3.8.41 \(\int e^{\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^m \, dx\) [741]

Optimal. Leaf size=80 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}} x^{1+m}}{m \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-\frac {c}{a^2 x^2}} x^{2+m}}{(1+m) \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 80, 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 = {6295, 6285, 45} \begin {gather*} \frac {x^{m+1} \sqrt {c-\frac {c}{a^2 x^2}}}{m \sqrt {1-a^2 x^2}}+\frac {a x^{m+2} \sqrt {c-\frac {c}{a^2 x^2}}}{(m+1) \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x^(1 + m))/(m*Sqrt[1 - a^2*x^2]) + (a*Sqrt[c - c/(a^2*x^2)]*x^(2 + m))/((1 + 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 6295

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 51, normalized size = 0.64 \begin {gather*} \frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (\frac {x^m}{m}+\frac {a x^{1+m}}{1+m}\right )}{\sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 53, normalized size = 0.66

method	result	size
gosper	\(\frac {x^{1+m} \left (a x m +m +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{\left (1+m \right ) m \sqrt {-a^{2} x^{2}+1}}\)	\(53\)
risch	\(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \sqrt {\frac {c \left (-a^{2} x^{2}+1\right )}{a^{2} x^{2}-1}}\, \left (a x m +m +1\right ) x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c}\, \left (1+m \right ) m}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.30, size = 30, normalized size = 0.38 \begin {gather*} \frac {\sqrt {c} x x^{m}}{i \, m + i} - \frac {i \, \sqrt {c} x^{m}}{a m} \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*(c-c/a^2/x^2)^(1/2),x, algorithm="maxima")

[Out]

sqrt(c)*x*x^m/(I*m + I) - I*sqrt(c)*x^m/(a*m)

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Fricas [A]
time = 0.37, size = 78, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a m x^{2} + {\left (m + 1\right )} x\right )} x^{m} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{{\left (a^{2} m^{2} + a^{2} m\right )} x^{2} - m^{2} - m} \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*(c-c/a^2/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{m} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\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*(c-c/a**2/x**2)**(1/2),x)

[Out]

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

[Out]

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

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Mupad [B]
time = 1.30, size = 45, normalized size = 0.56 \begin {gather*} \frac {x\,x^m\,\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (m+a\,m\,x+1\right )}{m\,\sqrt {1-a^2\,x^2}\,\left (m+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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