### 3.886 $$\int e^{2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} x^3 \, dx$$

Optimal. Leaf size=160 $\frac{x^2 (a x+1)^2 \sqrt{c-\frac{c}{a^2 x^2}}}{4 a^2}+\frac{x (a x+1)^2 \sqrt{c-\frac{c}{a^2 x^2}}}{6 a^3}+\frac{7 x (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}{24 a^3}+\frac{7 x \sqrt{c-\frac{c}{a^2 x^2}}}{8 a^3}-\frac{7 x \sqrt{c-\frac{c}{a^2 x^2}} \sin ^{-1}(a x)}{8 a^3 \sqrt{1-a x} \sqrt{a x+1}}$

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.557536, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.296, Rules used = {6167, 6159, 6129, 90, 80, 50, 41, 216} $\frac{x^2 (a x+1)^2 \sqrt{c-\frac{c}{a^2 x^2}}}{4 a^2}+\frac{x (a x+1)^2 \sqrt{c-\frac{c}{a^2 x^2}}}{6 a^3}+\frac{7 x (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}{24 a^3}+\frac{7 x \sqrt{c-\frac{c}{a^2 x^2}}}{8 a^3}-\frac{7 x \sqrt{c-\frac{c}{a^2 x^2}} \sin ^{-1}(a x)}{8 a^3 \sqrt{1-a x} \sqrt{a x+1}}$

Antiderivative was successfully veriﬁed.

[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 6167

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]

Rule 6159

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*(1 - a*x)^p*(1 + a*x)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d
, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &&  !GtQ[c, 0]

Rule 6129

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 90

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 80

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 50

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 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 216

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]

Rubi steps

\begin{align*} \int e^{2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} x^3 \, dx &=-\int e^{2 \tanh ^{-1}(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 \tanh ^{-1}(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 \sin ^{-1}(a x)}{8 a^3 \sqrt{1-a x} \sqrt{1+a x}}\\ \end{align*}

Mathematica [A]  time = 0.0911736, size = 93, normalized size = 0.58 $\frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (\sqrt{a^2 x^2-1} \left (6 a^3 x^3+16 a^2 x^2+21 a x+32\right )+21 \log \left (\sqrt{a^2 x^2-1}+a x\right )\right )}{24 a^3 \sqrt{a^2 x^2-1}}$

Warning: Unable to verify antiderivative.

[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]  time = 0.191, size = 196, normalized size = 1.2 \begin{align*} -{\frac{x}{24\,c{a}^{4}}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}} \left ( -6\,x \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}} \right ) ^{3/2}{a}^{4}-16\, \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}} \right ) ^{3/2}{a}^{3}-27\,\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}x{a}^{2}c+27\,{c}^{3/2}\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) -48\,{c}^{3/2}\ln \left ({\frac{1}{\sqrt{c}} \left ( \sqrt{c}\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}+cx \right ) } \right ) -48\,\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}ac \right ){\frac{1}{\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(-6*x*(c*(a^2*x^2-1)/a^2)^(3/2)*a^4-16*(c*(a^2*x^2-1)/a^2)^(3/2)*a^3-27*
(c*(a^2*x^2-1)/a^2)^(1/2)*x*a^2*c+27*c^(3/2)*ln(x*c^(1/2)+(c*(a^2*x^2-1)/a^2)^(1/2))-48*c^(3/2)*ln((c^(1/2)*((
a*x-1)*(a*x+1)*c/a^2)^(1/2)+c*x)/c^(1/2))-48*((a*x-1)*(a*x+1)*c/a^2)^(1/2)*a*c)/(c*(a^2*x^2-1)/a^2)^(1/2)/c/a^
4

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} \sqrt{c - \frac{c}{a^{2} x^{2}}} x^{3}}{a x - 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 1.6971, size = 487, normalized size = 3.04 \begin{align*} \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 ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]  time = 1.15859, size = 173, normalized size = 1.08 \begin{align*} \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 |} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)