\(\int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx\) [886]

Optimal result

Integrand size = 27, antiderivative size = 160 \[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x}{8 a^3}+\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)}{24 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2}{6 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}-\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x \arcsin (a x)}{8 a^3 \sqrt {1-a x} \sqrt {1+a x}} \]

[Out]

7/8*x*(c-c/a^2/x^2)^(1/2)/a^3+7/24*x*(a*x+1)*(c-c/a^2/x^2)^(1/2)/a^3+1/6*x*(a*x+1)^2*(c-c/a^2/x^2)^(1/2)/a^3+1
/4*x^2*(a*x+1)^2*(c-c/a^2/x^2)^(1/2)/a^2-7/8*x*arcsin(a*x)*(c-c/a^2/x^2)^(1/2)/a^3/(-a*x+1)^(1/2)/(a*x+1)^(1/2
)

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6302, 6294, 6264, 92, 81, 52, 41, 222} \[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {x^2 (a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^2}-\frac {7 x \arcsin (a x) \sqrt {c-\frac {c}{a^2 x^2}}}{8 a^3 \sqrt {1-a x} \sqrt {a x+1}}+\frac {x (a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}{6 a^3}+\frac {7 x (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}{24 a^3}+\frac {7 x \sqrt {c-\frac {c}{a^2 x^2}}}{8 a^3} \]

[In]

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

[Out]

(7*Sqrt[c - c/(a^2*x^2)]*x)/(8*a^3) + (7*Sqrt[c - c/(a^2*x^2)]*x*(1 + a*x))/(24*a^3) + (Sqrt[c - c/(a^2*x^2)]*
x*(1 + a*x)^2)/(6*a^3) + (Sqrt[c - c/(a^2*x^2)]*x^2*(1 + a*x)^2)/(4*a^2) - (7*Sqrt[c - c/(a^2*x^2)]*x*ArcSin[a
*x])/(8*a^3*Sqrt[1 - a*x]*Sqrt[1 + a*x])

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 222

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]

Rule 6264

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[u*(1 + d*(x/c))^
p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 6294

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[x^(2*p)*((c + d/x^2)^p/((
1 - a*x)^p*(1 + a*x)^p)), Int[(u/x^(2*p))*(1 - a*x)^p*(1 + a*x)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d
, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &&  !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)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx \\ & = -\frac {\left (\sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {1-a x} \sqrt {1+a x} \, dx}{\sqrt {1-a x} \sqrt {1+a x}} \\ & = -\frac {\left (\sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \frac {x^2 (1+a x)^{3/2}}{\sqrt {1-a x}} \, dx}{\sqrt {1-a x} \sqrt {1+a x}} \\ & = \frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}+\frac {\left (\sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \frac {(-1-2 a x) (1+a x)^{3/2}}{\sqrt {1-a x}} \, dx}{4 a^2 \sqrt {1-a x} \sqrt {1+a x}} \\ & = \frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2}{6 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}-\frac {\left (7 \sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \frac {(1+a x)^{3/2}}{\sqrt {1-a x}} \, dx}{12 a^2 \sqrt {1-a x} \sqrt {1+a x}} \\ & = \frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)}{24 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2}{6 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}-\frac {\left (7 \sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \frac {\sqrt {1+a x}}{\sqrt {1-a x}} \, dx}{8 a^2 \sqrt {1-a x} \sqrt {1+a x}} \\ & = \frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x}{8 a^3}+\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)}{24 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2}{6 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}-\frac {\left (7 \sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \frac {1}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{8 a^2 \sqrt {1-a x} \sqrt {1+a x}} \\ & = \frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x}{8 a^3}+\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)}{24 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2}{6 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}-\frac {\left (7 \sqrt {c-\frac {c}{a^2 x^2}} x\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2 \sqrt {1-a x} \sqrt {1+a x}} \\ & = \frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x}{8 a^3}+\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)}{24 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2}{6 a^3}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2 (1+a x)^2}{4 a^2}-\frac {7 \sqrt {c-\frac {c}{a^2 x^2}} x \arcsin (a x)}{8 a^3 \sqrt {1-a x} \sqrt {1+a x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.58 \[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (\sqrt {-1+a^2 x^2} \left (32+21 a x+16 a^2 x^2+6 a^3 x^3\right )+21 \log \left (a x+\sqrt {-1+a^2 x^2}\right )\right )}{24 a^3 \sqrt {-1+a^2 x^2}} \]

