3.999 \(\int e^{\tanh ^{-1}(a x)} x^m (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=174 \[ \frac{c x^{m+1} \sqrt{c-a^2 c x^2}}{(m+1) \sqrt{1-a^2 x^2}}+\frac{a c x^{m+2} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-a^2 x^2}}-\frac{a^2 c x^{m+3} \sqrt{c-a^2 c x^2}}{(m+3) \sqrt{1-a^2 x^2}}-\frac{a^3 c x^{m+4} \sqrt{c-a^2 c x^2}}{(m+4) \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.2033, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 75} \[ \frac{c x^{m+1} \sqrt{c-a^2 c x^2}}{(m+1) \sqrt{1-a^2 x^2}}+\frac{a c x^{m+2} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-a^2 x^2}}-\frac{a^2 c x^{m+3} \sqrt{c-a^2 c x^2}}{(m+3) \sqrt{1-a^2 x^2}}-\frac{a^3 c x^{m+4} \sqrt{c-a^2 c x^2}}{(m+4) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.0903083, size = 84, normalized size = 0.48 \[ \frac{c x^{m+1} \sqrt{c-a^2 c x^2} \left ((2 m+5) \left (\frac{a^2 x^2}{m+3}+\frac{2 a x}{m+2}+\frac{1}{m+1}\right )-(a x+1)^3\right )}{(m+4) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^(1 + m)*Sqrt[c - a^2*c*x^2]*(-(1 + a*x)^3 + (5 + 2*m)*((1 + m)^(-1) + (2*a*x)/(2 + m) + (a^2*x^2)/(3 + m)
)))/((4 + m)*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.03, size = 180, normalized size = 1. \begin{align*}{\frac{{x}^{1+m} \left ({a}^{3}{m}^{3}{x}^{3}+6\,{a}^{3}{m}^{2}{x}^{3}+11\,{a}^{3}m{x}^{3}+{a}^{2}{m}^{3}{x}^{2}+6\,{x}^{3}{a}^{3}+7\,{a}^{2}{m}^{2}{x}^{2}+14\,{a}^{2}m{x}^{2}-a{m}^{3}x+8\,{a}^{2}{x}^{2}-8\,a{m}^{2}x-19\,amx-{m}^{3}-12\,ax-9\,{m}^{2}-26\,m-24 \right ) }{ \left ( ax+1 \right ) \left ( ax-1 \right ) \left ( 4+m \right ) \left ( 3+m \right ) \left ( 2+m \right ) \left ( 1+m \right ) } \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1+m)*(a^3*m^3*x^3+6*a^3*m^2*x^3+11*a^3*m*x^3+a^2*m^3*x^2+6*a^3*x^3+7*a^2*m^2*x^2+14*a^2*m*x^2-a*m^3*x+8*a^2
*x^2-8*a*m^2*x-19*a*m*x-m^3-12*a*x-9*m^2-26*m-24)*(-a^2*c*x^2+c)^(3/2)/(4+m)/(3+m)/(2+m)/(1+m)/(a*x-1)/(a*x+1)
/(-a^2*x^2+1)^(1/2)

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Maxima [A]  time = 1.00073, size = 108, normalized size = 0.62 \begin{align*} -\frac{{\left (a^{2} c^{\frac{3}{2}}{\left (m + 2\right )} x^{4} - c^{\frac{3}{2}}{\left (m + 4\right )} x^{2}\right )} a x^{m}}{m^{2} + 6 \, m + 8} - \frac{{\left (a^{2} c^{\frac{3}{2}}{\left (m + 1\right )} x^{3} - c^{\frac{3}{2}}{\left (m + 3\right )} x\right )} x^{m}}{m^{2} + 4 \, m + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*c^(3/2)*(m + 2)*x^4 - c^(3/2)*(m + 4)*x^2)*a*x^m/(m^2 + 6*m + 8) - (a^2*c^(3/2)*(m + 1)*x^3 - c^(3/2)*(m
+ 3)*x)*x^m/(m^2 + 4*m + 3)

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Fricas [A]  time = 2.37073, size = 459, normalized size = 2.64 \begin{align*} -\frac{\sqrt{-a^{2} c x^{2} + c}{\left ({\left (a^{3} c m^{3} + 6 \, a^{3} c m^{2} + 11 \, a^{3} c m + 6 \, a^{3} c\right )} x^{4} +{\left (a^{2} c m^{3} + 7 \, a^{2} c m^{2} + 14 \, a^{2} c m + 8 \, a^{2} c\right )} x^{3} -{\left (a c m^{3} + 8 \, a c m^{2} + 19 \, a c m + 12 \, a c\right )} x^{2} -{\left (c m^{3} + 9 \, c m^{2} + 26 \, c m + 24 \, c\right )} x\right )} \sqrt{-a^{2} x^{2} + 1} x^{m}}{m^{4} + 10 \, m^{3} -{\left (a^{2} m^{4} + 10 \, a^{2} m^{3} + 35 \, a^{2} m^{2} + 50 \, a^{2} m + 24 \, a^{2}\right )} x^{2} + 35 \, m^{2} + 50 \, m + 24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-a^2*c*x^2 + c)*((a^3*c*m^3 + 6*a^3*c*m^2 + 11*a^3*c*m + 6*a^3*c)*x^4 + (a^2*c*m^3 + 7*a^2*c*m^2 + 14*a^
2*c*m + 8*a^2*c)*x^3 - (a*c*m^3 + 8*a*c*m^2 + 19*a*c*m + 12*a*c)*x^2 - (c*m^3 + 9*c*m^2 + 26*c*m + 24*c)*x)*sq
rt(-a^2*x^2 + 1)*x^m/(m^4 + 10*m^3 - (a^2*m^4 + 10*a^2*m^3 + 35*a^2*m^2 + 50*a^2*m + 24*a^2)*x^2 + 35*m^2 + 50
*m + 24)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}}{\left (a x + 1\right )} x^{m}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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