\(\int e^{-2 \coth ^{-1}(a x)} (c-a^2 c x^2)^{5/2} \, dx\) [651]

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}-\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}-\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {7 c^{5/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a} \]

[Out]

-7/24*c*x*(-a^2*c*x^2+c)^(3/2)-7/30*(-a^2*c*x^2+c)^(5/2)/a-1/6*(-a*x+1)*(-a^2*c*x^2+c)^(5/2)/a-7/16*c^(5/2)*ar
ctan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))/a-7/16*c^2*x*(-a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6302, 6277, 685, 655, 201, 223, 209} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {7 c^{5/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a}-\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}-\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a} \]

[In]

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

[Out]

(-7*c^2*x*Sqrt[c - a^2*c*x^2])/16 - (7*c*x*(c - a^2*c*x^2)^(3/2))/24 - (7*(c - a^2*c*x^2)^(5/2))/(30*a) - ((1
- a*x)*(c - a^2*c*x^2)^(5/2))/(6*a) - (7*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(16*a)

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*((m + p)/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rule 6277

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/c^(n/2), Int[(c + d*x^2)^(p
+ n/2)/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) && I
LtQ[n/2, 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)} \left (c-a^2 c x^2\right )^{5/2} \, dx \\ & = -\left (c \int (1-a x)^2 \left (c-a^2 c x^2\right )^{3/2} \, dx\right ) \\ & = -\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {1}{6} (7 c) \int (1-a x) \left (c-a^2 c x^2\right )^{3/2} \, dx \\ & = -\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {1}{6} (7 c) \int \left (c-a^2 c x^2\right )^{3/2} \, dx \\ & = -\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}-\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {1}{8} \left (7 c^2\right ) \int \sqrt {c-a^2 c x^2} \, dx \\ & = -\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}-\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}-\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {1}{16} \left (7 c^3\right ) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx \\ & = -\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}-\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}-\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {1}{16} \left (7 c^3\right ) \text {Subst}\left (\int \frac {1}{1+a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c-a^2 c x^2}}\right ) \\ & = -\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}-\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}-\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}-\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}-\frac {7 c^{5/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.04 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 \sqrt {c-a^2 c x^2} \left (-\sqrt {1+a x} \left (96+39 a x-327 a^2 x^2+202 a^3 x^3+86 a^4 x^4-136 a^5 x^5+40 a^6 x^6\right )+210 \sqrt {1-a x} \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{240 a \sqrt {1-a x} \sqrt {1-a^2 x^2}} \]

[In]

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

[Out]

(c^2*Sqrt[c - a^2*c*x^2]*(-(Sqrt[1 + a*x]*(96 + 39*a*x - 327*a^2*x^2 + 202*a^3*x^3 + 86*a^4*x^4 - 136*a^5*x^5
+ 40*a^6*x^6)) + 210*Sqrt[1 - a*x]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(240*a*Sqrt[1 - a*x]*Sqrt[1 - a^2*x^2])

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\left (40 a^{5} x^{5}-96 a^{4} x^{4}-10 a^{3} x^{3}+192 a^{2} x^{2}-135 a x -96\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{240 a \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {7 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c^{3}}{16 \sqrt {a^{2} c}}\)	\(106\)
default	\(\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )}{6}-\frac {2 \left (\frac {\left (-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}{5}+a c \left (-\frac {\left (-2 a^{2} c \left (x +\frac {1}{a}\right )+2 a c \right ) \left (-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}{8 a^{2} c}+\frac {3 c \left (-\frac {\left (-2 a^{2} c \left (x +\frac {1}{a}\right )+2 a c \right ) \sqrt {-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c}}{4 a^{2} c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}\right )\right )}{a}\)	\(275\)

[In]

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

[Out]

-1/240*(40*a^5*x^5-96*a^4*x^4-10*a^3*x^3+192*a^2*x^2-135*a*x-96)*(a^2*x^2-1)/a/(-c*(a^2*x^2-1))^(1/2)*c^3-7/16
/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))*c^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.84 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\left [\frac {105 \, \sqrt {-c} c^{2} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + 2 \, {\left (40 \, a^{5} c^{2} x^{5} - 96 \, a^{4} c^{2} x^{4} - 10 \, a^{3} c^{2} x^{3} + 192 \, a^{2} c^{2} x^{2} - 135 \, a c^{2} x - 96 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{480 \, a}, \frac {105 \, c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + {\left (40 \, a^{5} c^{2} x^{5} - 96 \, a^{4} c^{2} x^{4} - 10 \, a^{3} c^{2} x^{3} + 192 \, a^{2} c^{2} x^{2} - 135 \, a c^{2} x - 96 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{240 \, a}\right ] \]

[In]

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

[Out]

[1/480*(105*sqrt(-c)*c^2*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) + 2*(40*a^5*c^2*x^5 - 96*a
^4*c^2*x^4 - 10*a^3*c^2*x^3 + 192*a^2*c^2*x^2 - 135*a*c^2*x - 96*c^2)*sqrt(-a^2*c*x^2 + c))/a, 1/240*(105*c^(5
/2)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + (40*a^5*c^2*x^5 - 96*a^4*c^2*x^4 - 10*a^3*c^2*x
^3 + 192*a^2*c^2*x^2 - 135*a*c^2*x - 96*c^2)*sqrt(-a^2*c*x^2 + c))/a]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (119) = 238\).

