Integrand size = 10, antiderivative size = 84 \[ \int e^{\text {arctanh}(a+b x)} x \, dx=-\frac {(1-2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2}}{2 b^2}+\frac {(1-2 a) \arcsin (a+b x)}{2 b^2} \]
1/2*(1-2*a)*arcsin(b*x+a)/b^2-1/2*(b*x+a+1)^(3/2)*(-b*x-a+1)^(1/2)/b^2-1/2 *(1-2*a)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^2
Time = 0.06 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.55 \[ \int e^{\text {arctanh}(a+b x)} x \, dx=\frac {\sqrt {b} (-2+a-b x) \sqrt {1-a^2-2 a b x-b^2 x^2}+2 \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )+4 a \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{2 b^{5/2}} \]
(Sqrt[b]*(-2 + a - b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] + 2*Sqrt[-b]*Arc Sinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])] + 4*a*Sqrt[-b]*ArcSin h[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(2*b^(5/2))
Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6713, 90, 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.
\(\displaystyle \int x e^{\text {arctanh}(a+b x)} \, dx\) |
\(\Big \downarrow \) 6713 |
\(\displaystyle \int \frac {x \sqrt {a+b x+1}}{\sqrt {-a-b x+1}}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(1-2 a) \int \frac {\sqrt {a+b x+1}}{\sqrt {-a-b x+1}}dx}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(1-2 a) \left (\int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}\) |
\(\Big \downarrow \) 62 |
\(\displaystyle \frac {(1-2 a) \left (\int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {(1-2 a) \left (-\frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b^2}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {(1-2 a) \left (-\frac {\arcsin \left (\frac {-2 a b-2 b^2 x}{2 b}\right )}{b}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b^2}\) |
-1/2*(Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2))/b^2 + ((1 - 2*a)*(-((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/b) - ArcSin[(-2*a*b - 2*b^2*x)/(2*b)]/b))/(2* b)
3.9.20.3.1 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 + 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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 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_.))*((d_.) + (e_.)*(x_))^(m_.) , x_Symbol] :> Int[(d + e*x)^m*((1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^( n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.24
method | result | size |
risch | \(-\frac {\left (-b x +a -2\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\left (-1+2 a \right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b \sqrt {b^{2}}}\) | \(104\) |
default | \(b \left (-\frac {x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 b^{2}}-\frac {3 a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b^{2}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}\right )}{2 b}+\frac {\left (-a^{2}+1\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b^{2} \sqrt {b^{2}}}\right )+\left (1+a \right ) \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b^{2}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}\right )\) | \(246\) |
-1/2*(-b*x+a-2)*(b^2*x^2+2*a*b*x+a^2-1)/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2) -1/2/b*(-1+2*a)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a ^2+1)^(1/2))
Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.10 \[ \int e^{\text {arctanh}(a+b x)} x \, dx=\frac {{\left (2 \, a - 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x - a + 2\right )}}{2 \, b^{2}} \]
1/2*((2*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 + 2) )/b^2
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (68) = 136\).
Time = 1.59 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.19 \[ \int e^{\text {arctanh}(a+b x)} x \, dx=\begin {cases} \left (- \frac {x}{2 b} - \frac {1 - \frac {a}{2}}{b^{2}}\right ) \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} + \frac {\left (- \frac {a \left (1 - \frac {a}{2}\right )}{b} + \frac {1 - a^{2}}{2 b}\right ) \log {\left (- 2 a b - 2 b^{2} x + 2 \sqrt {- b^{2}} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \right )}}{\sqrt {- b^{2}}} & \text {for}\: b^{2} \neq 0 \\- \frac {- \frac {a^{2} \sqrt {- a^{2} - 2 a b x + 1} + \frac {\left (- a^{2} - 2 a b x + 1\right )^{\frac {3}{2}}}{3} - \sqrt {- a^{2} - 2 a b x + 1}}{b} - \frac {a^{2} \sqrt {- a^{2} - 2 a b x + 1} + \frac {\left (- a^{2} - 2 a b x + 1\right )^{\frac {3}{2}}}{3} - \sqrt {- a^{2} - 2 a b x + 1}}{a b} + \frac {a^{4} \sqrt {- a^{2} - 2 a b x + 1} - 2 a^{2} \sqrt {- a^{2} - 2 a b x + 1} + \frac {\left (2 a^{2} - 2\right ) \left (- a^{2} - 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (- a^{2} - 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \sqrt {- a^{2} - 2 a b x + 1}}{2 a^{2} b}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {\frac {a x^{2}}{2} + \frac {b x^{3}}{3} + \frac {x^{2}}{2}}{\sqrt {1 - a^{2}}} & \text {otherwise} \end {cases} \]
Piecewise(((-x/(2*b) - (1 - a/2)/b**2)*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1) + (-a*(1 - a/2)/b + (1 - a**2)/(2*b))*log(-2*a*b - 2*b**2*x + 2*sqrt(-b **2)*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1))/sqrt(-b**2), Ne(b**2, 0)), (-( -(a**2*sqrt(-a**2 - 2*a*b*x + 1) + (-a**2 - 2*a*b*x + 1)**(3/2)/3 - sqrt(- a**2 - 2*a*b*x + 1))/b - (a**2*sqrt(-a**2 - 2*a*b*x + 1) + (-a**2 - 2*a*b* x + 1)**(3/2)/3 - sqrt(-a**2 - 2*a*b*x + 1))/(a*b) + (a**4*sqrt(-a**2 - 2* a*b*x + 1) - 2*a**2*sqrt(-a**2 - 2*a*b*x + 1) + (2*a**2 - 2)*(-a**2 - 2*a* b*x + 1)**(3/2)/3 + (-a**2 - 2*a*b*x + 1)**(5/2)/5 + sqrt(-a**2 - 2*a*b*x + 1))/(2*a**2*b))/(2*a*b), Ne(a*b, 0)), ((a*x**2/2 + b*x**3/3 + x**2/2)/sq rt(1 - a**2), True))
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (70) = 140\).
Time = 0.30 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.49 \[ \int e^{\text {arctanh}(a+b x)} x \, dx=\frac {{\left (a + 1\right )} a \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{2}} - \frac {3 \, a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{2 \, b} + \frac {{\left (a^{2} - 1\right )} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a + 1\right )}}{b^{2}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{2 \, b^{2}} \]
(a + 1)*a*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^2 - 3/2*a ^2*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^2 - 1/2*sqrt(-b^ 2*x^2 - 2*a*b*x - a^2 + 1)*x/b + 1/2*(a^2 - 1)*arcsin(-(b^2*x + a*b)/sqrt( a^2*b^2 - (a^2 - 1)*b^2))/b^2 - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a + 1) /b^2 + 3/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a/b^2
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int e^{\text {arctanh}(a+b x)} x \, dx=-\frac {1}{2} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\frac {x}{b} - \frac {a b - 2 \, b}{b^{3}}\right )} + \frac {{\left (2 \, a - 1\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b {\left | b \right |}} \]
-1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x/b - (a*b - 2*b)/b^3) + 1/2*(2*a - 1)*arcsin(-b*x - a)*sgn(b)/(b*abs(b))
Timed out. \[ \int e^{\text {arctanh}(a+b x)} x \, dx=\int \frac {x\,\left (a+b\,x+1\right )}{\sqrt {1-{\left (a+b\,x\right )}^2}} \,d x \]