Integrand size = 14, antiderivative size = 110 \[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=-\frac {b \coth (c+d x) \sqrt {b \coth ^4(c+d x)}}{3 d}-\frac {b \coth ^3(c+d x) \sqrt {b \coth ^4(c+d x)}}{5 d}-\frac {b \sqrt {b \coth ^4(c+d x)} \tanh (c+d x)}{d}+b x \sqrt {b \coth ^4(c+d x)} \tanh ^2(c+d x) \]
-1/3*b*coth(d*x+c)*(b*coth(d*x+c)^4)^(1/2)/d-1/5*b*coth(d*x+c)^3*(b*coth(d *x+c)^4)^(1/2)/d-b*(b*coth(d*x+c)^4)^(1/2)*tanh(d*x+c)/d+b*x*(b*coth(d*x+c )^4)^(1/2)*tanh(d*x+c)^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.39 \[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=-\frac {\left (b \coth ^4(c+d x)\right )^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\tanh ^2(c+d x)\right ) \tanh (c+d x)}{5 d} \]
-1/5*((b*Coth[c + d*x]^4)^(3/2)*Hypergeometric2F1[-5/2, 1, -3/2, Tanh[c + d*x]^2]*Tanh[c + d*x])/d
Time = 0.42 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, 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.
\(\displaystyle \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan \left (i c+i d x+\frac {\pi }{2}\right )^4\right )^{3/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \int \coth ^6(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \int -\tan \left (i c+i d x+\frac {\pi }{2}\right )^6dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^6dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (\frac {\coth ^5(c+d x)}{5 d}-\int \coth ^4(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (\frac {\coth ^5(c+d x)}{5 d}-\int \tan \left (i c+i d x+\frac {\pi }{2}\right )^4dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (\int -\coth ^2(c+d x)dx+\frac {\coth ^5(c+d x)}{5 d}+\frac {\coth ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (-\int \coth ^2(c+d x)dx+\frac {\coth ^5(c+d x)}{5 d}+\frac {\coth ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (-\int -\tan \left (i c+i d x+\frac {\pi }{2}\right )^2dx+\frac {\coth ^5(c+d x)}{5 d}+\frac {\coth ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (\int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^2dx+\frac {\coth ^5(c+d x)}{5 d}+\frac {\coth ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -b \tanh ^2(c+d x) \sqrt {b \coth ^4(c+d x)} \left (-\int 1dx+\frac {\coth ^5(c+d x)}{5 d}+\frac {\coth ^3(c+d x)}{3 d}+\frac {\coth (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -b \tanh ^2(c+d x) \left (\frac {\coth ^5(c+d x)}{5 d}+\frac {\coth ^3(c+d x)}{3 d}+\frac {\coth (c+d x)}{d}-x\right ) \sqrt {b \coth ^4(c+d x)}\) |
-(b*Sqrt[b*Coth[c + d*x]^4]*(-x + Coth[c + d*x]/d + Coth[c + d*x]^3/(3*d) + Coth[c + d*x]^5/(5*d))*Tanh[c + d*x]^2)
3.1.40.3.1 Defintions of rubi rules used
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 = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(-\frac {\left (b \coth \left (d x +c \right )^{4}\right )^{\frac {3}{2}} \left (6 \coth \left (d x +c \right )^{5}+10 \coth \left (d x +c \right )^{3}+15 \ln \left (\coth \left (d x +c \right )-1\right )-15 \ln \left (\coth \left (d x +c \right )+1\right )+30 \coth \left (d x +c \right )\right )}{30 d \coth \left (d x +c \right )^{6}}\) | \(77\) |
default | \(-\frac {\left (b \coth \left (d x +c \right )^{4}\right )^{\frac {3}{2}} \left (6 \coth \left (d x +c \right )^{5}+10 \coth \left (d x +c \right )^{3}+15 \ln \left (\coth \left (d x +c \right )-1\right )-15 \ln \left (\coth \left (d x +c \right )+1\right )+30 \coth \left (d x +c \right )\right )}{30 d \coth \left (d x +c \right )^{6}}\) | \(77\) |
risch | \(\frac {b \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}-\frac {2 b \sqrt {\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}}\, \left (45 \,{\mathrm e}^{8 d x +8 c}-90 \,{\mathrm e}^{6 d x +6 c}+140 \,{\mathrm e}^{4 d x +4 c}-70 \,{\mathrm e}^{2 d x +2 c}+23\right )}{15 \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} d}\) | \(169\) |
-1/30/d*(b*coth(d*x+c)^4)^(3/2)*(6*coth(d*x+c)^5+10*coth(d*x+c)^3+15*ln(co th(d*x+c)-1)-15*ln(coth(d*x+c)+1)+30*coth(d*x+c))/coth(d*x+c)^6
Leaf count of result is larger than twice the leaf count of optimal. 3421 vs. \(2 (98) = 196\).