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*(Sqrt[-1 + a^2*x^2]*(32 + 21*a*x + 16*a^2*x^2 + 6*a^3*x^3) + 21*Log[a*x + Sqrt[-1 + a
^2*x^2]]))/(24*a^3*Sqrt[-1 + a^2*x^2])

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {\left (6 a^{3} x^{3}+16 a^{2} x^{2}+21 a x +32\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x}{24 a^{3}}+\frac {7 \ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c}\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {c \left (a^{2} x^{2}-1\right )}\, x}{8 a^{2} \sqrt {a^{2} c}\, \left (a^{2} x^{2}-1\right )}\)	\(134\)
default	\(-\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \left (-6 x {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{4}-16 {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{3}-27 \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2} c x +27 c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right )-48 c^{\frac {3}{2}} \ln \left (\frac {\sqrt {c}\, \sqrt {\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}}+c x}{\sqrt {c}}\right )-48 \sqrt {\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}}\, a c \right )}{24 \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, c \,a^{4}}\)	\(196\)

[In]

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

[Out]

1/24*(6*a^3*x^3+16*a^2*x^2+21*a*x+32)/a^3*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x+7/8/a^2*ln(a^2*c*x/(a^2*c)^(1/2)+(a^
2*c*x^2-c)^(1/2))/(a^2*c)^(1/2)*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*(c*(a^2*x^2-1))^(1/2)/(a^2*x^2-1)*x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.39 \[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\left [\frac {2 \, {\left (6 \, a^{4} x^{4} + 16 \, a^{3} x^{3} + 21 \, a^{2} x^{2} + 32 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} + 21 \, \sqrt {c} \log \left (2 \, a^{2} c x^{2} + 2 \, a^{2} \sqrt {c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right )}{48 \, a^{4}}, \frac {{\left (6 \, a^{4} x^{4} + 16 \, a^{3} x^{3} + 21 \, a^{2} x^{2} + 32 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 21 \, \sqrt {-c} \arctan \left (\frac {a^{2} \sqrt {-c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right )}{24 \, a^{4}}\right ] \]

[In]

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

[Out]

[1/48*(2*(6*a^4*x^4 + 16*a^3*x^3 + 21*a^2*x^2 + 32*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) + 21*sqrt(c)*log(2*a^2
*c*x^2 + 2*a^2*sqrt(c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - c))/a^4, 1/24*((6*a^4*x^4 + 16*a^3*x^3 + 21*a^2*x
^2 + 32*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 21*sqrt(-c)*arctan(a^2*sqrt(-c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x
^2))/(a^2*c*x^2 - c)))/a^4]

Sympy [F]

\[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int \frac {x^{3} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{a x - 1}\, dx \]

[In]

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

[Out]

Integral(x**3*sqrt(-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))*(a*x + 1)/(a*x - 1), x)

Maxima [F]

\[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a^{2} x^{2}}} x^{3}}{a x - 1} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {1}{48} \, {\left (2 \, \sqrt {a^{2} c x^{2} - c} {\left ({\left (2 \, x {\left (\frac {3 \, x \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {8 \, \mathrm {sgn}\left (x\right )}{a^{3}}\right )} + \frac {21 \, \mathrm {sgn}\left (x\right )}{a^{4}}\right )} x + \frac {32 \, \mathrm {sgn}\left (x\right )}{a^{5}}\right )} - \frac {42 \, \sqrt {c} \log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} - c} \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{4} {\left | a \right |}} + \frac {{\left (21 \, a \sqrt {c} \log \left ({\left | c \right |}\right ) - 64 \, \sqrt {-c} {\left | a \right |}\right )} \mathrm {sgn}\left (x\right )}{a^{5} {\left | a \right |}}\right )} {\left | a \right |} \]

[In]

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

[Out]

1/48*(2*sqrt(a^2*c*x^2 - c)*((2*x*(3*x*sgn(x)/a^2 + 8*sgn(x)/a^3) + 21*sgn(x)/a^4)*x + 32*sgn(x)/a^5) - 42*sqr
t(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 - c)))*sgn(x)/(a^4*abs(a)) + (21*a*sqrt(c)*log(abs(c)) - 64*sqrt(
-c)*abs(a))*sgn(x)/(a^5*abs(a)))*abs(a)

Mupad [F(-1)]

Timed out. \[ \int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int \frac {x^3\,\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (a\,x+1\right )}{a\,x-1} \,d x \]

[In]

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

[Out]

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