3.9.80 \(\int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^m \, dx\) [880]

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

[Out]

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

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Rubi [A]
time = 0.18, 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 = {6332, 6328, 45} \begin {gather*} \frac {x^{m+1} \sqrt {c-\frac {c}{a^2 x^2}}}{(m+1) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^m \sqrt {c-\frac {c}{a^2 x^2}}}{a m \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x^m)/(a*m*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[c - c/(a^2*x^2)]*x^(1 + m))/((1 + m)*Sqrt[1 -
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 6328

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

Rule 6332

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 53, normalized size = 0.66 \begin {gather*} \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {x^m}{m}+\frac {a x^{1+m}}{1+m}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.03, size = 63, normalized size = 0.79

method	result	size
gosper	\(\frac {x^{1+m} \left (a x m +m +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{\left (1+m \right ) m \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(63\)
risch	\(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{\left (a x +1\right ) \left (a x -1\right )}}\, \left (a x -1\right ) \left (a x m +m +1\right ) x^{m}}{\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {c}\, \left (a^{2} x^{2}-1\right ) \left (1+m \right ) m}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]
time = 0.30, size = 44, normalized size = 0.55 \begin {gather*} \frac {{\left (a \sqrt {c} m x + \sqrt {c} {\left (m + 1\right )}\right )} {\left (a x + 1\right )} x^{m}}{{\left (m^{2} + m\right )} a^{2} x + {\left (m^{2} + m\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 72, normalized size = 0.90 \begin {gather*} -\frac {{\left (a m x^{2} + {\left (m + 1\right )} x\right )} x^{m} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{m^{2} - {\left (a m^{2} + a m\right )} x + m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m\,\sqrt {c-\frac {c}{a^2\,x^2}}}{\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \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)/(a*x + 1))^(1/2),x)

[Out]

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

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