\(\int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx\) [262]

Optimal result

Integrand size = 20, antiderivative size = 116 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}+\frac {16 \sqrt {2} c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

[Out]

-8/3*c*(-a*c*x+c)^(3/2)/a-4/5*(-a*c*x+c)^(5/2)/a-2/7*(-a*c*x+c)^(7/2)/a/c+16*c^(5/2)*arctanh(1/2*(-a*c*x+c)^(1
/2)*2^(1/2)/c^(1/2))*2^(1/2)/a-16*c^2*(-a*c*x+c)^(1/2)/a

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6302, 6265, 21, 52, 65, 212} \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {16 \sqrt {2} c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a}-\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {8 c (c-a c x)^{3/2}}{3 a} \]

[In]

Int[(c - a*c*x)^(5/2)/E^(2*ArcCoth[a*x]),x]

[Out]

(-16*c^2*Sqrt[c - a*c*x])/a - (8*c*(c - a*c*x)^(3/2))/(3*a) - (4*(c - a*c*x)^(5/2))/(5*a) - (2*(c - a*c*x)^(7/
2))/(7*a*c) + (16*Sqrt[2]*c^(5/2)*ArcTanh[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)^{5/2} \, dx \\ & = -\int \frac {(1-a x) (c-a c x)^{5/2}}{1+a x} \, dx \\ & = -\frac {\int \frac {(c-a c x)^{7/2}}{1+a x} \, dx}{c} \\ & = -\frac {2 (c-a c x)^{7/2}}{7 a c}-2 \int \frac {(c-a c x)^{5/2}}{1+a x} \, dx \\ & = -\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-(4 c) \int \frac {(c-a c x)^{3/2}}{1+a x} \, dx \\ & = -\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\left (8 c^2\right ) \int \frac {\sqrt {c-a c x}}{1+a x} \, dx \\ & = -\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}-\left (16 c^3\right ) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx \\ & = -\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}+\frac {\left (32 c^2\right ) \text {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{a} \\ & = -\frac {16 c^2 \sqrt {c-a c x}}{a}-\frac {8 c (c-a c x)^{3/2}}{3 a}-\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c}+\frac {16 \sqrt {2} c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 c^2 \left (\sqrt {c-a c x} \left (-1037+269 a x-87 a^2 x^2+15 a^3 x^3\right )+840 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )\right )}{105 a} \]

[In]

Integrate[(c - a*c*x)^(5/2)/E^(2*ArcCoth[a*x]),x]

[Out]

(2*c^2*(Sqrt[c - a*c*x]*(-1037 + 269*a*x - 87*a^2*x^2 + 15*a^3*x^3) + 840*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c
*x]/(Sqrt[2]*Sqrt[c])]))/(105*a)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(\frac {2 \left (56 \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+\frac {\left (15 a^{3} x^{3}-87 a^{2} x^{2}+269 a x -1037\right ) \sqrt {-c \left (a x -1\right )}}{15}\right ) c^{2}}{7 a}\)	\(71\)
risch	\(-\frac {2 \left (15 a^{3} x^{3}-87 a^{2} x^{2}+269 a x -1037\right ) \left (a x -1\right ) c^{3}}{105 a \sqrt {-c \left (a x -1\right )}}+\frac {16 c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}}{a}\)	\(76\)
derivativedivides	\(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {2 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {4 c^{2} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+8 c^{3} \sqrt {-a c x +c}-8 c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{c a}\)	\(87\)
default	\(\frac {-\frac {2 \left (-a c x +c \right )^{\frac {7}{2}}}{7}-\frac {4 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {8 c^{2} \left (-a c x +c \right )^{\frac {3}{2}}}{3}-16 c^{3} \sqrt {-a c x +c}+16 c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a c}\)	\(87\)

[In]

int((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x,method=_RETURNVERBOSE)

[Out]

2/7*(56*c^(1/2)*2^(1/2)*arctanh(1/2*(-c*(a*x-1))^(1/2)*2^(1/2)/c^(1/2))+1/15*(15*a^3*x^3-87*a^2*x^2+269*a*x-10
37)*(-c*(a*x-1))^(1/2))*c^2/a

