3.1201 \(\int e^{-\tanh ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=83 \[ \frac {x^{m+1} \sqrt {c-a^2 c x^2}}{(m+1) \sqrt {1-a^2 x^2}}-\frac {a x^{m+2} \sqrt {c-a^2 c x^2}}{(m+2) \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6153, 6150, 43} \[ \frac {x^{m+1} \sqrt {c-a^2 c x^2}}{(m+1) \sqrt {1-a^2 x^2}}-\frac {a x^{m+2} \sqrt {c-a^2 c x^2}}{(m+2) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Sqrt[c - a^2*c*x^2])/((1 + m)*Sqrt[1 - a^2*x^2]) - (a*x^(2 + m)*Sqrt[c - a^2*c*x^2])/((2 + m)*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^m \sqrt {c-a^2 c x^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int e^{-\tanh ^{-1}(a x)} x^m \sqrt {1-a^2 x^2} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int x^m (1-a x) \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \left (x^m-a x^{1+m}\right ) \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {x^{1+m} \sqrt {c-a^2 c x^2}}{(1+m) \sqrt {1-a^2 x^2}}-\frac {a x^{2+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 50, normalized size = 0.60 \[ \frac {x^{m+1} \sqrt {c-a^2 c x^2} \left (\frac {1}{m+1}-\frac {a x}{m+2}\right )}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 80, normalized size = 0.96 \[ \frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (a m + a\right )} x^{2} - {\left (m + 2\right )} x\right )} x^{m}}{{\left (a^{2} m^{2} + 3 \, a^{2} m + 2 \, a^{2}\right )} x^{2} - m^{2} - 3 \, m - 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{m}}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 68, normalized size = 0.82 \[ \frac {x^{1+m} \left (a m x +a x -m -2\right ) \sqrt {-a^{2} c \,x^{2}+c}\, \sqrt {-a^{2} x^{2}+1}}{\left (2+m \right ) \left (1+m \right ) \left (a x -1\right ) \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 63, normalized size = 0.76 \[ -\frac {{\left (a \sqrt {c} {\left (m + 1\right )} x^{2} - \sqrt {c} {\left (m + 2\right )} x\right )} {\left (a x + 1\right )} {\left (a x - 1\right )} x^{m}}{{\left (m^{2} + 3 \, m + 2\right )} a^{2} x^{2} - m^{2} - 3 \, m - 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.23, size = 52, normalized size = 0.63 \[ \frac {x\,x^m\,\sqrt {c-a^2\,c\,x^2}\,\left (m-a\,x-a\,m\,x+2\right )}{\sqrt {1-a^2\,x^2}\,\left (m^2+3\,m+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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