### 3.674 $$\int e^{2 \coth ^{-1}(a x)} x^2 \sqrt{c-a^2 c x^2} \, dx$$

Optimal. Leaf size=112 $\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}+\frac{2 x^2 \sqrt{c-a^2 c x^2}}{3 a}+\frac{(21 a x+32) \sqrt{c-a^2 c x^2}}{24 a^3}-\frac{7 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{8 a^3}$

[Out]

(2*x^2*Sqrt[c - a^2*c*x^2])/(3*a) + (x^3*Sqrt[c - a^2*c*x^2])/4 + ((32 + 21*a*x)*Sqrt[c - a^2*c*x^2])/(24*a^3)
- (7*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(8*a^3)

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Rubi [A]  time = 0.387016, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.259, Rules used = {6167, 6151, 1809, 833, 780, 217, 203} $\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}+\frac{2 x^2 \sqrt{c-a^2 c x^2}}{3 a}+\frac{(21 a x+32) \sqrt{c-a^2 c x^2}}{24 a^3}-\frac{7 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{8 a^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*x^2*Sqrt[c - a^2*c*x^2])/(3*a) + (x^3*Sqrt[c - a^2*c*x^2])/4 + ((32 + 21*a*x)*Sqrt[c - a^2*c*x^2])/(24*a^3)
- (7*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(8*a^3)

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 6151

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 833

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 217

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 203

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

Rubi steps

\begin{align*} \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt{c-a^2 c x^2} \, dx &=-\int e^{2 \tanh ^{-1}(a x)} x^2 \sqrt{c-a^2 c x^2} \, dx\\ &=-\left (c \int \frac{x^2 (1+a x)^2}{\sqrt{c-a^2 c x^2}} \, dx\right )\\ &=\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}+\frac{\int \frac{x^2 \left (-7 a^2 c-8 a^3 c x\right )}{\sqrt{c-a^2 c x^2}} \, dx}{4 a^2}\\ &=\frac{2 x^2 \sqrt{c-a^2 c x^2}}{3 a}+\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}-\frac{\int \frac{x \left (16 a^3 c^2+21 a^4 c^2 x\right )}{\sqrt{c-a^2 c x^2}} \, dx}{12 a^4 c}\\ &=\frac{2 x^2 \sqrt{c-a^2 c x^2}}{3 a}+\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}+\frac{(32+21 a x) \sqrt{c-a^2 c x^2}}{24 a^3}-\frac{(7 c) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{8 a^2}\\ &=\frac{2 x^2 \sqrt{c-a^2 c x^2}}{3 a}+\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}+\frac{(32+21 a x) \sqrt{c-a^2 c x^2}}{24 a^3}-\frac{(7 c) \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )}{8 a^2}\\ &=\frac{2 x^2 \sqrt{c-a^2 c x^2}}{3 a}+\frac{1}{4} x^3 \sqrt{c-a^2 c x^2}+\frac{(32+21 a x) \sqrt{c-a^2 c x^2}}{24 a^3}-\frac{7 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{8 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0953186, size = 88, normalized size = 0.79 $\frac{\left (6 a^3 x^3+16 a^2 x^2+21 a x+32\right ) \sqrt{c-a^2 c x^2}+21 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{24 a^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.052, size = 186, normalized size = 1.7 \begin{align*} -{\frac{x}{4\,{a}^{2}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{9\,x}{8\,{a}^{2}}\sqrt{-{a}^{2}c{x}^{2}+c}}+{\frac{9\,c}{8\,{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-{\frac{2}{3\,{a}^{3}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{1}{{a}^{3}}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}-2\,{\frac{c}{{a}^{2}\sqrt{{a}^{2}c}}\arctan \left ({\sqrt{{a}^{2}c}x{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*(-a^2*c*x^2+c)^(3/2)/a^2/c+9/8/a^2*x*(-a^2*c*x^2+c)^(1/2)+9/8/a^2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*
x/(-a^2*c*x^2+c)^(1/2))-2/3/a^3*(-a^2*c*x^2+c)^(3/2)/c+2/a^3*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)-2/a^2*c/(a
^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68948, size = 396, normalized size = 3.54 \begin{align*} \left [\frac{2 \,{\left (6 \, a^{3} x^{3} + 16 \, a^{2} x^{2} + 21 \, a x + 32\right )} \sqrt{-a^{2} c x^{2} + c} + 21 \, \sqrt{-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right )}{48 \, a^{3}}, \frac{{\left (6 \, a^{3} x^{3} + 16 \, a^{2} x^{2} + 21 \, a x + 32\right )} \sqrt{-a^{2} c x^{2} + c} + 21 \, \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right )}{24 \, a^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(x**2*sqrt(-c*(a*x - 1)*(a*x + 1))*(a*x + 1)/(a*x - 1), x)

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Giac [A]  time = 1.13488, size = 113, normalized size = 1.01 \begin{align*} \frac{1}{24} \, \sqrt{-a^{2} c x^{2} + c}{\left ({\left (2 \,{\left (3 \, x + \frac{8}{a}\right )} x + \frac{21}{a^{2}}\right )} x + \frac{32}{a^{3}}\right )} + \frac{7 \, c \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{8 \, a^{2} \sqrt{-c}{\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^2*(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(-a^2*c*x^2 + c)*((2*(3*x + 8/a)*x + 21/a^2)*x + 32/a^3) + 7/8*c*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*
c*x^2 + c)))/(a^2*sqrt(-c)*abs(a))