5(x)\right ) \left (a \tanh ^4(x)\right )^{3/2} \] Input:
Integrate[(a*Tanh[x]^4)^(3/2),x]
Output:
(Coth[x]*(-3 - 5*Coth[x]^2 - 15*Coth[x]^4 + 15*ArcTanh[Tanh[x]]*Coth[x]^5) *(a*Tanh[x]^4)^(3/2))/15
Time = 0.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {3042, 4141, 3042, 25, 3954, 3042, 3954, 25, 3042, 25, 3954, 24}
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.
\(\displaystyle \int \left (a \tanh ^4(x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \tan (i x)^4\right )^{3/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle a \coth ^2(x) \sqrt {a \tanh ^4(x)} \int \tanh ^6(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \coth ^2(x) \sqrt {a \tanh ^4(x)} \int -\tan (i x)^6dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \int \tan (i x)^6dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (\frac {\tanh ^5(x)}{5}-\int \tanh ^4(x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (\frac {\tanh ^5(x)}{5}-\int \tan (i x)^4dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (\int -\tanh ^2(x)dx+\frac {\tanh ^5(x)}{5}+\frac {\tanh ^3(x)}{3}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (-\int \tanh ^2(x)dx+\frac {\tanh ^5(x)}{5}+\frac {\tanh ^3(x)}{3}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (-\int -\tan (i x)^2dx+\frac {\tanh ^5(x)}{5}+\frac {\tanh ^3(x)}{3}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (\int \tan (i x)^2dx+\frac {\tanh ^5(x)}{5}+\frac {\tanh ^3(x)}{3}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a \coth ^2(x) \sqrt {a \tanh ^4(x)} \left (-\int 1dx+\frac {\tanh ^5(x)}{5}+\frac {\tanh ^3(x)}{3}+\tanh (x)\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -a \left (-x+\frac {\tanh ^5(x)}{5}+\frac {\tanh ^3(x)}{3}+\tanh (x)\right ) \coth ^2(x) \sqrt {a \tanh ^4(x)}\) |
Input:
Int[(a*Tanh[x]^4)^(3/2),x]
Output:
-(a*Coth[x]^2*Sqrt[a*Tanh[x]^4]*(-x + Tanh[x] + Tanh[x]^3/3 + Tanh[x]^5/5) )
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(-\frac {\left (a \tanh \left (x \right )^{4}\right )^{\frac {3}{2}} \left (6 \tanh \left (x \right )^{5}+10 \tanh \left (x \right )^{3}+15 \ln \left (\tanh \left (x \right )-1\right )-15 \ln \left (1+\tanh \left (x \right )\right )+30 \tanh \left (x \right )\right )}{30 \tanh \left (x \right )^{6}}\) | \(46\) |
default | \(-\frac {\left (a \tanh \left (x \right )^{4}\right )^{\frac {3}{2}} \left (6 \tanh \left (x \right )^{5}+10 \tanh \left (x \right )^{3}+15 \ln \left (\tanh \left (x \right )-1\right )-15 \ln \left (1+\tanh \left (x \right )\right )+30 \tanh \left (x \right )\right )}{30 \tanh \left (x \right )^{6}}\) | \(46\) |
risch | \(\frac {a \left ({\mathrm e}^{2 x}+1\right )^{2} \sqrt {\frac {a \left ({\mathrm e}^{2 x}-1\right )^{4}}{\left ({\mathrm e}^{2 x}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 x}-1\right )^{2}}+\frac {2 a \sqrt {\frac {a \left ({\mathrm e}^{2 x}-1\right )^{4}}{\left ({\mathrm e}^{2 x}+1\right )^{4}}}\, \left (45 \,{\mathrm e}^{8 x}+90 \,{\mathrm e}^{6 x}+140 \,{\mathrm e}^{4 x}+70 \,{\mathrm e}^{2 x}+23\right )}{15 \left ({\mathrm e}^{2 x}-1\right )^{2} \left ({\mathrm e}^{2 x}+1\right )^{3}}\) | \(106\) |
Input:
int((a*tanh(x)^4)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/30*(a*tanh(x)^4)^(3/2)*(6*tanh(x)^5+10*tanh(x)^3+15*ln(tanh(x)-1)-15*ln (1+tanh(x))+30*tanh(x))/tanh(x)^6
Leaf count of result is larger than twice the leaf count of optimal. 2114 vs. \(2 (57) = 114\).
