3.784 \(\int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx\)

Optimal. Leaf size=67 \[ -\frac {\left (a^2-b^2 x^2\right )^{3/2}}{15 a^2 b (a+b x)^3}-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4} \]

[Out]

-1/5*(-b^2*x^2+a^2)^(3/2)/a/b/(b*x+a)^4-1/15*(-b^2*x^2+a^2)^(3/2)/a^2/b/(b*x+a)^3

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {659, 651} \[ -\frac {\left (a^2-b^2 x^2\right )^{3/2}}{15 a^2 b (a+b x)^3}-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 - b^2*x^2]/(a + b*x)^4,x]

[Out]

-(a^2 - b^2*x^2)^(3/2)/(5*a*b*(a + b*x)^4) - (a^2 - b^2*x^2)^(3/2)/(15*a^2*b*(a + b*x)^3)

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx &=-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4}+\frac {\int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^3} \, dx}{5 a}\\ &=-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4}-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{15 a^2 b (a+b x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 51, normalized size = 0.76 \[ \frac {\sqrt {a^2-b^2 x^2} \left (-4 a^2+3 a b x+b^2 x^2\right )}{15 a^2 b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 - b^2*x^2]/(a + b*x)^4,x]

[Out]

(Sqrt[a^2 - b^2*x^2]*(-4*a^2 + 3*a*b*x + b^2*x^2))/(15*a^2*b*(a + b*x)^3)

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fricas [A]  time = 0.90, size = 104, normalized size = 1.55 \[ -\frac {4 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 12 \, a^{2} b x + 4 \, a^{3} - {\left (b^{2} x^{2} + 3 \, a b x - 4 \, a^{2}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{15 \, {\left (a^{2} b^{4} x^{3} + 3 \, a^{3} b^{3} x^{2} + 3 \, a^{4} b^{2} x + a^{5} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(1/2)/(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/15*(4*b^3*x^3 + 12*a*b^2*x^2 + 12*a^2*b*x + 4*a^3 - (b^2*x^2 + 3*a*b*x - 4*a^2)*sqrt(-b^2*x^2 + a^2))/(a^2*
b^4*x^3 + 3*a^3*b^3*x^2 + 3*a^4*b^2*x + a^5*b)

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giac [B]  time = 0.23, size = 165, normalized size = 2.46 \[ \frac {2 \, {\left (\frac {5 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}}{b^{2} x} + \frac {25 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}^{2}}{b^{4} x^{2}} + \frac {15 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}^{3}}{b^{6} x^{3}} + \frac {15 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}^{4}}{b^{8} x^{4}} + 4\right )}}{15 \, a^{2} {\left (\frac {a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}}{b^{2} x} + 1\right )}^{5} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(1/2)/(b*x+a)^4,x, algorithm="giac")

[Out]

2/15*(5*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))/(b^2*x) + 25*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^2/(b^4*x^2) + 15*
(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^3/(b^6*x^3) + 15*(a*b + sqrt(-b^2*x^2 + a^2)*abs(b))^4/(b^8*x^4) + 4)/(a^2
*((a*b + sqrt(-b^2*x^2 + a^2)*abs(b))/(b^2*x) + 1)^5*abs(b))

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maple [A]  time = 0.04, size = 43, normalized size = 0.64 \[ -\frac {\left (-b x +a \right ) \left (b x +4 a \right ) \sqrt {-b^{2} x^{2}+a^{2}}}{15 \left (b x +a \right )^{3} a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b^2*x^2+a^2)^(1/2)/(b*x+a)^4,x)

[Out]

-1/15*(-b*x+a)*(b*x+4*a)*(-b^2*x^2+a^2)^(1/2)/(b*x+a)^3/a^2/b

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maxima [B]  time = 1.35, size = 123, normalized size = 1.84 \[ -\frac {2 \, \sqrt {-b^{2} x^{2} + a^{2}}}{5 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{15 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{15 \, {\left (a^{2} b^{2} x + a^{3} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(1/2)/(b*x+a)^4,x, algorithm="maxima")

[Out]

-2/5*sqrt(-b^2*x^2 + a^2)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) + 1/15*sqrt(-b^2*x^2 + a^2)/(a*b^3*x^2
+ 2*a^2*b^2*x + a^3*b) + 1/15*sqrt(-b^2*x^2 + a^2)/(a^2*b^2*x + a^3*b)

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mupad [B]  time = 0.62, size = 47, normalized size = 0.70 \[ \frac {\sqrt {a^2-b^2\,x^2}\,\left (-4\,a^2+3\,a\,b\,x+b^2\,x^2\right )}{15\,a^2\,b\,{\left (a+b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^(1/2)/(a + b*x)^4,x)

[Out]

((a^2 - b^2*x^2)^(1/2)*(b^2*x^2 - 4*a^2 + 3*a*b*x))/(15*a^2*b*(a + b*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (- a + b x\right ) \left (a + b x\right )}}{\left (a + b x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*x**2+a**2)**(1/2)/(b*x+a)**4,x)

[Out]

Integral(sqrt(-(-a + b*x)*(a + b*x))/(a + b*x)**4, x)

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