\(\int \frac {(a^2-b^2 x^2)^{3/2}}{(a+b x)^4} \, dx\) [794]

Optimal result

Integrand size = 24, antiderivative size = 83 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {\arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {677, 223, 209} \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {\arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)} \]

[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 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]

Rubi steps \begin{align*} \text {integral}& = -\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}+\text {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*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {4 (a+2 b x) \sqrt {a^2-b^2 x^2}}{3 b (a+b x)^2}-\frac {2 \arctan \left (\frac {b x}{\sqrt {a^2}-\sqrt {a^2-b^2 x^2}}\right )}{b} \]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(290\) vs. \(2(75)=150\).

Time = 2.34 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.51

method	result	size
default	\(\frac {-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{3 a b \left (x +\frac {a}{b}\right )^{4}}-\frac {b \left (-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{3}}-\frac {2 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{2}}+\frac {3 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{3}+a b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a}{b}\right )+2 a b \right ) \sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}{4 b^{2}}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )}{a}\right )}{a}\right )}{3 a}}{b^{4}}\)	\(291\)

[In]

int((-b^2*x^2+a^2)^(3/2)/(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {2 \, {\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + 2 \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (2 \, b x + a\right )}\right )}}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\int \frac {\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{\frac {3}{2}}}{\left (a + b x\right )^{4}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=-\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac {2 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {\arcsin \left (\frac {b x}{a}\right )}{b} + \frac {7 \, \sqrt {-b^{2} x^{2} + a^{2}}}{3 \, {\left (b^{2} x + a b\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {\arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{{\left | b \right |}} - \frac {8 \, {\left (\frac {3 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}}{b^{2} x} + 1\right )}}{3 \, {\left (\frac {a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}}{b^{2} x} + 1\right )}^{3} {\left | b \right |}} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\int \frac {{\left (a^2-b^2\,x^2\right )}^{3/2}}{{\left (a+b\,x\right )}^4} \,d x \]

[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)