3.8.93 \(\int \frac {(a^2-b^2 x^2)^{3/2}}{(a+b x)^3} \, dx\) [793]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, 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 = {677, 679, 223, 209} \begin {gather*} -\frac {3 a \text {ArcTan}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b}-\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} \end {gather*}

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 209

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

Rule 223

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 677

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 679

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)^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) \text {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*}

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Mathematica [A]
time = 0.26, size = 78, normalized size = 1.03 \begin {gather*} \frac {(-5 a-b x) \sqrt {a^2-b^2 x^2}}{b (a+b x)}+\frac {3 a \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{\sqrt {-b^2}} \end {gather*}

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(238\) vs. \(2(70)=140\).
time = 0.51, size = 239, normalized size = 3.14

method	result	size
risch	\(-\frac {\sqrt {-b^{2} x^{2}+a^{2}}}{b}-\frac {3 a \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{\sqrt {b^{2}}}-\frac {4 a \sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}{b^{2} \left (x +\frac {a}{b}\right )}\)	\(94\)
default	\(\frac {-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{3}}-\frac {2 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{2}}+\frac {3 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{3}+a b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a}{b}\right )+2 a b \right ) \sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}{4 b^{2}}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )}{a}\right )}{a}}{b^{3}}\)	\(239\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 79, normalized size = 1.04 \begin {gather*} -\frac {3 \, a \arcsin \left (\frac {b x}{a}\right )}{b} + \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b} - \frac {6 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{b^{2} x + a 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)^3,x, algorithm="maxima")

[Out]

-3*a*arcsin(b*x/a)/b + (-b^2*x^2 + a^2)^(3/2)/(b^3*x^2 + 2*a*b^2*x + a^2*b) - 6*sqrt(-b^2*x^2 + a^2)*a/(b^2*x
+ a*b)

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Fricas [A]
time = 2.35, size = 83, normalized size = 1.09 \begin {gather*} -\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 {gather*}

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 = 0.00, 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 )^{3}}\, 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)**3,x)

[Out]

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

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Giac [A]
time = 1.44, size = 77, normalized size = 1.01 \begin {gather*} -\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 {gather*}

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))

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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 )}^3} \,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)^3,x)

[Out]

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

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