3.7.73 \(\int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx\) [673]

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

[Out]

-3/4*arctan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))*c^(1/2)/a^4+3/5*x^2*(-a^2*c*x^2+c)^(1/2)/a^2+1/2*x^3*(-a^2*c*x^2
+c)^(1/2)/a+1/5*x^4*(-a^2*c*x^2+c)^(1/2)+3/20*(5*a*x+8)*(-a^2*c*x^2+c)^(1/2)/a^4

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Rubi [A]
time = 0.29, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {6302, 6286, 1823, 847, 794, 223, 209} \begin {gather*} \frac {3 x^2 \sqrt {c-a^2 c x^2}}{5 a^2}+\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 \sqrt {c} \text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^4}+\frac {3 (5 a x+8) \sqrt {c-a^2 c x^2}}{20 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])*x^3*Sqrt[c - a^2*c*x^2],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 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 794

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 847

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 1823

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 6286

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 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*} \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) \text {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*}

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Mathematica [A]
time = 0.15, size = 96, normalized size = 0.70 \begin {gather*} \frac {\sqrt {c-a^2 c x^2} \left (24+15 a x+12 a^2 x^2+10 a^3 x^3+4 a^4 x^4\right )+15 \sqrt {c} \text {ArcTan}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{20 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])*x^3*Sqrt[c - a^2*c*x^2],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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(269\) vs. \(2(113)=226\).
time = 0.21, size = 270, normalized size = 1.97

method	result	size
risch	\(-\frac {\left (4 a^{4} x^{4}+10 a^{3} x^{3}+12 a^{2} x^{2}+15 a x +24\right ) \left (a^{2} x^{2}-1\right ) c}{20 a^{4} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c}{4 a^{3} \sqrt {a^{2} c}}\)	\(97\)
default	\(-\frac {x^{2} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{5 a^{2} c}-\frac {4 \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{5 c \,a^{4}}+\frac {-\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 a^{2} c}+\frac {\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}}{2 a^{2}}}{a}+\frac {x \sqrt {-a^{2} c \,x^{2}+c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}}{a^{3}}+\frac {2 \sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}-\frac {2 a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}}{a^{4}}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)+2/a*(-1/4*x*(-a^2*c*x^2+c)^(3/2)/a^2/c+1/4/
a^2*(1/2*x*(-a^2*c*x^2+c)^(1/2)+1/2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))))+2/a^3*(1/2*
x*(-a^2*c*x^2+c)^(1/2)+1/2*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)-a*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 [A]
time = 0.47, size = 117, normalized size = 0.85 \begin {gather*} -\frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{2}}{5 \, a^{2} c} + \frac {5 \, \sqrt {-a^{2} c x^{2} + c} x}{4 \, a^{3}} - \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{2 \, a^{3} c} - \frac {3 \, \sqrt {c} \arcsin \left (a x\right )}{4 \, a^{4}} + \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{4}} - \frac {4 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{5 \, a^{4} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 184, normalized size = 1.34 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*x^3*(-a^2*c*x^2+c)^(1/2),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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*x**3*(-a**2*c*x**2+c)**(1/2),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 [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\sqrt {c-a^2\,c\,x^2}\,\left (a\,x+1\right )}{a\,x-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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