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]

-(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)

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Rubi [A]  time = 0.0208856, antiderivative size = 67, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0374707, 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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Maple [A]  time = 0.044, size = 43, normalized size = 0.6 \begin{align*} -{\frac{ \left ( bx+4\,a \right ) \left ( -bx+a \right ) }{15\, \left ( bx+a \right ) ^{3}b{a}^{2}}\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}} \end{align*}

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/b/a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.87622, size = 213, normalized size = 3.18 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (- a + b x\right ) \left (a + b x\right )}}{\left (a + b x\right )^{4}}\, dx \end{align*}

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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Giac [B]  time = 1.2234, size = 223, normalized size = 3.33 \begin{align*} \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 |}} \end{align*}

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))