Integrand size = 14, antiderivative size = 61 \[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=-\frac {b \coth (c+d x) \sqrt {b \coth ^2(c+d x)}}{2 d}+\frac {b \sqrt {b \coth ^2(c+d x)} \log (\sinh (c+d x)) \tanh (c+d x)}{d} \]
-1/2*b*coth(d*x+c)*(b*coth(d*x+c)^2)^(1/2)/d+b*ln(sinh(d*x+c))*(b*coth(d*x +c)^2)^(1/2)*tanh(d*x+c)/d
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=-\frac {\left (b \coth ^2(c+d x)\right )^{3/2} \left (\coth ^2(c+d x)-2 \log (\cosh (c+d x))-2 \log (\tanh (c+d x))\right ) \tanh ^3(c+d x)}{2 d} \]
-1/2*((b*Coth[c + d*x]^2)^(3/2)*(Coth[c + d*x]^2 - 2*Log[Cosh[c + d*x]] - 2*Log[Tanh[c + d*x]])*Tanh[c + d*x]^3)/d
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4141, 3042, 26, 3954, 26, 3042, 26, 3956}
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.
\(\displaystyle \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-b \tan \left (i c+i d x+\frac {\pi }{2}\right )^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \int \coth ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \int i \tan \left (i c+i d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle i b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \left (\frac {i \coth ^2(c+d x)}{2 d}-\int i \coth (c+d x)dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \left (\frac {i \coth ^2(c+d x)}{2 d}-i \int \coth (c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \left (\frac {i \coth ^2(c+d x)}{2 d}-i \int -i \tan \left (i c+i d x+\frac {\pi }{2}\right )dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \left (\frac {i \coth ^2(c+d x)}{2 d}-\int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )dx\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle i b \tanh (c+d x) \sqrt {b \coth ^2(c+d x)} \left (\frac {i \coth ^2(c+d x)}{2 d}-\frac {i \log (-i \sinh (c+d x))}{d}\right )\) |
I*b*Sqrt[b*Coth[c + d*x]^2]*(((I/2)*Coth[c + d*x]^2)/d - (I*Log[(-I)*Sinh[ c + d*x]])/d)*Tanh[c + d*x]
3.1.18.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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 = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(-\frac {\left (\coth \left (d x +c \right )^{2} b \right )^{\frac {3}{2}} \left (\coth \left (d x +c \right )^{2}+\ln \left (\coth \left (d x +c \right )-1\right )+\ln \left (\coth \left (d x +c \right )+1\right )\right )}{2 d \coth \left (d x +c \right )^{3}}\) | \(53\) |
default | \(-\frac {\left (\coth \left (d x +c \right )^{2} b \right )^{\frac {3}{2}} \left (\coth \left (d x +c \right )^{2}+\ln \left (\coth \left (d x +c \right )-1\right )+\ln \left (\coth \left (d x +c \right )+1\right )\right )}{2 d \coth \left (d x +c \right )^{3}}\) | \(53\) |
risch | \(\frac {b \sqrt {\frac {\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2} b}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}}\, \left (-{\mathrm e}^{4 d x +4 c} d x +{\mathrm e}^{4 d x +4 c} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )-2 \,{\mathrm e}^{4 d x +4 c} c +2 \,{\mathrm e}^{2 d x +2 c} d x -2 \,{\mathrm e}^{2 d x +2 c} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )+4 \,{\mathrm e}^{2 d x +2 c} c -d x -2 \,{\mathrm e}^{2 d x +2 c}+\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )-2 c \right )}{\left ({\mathrm e}^{2 d x +2 c}+1\right ) \left ({\mathrm e}^{2 d x +2 c}-1\right ) d}\) | \(188\) |
-1/2/d*(coth(d*x+c)^2*b)^(3/2)*(coth(d*x+c)^2+ln(coth(d*x+c)-1)+ln(coth(d* x+c)+1))/coth(d*x+c)^3
Leaf count of result is larger than twice the leaf count of optimal. 823 vs. \(2 (55) = 110\).
