### 3.793 $$\int \frac{(a^2-b^2 x^2)^{3/2}}{(a+b x)^3} \, dx$$

Optimal. Leaf size=76 $-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-\frac{3 \sqrt{a^2-b^2 x^2}}{b}-\frac{3 a \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.0277218, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {663, 665, 217, 203} $-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-\frac{3 \sqrt{a^2-b^2 x^2}}{b}-\frac{3 a \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^3,x]

[Out]

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

Rule 663

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

Rule 665

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx &=-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-3 \int \frac{\sqrt{a^2-b^2 x^2}}{a+b x} \, dx\\ &=-\frac{3 \sqrt{a^2-b^2 x^2}}{b}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-(3 a) \int \frac{1}{\sqrt{a^2-b^2 x^2}} \, dx\\ &=-\frac{3 \sqrt{a^2-b^2 x^2}}{b}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-(3 a) \operatorname{Subst}\left (\int \frac{1}{1+b^2 x^2} \, dx,x,\frac{x}{\sqrt{a^2-b^2 x^2}}\right )\\ &=-\frac{3 \sqrt{a^2-b^2 x^2}}{b}-\frac{2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-\frac{3 a \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0853772, size = 60, normalized size = 0.79 $-\frac{\frac{\sqrt{a^2-b^2 x^2} (5 a+b x)}{a+b x}+3 a \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^3,x]

[Out]

-((((5*a + b*x)*Sqrt[a^2 - b^2*x^2])/(a + b*x) + 3*a*ArcTan[(b*x)/Sqrt[a^2 - b^2*x^2]])/b)

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Maple [B]  time = 0.047, size = 206, normalized size = 2.7 \begin{align*} -{\frac{1}{{b}^{4}a} \left ( - \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab \right ) ^{{\frac{5}{2}}} \left ( x+{\frac{a}{b}} \right ) ^{-3}}-2\,{\frac{1}{{b}^{3}{a}^{2}} \left ( - \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab \right ) ^{5/2} \left ( x+{\frac{a}{b}} \right ) ^{-2}}-2\,{\frac{1}{b{a}^{2}} \left ( - \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab \right ) ^{3/2}}-3\,{\frac{x}{a}\sqrt{- \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab}}-3\,{\frac{a}{\sqrt{{b}^{2}}}\arctan \left ({\sqrt{{b}^{2}}x{\frac{1}{\sqrt{- \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((-b^2*x^2+a^2)^(3/2)/(b*x+a)^3,x)

[Out]

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

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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)^(3/2)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.15592, size = 174, normalized size = 2.29 \begin{align*} -\frac{5 \, a b x + 5 \, a^{2} - 6 \,{\left (a b x + a^{2}\right )} \arctan \left (-\frac{a - \sqrt{-b^{2} x^{2} + a^{2}}}{b x}\right ) + \sqrt{-b^{2} x^{2} + a^{2}}{\left (b x + 5 \, a\right )}}{b^{2} x + a b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a)^3,x, algorithm="fricas")

[Out]

-(5*a*b*x + 5*a^2 - 6*(a*b*x + a^2)*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + sqrt(-b^2*x^2 + a^2)*(b*x + 5*
a))/(b^2*x + a*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{\frac{3}{2}}}{\left (a + b x\right )^{3}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*x**2+a**2)**(3/2)/(b*x+a)**3,x)

[Out]

Integral((-(-a + b*x)*(a + b*x))**(3/2)/(a + b*x)**3, x)

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Giac [A]  time = 1.2186, size = 104, normalized size = 1.37 \begin{align*} -\frac{3 \, a \arcsin \left (\frac{b x}{a}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )}{{\left | b \right |}} - \frac{\sqrt{-b^{2} x^{2} + a^{2}}}{b} + \frac{8 \, a}{{\left (\frac{a b + \sqrt{-b^{2} x^{2} + a^{2}}{\left | b \right |}}{b^{2} x} + 1\right )}{\left | b \right |}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a)^3,x, algorithm="giac")

[Out]

-3*a*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - sqrt(-b^2*x^2 + a^2)/b + 8*a/(((a*b + sqrt(-b^2*x^2 + a^2)*abs(b))/(
b^2*x) + 1)*abs(b))