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

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

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Rubi [A]  time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 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])

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 60, normalized size = 0.71 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.33, size = 81, normalized size = 0.95 \begin {gather*} \frac {(4 a-b x) \sqrt {a^2-b^2 x^2}}{2 b}+\frac {3 a^2 \sqrt {-b^2} \log \left (\sqrt {a^2-b^2 x^2}-\sqrt {-b^2} x\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 60, normalized size = 0.71 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.26, size = 121, normalized size = 1.42 \begin {gather*} -\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}\relax (b) - \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}\relax (b) + 3 \, a^{3} b^{3} \sqrt {\frac {2 \, a}{b x + a} - 1} \mathrm {sgn}\left (\frac {1}{b x + a}\right ) \mathrm {sgn}\relax (b)\right )} {\left (b x + a\right )}^{2}}{a^{2}}\right )} {\left | b \right |}}{4 \, a b^{5}} \end {gather*}

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)

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maple [B]  time = 0.05, size = 158, normalized size = 1.86 \begin {gather*} \frac {3 a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}}}\right )}{2 \sqrt {b^{2}}}+\frac {3 \sqrt {2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}}\, x}{2}+\frac {\left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {3}{2}}}{a b}+\frac {\left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {5}{2}}}{\left (x +\frac {a}{b}\right )^{2} a \,b^{3}} \end {gather*}

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

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maxima [A]  time = 3.04, size = 63, normalized size = 0.74 \begin {gather*} \frac {3 \, a^{2} \arcsin \left (\frac {b x}{a}\right )}{2 \, b} + \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{2 \, {\left (b^{2} x + a b\right )}} + \frac {3 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{2 \, b} \end {gather*}

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2-b^2\,x^2\right )}^{3/2}}{{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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