3.712 \(\int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=137 \[ \frac{1}{5} x^4 \sqrt{c-a^2 c x^2}-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{3 x^2 \sqrt{c-a^2 c x^2}}{5 a^2}+\frac{3 (8-5 a x) \sqrt{c-a^2 c x^2}}{20 a^4}+\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{4 a^4} \]

[Out]

(3*x^2*Sqrt[c - a^2*c*x^2])/(5*a^2) - (x^3*Sqrt[c - a^2*c*x^2])/(2*a) + (x^4*Sqrt[c - a^2*c*x^2])/5 + (3*(8 -
5*a*x)*Sqrt[c - a^2*c*x^2])/(20*a^4) + (3*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(4*a^4)

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Rubi [A]  time = 0.429526, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {6167, 6152, 1809, 833, 780, 217, 203} \[ \frac{1}{5} x^4 \sqrt{c-a^2 c x^2}-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{3 x^2 \sqrt{c-a^2 c x^2}}{5 a^2}+\frac{3 (8-5 a x) \sqrt{c-a^2 c x^2}}{20 a^4}+\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{4 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x^2*Sqrt[c - a^2*c*x^2])/(5*a^2) - (x^3*Sqrt[c - a^2*c*x^2])/(2*a) + (x^4*Sqrt[c - a^2*c*x^2])/5 + (3*(8 -
5*a*x)*Sqrt[c - a^2*c*x^2])/(20*a^4) + (3*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(4*a^4)

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 6152

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/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]) && ILtQ[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^3 \sqrt{c-a^2 c x^2} \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} x^3 \sqrt{c-a^2 c x^2} \, dx\\ &=-\left (c \int \frac{x^3 (1-a x)^2}{\sqrt{c-a^2 c x^2}} \, dx\right )\\ &=\frac{1}{5} x^4 \sqrt{c-a^2 c x^2}+\frac{\int \frac{x^3 \left (-9 a^2 c+10 a^3 c x\right )}{\sqrt{c-a^2 c x^2}} \, dx}{5 a^2}\\ &=-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{1}{5} x^4 \sqrt{c-a^2 c x^2}-\frac{\int \frac{x^2 \left (-30 a^3 c^2+36 a^4 c^2 x\right )}{\sqrt{c-a^2 c x^2}} \, dx}{20 a^4 c}\\ &=\frac{3 x^2 \sqrt{c-a^2 c x^2}}{5 a^2}-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{1}{5} x^4 \sqrt{c-a^2 c x^2}+\frac{\int \frac{x \left (-72 a^4 c^3+90 a^5 c^3 x\right )}{\sqrt{c-a^2 c x^2}} \, dx}{60 a^6 c^2}\\ &=\frac{3 x^2 \sqrt{c-a^2 c x^2}}{5 a^2}-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{1}{5} x^4 \sqrt{c-a^2 c x^2}+\frac{3 (8-5 a x) \sqrt{c-a^2 c x^2}}{20 a^4}+\frac{(3 c) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{4 a^3}\\ &=\frac{3 x^2 \sqrt{c-a^2 c x^2}}{5 a^2}-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{1}{5} x^4 \sqrt{c-a^2 c x^2}+\frac{3 (8-5 a x) \sqrt{c-a^2 c x^2}}{20 a^4}+\frac{(3 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 )}{4 a^3}\\ &=\frac{3 x^2 \sqrt{c-a^2 c x^2}}{5 a^2}-\frac{x^3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{1}{5} x^4 \sqrt{c-a^2 c x^2}+\frac{3 (8-5 a x) \sqrt{c-a^2 c x^2}}{20 a^4}+\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{4 a^4}\\ \end{align*}

Mathematica [A]  time = 0.127231, size = 96, normalized size = 0.7 \[ \frac{\left (4 a^4 x^4-10 a^3 x^3+12 a^2 x^2-15 a x+24\right ) \sqrt{c-a^2 c x^2}-15 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{20 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*(24 - 15*a*x + 12*a^2*x^2 - 10*a^3*x^3 + 4*a^4*x^4) - 15*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2
*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(20*a^4)

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Maple [A]  time = 0.055, size = 202, normalized size = 1.5 \begin{align*} -{\frac{{x}^{2}}{5\,{a}^{2}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{4}{5\,c{a}^{4}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{x}{2\,{a}^{3}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{5\,x}{4\,{a}^{3}}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{5\,c}{4\,{a}^{3}}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}+2\,{\frac{\sqrt{-{a}^{2}c \left ( x+{a}^{-1} \right ) ^{2}+2\, \left ( x+{a}^{-1} \right ) ac}}{{a}^{4}}}+2\,{\frac{c}{{a}^{3}\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-{a}^{2}c \left ( x+{a}^{-1} \right ) ^{2}+2\, \left ( x+{a}^{-1} \right ) ac}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*x^2*(-a^2*c*x^2+c)^(3/2)/a^2/c-4/5/c/a^4*(-a^2*c*x^2+c)^(3/2)+1/2/a^3*x*(-a^2*c*x^2+c)^(3/2)/c-5/4/a^3*x*
(-a^2*c*x^2+c)^(1/2)-5/4/a^3*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+2/a^4*(-a^2*c*(x+1/a
)^2+2*(x+1/a)*a*c)^(1/2)+2/a^3*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*(x+1/a)^2+2*(x+1/a)*a*c)^(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(x^3*(-a^2*c*x^2+c)^(1/2)*(a*x-1)/(a*x+1),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77678, size = 431, normalized size = 3.15 \begin{align*} \left [\frac{2 \,{\left (4 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 15 \, a x + 24\right )} \sqrt{-a^{2} c x^{2} + c} + 15 \, \sqrt{-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right )}{40 \, a^{4}}, \frac{{\left (4 \, a^{4} x^{4} - 10 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 15 \, a x + 24\right )} \sqrt{-a^{2} c x^{2} + c} - 15 \, \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right )}{20 \, a^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40*(2*(4*a^4*x^4 - 10*a^3*x^3 + 12*a^2*x^2 - 15*a*x + 24)*sqrt(-a^2*c*x^2 + c) + 15*sqrt(-c)*log(2*a^2*c*x^
2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c))/a^4, 1/20*((4*a^4*x^4 - 10*a^3*x^3 + 12*a^2*x^2 - 15*a*x + 24)*s
qrt(-a^2*c*x^2 + c) - 15*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)))/a^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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(x**3*(-a**2*c*x**2+c)**(1/2)*(a*x-1)/(a*x+1),x)

[Out]

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

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Giac [A]  time = 1.16753, size = 124, normalized size = 0.91 \begin{align*} \frac{1}{20} \, \sqrt{-a^{2} c x^{2} + c}{\left ({\left (2 \,{\left ({\left (2 \, x - \frac{5}{a}\right )} x + \frac{6}{a^{2}}\right )} x - \frac{15}{a^{3}}\right )} x + \frac{24}{a^{4}}\right )} - \frac{3 \, c \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{4 \, a^{3} \sqrt{-c}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*sqrt(-a^2*c*x^2 + c)*((2*((2*x - 5/a)*x + 6/a^2)*x - 15/a^3)*x + 24/a^4) - 3/4*c*log(abs(-sqrt(-a^2*c)*x
+ sqrt(-a^2*c*x^2 + c)))/(a^3*sqrt(-c)*abs(a))