Integrand size = 14, antiderivative size = 118 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\coth (c+d x)}{b d \sqrt {b \coth ^4(c+d x)}}+\frac {x \coth ^2(c+d x)}{b \sqrt {b \coth ^4(c+d x)}}-\frac {\tanh (c+d x)}{3 b d \sqrt {b \coth ^4(c+d x)}}-\frac {\tanh ^3(c+d x)}{5 b d \sqrt {b \coth ^4(c+d x)}} \]
-coth(d*x+c)/b/d/(b*coth(d*x+c)^4)^(1/2)+x*coth(d*x+c)^2/b/(b*coth(d*x+c)^ 4)^(1/2)-1/3*tanh(d*x+c)/b/d/(b*coth(d*x+c)^4)^(1/2)-1/5*tanh(d*x+c)^3/b/d /(b*coth(d*x+c)^4)^(1/2)
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=\frac {-15 \coth (c+d x)+15 \text {arctanh}(\tanh (c+d x)) \coth ^2(c+d x)-5 \tanh (c+d x)-3 \tanh ^3(c+d x)}{15 b d \sqrt {b \coth ^4(c+d x)}} \]
(-15*Coth[c + d*x] + 15*ArcTanh[Tanh[c + d*x]]*Coth[c + d*x]^2 - 5*Tanh[c + d*x] - 3*Tanh[c + d*x]^3)/(15*b*d*Sqrt[b*Coth[c + d*x]^4])
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.60, 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.
\(\displaystyle \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (b \tan \left (i c+i d x+\frac {\pi }{2}\right )^4\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \frac {\coth ^2(c+d x) \int \tanh ^6(c+d x)dx}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\coth ^2(c+d x) \int -\tan (i c+i d x)^6dx}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\coth ^2(c+d x) \int \tan (i c+i d x)^6dx}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (\frac {\tanh ^5(c+d x)}{5 d}-\int \tanh ^4(c+d x)dx\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (\frac {\tanh ^5(c+d x)}{5 d}-\int \tan (i c+i d x)^4dx\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (\int -\tanh ^2(c+d x)dx+\frac {\tanh ^5(c+d x)}{5 d}+\frac {\tanh ^3(c+d x)}{3 d}\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (-\int \tanh ^2(c+d x)dx+\frac {\tanh ^5(c+d x)}{5 d}+\frac {\tanh ^3(c+d x)}{3 d}\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (-\int -\tan (i c+i d x)^2dx+\frac {\tanh ^5(c+d x)}{5 d}+\frac {\tanh ^3(c+d x)}{3 d}\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (\int \tan (i c+i d x)^2dx+\frac {\tanh ^5(c+d x)}{5 d}+\frac {\tanh ^3(c+d x)}{3 d}\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -\frac {\coth ^2(c+d x) \left (-\int 1dx+\frac {\tanh ^5(c+d x)}{5 d}+\frac {\tanh ^3(c+d x)}{3 d}+\frac {\tanh (c+d x)}{d}\right )}{b \sqrt {b \coth ^4(c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {\left (\frac {\tanh ^5(c+d x)}{5 d}+\frac {\tanh ^3(c+d x)}{3 d}+\frac {\tanh (c+d x)}{d}-x\right ) \coth ^2(c+d x)}{b \sqrt {b \coth ^4(c+d x)}}\) |
-((Coth[c + d*x]^2*(-x + Tanh[c + d*x]/d + Tanh[c + d*x]^3/(3*d) + Tanh[c + d*x]^5/(5*d)))/(b*Sqrt[b*Coth[c + d*x]^4]))
3.1.43.3.1 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 = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(-\frac {\coth \left (d x +c \right ) \left (15 \ln \left (\coth \left (d x +c \right )-1\right ) \coth \left (d x +c \right )^{5}-15 \ln \left (\coth \left (d x +c \right )+1\right ) \coth \left (d x +c \right )^{5}+30 \coth \left (d x +c \right )^{4}+10 \coth \left (d x +c \right )^{2}+6\right )}{30 d \left (b \coth \left (d x +c \right )^{4}\right )^{\frac {3}{2}}}\) | \(84\) |
default | \(-\frac {\coth \left (d x +c \right ) \left (15 \ln \left (\coth \left (d x +c \right )-1\right ) \coth \left (d x +c \right )^{5}-15 \ln \left (\coth \left (d x +c \right )+1\right ) \coth \left (d x +c \right )^{5}+30 \coth \left (d x +c \right )^{4}+10 \coth \left (d x +c \right )^{2}+6\right )}{30 d \left (b \coth \left (d x +c \right )^{4}\right )^{\frac {3}{2}}}\) | \(84\) |
risch | \(\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2} x}{b \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}}}+\frac {6 \,{\mathrm e}^{8 d x +8 c}+12 \,{\mathrm e}^{6 d x +6 c}+\frac {56 \,{\mathrm e}^{4 d x +4 c}}{3}+\frac {28 \,{\mathrm e}^{2 d x +2 c}}{3}+\frac {46}{15}}{b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{4}}}\, d}\) | \(173\) |
-1/30/d*coth(d*x+c)*(15*ln(coth(d*x+c)-1)*coth(d*x+c)^5-15*ln(coth(d*x+c)+ 1)*coth(d*x+c)^5+30*coth(d*x+c)^4+10*coth(d*x+c)^2+6)/(b*coth(d*x+c)^4)^(3 /2)
Leaf count of result is larger than twice the leaf count of optimal. 3473 vs. \(2 (106) = 212\).
