Integrand size = 12, antiderivative size = 58 \[ \int \sqrt {b \tanh (c+d x)} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b \tanh (c+d x)}}{\sqrt {b}}\right )}{d} \]
-arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))*b^(1/2)/d+arctanh((b*tanh(d*x+c))^( 1/2)/b^(1/2))*b^(1/2)/d
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int \sqrt {b \tanh (c+d x)} \, dx=-\frac {\left (\arctan \left (\sqrt {\tanh (c+d x)}\right )-\text {arctanh}\left (\sqrt {\tanh (c+d x)}\right )\right ) \sqrt {b \tanh (c+d x)}}{d \sqrt {\tanh (c+d x)}} \]
-(((ArcTan[Sqrt[Tanh[c + d*x]]] - ArcTanh[Sqrt[Tanh[c + d*x]]])*Sqrt[b*Tan h[c + d*x]])/(d*Sqrt[Tanh[c + d*x]]))
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {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.
\(\displaystyle \int \sqrt {b \tanh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {-i b \tan (i c+i d x)}dx\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle -\frac {b \int -\frac {\sqrt {b \tanh (c+d x)}}{b^2-b^2 \tanh ^2(c+d x)}d(b \tanh (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \int \frac {\sqrt {b \tanh (c+d x)}}{b^2-b^2 \tanh ^2(c+d x)}d(b \tanh (c+d x))}{d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b \int \frac {b^2 \tanh ^2(c+d x)}{b^2-b^4 \tanh ^4(c+d x)}d\sqrt {b \tanh (c+d x)}}{d}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}-\frac {1}{2} \int \frac {1}{b^2 \tanh ^2(c+d x)+b}d\sqrt {b \tanh (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \int \frac {1}{b-b^2 \tanh ^2(c+d x)}d\sqrt {b \tanh (c+d x)}-\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 b \left (\frac {\text {arctanh}\left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \tanh (c+d x)\right )}{2 \sqrt {b}}\right )}{d}\) |
(2*b*(-1/2*ArcTan[Sqrt[b]*Tanh[c + d*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Tanh[c + d*x]]/(2*Sqrt[b])))/d
3.1.16.3.1 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/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 = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(-\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right ) \sqrt {b}}{d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right ) \sqrt {b}}{d}\) | \(47\) |
default | \(-\frac {\arctan \left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right ) \sqrt {b}}{d}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \tanh \left (d x +c \right )}}{\sqrt {b}}\right ) \sqrt {b}}{d}\) | \(47\) |
-arctan((b*tanh(d*x+c))^(1/2)/b^(1/2))*b^(1/2)/d+arctanh((b*tanh(d*x+c))^( 1/2)/b^(1/2))*b^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (46) = 92\).
Time = 0.29 (sec) , antiderivative size = 593, normalized size of antiderivative = 10.22 \[ \int \sqrt {b \tanh (c+d x)} \, dx=\left [-\frac {2 \, \sqrt {-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 \sinh \left (d x + c\right )}{\cosh \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} \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 \sinh \left (d x + c\right )}{\cosh \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 )}{4 \, d}, \frac {2 \, \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {\frac {b \sinh \left (d x + c\right )}{\cosh \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} \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 \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )}} - b\right )}{4 \, d}\right ] \]
[-1/4*(2*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)) - sqr t(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*co sh(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)))/d, 1/4*(2*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)) + 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)^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*cos h(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c))*sqrt(b)*sqrt(b*sinh(d*x + c)/ cosh(d*x + c)) - b))/d]
\[ \int \sqrt {b \tanh (c+d x)} \, dx=\int \sqrt {b \tanh {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt {b \tanh (c+d x)} \, dx=\int { \sqrt {b \tanh \left (d x + c\right )} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.52 \[ \int \sqrt {b \tanh (c+d x)} \, dx=-\frac {2 \, \sqrt {b} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right ) + \sqrt {b} \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{2 \, d} \]
-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)) + sqrt(b)*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqrt(b*e^(4*d* x + 4*c) - b))))/d
Time = 1.72 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \sqrt {b \tanh (c+d x)} \, dx=-\frac {\sqrt {b}\,\left (\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )-\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {tanh}\left (c+d\,x\right )}}{\sqrt {b}}\right )\right )}{d} \]