Integrand size = 12, antiderivative size = 75 \[ \int (b \coth (c+d x))^{3/2} \, dx=\frac {b^{3/2} \arctan \left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d} \]
b^(3/2)*arctan((b*coth(d*x+c))^(1/2)/b^(1/2))/d+b^(3/2)*arctanh((b*coth(d* x+c))^(1/2)/b^(1/2))/d-2*b*(b*coth(d*x+c))^(1/2)/d
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int (b \coth (c+d x))^{3/2} \, dx=-\frac {\left (-\arctan \left (\sqrt {\coth (c+d x)}\right )-\text {arctanh}\left (\sqrt {\coth (c+d x)}\right )+2 \sqrt {\coth (c+d x)}\right ) (b \coth (c+d x))^{3/2}}{d \coth ^{\frac {3}{2}}(c+d x)} \]
-(((-ArcTan[Sqrt[Coth[c + d*x]]] - ArcTanh[Sqrt[Coth[c + d*x]]] + 2*Sqrt[C oth[c + d*x]])*(b*Coth[c + d*x])^(3/2))/(d*Coth[c + d*x]^(3/2)))
Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3954, 3042, 3957, 25, 266, 756, 216, 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.
\(\displaystyle \int (b \coth (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-i b \tan \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b^2 \int \frac {1}{\sqrt {b \coth (c+d x)}}dx-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b \sqrt {b \coth (c+d x)}}{d}+b^2 \int \frac {1}{\sqrt {-i b \tan \left (i c+i d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle -\frac {b^3 \int -\frac {1}{\sqrt {b \coth (c+d x)} \left (b^2-b^2 \coth ^2(c+d x)\right )}d(b \coth (c+d x))}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \int \frac {1}{\sqrt {b \coth (c+d x)} \left (b^2-b^2 \coth ^2(c+d x)\right )}d(b \coth (c+d x))}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b^3 \int \frac {1}{b^2-b^4 \coth ^4(c+d x)}d\sqrt {b \coth (c+d x)}}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {2 b^3 \left (\frac {\int \frac {1}{b-b^2 \coth ^2(c+d x)}d\sqrt {b \coth (c+d x)}}{2 b}+\frac {\int \frac {1}{b^2 \coth ^2(c+d x)+b}d\sqrt {b \coth (c+d x)}}{2 b}\right )}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 b^3 \left (\frac {\int \frac {1}{b-b^2 \coth ^2(c+d x)}d\sqrt {b \coth (c+d x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \coth (c+d x)\right )}{2 b^{3/2}}\right )}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \coth (c+d x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \coth (c+d x)\right )}{2 b^{3/2}}\right )}{d}-\frac {2 b \sqrt {b \coth (c+d x)}}{d}\) |
(2*b^3*(ArcTan[Sqrt[b]*Coth[c + d*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Coth[c + d*x]]/(2*b^(3/2))))/d - (2*b*Sqrt[b*Coth[c + d*x]])/d
3.1.3.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
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[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{d}+\frac {b^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{d}-\frac {2 b \sqrt {b \coth \left (d x +c \right )}}{d}\) | \(62\) |
default | \(\frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{d}+\frac {b^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{d}-\frac {2 b \sqrt {b \coth \left (d x +c \right )}}{d}\) | \(62\) |
b^(3/2)*arctan((b*coth(d*x+c))^(1/2)/b^(1/2))/d+b^(3/2)*arctanh((b*coth(d* x+c))^(1/2)/b^(1/2))/d-2*b*(b*coth(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (61) = 122\).
Time = 0.27 (sec) , antiderivative size = 637, normalized size of antiderivative = 8.49 \[ \int (b \coth (c+d x))^{3/2} \, dx=\left [-\frac {2 \, \sqrt {-b} b \arctan \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}\right )} \sqrt {-b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b}\right ) - \sqrt {-b} b \log \left (-\frac {b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} - 2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {-b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}} - 2 \, b}{\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4}}\right ) + 8 \, b \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{4 \, d}, \frac {2 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b}\right ) + b^{\frac {3}{2}} \log \left (2 \, b \cosh \left (d x + c\right )^{4} + 8 \, b \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 12 \, b \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 8 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, b \sinh \left (d x + c\right )^{4} + 2 \, {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} + {\left (6 \, \cosh \left (d x + c\right )^{2} - 1\right )} \sinh \left (d x + c\right )^{2} - \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cosh \left (d x + c\right )^{3} - \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b} \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}} - b\right ) - 8 \, b \sqrt {\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}}}{4 \, d}\right ] \]
[-1/4*(2*sqrt(-b)*b*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c ) + sinh(d*x + c)^2)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh( d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) - s qrt(-b)*b*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6* b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b* sinh(d*x + c)^4 - 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sin h(d*x + c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - 2*b)/(cos h(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d* x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)) + 8*b*sqrt( b*cosh(d*x + c)/sinh(d*x + c)))/d, 1/4*(2*b^(3/2)*arctan(sqrt(b)*sqrt(b*co sh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) + b^(3/2)*log(2*b*cosh(d*x + c)^4 + 8*b*co sh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*c osh(d*x + c)*sinh(d*x + c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*cosh(d*x + c)^2 - 1 )*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(2*cosh(d*x + c)^3 - cosh(d*x + c) )*sinh(d*x + c))*sqrt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - b) - 8*b*sq rt(b*cosh(d*x + c)/sinh(d*x + c)))/d]
\[ \int (b \coth (c+d x))^{3/2} \, dx=\int \left (b \coth {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
\[ \int (b \coth (c+d x))^{3/2} \, dx=\int { \left (b \coth \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (61) = 122\).
Time = 0.34 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.24 \[ \int (b \coth (c+d x))^{3/2} \, dx=-\frac {{\left (2 \, \sqrt {b} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right ) \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) + \sqrt {b} \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) - \frac {8 \, b \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} - \sqrt {b}}\right )} b}{2 \, d} \]
-1/2*(2*sqrt(b)*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b))*sgn(e^(2*d*x + 2*c) - 1) + sqrt(b)*log(abs(-sqrt(b)*e^(2*d* x + 2*c) + sqrt(b*e^(4*d*x + 4*c) - b)))*sgn(e^(2*d*x + 2*c) - 1) - 8*b*sg n(e^(2*d*x + 2*c) - 1)/(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b) - sqrt(b)))*b/d
Time = 2.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int (b \coth (c+d x))^{3/2} \, dx=\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {2\,b\,\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{d}+\frac {b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d} \]