3.7.35 \(\int \frac {(a^2-b^2 x^2)^{3/2}}{(a+b x)^4} \, dx\)

Optimal. Leaf size=83 \[ -\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}+\frac {\tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \]

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Rubi [A]  time = 0.02, antiderivative size = 83, 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 = {663, 217, 203} \begin {gather*} -\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}+\frac {\tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^4,x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx &=-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}-\int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^2} \, dx\\ &=\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\int \frac {1}{\sqrt {a^2-b^2 x^2}} \, dx\\ &=\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\operatorname {Subst}\left (\int \frac {1}{1+b^2 x^2} \, dx,x,\frac {x}{\sqrt {a^2-b^2 x^2}}\right )\\ &=\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {\tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 61, normalized size = 0.73 \begin {gather*} \frac {\frac {4 \sqrt {a^2-b^2 x^2} (a+2 b x)}{(a+b x)^2}+3 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.43, size = 80, normalized size = 0.96 \begin {gather*} \frac {4 \sqrt {a^2-b^2 x^2} (a+2 b x)}{3 b (a+b x)^2}+\frac {\sqrt {-b^2} \log \left (\sqrt {a^2-b^2 x^2}-\sqrt {-b^2} x\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 - b^2*x^2)^(3/2)/(a + b*x)^4,x]

[Out]

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

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fricas [A]  time = 0.42, size = 110, normalized size = 1.33 \begin {gather*} \frac {2 \, {\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + 2 \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (2 \, b x + a\right )}\right )}}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} 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)^4,x, algorithm="fricas")

[Out]

2/3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 - 3*(b^2*x^2 + 2*a*b*x + a^2)*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + 2*s
qrt(-b^2*x^2 + a^2)*(2*b*x + a))/(b^3*x^2 + 2*a*b^2*x + a^2*b)

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giac [A]  time = 0.22, size = 86, normalized size = 1.04 \begin {gather*} \frac {\arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (b)}{{\left | b \right |}} - \frac {8 \, {\left (\frac {3 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}}{b^{2} x} + 1\right )}}{3 \, {\left (\frac {a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}}{b^{2} x} + 1\right )}^{3} {\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)^4,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 248, normalized size = 2.99 \begin {gather*} \frac {\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 )}{\sqrt {b^{2}}}+\frac {\sqrt {2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}}\, x}{a^{2}}+\frac {2 \left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {3}{2}}}{3 a^{3} 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}}}{3 \left (x +\frac {a}{b}\right )^{4} a \,b^{5}}+\frac {\left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {5}{2}}}{3 \left (x +\frac {a}{b}\right )^{3} a^{2} b^{4}}+\frac {2 \left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {5}{2}}}{3 \left (x +\frac {a}{b}\right )^{2} a^{3} 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)^4,x)

[Out]

-1/3/b^5/a/(x+a/b)^4*(2*(x+a/b)*a*b-(x+a/b)^2*b^2)^(5/2)+1/3/b^4/a^2/(x+a/b)^3*(2*(x+a/b)*a*b-(x+a/b)^2*b^2)^(
5/2)+2/3/b^3/a^3/(x+a/b)^2*(2*(x+a/b)*a*b-(x+a/b)^2*b^2)^(5/2)+2/3/b/a^3*(2*(x+a/b)*a*b-(x+a/b)^2*b^2)^(3/2)+1
/a^2*(2*(x+a/b)*a*b-(x+a/b)^2*b^2)^(1/2)*x+1/(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 = 3.08, size = 127, normalized size = 1.53 \begin {gather*} -\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac {2 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {\arcsin \left (\frac {b x}{a}\right )}{b} + \frac {7 \, \sqrt {-b^{2} x^{2} + a^{2}}}{3 \, {\left (b^{2} x + a 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)^4,x, algorithm="maxima")

[Out]

-1/3*(-b^2*x^2 + a^2)^(3/2)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) - 2/3*sqrt(-b^2*x^2 + a^2)*a/(b^3*x^
2 + 2*a*b^2*x + a^2*b) + arcsin(b*x/a)/b + 7/3*sqrt(-b^2*x^2 + a^2)/(b^2*x + 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 )}^4} \,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)^4,x)

[Out]

int((a^2 - b^2*x^2)^(3/2)/(a + b*x)^4, 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 )^{4}}\, 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)**4,x)

[Out]

Integral((-(-a + b*x)*(a + b*x))**(3/2)/(a + b*x)**4, x)

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