Time = 0.32 (sec) , antiderivative size = 3473, normalized size of antiderivative = 29.43 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=\text {Too large to display} \]
1/15*(15*d*x*cosh(d*x + c)^10 + 15*(d*x*e^(4*d*x + 4*c) - 2*d*x*e^(2*d*x + 2*c) + d*x)*sinh(d*x + c)^10 + 150*(d*x*cosh(d*x + c)*e^(4*d*x + 4*c) - 2 *d*x*cosh(d*x + c)*e^(2*d*x + 2*c) + d*x*cosh(d*x + c))*sinh(d*x + c)^9 + 15*(5*d*x + 6)*cosh(d*x + c)^8 + 15*(45*d*x*cosh(d*x + c)^2 + 5*d*x + (45* d*x*cosh(d*x + c)^2 + 5*d*x + 6)*e^(4*d*x + 4*c) - 2*(45*d*x*cosh(d*x + c) ^2 + 5*d*x + 6)*e^(2*d*x + 2*c) + 6)*sinh(d*x + c)^8 + 120*(15*d*x*cosh(d* x + c)^3 + (5*d*x + 6)*cosh(d*x + c) + (15*d*x*cosh(d*x + c)^3 + (5*d*x + 6)*cosh(d*x + c))*e^(4*d*x + 4*c) - 2*(15*d*x*cosh(d*x + c)^3 + (5*d*x + 6 )*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c)^7 + 30*(5*d*x + 6)*cosh(d* x + c)^6 + 30*(105*d*x*cosh(d*x + c)^4 + 14*(5*d*x + 6)*cosh(d*x + c)^2 + 5*d*x + (105*d*x*cosh(d*x + c)^4 + 14*(5*d*x + 6)*cosh(d*x + c)^2 + 5*d*x + 6)*e^(4*d*x + 4*c) - 2*(105*d*x*cosh(d*x + c)^4 + 14*(5*d*x + 6)*cosh(d* x + c)^2 + 5*d*x + 6)*e^(2*d*x + 2*c) + 6)*sinh(d*x + c)^6 + 60*(63*d*x*co sh(d*x + c)^5 + 14*(5*d*x + 6)*cosh(d*x + c)^3 + 3*(5*d*x + 6)*cosh(d*x + c) + (63*d*x*cosh(d*x + c)^5 + 14*(5*d*x + 6)*cosh(d*x + c)^3 + 3*(5*d*x + 6)*cosh(d*x + c))*e^(4*d*x + 4*c) - 2*(63*d*x*cosh(d*x + c)^5 + 14*(5*d*x + 6)*cosh(d*x + c)^3 + 3*(5*d*x + 6)*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh (d*x + c)^5 + 10*(15*d*x + 28)*cosh(d*x + c)^4 + 10*(315*d*x*cosh(d*x + c) ^6 + 105*(5*d*x + 6)*cosh(d*x + c)^4 + 45*(5*d*x + 6)*cosh(d*x + c)^2 + 15 *d*x + (315*d*x*cosh(d*x + c)^6 + 105*(5*d*x + 6)*cosh(d*x + c)^4 + 45*...
\[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \coth ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=-\frac {2 \, {\left (70 \, \sqrt {b} e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, \sqrt {b} e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, \sqrt {b} e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, \sqrt {b} e^{\left (-8 \, d x - 8 \, c\right )} + 23 \, \sqrt {b}\right )}}{15 \, {\left (5 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, b^{2} e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, b^{2} e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, b^{2} e^{\left (-8 \, d x - 8 \, c\right )} + b^{2} e^{\left (-10 \, d x - 10 \, c\right )} + b^{2}\right )} d} + \frac {d x + c}{b^{\frac {3}{2}} d} \]
-2/15*(70*sqrt(b)*e^(-2*d*x - 2*c) + 140*sqrt(b)*e^(-4*d*x - 4*c) + 90*sqr t(b)*e^(-6*d*x - 6*c) + 45*sqrt(b)*e^(-8*d*x - 8*c) + 23*sqrt(b))/((5*b^2* e^(-2*d*x - 2*c) + 10*b^2*e^(-4*d*x - 4*c) + 10*b^2*e^(-6*d*x - 6*c) + 5*b ^2*e^(-8*d*x - 8*c) + b^2*e^(-10*d*x - 10*c) + b^2)*d) + (d*x + c)/(b^(3/2 )*d)
Time = 0.34 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=\frac {\frac {15 \, {\left (d x + c\right )}}{\sqrt {b}} + \frac {2 \, {\left (45 \, e^{\left (8 \, d x + 8 \, c\right )} + 90 \, e^{\left (6 \, d x + 6 \, c\right )} + 140 \, e^{\left (4 \, d x + 4 \, c\right )} + 70 \, e^{\left (2 \, d x + 2 \, c\right )} + 23\right )}}{\sqrt {b} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, b d} \]
1/15*(15*(d*x + c)/sqrt(b) + 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)/(sqrt(b)*(e^(2*d*x + 2*c) + 1)^5))/(b*d)
Timed out. \[ \int \frac {1}{\left (b \coth ^4(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^4\right )}^{3/2}} \,d x \]