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.57 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\left [\frac {2 \, {\left (420 \, \sqrt {2} c^{\frac {5}{2}} \log \left (\frac {a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + {\left (15 \, a^{3} c^{2} x^{3} - 87 \, a^{2} c^{2} x^{2} + 269 \, a c^{2} x - 1037 \, c^{2}\right )} \sqrt {-a c x + c}\right )}}{105 \, a}, -\frac {2 \, {\left (840 \, \sqrt {2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) - {\left (15 \, a^{3} c^{2} x^{3} - 87 \, a^{2} c^{2} x^{2} + 269 \, a c^{2} x - 1037 \, c^{2}\right )} \sqrt {-a c x + c}\right )}}{105 \, a}\right ] \]

[In]

integrate((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x, algorithm="fricas")

[Out]

[2/105*(420*sqrt(2)*c^(5/2)*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (15*a^3*c^2*x^
3 - 87*a^2*c^2*x^2 + 269*a*c^2*x - 1037*c^2)*sqrt(-a*c*x + c))/a, -2/105*(840*sqrt(2)*sqrt(-c)*c^2*arctan(1/2*
sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) - (15*a^3*c^2*x^3 - 87*a^2*c^2*x^2 + 269*a*c^2*x - 1037*c^2)*sqrt(-a*c*x
+ c))/a]

Sympy [A] (verification not implemented)

Time = 2.75 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.15 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\begin {cases} - \frac {2 \cdot \left (\frac {8 \sqrt {2} c^{4} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} + 8 c^{3} \sqrt {- a c x + c} + \frac {4 c^{2} \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {2 c \left (- a c x + c\right )^{\frac {5}{2}}}{5} + \frac {\left (- a c x + c\right )^{\frac {7}{2}}}{7}\right )}{a c} & \text {for}\: a c \neq 0 \\c^{\frac {5}{2}} \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {a x - 2 \log {\left (a x + 1 \right )} + 1}{a} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((-a*c*x+c)**(5/2)*(a*x-1)/(a*x+1),x)

[Out]

Piecewise((-2*(8*sqrt(2)*c**4*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/sqrt(-c) + 8*c**3*sqrt(-a*c*x + c) +
4*c**2*(-a*c*x + c)**(3/2)/3 + 2*c*(-a*c*x + c)**(5/2)/5 + (-a*c*x + c)**(7/2)/7)/(a*c), Ne(a*c, 0)), (c**(5/
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 = 109, normalized size of antiderivative = 0.94 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (420 \, \sqrt {2} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 15 \, {\left (-a c x + c\right )}^{\frac {7}{2}} + 42 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c + 140 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 840 \, \sqrt {-a c x + c} c^{3}\right )}}{105 \, a c} \]

[In]

integrate((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

-2/105*(420*sqrt(2)*c^(7/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c))) +
15*(-a*c*x + c)^(7/2) + 42*(-a*c*x + c)^(5/2)*c + 140*(-a*c*x + c)^(3/2)*c^2 + 840*sqrt(-a*c*x + c)*c^3)/(a*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.16 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {16 \, \sqrt {2} c^{3} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {2 \, {\left (15 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} a^{6} c^{6} - 42 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{6} c^{7} - 140 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{6} c^{8} - 840 \, \sqrt {-a c x + c} a^{6} c^{9}\right )}}{105 \, a^{7} c^{7}} \]

[In]

integrate((-a*c*x+c)^(5/2)*(a*x-1)/(a*x+1),x, algorithm="giac")

[Out]

-16*sqrt(2)*c^3*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a*sqrt(-c)) + 2/105*(15*(a*c*x - c)^3*sqrt(-a*c
*x + c)*a^6*c^6 - 42*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^6*c^7 - 140*(-a*c*x + c)^(3/2)*a^6*c^8 - 840*sqrt(-a*c*x
+ c)*a^6*c^9)/(a^7*c^7)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.82 \[ \int e^{-2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {4\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a}-\frac {8\,c\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a}-\frac {16\,c^2\,\sqrt {c-a\,c\,x}}{a}-\frac {2\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a\,c}-\frac {\sqrt {2}\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,16{}\mathrm {i}}{a} \]

[In]

int(((c - a*c*x)^(5/2)*(a*x - 1))/(a*x + 1),x)

[Out]

- (4*(c - a*c*x)^(5/2))/(5*a) - (8*c*(c - a*c*x)^(3/2))/(3*a) - (16*c^2*(c - a*c*x)^(1/2))/a - (2*(c - a*c*x)^
(7/2))/(7*a*c) - (2^(1/2)*c^(5/2)*atan((2^(1/2)*(c - a*c*x)^(1/2)*1i)/(2*c^(1/2)))*16i)/a