### 3.785 $$\int \frac{\sqrt{a^2-b^2 x^2}}{(a+b x)^5} \, dx$$

Optimal. Leaf size=100 $-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{105 a^3 b (a+b x)^3}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{35 a^2 b (a+b x)^4}-\frac{\left (a^2-b^2 x^2\right )^{3/2}}{7 a b (a+b x)^5}$

[Out]

-(a^2 - b^2*x^2)^(3/2)/(7*a*b*(a + b*x)^5) - (2*(a^2 - b^2*x^2)^(3/2))/(35*a^2*b*(a + b*x)^4) - (2*(a^2 - b^2*
x^2)^(3/2))/(105*a^3*b*(a + b*x)^3)

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Rubi [A]  time = 0.0357099, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {659, 651} $-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{105 a^3 b (a+b x)^3}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{35 a^2 b (a+b x)^4}-\frac{\left (a^2-b^2 x^2\right )^{3/2}}{7 a b (a+b x)^5}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[a^2 - b^2*x^2]/(a + b*x)^5,x]

[Out]

-(a^2 - b^2*x^2)^(3/2)/(7*a*b*(a + b*x)^5) - (2*(a^2 - b^2*x^2)^(3/2))/(35*a^2*b*(a + b*x)^4) - (2*(a^2 - b^2*
x^2)^(3/2))/(105*a^3*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)^5} \, dx &=-\frac{\left (a^2-b^2 x^2\right )^{3/2}}{7 a b (a+b x)^5}+\frac{2 \int \frac{\sqrt{a^2-b^2 x^2}}{(a+b x)^4} \, dx}{7 a}\\ &=-\frac{\left (a^2-b^2 x^2\right )^{3/2}}{7 a b (a+b x)^5}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{35 a^2 b (a+b x)^4}+\frac{2 \int \frac{\sqrt{a^2-b^2 x^2}}{(a+b x)^3} \, dx}{35 a^2}\\ &=-\frac{\left (a^2-b^2 x^2\right )^{3/2}}{7 a b (a+b x)^5}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{35 a^2 b (a+b x)^4}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{105 a^3 b (a+b x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0409247, size = 63, normalized size = 0.63 $\frac{\sqrt{a^2-b^2 x^2} \left (13 a^2 b x-23 a^3+8 a b^2 x^2+2 b^3 x^3\right )}{105 a^3 b (a+b x)^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[a^2 - b^2*x^2]/(a + b*x)^5,x]

[Out]

(Sqrt[a^2 - b^2*x^2]*(-23*a^3 + 13*a^2*b*x + 8*a*b^2*x^2 + 2*b^3*x^3))/(105*a^3*b*(a + b*x)^4)

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Maple [A]  time = 0.043, size = 55, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,{b}^{2}{x}^{2}+10\,abx+23\,{a}^{2} \right ) \left ( -bx+a \right ) }{105\, \left ( bx+a \right ) ^{4}{a}^{3}b}\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)^5,x)

[Out]

-1/105*(-b*x+a)*(2*b^2*x^2+10*a*b*x+23*a^2)*(-b^2*x^2+a^2)^(1/2)/(b*x+a)^4/a^3/b

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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)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91067, size = 290, normalized size = 2.9 \begin{align*} -\frac{23 \, b^{4} x^{4} + 92 \, a b^{3} x^{3} + 138 \, a^{2} b^{2} x^{2} + 92 \, a^{3} b x + 23 \, a^{4} -{\left (2 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + 13 \, a^{2} b x - 23 \, a^{3}\right )} \sqrt{-b^{2} x^{2} + a^{2}}}{105 \,{\left (a^{3} b^{5} x^{4} + 4 \, a^{4} b^{4} x^{3} + 6 \, a^{5} b^{3} x^{2} + 4 \, a^{6} b^{2} x + a^{7} 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)^5,x, algorithm="fricas")

[Out]

-1/105*(23*b^4*x^4 + 92*a*b^3*x^3 + 138*a^2*b^2*x^2 + 92*a^3*b*x + 23*a^4 - (2*b^3*x^3 + 8*a*b^2*x^2 + 13*a^2*
b*x - 23*a^3)*sqrt(-b^2*x^2 + a^2))/(a^3*b^5*x^4 + 4*a^4*b^4*x^3 + 6*a^5*b^3*x^2 + 4*a^6*b^2*x + a^7*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 )^{5}}\, 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)**5,x)

[Out]

Integral(sqrt(-(-a + b*x)*(a + b*x))/(a + b*x)**5, x)

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Giac [C]  time = 1.21159, size = 120, normalized size = 1.2 \begin{align*} -\frac{1}{420} \,{\left (\frac{{\left (15 \,{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{7}{2}} + 42 \,{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{5}{2}} + 35 \,{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{3}{2}}\right )} \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right )}{a^{3} b^{2}} + \frac{8 i \, \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right )}{a^{3} b^{2}}\right )}{\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)^5,x, algorithm="giac")

[Out]

-1/420*((15*(2*a/(b*x + a) - 1)^(7/2) + 42*(2*a/(b*x + a) - 1)^(5/2) + 35*(2*a/(b*x + a) - 1)^(3/2))*sgn(1/(b*
x + a))*sgn(b)/(a^3*b^2) + 8*I*sgn(1/(b*x + a))*sgn(b)/(a^3*b^2))*abs(b)