Integrand size = 10, antiderivative size = 68 \[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=-\frac {2 (1-a-b x)^{3/2}}{b \sqrt {1+a+b x}}-\frac {3 \sqrt {1-a-b x} \sqrt {1+a+b x}}{b}-\frac {3 \arcsin (a+b x)}{b} \]
-3*arcsin(b*x+a)/b-2*(-b*x-a+1)^(3/2)/b/(b*x+a+1)^(1/2)-3*(-b*x-a+1)^(1/2) *(b*x+a+1)^(1/2)/b
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63 \[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=\frac {\sqrt {1-(a+b x)^2} \left (-1-\frac {4}{1+a+b x}\right )}{b}-\frac {3 \arcsin (a+b x)}{b} \]
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6711, 57, 60, 62, 1090, 223}
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.
\(\displaystyle \int e^{-3 \text {arctanh}(a+b x)} \, dx\) |
\(\Big \downarrow \) 6711 |
\(\displaystyle \int \frac {(-a-b x+1)^{3/2}}{(a+b x+1)^{3/2}}dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle -3 \int \frac {\sqrt {-a-b x+1}}{\sqrt {a+b x+1}}dx-\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -3 \left (\int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx+\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )-\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}\) |
\(\Big \downarrow \) 62 |
\(\displaystyle -3 \left (\int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx+\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )-\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle -3 \left (\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}-\frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b^2}\right )-\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -3 \left (\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}-\frac {\arcsin \left (\frac {-2 a b-2 b^2 x}{2 b}\right )}{b}\right )-\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}\) |
(-2*(1 - a - b*x)^(3/2))/(b*Sqrt[1 + a + b*x]) - 3*((Sqrt[1 - a - b*x]*Sqr t[1 + a + b*x])/b - ArcSin[(-2*a*b - 2*b^2*x)/(2*b)]/b)
3.9.63.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, n}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(134\) vs. \(2(60)=120\).
Time = 0.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.99
method | result | size |
risch | \(\frac {b^{2} x^{2}+2 a b x +a^{2}-1}{b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}-\frac {4 \sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{b^{2} \left (x +\frac {1+a}{b}\right )}\) | \(135\) |
default | \(\frac {-\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x +\frac {1+a}{b}\right )^{3}}-2 b \left (\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x +\frac {1+a}{b}\right )^{2}}+3 b \left (\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {3}{2}}}{3}+b \left (-\frac {\left (-2 b^{2} \left (x +\frac {1+a}{b}\right )+2 b \right ) \sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{4 b^{2}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )}{b^{3}}\) | \(256\) |
1/b*(b^2*x^2+2*a*b*x+a^2-1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3/(b^2)^(1/2)*a rctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))-4/b^2/(x+(1+a)/b )*(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(1/2)
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.49 \[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=\frac {3 \, {\left (b x + a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a + 5\right )}}{b^{2} x + {\left (a + 1\right )} b} \]
(3*(b*x + a + 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2* x^2 + 2*a*b*x + a^2 - 1)) - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a + 5))/(b^2*x + (a + 1)*b)
\[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=\int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{\left (a + b x + 1\right )^{3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.53 \[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + 2 \, b^{2} x + 2 \, a b + b} - \frac {3 \, \arcsin \left (b x + a\right )}{b} - \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{2} x + a b + b} \]
(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/(b^3*x^2 + 2*a*b^2*x + a^2*b + 2*b^2* x + 2*a*b + b) - 3*arcsin(b*x + a)/b - 6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1 )/(b^2*x + a*b + b)
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.38 \[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=\frac {3 \, \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{{\left | b \right |}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b} + \frac {8}{{\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} + 1\right )} {\left | b \right |}} \]
3*arcsin(-b*x - a)*sgn(b)/abs(b) - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/b + 8/(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) + 1)*abs (b))
Timed out. \[ \int e^{-3 \text {arctanh}(a+b x)} \, dx=\int \frac {{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}}{{\left (a+b\,x+1\right )}^3} \,d x \]