### 3.792 $$\int \frac{(a^2-b^2 x^2)^{3/2}}{(a+b x)^2} \, dx$$

Optimal. Leaf size=85 $\frac{3 a \sqrt{a^2-b^2 x^2}}{2 b}+\frac{\left (a^2-b^2 x^2\right )^{3/2}}{2 b (a+b x)}+\frac{3 a^2 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{2 b}$

[Out]

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

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Rubi [A]  time = 0.0293329, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {665, 217, 203} $\frac{3 a \sqrt{a^2-b^2 x^2}}{2 b}+\frac{\left (a^2-b^2 x^2\right )^{3/2}}{2 b (a+b x)}+\frac{3 a^2 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0561479, size = 60, normalized size = 0.71 $\left (\frac{2 a}{b}-\frac{x}{2}\right ) \sqrt{a^2-b^2 x^2}+\frac{3 a^2 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.05, size = 158, normalized size = 1.9 \begin{align*}{\frac{1}{a{b}^{3}} \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 ) ^{-2}}+{\frac{1}{ab} \left ( - \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab \right ) ^{{\frac{3}{2}}}}+{\frac{3\,x}{2}\sqrt{- \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab}}+{\frac{3\,{a}^{2}}{2}\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{- \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \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)^2,x)

[Out]

1/b^3/a/(x+1/b*a)^2*(-(x+1/b*a)^2*b^2+2*(x+1/b*a)*a*b)^(5/2)+1/b/a*(-(x+1/b*a)^2*b^2+2*(x+1/b*a)*a*b)^(3/2)+3/
2*(-(x+1/b*a)^2*b^2+2*(x+1/b*a)*a*b)^(1/2)*x+3/2*a^2/(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08218, size = 126, normalized size = 1.48 \begin{align*} -\frac{6 \, a^{2} \arctan \left (-\frac{a - \sqrt{-b^{2} x^{2} + a^{2}}}{b x}\right ) + \sqrt{-b^{2} x^{2} + a^{2}}{\left (b x - 4 \, a\right )}}{2 \, 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)^2,x, algorithm="fricas")

[Out]

-1/2*(6*a^2*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + sqrt(-b^2*x^2 + a^2)*(b*x - 4*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 )^{2}}\, 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)**2,x)

[Out]

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

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Giac [A]  time = 1.26223, size = 163, normalized size = 1.92 \begin{align*} -\frac{{\left (12 \, a^{3} b^{3} \arctan \left (\sqrt{\frac{2 \, a}{b x + a} - 1}\right ) \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right ) - \frac{{\left (5 \, a^{3} b^{3}{\left (\frac{2 \, a}{b x + a} - 1\right )}^{\frac{3}{2}} \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right ) + 3 \, a^{3} b^{3} \sqrt{\frac{2 \, a}{b x + a} - 1} \mathrm{sgn}\left (\frac{1}{b x + a}\right ) \mathrm{sgn}\left (b\right )\right )}{\left (b x + a\right )}^{2}}{a^{2}}\right )}{\left | b \right |}}{4 \, a b^{5}} \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)^2,x, algorithm="giac")

[Out]

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