Integrand size = 12, antiderivative size = 78 \[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}} \]
-arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(3/2)/d+arctanh((b*tanh(d*x+c))^( 1/2)/b^(1/2))/b^(3/2)/d-2/b/d/(b*tanh(d*x+c))^(1/2)
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=\frac {-2-\arctan \left (\sqrt [4]{\tanh ^2(c+d x)}\right ) \sqrt [4]{\tanh ^2(c+d x)}+\text {arctanh}\left (\sqrt [4]{\tanh ^2(c+d x)}\right ) \sqrt [4]{\tanh ^2(c+d x)}}{b d \sqrt {b \tanh (c+d x)}} \]
(-2 - ArcTan[(Tanh[c + d*x]^2)^(1/4)]*(Tanh[c + d*x]^2)^(1/4) + ArcTanh[(T anh[c + d*x]^2)^(1/4)]*(Tanh[c + d*x]^2)^(1/4))/(b*d*Sqrt[b*Tanh[c + d*x]] )
Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3955, 3042, 3957, 25, 266, 827, 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.
\(\displaystyle \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(-i b \tan (i c+i d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3955 |
\(\displaystyle \frac {\int \sqrt {b \tanh (c+d x)}dx}{b^2}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{b d \sqrt {b \tanh (c+d x)}}+\frac {\int \sqrt {-i b \tan (i c+i d x)}dx}{b^2}\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle -\frac {\int -\frac {\sqrt {b \tanh (c+d x)}}{b^2-b^2 \tanh ^2(c+d x)}d(b \tanh (c+d x))}{b d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sqrt {b \tanh (c+d x)}}{b^2-b^2 \tanh ^2(c+d x)}d(b \tanh (c+d x))}{b d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 \int \frac {b^2 \tanh ^2(c+d x)}{b^2-b^4 \tanh ^4(c+d x)}d\sqrt {b \tanh (c+d x)}}{b d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}-\frac {1}{2} \int \frac {1}{b^2 \tanh ^2(c+d x)+b}d\sqrt {b \tanh (c+d x)}\right )}{b d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}-\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}\right )}{b d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \left (\frac {\text {arctanh}\left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}\right )}{b d}-\frac {2}{b d \sqrt {b \tanh (c+d x)}}\) |
(2*(-1/2*ArcTan[Sqrt[b]*Tanh[c + d*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Tanh[c + d*x]]/(2*Sqrt[b])))/(b*d) - 2/(b*d*Sqrt[b*Tanh[c + d*x]])
3.1.18.3.1 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) 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*Tan[c + d*x] )^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2 Int[(b*Tan[c + d*x])^(n + 2), x] , x] /; FreeQ[{b, c, d}, x] && LtQ[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 = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(-\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}-\frac {2}{b d \sqrt {b \tanh \left (d x +c \right )}}\) | \(65\) |
default | \(-\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}} d}-\frac {2}{b d \sqrt {b \tanh \left (d x +c \right )}}\) | \(65\) |
-arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))/b^(3/2)/d+arctanh((b*tanh(d*x+c))^( 1/2)/b^(1/2))/b^(3/2)/d-2/b/d/(b*tanh(d*x+c))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (64) = 128\).
Time = 0.28 (sec) , antiderivative size = 924, normalized size of antiderivative = 11.85 \[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
[-1/4*(2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^ 2 - 1)*sqrt(-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*sinh(d*x + c)/cosh(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)) + (cosh (d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(-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) + sinh(d*x + c)^2 + 1)*sqrt(-b)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)) - 2*b)/(cosh(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*(cosh(d*x + c)^ 2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 + 1)*sqrt(b*sinh(d*x + c)/cosh(d*x + c)))/(b^2*d*cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x + c)*sinh(d* x + c) + b^2*d*sinh(d*x + c)^2 - b^2*d), 1/4*(2*(cosh(d*x + c)^2 + 2*cosh( d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(b)*arctan(sqrt(b)*sqrt( b*sinh(d*x + c)/cosh(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)) + (cosh(d*x + c)^2 + 2*cosh(d*x + c)*s inh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(b)*log(2*b*cosh(d*x + c)^4 + 8*b* cosh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b *cosh(d*x + c)*sinh(d*x + c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)...
\[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (b \tanh {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=\int { \frac {1}{\left (b \tanh \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=\frac {4}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} - \sqrt {b}\right )} b d} \]
Time = 1.84 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(b \tanh (c+d x))^{3/2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{3/2}\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{3/2}\,d}-\frac {2}{b\,d\,\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}} \]