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

Optimal result

Integrand size = 24, antiderivative size = 76 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx=-\frac {3 \sqrt {a^2-b^2 x^2}}{b}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-\frac {3 a \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {677, 679, 223, 209} \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx=-\frac {3 a \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{b (a+b x)^2}-\frac {3 \sqrt {a^2-b^2 x^2}}{b} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx=\frac {(-5 a-b x) \sqrt {a^2-b^2 x^2}}{b (a+b x)}+\frac {3 a \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{\sqrt {-b^2}} \]

[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]

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.24

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	\(\frac {-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{3}}-\frac {2 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{2}}+\frac {3 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{3}+a b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a}{b}\right )+2 a b \right ) \sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}{4 b^{2}}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )}{a}\right )}{a}}{b^{3}}\)	\(239\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx=-\frac {5 \, a b x + 5 \, a^{2} - 6 \, {\left (a b x + a^{2}\right )} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + \sqrt {-b^{2} x^{2} + a^{2}} {\left (b x + 5 \, a\right )}}{b^{2} x + a b} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx=-\frac {3 \, a \arcsin \left (\frac {b x}{a}\right )}{b} + \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b} - \frac {6 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{b^{2} x + a b} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^3} \, dx=-\frac {3 \, a \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{{\left | b \right |}} - \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{b} + \frac {8 \, a}{{\left (\frac {a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}}{b^{2} x} + 1\right )} {\left | b \right |}} \]

[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))

Mupad [F(-1)]

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

[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)