3.595 \(\int \frac {\sqrt {b-\frac {a}{x^2}} x^m}{\sqrt {a-b x^2}} \, dx\)

Optimal. Leaf size=33 \[ \frac {x^{m+1} \sqrt {b-\frac {a}{x^2}}}{m \sqrt {a-b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {515, 23, 30} \[ \frac {x^{m+1} \sqrt {b-\frac {a}{x^2}}}{m \sqrt {a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[b - a/x^2]*x^m)/Sqrt[a - b*x^2],x]

[Out]

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

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 515

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Dist[(x^(n*FracPa
rt[q])*(c + d/x^n)^FracPart[q])/(d + c*x^n)^FracPart[q], Int[x^(m - n*q)*(a + b*x^n)^p*(d + c*x^n)^q, x], x] /
; FreeQ[{a, b, c, d, m, n, p, q}, x] && EqQ[mn, -n] &&  !IntegerQ[q] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.00 \[ \frac {x^{m+1} \sqrt {b-\frac {a}{x^2}}}{m \sqrt {a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[b - a/x^2]*x^m)/Sqrt[a - b*x^2],x]

[Out]

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

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fricas [A]  time = 0.42, size = 44, normalized size = 1.33 \[ -\frac {\sqrt {-b x^{2} + a} x x^{m} \sqrt {\frac {b x^{2} - a}{x^{2}}}}{b m x^{2} - a m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 35, normalized size = 1.06 \[ \frac {\sqrt {-\frac {-b \,x^{2}+a}{x^{2}}}\, x^{m +1}}{\sqrt {-b \,x^{2}+a}\, m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.07, size = 8, normalized size = 0.24 \[ -\frac {i \, x^{m}}{m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x^m/m

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mupad [B]  time = 3.36, size = 29, normalized size = 0.88 \[ \frac {x^{m+1}\,\sqrt {b-\frac {a}{x^2}}}{m\,\sqrt {a-b\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \sqrt {- \frac {a}{x^{2}} + b}}{\sqrt {a - b x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**m*sqrt(-a/x**2 + b)/sqrt(a - b*x**2), x)

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