Time = 0.33 (sec) , antiderivative size = 3421, normalized size of antiderivative = 31.10 \[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=\text {Too large to display} \]
1/15*(15*b*d*x*cosh(d*x + c)^10 + 15*(b*d*x*e^(4*d*x + 4*c) - 2*b*d*x*e^(2 *d*x + 2*c) + b*d*x)*sinh(d*x + c)^10 + 150*(b*d*x*cosh(d*x + c)*e^(4*d*x + 4*c) - 2*b*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) + b*d*x*cosh(d*x + c))*sinh (d*x + c)^9 - 15*(5*b*d*x + 6*b)*cosh(d*x + c)^8 + 15*(45*b*d*x*cosh(d*x + c)^2 - 5*b*d*x + (45*b*d*x*cosh(d*x + c)^2 - 5*b*d*x - 6*b)*e^(4*d*x + 4* c) - 2*(45*b*d*x*cosh(d*x + c)^2 - 5*b*d*x - 6*b)*e^(2*d*x + 2*c) - 6*b)*s inh(d*x + c)^8 + 120*(15*b*d*x*cosh(d*x + c)^3 - (5*b*d*x + 6*b)*cosh(d*x + c) + (15*b*d*x*cosh(d*x + c)^3 - (5*b*d*x + 6*b)*cosh(d*x + c))*e^(4*d*x + 4*c) - 2*(15*b*d*x*cosh(d*x + c)^3 - (5*b*d*x + 6*b)*cosh(d*x + c))*e^( 2*d*x + 2*c))*sinh(d*x + c)^7 + 30*(5*b*d*x + 6*b)*cosh(d*x + c)^6 + 30*(1 05*b*d*x*cosh(d*x + c)^4 + 5*b*d*x - 14*(5*b*d*x + 6*b)*cosh(d*x + c)^2 + (105*b*d*x*cosh(d*x + c)^4 + 5*b*d*x - 14*(5*b*d*x + 6*b)*cosh(d*x + c)^2 + 6*b)*e^(4*d*x + 4*c) - 2*(105*b*d*x*cosh(d*x + c)^4 + 5*b*d*x - 14*(5*b* d*x + 6*b)*cosh(d*x + c)^2 + 6*b)*e^(2*d*x + 2*c) + 6*b)*sinh(d*x + c)^6 + 60*(63*b*d*x*cosh(d*x + c)^5 - 14*(5*b*d*x + 6*b)*cosh(d*x + c)^3 + 3*(5* b*d*x + 6*b)*cosh(d*x + c) + (63*b*d*x*cosh(d*x + c)^5 - 14*(5*b*d*x + 6*b )*cosh(d*x + c)^3 + 3*(5*b*d*x + 6*b)*cosh(d*x + c))*e^(4*d*x + 4*c) - 2*( 63*b*d*x*cosh(d*x + c)^5 - 14*(5*b*d*x + 6*b)*cosh(d*x + c)^3 + 3*(5*b*d*x + 6*b)*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c)^5 - 10*(15*b*d*x + 2 8*b)*cosh(d*x + c)^4 + 10*(315*b*d*x*cosh(d*x + c)^6 - 105*(5*b*d*x + 6...
\[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=\int \left (b \coth ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.25 \[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=\frac {{\left (d x + c\right )} b^{\frac {3}{2}}}{d} - \frac {2 \, {\left (70 \, b^{\frac {3}{2}} e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, b^{\frac {3}{2}} e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, b^{\frac {3}{2}} e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, b^{\frac {3}{2}} e^{\left (-8 \, d x - 8 \, c\right )} - 23 \, b^{\frac {3}{2}}\right )}}{15 \, d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} \]
(d*x + c)*b^(3/2)/d - 2/15*(70*b^(3/2)*e^(-2*d*x - 2*c) - 140*b^(3/2)*e^(- 4*d*x - 4*c) + 90*b^(3/2)*e^(-6*d*x - 6*c) - 45*b^(3/2)*e^(-8*d*x - 8*c) - 23*b^(3/2))/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1))
Time = 0.32 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70 \[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=\frac {{\left (15 \, d x + 15 \, c - \frac {2 \, {\left (45 \, e^{\left (8 \, d x + 8 \, c\right )} - 90 \, e^{\left (6 \, d x + 6 \, c\right )} + 140 \, e^{\left (4 \, d x + 4 \, c\right )} - 70 \, e^{\left (2 \, d x + 2 \, c\right )} + 23\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}\right )} b^{\frac {3}{2}}}{15 \, d} \]
1/15*(15*d*x + 15*c - 2*(45*e^(8*d*x + 8*c) - 90*e^(6*d*x + 6*c) + 140*e^( 4*d*x + 4*c) - 70*e^(2*d*x + 2*c) + 23)/(e^(2*d*x + 2*c) - 1)^5)*b^(3/2)/d
Timed out. \[ \int \left (b \coth ^4(c+d x)\right )^{3/2} \, dx=\int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^4\right )}^{3/2} \,d x \]