Time = 2.74 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.40 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\begin {cases} \frac {2 c^{2} \left (\begin {cases} \left (\frac {a^{2} x^{2}}{3} - \frac {1}{3}\right ) \sqrt {- a^{2} c x^{2} + c} & \text {for}\: c \neq 0 \\\frac {a^{2} \sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) - 2 c^{2} \left (\begin {cases} \sqrt {- a^{2} c x^{2} + c} \left (\frac {a^{4} x^{4}}{5} - \frac {a^{2} x^{2}}{15} - \frac {2}{15}\right ) & \text {for}\: c \neq 0 \\\frac {a^{4} \sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a c x + 2 \sqrt {- c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- c}} & \text {for}\: c \neq 0 \\\frac {a x \log {\left (a x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{16} + \sqrt {- a^{2} c x^{2} + c} \left (\frac {a^{5} x^{5}}{6} - \frac {a^{3} x^{3}}{24} - \frac {a x}{16}\right ) & \text {for}\: c \neq 0 \\\frac {a^{5} \sqrt {c} x^{5}}{5} & \text {otherwise} \end {cases}\right ) - c^{2} \left (\begin {cases} \frac {a x \sqrt {- a^{2} c x^{2} + c}}{2} + \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a c x + 2 \sqrt {- c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- c}} & \text {for}\: c \neq 0 \\\frac {a x \log {\left (a x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{2} & \text {for}\: c \neq 0 \\a \sqrt {c} x & \text {otherwise} \end {cases}\right )}{a} & \text {for}\: a \neq 0 \\- c^{\frac {5}{2}} x & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((2*c**2*Piecewise(((a**2*x**2/3 - 1/3)*sqrt(-a**2*c*x**2 + c), Ne(c, 0)), (a**2*sqrt(c)*x**2/2, Tru
e)) - 2*c**2*Piecewise((sqrt(-a**2*c*x**2 + c)*(a**4*x**4/5 - a**2*x**2/15 - 2/15), Ne(c, 0)), (a**4*sqrt(c)*x
**4/4, True)) + c**2*Piecewise((c*Piecewise((log(-2*a*c*x + 2*sqrt(-c)*sqrt(-a**2*c*x**2 + c))/sqrt(-c), Ne(c,
0)), (a*x*log(a*x)/sqrt(-a**2*c*x**2), True))/16 + sqrt(-a**2*c*x**2 + c)*(a**5*x**5/6 - a**3*x**3/24 - a*x/1
6), Ne(c, 0)), (a**5*sqrt(c)*x**5/5, True)) - c**2*Piecewise((a*x*sqrt(-a**2*c*x**2 + c)/2 + c*Piecewise((log(
-2*a*c*x + 2*sqrt(-c)*sqrt(-a**2*c*x**2 + c))/sqrt(-c), Ne(c, 0)), (a*x*log(a*x)/sqrt(-a**2*c*x**2), True))/2,
Ne(c, 0)), (a*sqrt(c)*x, True)))/a, Ne(a, 0)), (-c**(5/2)*x, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.18 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x - \frac {7}{24} \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c x - \frac {3}{4} \, \sqrt {a^{2} c x^{2} + 4 \, a c x + 3 \, c} c^{2} x + \frac {5}{16} \, \sqrt {-a^{2} c x^{2} + c} c^{2} x + \frac {3 \, c^{4} \arcsin \left (a x + 2\right )}{4 \, a \left (-c\right )^{\frac {3}{2}}} + \frac {5 \, c^{\frac {5}{2}} \arcsin \left (a x\right )}{16 \, a} - \frac {2 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{5 \, a} - \frac {3 \, \sqrt {a^{2} c x^{2} + 4 \, a c x + 3 \, c} c^{2}}{2 \, a} \]

[In]

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

[Out]

1/6*(-a^2*c*x^2 + c)^(5/2)*x - 7/24*(-a^2*c*x^2 + c)^(3/2)*c*x - 3/4*sqrt(a^2*c*x^2 + 4*a*c*x + 3*c)*c^2*x + 5
/16*sqrt(-a^2*c*x^2 + c)*c^2*x + 3/4*c^4*arcsin(a*x + 2)/(a*(-c)^(3/2)) + 5/16*c^(5/2)*arcsin(a*x)/a - 2/5*(-a
^2*c*x^2 + c)^(5/2)/a - 3/2*sqrt(a^2*c*x^2 + 4*a*c*x + 3*c)*c^2/a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {7 \, c^{3} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{16 \, \sqrt {-c} {\left | a \right |}} - \frac {1}{240} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (135 \, c^{2} - 2 \, {\left (96 \, a c^{2} - {\left (5 \, a^{2} c^{2} - 4 \, {\left (5 \, a^{4} c^{2} x - 12 \, a^{3} c^{2}\right )} x\right )} x\right )} x\right )} x + \frac {96 \, c^{2}}{a}\right )} \]

[In]

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

[Out]

7/16*c^3*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(sqrt(-c)*abs(a)) - 1/240*sqrt(-a^2*c*x^2 + c)*((135
*c^2 - 2*(96*a*c^2 - (5*a^2*c^2 - 4*(5*a^4*c^2*x - 12*a^3*c^2)*x)*x)*x)*x + 96*c^2/a)

Mupad [F(-1)]

Timed out. \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\left (a\,x-1\right )}{a\,x+1} \,d x \]

[In]

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

[Out]

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