3.949 \(\int e^{\tanh ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=74 \[ \frac {a x^4 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-a^2 x^2}}+\frac {x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6153, 6150, 43} \[ \frac {a x^4 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-a^2 x^2}}+\frac {x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^2*Sqrt[c - a^2*c*x^2],x]

[Out]

(x^3*Sqrt[c - a^2*c*x^2])/(3*Sqrt[1 - a^2*x^2]) + (a*x^4*Sqrt[c - a^2*c*x^2])/(4*Sqrt[1 - a^2*x^2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6150

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

Rule 6153

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int e^{\tanh ^{-1}(a x)} x^2 \sqrt {1-a^2 x^2} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int x^2 (1+a x) \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \left (x^2+a x^3\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}}+\frac {a x^4 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-a^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 42, normalized size = 0.57 \[ \frac {x^3 (3 a x+4) \sqrt {c-a^2 c x^2}}{12 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^2*Sqrt[c - a^2*c*x^2],x]

[Out]

(x^3*(4 + 3*a*x)*Sqrt[c - a^2*c*x^2])/(12*Sqrt[1 - a^2*x^2])

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 50, normalized size = 0.68 \[ -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (3 \, a x^{4} + 4 \, x^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{12 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(-a^2*c*x^2 + c)*(3*a*x^4 + 4*x^3)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2*(-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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.03, size = 37, normalized size = 0.50 \[ \frac {x^{3} \left (3 a x +4\right ) \sqrt {-a^{2} c \,x^{2}+c}}{12 \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^3*(3*a*x+4)*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

mupad [B]  time = 1.05, size = 38, normalized size = 0.51 \[ \frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {a\,x^4}{4}+\frac {x^3}{3}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________