Integrand size = 20, antiderivative size = 95 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\frac {8 c \sqrt {c-a c x}}{a}+\frac {4 (c-a c x)^{3/2}}{3 a}+\frac {2 (c-a c x)^{5/2}}{5 a c}-\frac {8 \sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a} \]
4/3*(-a*c*x+c)^(3/2)/a+2/5*(-a*c*x+c)^(5/2)/a/c-8*c^(3/2)*arctanh(1/2*(-a* c*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a+8*c*(-a*c*x+c)^(1/2)/a
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\frac {2 c \sqrt {c-a c x} \left (73-16 a x+3 a^2 x^2\right )-120 \sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{15 a} \]
(2*c*Sqrt[c - a*c*x]*(73 - 16*a*x + 3*a^2*x^2) - 120*Sqrt[2]*c^(3/2)*ArcTa nh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/(15*a)
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6680, 35, 60, 60, 60, 73, 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.
\(\displaystyle \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {(1-a x) (c-a c x)^{3/2}}{a x+1}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle \frac {\int \frac {(c-a c x)^{5/2}}{a x+1}dx}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c \int \frac {(c-a c x)^{3/2}}{a x+1}dx+\frac {2 (c-a c x)^{5/2}}{5 a}}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c \left (2 c \int \frac {\sqrt {c-a c x}}{a x+1}dx+\frac {2 (c-a c x)^{3/2}}{3 a}\right )+\frac {2 (c-a c x)^{5/2}}{5 a}}{c}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 c \left (2 c \left (2 c \int \frac {1}{(a x+1) \sqrt {c-a c x}}dx+\frac {2 \sqrt {c-a c x}}{a}\right )+\frac {2 (c-a c x)^{3/2}}{3 a}\right )+\frac {2 (c-a c x)^{5/2}}{5 a}}{c}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 c \left (2 c \left (\frac {2 \sqrt {c-a c x}}{a}-\frac {4 \int \frac {1}{2-\frac {c-a c x}{c}}d\sqrt {c-a c x}}{a}\right )+\frac {2 (c-a c x)^{3/2}}{3 a}\right )+\frac {2 (c-a c x)^{5/2}}{5 a}}{c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 c \left (2 c \left (\frac {2 \sqrt {c-a c x}}{a}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}\right )+\frac {2 (c-a c x)^{3/2}}{3 a}\right )+\frac {2 (c-a c x)^{5/2}}{5 a}}{c}\) |
((2*(c - a*c*x)^(5/2))/(5*a) + 2*c*((2*(c - a*c*x)^(3/2))/(3*a) + 2*c*((2* Sqrt[c - a*c*x])/a - (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*S qrt[c])])/a)))/c
3.3.63.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Int[u*(c + d*x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c , d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.63
method | result | size |
pseudoelliptic | \(-\frac {8 \left (\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-\frac {\sqrt {-c \left (a x -1\right )}\, \left (3 a^{2} x^{2}-16 a x +73\right )}{60}\right ) c}{a}\) | \(60\) |
risch | \(-\frac {2 \left (3 a^{2} x^{2}-16 a x +73\right ) \left (a x -1\right ) c^{2}}{15 a \sqrt {-c \left (a x -1\right )}}-\frac {8 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}}{a}\) | \(68\) |
derivativedivides | \(\frac {\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {4 c \left (-a c x +c \right )^{\frac {3}{2}}}{3}+8 c^{2} \sqrt {-a c x +c}-8 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a c}\) | \(73\) |
default | \(-\frac {2 \left (-\frac {\left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {2 c \left (-a c x +c \right )^{\frac {3}{2}}}{3}-4 c^{2} \sqrt {-a c x +c}+4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{c a}\) | \(73\) |
-8*(c^(1/2)*2^(1/2)*arctanh(1/2*(-c*(a*x-1))^(1/2)*2^(1/2)/c^(1/2))-1/60*( -c*(a*x-1))^(1/2)*(3*a^2*x^2-16*a*x+73))*c/a
Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.53 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\left [\frac {2 \, {\left (30 \, \sqrt {2} c^{\frac {3}{2}} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + {\left (3 \, a^{2} c x^{2} - 16 \, a c x + 73 \, c\right )} \sqrt {-a c x + c}\right )}}{15 \, a}, \frac {2 \, {\left (60 \, \sqrt {2} \sqrt {-c} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) + {\left (3 \, a^{2} c x^{2} - 16 \, a c x + 73 \, c\right )} \sqrt {-a c x + c}\right )}}{15 \, a}\right ] \]
[2/15*(30*sqrt(2)*c^(3/2)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (3*a^2*c*x^2 - 16*a*c*x + 73*c)*sqrt(-a*c*x + c))/a, 2 /15*(60*sqrt(2)*sqrt(-c)*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) + (3*a^2*c*x^2 - 16*a*c*x + 73*c)*sqrt(-a*c*x + c))/a]
Time = 5.46 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.18 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\begin {cases} - \frac {2 \left (- \frac {4 \sqrt {2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} - 4 c^{2} \sqrt {- a c x + c} - \frac {2 c \left (- a c x + c\right )^{\frac {3}{2}}}{3} - \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a c} & \text {for}\: a c \neq 0 \\c^{\frac {3}{2}} \left (- x + 2 \left (\begin {cases} x & \text {for}\: a = 0 \\\frac {\log {\left (a x + 1 \right )}}{a} & \text {otherwise} \end {cases}\right )\right ) & \text {otherwise} \end {cases} \]
Piecewise((-2*(-4*sqrt(2)*c**3*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c))) /sqrt(-c) - 4*c**2*sqrt(-a*c*x + c) - 2*c*(-a*c*x + c)**(3/2)/3 - (-a*c*x + c)**(5/2)/5)/(a*c), Ne(a*c, 0)), (c**(3/2)*(-x + 2*Piecewise((x, Eq(a, 0 )), (log(a*x + 1)/a, True))), True))
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\frac {2 \, {\left (30 \, \sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 3 \, {\left (-a c x + c\right )}^{\frac {5}{2}} + 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 60 \, \sqrt {-a c x + c} c^{2}\right )}}{15 \, a c} \]
2/15*(30*sqrt(2)*c^(5/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2 )*sqrt(c) + sqrt(-a*c*x + c))) + 3*(-a*c*x + c)^(5/2) + 10*(-a*c*x + c)^(3 /2)*c + 60*sqrt(-a*c*x + c)*c^2)/(a*c)
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\frac {8 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {2 \, {\left (3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{4} c^{4} + 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{4} c^{5} + 60 \, \sqrt {-a c x + c} a^{4} c^{6}\right )}}{15 \, a^{5} c^{5}} \]
8*sqrt(2)*c^2*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/15*(3*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^4*c^4 + 10*(-a*c*x + c)^(3/2)*a^ 4*c^5 + 60*sqrt(-a*c*x + c)*a^4*c^6)/(a^5*c^5)
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx=\frac {4\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a}+\frac {8\,c\,\sqrt {c-a\,c\,x}}{a}+\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a\,c}+\frac {\sqrt {2}\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,8{}\mathrm {i}}{a} \]