3.10.100 \(\int e^{\tanh ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx\) [1000]

Optimal. Leaf size=82 \[ \frac {x^{1+m} \sqrt {c-a^2 c x^2}}{(1+m) \sqrt {1-a^2 x^2}}+\frac {a x^{2+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-a^2 x^2}} \]

[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 = 0.12, antiderivative size = 82, 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 = {6288, 6285, 45} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^m*Sqrt[c - a^2*c*x^2],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 45

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 6285

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 6288

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 = 0.02, size = 49, normalized size = 0.60 \begin {gather*} \frac {x^{1+m} \left (\frac {1}{1+m}+\frac {a x}{2+m}\right ) \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^m*Sqrt[c - a^2*c*x^2],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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Maple [A]
time = 0.06, size = 52, normalized size = 0.63

method	result	size
gosper	\(\frac {x^{1+m} \left (a m x +a x +m +2\right ) \sqrt {-a^{2} c \,x^{2}+c}}{\left (2+m \right ) \left (1+m \right ) \sqrt {-a^{2} x^{2}+1}}\)	\(52\)
risch	\(-\frac {\sqrt {-\frac {c \left (-a^{2} x^{2}+1\right )}{a^{2} x^{2}-1}}\, \left (a^{2} x^{2}-1\right ) \sqrt {c}\, \left (a m x +a x +m +2\right ) x \,x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (2+m \right ) \left (1+m \right )}\)	\(91\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 30, normalized size = 0.37 \begin {gather*} \frac {a \sqrt {c} x^{2} x^{m}}{m + 2} + \frac {\sqrt {c} x x^{m}}{m + 1} \end {gather*}

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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 80, normalized size = 0.98 \begin {gather*} -\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} \end {gather*}

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)^(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{m} \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 \end {gather*}

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)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.07, size = 51, normalized size = 0.62 \begin {gather*} \frac {x^{m+1}\,\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^(m + 1)*(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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