Time = 0.28 (sec) , antiderivative size = 823, normalized size of antiderivative = 13.49 \[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=\text {Too large to display} \]
(b*d*x*cosh(d*x + c)^4 - (b*d*x*e^(2*d*x + 2*c) - b*d*x)*sinh(d*x + c)^4 - 4*(b*d*x*cosh(d*x + c)*e^(2*d*x + 2*c) - b*d*x*cosh(d*x + c))*sinh(d*x + c)^3 + b*d*x - 2*(b*d*x - b)*cosh(d*x + c)^2 + 2*(3*b*d*x*cosh(d*x + c)^2 - b*d*x - (3*b*d*x*cosh(d*x + c)^2 - b*d*x + b)*e^(2*d*x + 2*c) + b)*sinh( d*x + c)^2 - (b*d*x*cosh(d*x + c)^4 + b*d*x - 2*(b*d*x - b)*cosh(d*x + c)^ 2)*e^(2*d*x + 2*c) - (b*cosh(d*x + c)^4 - (b*e^(2*d*x + 2*c) - b)*sinh(d*x + c)^4 - 4*(b*cosh(d*x + c)*e^(2*d*x + 2*c) - b*cosh(d*x + c))*sinh(d*x + c)^3 - 2*b*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - (3*b*cosh(d*x + c)^ 2 - b)*e^(2*d*x + 2*c) - b)*sinh(d*x + c)^2 - (b*cosh(d*x + c)^4 - 2*b*cos h(d*x + c)^2 + b)*e^(2*d*x + 2*c) + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x + c) - (b*cosh(d*x + c)^3 - b*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c) + b)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(b*d*x*cosh(d* x + c)^3 - (b*d*x - b)*cosh(d*x + c) - (b*d*x*cosh(d*x + c)^3 - (b*d*x - b )*cosh(d*x + c))*e^(2*d*x + 2*c))*sinh(d*x + c))*sqrt((b*e^(4*d*x + 4*c) + 2*b*e^(2*d*x + 2*c) + b)/(e^(4*d*x + 4*c) - 2*e^(2*d*x + 2*c) + 1))/(d*co sh(d*x + c)^4 + (d*e^(2*d*x + 2*c) + d)*sinh(d*x + c)^4 + 4*(d*cosh(d*x + c)*e^(2*d*x + 2*c) + d*cosh(d*x + c))*sinh(d*x + c)^3 - 2*d*cosh(d*x + c)^ 2 + 2*(3*d*cosh(d*x + c)^2 + (3*d*cosh(d*x + c)^2 - d)*e^(2*d*x + 2*c) - d )*sinh(d*x + c)^2 + (d*cosh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + d)*e^(2*d*x + 2*c) + 4*(d*cosh(d*x + c)^3 - d*cosh(d*x + c) + (d*cosh(d*x + c)^3 -...
\[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=\int \left (b \coth ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.59 \[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=-\frac {{\left (d x + c\right )} b^{\frac {3}{2}}}{d} - \frac {b^{\frac {3}{2}} \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {b^{\frac {3}{2}} \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {2 \, b^{\frac {3}{2}} e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \]
-(d*x + c)*b^(3/2)/d - b^(3/2)*log(e^(-d*x - c) + 1)/d - b^(3/2)*log(e^(-d *x - c) - 1)/d - 2*b^(3/2)*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - e^(-4 *d*x - 4*c) - 1))
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.48 \[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=-\frac {{\left ({\left (d x + c\right )} \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) - \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right ) + \frac {2 \, e^{\left (2 \, d x + 2 \, c\right )} \mathrm {sgn}\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}\right )} b^{\frac {3}{2}}}{d} \]
-((d*x + c)*sgn(e^(4*d*x + 4*c) - 1) - log(abs(e^(2*d*x + 2*c) - 1))*sgn(e ^(4*d*x + 4*c) - 1) + 2*e^(2*d*x + 2*c)*sgn(e^(4*d*x + 4*c) - 1)/(e^(2*d*x + 2*c) - 1)^2)*b^(3/2)/d
Timed out. \[ \int \left (b \coth ^2(c+d x)\right )^{3/2} \, dx=\int {\left (b\,{\mathrm {coth}\left (c+d\,x\right )}^2\right )}^{3/2} \,d x \]