Time = 0.17 (sec) , antiderivative size = 2114, normalized size of antiderivative = 30.64 \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate((a*tanh(x)^4)^(3/2),x, algorithm="fricas")
Output:
1/15*(15*a*x*cosh(x)^10 + 15*(a*x*e^(4*x) + 2*a*x*e^(2*x) + a*x)*sinh(x)^1 0 + 150*(a*x*cosh(x)*e^(4*x) + 2*a*x*cosh(x)*e^(2*x) + a*x*cosh(x))*sinh(x )^9 + 15*(5*a*x + 6*a)*cosh(x)^8 + 15*(45*a*x*cosh(x)^2 + 5*a*x + (45*a*x* cosh(x)^2 + 5*a*x + 6*a)*e^(4*x) + 2*(45*a*x*cosh(x)^2 + 5*a*x + 6*a)*e^(2 *x) + 6*a)*sinh(x)^8 + 120*(15*a*x*cosh(x)^3 + (5*a*x + 6*a)*cosh(x) + (15 *a*x*cosh(x)^3 + (5*a*x + 6*a)*cosh(x))*e^(4*x) + 2*(15*a*x*cosh(x)^3 + (5 *a*x + 6*a)*cosh(x))*e^(2*x))*sinh(x)^7 + 30*(5*a*x + 6*a)*cosh(x)^6 + 30* (105*a*x*cosh(x)^4 + 14*(5*a*x + 6*a)*cosh(x)^2 + 5*a*x + (105*a*x*cosh(x) ^4 + 14*(5*a*x + 6*a)*cosh(x)^2 + 5*a*x + 6*a)*e^(4*x) + 2*(105*a*x*cosh(x )^4 + 14*(5*a*x + 6*a)*cosh(x)^2 + 5*a*x + 6*a)*e^(2*x) + 6*a)*sinh(x)^6 + 60*(63*a*x*cosh(x)^5 + 14*(5*a*x + 6*a)*cosh(x)^3 + 3*(5*a*x + 6*a)*cosh( x) + (63*a*x*cosh(x)^5 + 14*(5*a*x + 6*a)*cosh(x)^3 + 3*(5*a*x + 6*a)*cosh (x))*e^(4*x) + 2*(63*a*x*cosh(x)^5 + 14*(5*a*x + 6*a)*cosh(x)^3 + 3*(5*a*x + 6*a)*cosh(x))*e^(2*x))*sinh(x)^5 + 10*(15*a*x + 28*a)*cosh(x)^4 + 10*(3 15*a*x*cosh(x)^6 + 105*(5*a*x + 6*a)*cosh(x)^4 + 45*(5*a*x + 6*a)*cosh(x)^ 2 + 15*a*x + (315*a*x*cosh(x)^6 + 105*(5*a*x + 6*a)*cosh(x)^4 + 45*(5*a*x + 6*a)*cosh(x)^2 + 15*a*x + 28*a)*e^(4*x) + 2*(315*a*x*cosh(x)^6 + 105*(5* a*x + 6*a)*cosh(x)^4 + 45*(5*a*x + 6*a)*cosh(x)^2 + 15*a*x + 28*a)*e^(2*x) + 28*a)*sinh(x)^4 + 40*(45*a*x*cosh(x)^7 + 21*(5*a*x + 6*a)*cosh(x)^5 + 1 5*(5*a*x + 6*a)*cosh(x)^3 + (15*a*x + 28*a)*cosh(x) + (45*a*x*cosh(x)^7...
\[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=\int \left (a \tanh ^{4}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a*tanh(x)**4)**(3/2),x)
Output:
Integral((a*tanh(x)**4)**(3/2), x)
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=a^{\frac {3}{2}} x - \frac {2 \, {\left (70 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} + 140 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 90 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} + 45 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + 23 \, a^{\frac {3}{2}}\right )}}{15 \, {\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} \] Input:
integrate((a*tanh(x)^4)^(3/2),x, algorithm="maxima")
Output:
a^(3/2)*x - 2/15*(70*a^(3/2)*e^(-2*x) + 140*a^(3/2)*e^(-4*x) + 90*a^(3/2)* e^(-6*x) + 45*a^(3/2)*e^(-8*x) + 23*a^(3/2))/(5*e^(-2*x) + 10*e^(-4*x) + 1 0*e^(-6*x) + 5*e^(-8*x) + e^(-10*x) + 1)
Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.65 \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=\frac {1}{15} \, a^{\frac {3}{2}} {\left (15 \, x + \frac {2 \, {\left (45 \, e^{\left (8 \, x\right )} + 90 \, e^{\left (6 \, x\right )} + 140 \, e^{\left (4 \, x\right )} + 70 \, e^{\left (2 \, x\right )} + 23\right )}}{{\left (e^{\left (2 \, x\right )} + 1\right )}^{5}}\right )} \] Input:
integrate((a*tanh(x)^4)^(3/2),x, algorithm="giac")
Output:
1/15*a^(3/2)*(15*x + 2*(45*e^(8*x) + 90*e^(6*x) + 140*e^(4*x) + 70*e^(2*x) + 23)/(e^(2*x) + 1)^5)
Timed out. \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=\int {\left (a\,{\mathrm {tanh}\left (x\right )}^4\right )}^{3/2} \,d x \] Input:
int((a*tanh(x)^4)^(3/2),x)
Output:
int((a*tanh(x)^4)^(3/2), x)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.36 \[ \int \left (a \tanh ^4(x)\right )^{3/2} \, dx=\frac {\sqrt {a}\, a \left (-3 \tanh \left (x \right )^{5}-5 \tanh \left (x \right )^{3}-15 \tanh \left (x \right )+15 x \right )}{15} \] Input:
int((a*tanh(x)^4)^(3/2),x)
Output:
(sqrt(a)*a*( - 3*tanh(x)**5 - 5*tanh(x)**3 - 15*tanh(x) + 15*x))/15