\(\int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx\) [263]

Optimal result

Integrand size = 20, antiderivative size = 95 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {8 c \sqrt {c-a c x}}{a}-\frac {4 (c-a c x)^{3/2}}{3 a}-\frac {2 (c-a c x)^{5/2}}{5 a c}+\frac {8 \sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6302, 6265, 21, 52, 65, 212} \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\frac {8 \sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}-\frac {2 (c-a c x)^{5/2}}{5 a c}-\frac {4 (c-a c x)^{3/2}}{3 a}-\frac {8 c \sqrt {c-a c x}}{a} \]

[In]

Int[(c - a*c*x)^(3/2)/E^(2*ArcCoth[a*x]),x]

[Out]

(-8*c*Sqrt[c - a*c*x])/a - (4*(c - a*c*x)^(3/2))/(3*a) - (2*(c - a*c*x)^(5/2))/(5*a*c) + (8*Sqrt[2]*c^(3/2)*Ar
cTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6265

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]
)

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps \begin{align*} \text {integral}& = -\int e^{-2 \text {arctanh}(a x)} (c-a c x)^{3/2} \, dx \\ & = -\int \frac {(1-a x) (c-a c x)^{3/2}}{1+a x} \, dx \\ & = -\frac {\int \frac {(c-a c x)^{5/2}}{1+a x} \, dx}{c} \\ & = -\frac {2 (c-a c x)^{5/2}}{5 a c}-2 \int \frac {(c-a c x)^{3/2}}{1+a x} \, dx \\ & = -\frac {4 (c-a c x)^{3/2}}{3 a}-\frac {2 (c-a c x)^{5/2}}{5 a c}-(4 c) \int \frac {\sqrt {c-a c x}}{1+a x} \, 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}-\left (8 c^2\right ) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, 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 {(16 c) \text {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{a} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.75 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\frac {-2 c \sqrt {c-a c x} \left (73-16 a x+3 a^2 x^2\right )+120 \sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{15 a} \]

[In]

Integrate[(c - a*c*x)^(3/2)/E^(2*ArcCoth[a*x]),x]

[Out]

(-2*c*Sqrt[c - a*c*x]*(73 - 16*a*x + 3*a^2*x^2) + 120*Sqrt[2]*c^(3/2)*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c]
)])/(15*a)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(-\frac {2 \left (-20 \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+\frac {\left (3 a^{2} x^{2}-16 a x +73\right ) \sqrt {-c \left (a x -1\right )}}{3}\right ) c}{5 a}\)	\(61\)
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	\(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {2 c \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c^{2} \sqrt {-a c x +c}-4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{c a}\)	\(73\)
default	\(\frac {-\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {4 c \left (-a c x +c \right )^{\frac {3}{2}}}{3}-8 c^{2} \sqrt {-a c x +c}+8 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a c}\)	\(73\)

[In]

int((-a*c*x+c)^(3/2)*(a*x-1)/(a*x+1),x,method=_RETURNVERBOSE)

[Out]

-2/5*(-20*c^(1/2)*2^(1/2)*arctanh(1/2*(-c*(a*x-1))^(1/2)*2^(1/2)/c^(1/2))+1/3*(3*a^2*x^2-16*a*x+73)*(-c*(a*x-1
))^(1/2))*c/a

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.54 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\left [\frac {2 \, {\left (30 \, \sqrt {2} c^{\frac {3}{2}} \log \left (\frac {a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) - {\left (3 \, a^{2} c x^{2} - 16 \, a c x + 73 \, c\right )} \sqrt {-a c x + c}\right )}}{15 \, a}, -\frac {2 \, {\left (60 \, \sqrt {2} \sqrt {-c} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) + {\left (3 \, a^{2} c x^{2} - 16 \, a c x + 73 \, c\right )} \sqrt {-a c x + c}\right )}}{15 \, a}\right ] \]

[In]

integrate((-a*c*x+c)^(3/2)*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

[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 - 1
6*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]

Sympy [A] (verification not implemented)

Time = 2.61 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.22 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\begin {cases} - \frac {2 \cdot \left (\frac {4 \sqrt {2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} + 4 c^{2} \sqrt {- a c x + c} + \frac {2 c \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a c} & \text {for}\: a c \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {a x - 2 \log {\left (a x + 1 \right )} + 1}{a} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((-a*c*x+c)**(3/2)*(a*x-1)/(a*x+1),x)

[Out]

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)*Piecewise((-x, Eq(a, 0)), ((
a*x - 2*log(a*x + 1) + 1)/a, True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {2 \, {\left (30 \, \sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 3 \, {\left (-a c x + c\right )}^{\frac {5}{2}} + 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 60 \, \sqrt {-a c x + c} c^{2}\right )}}{15 \, a c} \]

[In]

integrate((-a*c*x+c)^(3/2)*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {8 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} - \frac {2 \, {\left (3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{4} c^{4} + 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{4} c^{5} + 60 \, \sqrt {-a c x + c} a^{4} c^{6}\right )}}{15 \, a^{5} c^{5}} \]

[In]

integrate((-a*c*x+c)^(3/2)*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 4.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {4\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a}-\frac {8\,c\,\sqrt {c-a\,c\,x}}{a}-\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a\,c}-\frac {\sqrt {2}\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,8{}\mathrm {i}}{a} \]

[In]

int(((c - a*c*x)^(3/2)*(a*x - 1))/(a*x + 1),x)

[Out]

- (4*(c - a*c*x)^(3/2))/(3*a) - (8*c*(c - a*c*x)^(1/2))/a - (2*(c - a*c*x)^(5/2))/(5*a*c) - (2^(1/2)*c^(3/2)*a
tan((2^(1/2)*(c - a*c*x)^(1/2)*1i)/(2*c^(1/2)))*8i)/a