\(\int \frac {(a^2-b^2 x^2)^{3/2}}{a+b x} \, dx\) [791]

Optimal result

Integrand size = 24, antiderivative size = 76 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=\frac {1}{2} a x \sqrt {a^2-b^2 x^2}+\frac {\left (a^2-b^2 x^2\right )^{3/2}}{3 b}+\frac {a^3 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (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 = {679, 201, 223, 209} \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=\frac {1}{2} a x \sqrt {a^2-b^2 x^2}+\frac {\left (a^2-b^2 x^2\right )^{3/2}}{3 b}+\frac {a^3 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{2 b} \]

[In]

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

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=\frac {\left (2 a^2+3 a b x-2 b^2 x^2\right ) \sqrt {a^2-b^2 x^2}}{6 b}-\frac {a^3 \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{2 \sqrt {-b^2}} \]

[In]

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

[Out]

((2*a^2 + 3*a*b*x - 2*b^2*x^2)*Sqrt[a^2 - b^2*x^2])/(6*b) - (a^3*Log[-(Sqrt[-b^2]*x) + Sqrt[a^2 - b^2*x^2]])/(
2*Sqrt[-b^2])

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {\left (-2 b^{2} x^{2}+3 a b x +2 a^{2}\right ) \sqrt {-b^{2} x^{2}+a^{2}}}{6 b}+\frac {a^{3} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\)	\(72\)
default	\(\frac {\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{3}+a b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a}{b}\right )+2 a b \right ) \sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}{4 b^{2}}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )}{b}\)	\(136\)

[In]

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

[Out]

1/6*(-2*b^2*x^2+3*a*b*x+2*a^2)/b*(-b^2*x^2+a^2)^(1/2)+1/2*a^3/(b^2)^(1/2)*arctan((b^2)^(1/2)*x/(-b^2*x^2+a^2)^
(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=-\frac {6 \, a^{3} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + {\left (2 \, b^{2} x^{2} - 3 \, a b x - 2 \, a^{2}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{6 \, b} \]

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a),x, algorithm="fricas")

[Out]

-1/6*(6*a^3*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + (2*b^2*x^2 - 3*a*b*x - 2*a^2)*sqrt(-b^2*x^2 + a^2))/b

Sympy [A] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=a \left (\begin {cases} \frac {a^{2} \left (\begin {cases} \frac {\log {\left (- 2 b^{2} x + 2 \sqrt {- b^{2}} \sqrt {a^{2} - b^{2} x^{2}} \right )}}{\sqrt {- b^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- b^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a^{2} - b^{2} x^{2}}}{2} & \text {for}\: b^{2} \neq 0 \\x \sqrt {a^{2}} & \text {otherwise} \end {cases}\right ) - b \left (\begin {cases} \sqrt {a^{2} - b^{2} x^{2}} \left (- \frac {a^{2}}{3 b^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a*Piecewise((a**2*Piecewise((log(-2*b**2*x + 2*sqrt(-b**2)*sqrt(a**2 - b**2*x**2))/sqrt(-b**2), Ne(a**2, 0)),
(x*log(x)/sqrt(-b**2*x**2), True))/2 + x*sqrt(a**2 - b**2*x**2)/2, Ne(b**2, 0)), (x*sqrt(a**2), True)) - b*Pie
cewise((sqrt(a**2 - b**2*x**2)*(-a**2/(3*b**2) + x**2/3), Ne(b**2, 0)), (x**2*sqrt(a**2)/2, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=-\frac {i \, a^{3} \arcsin \left (\frac {b x}{a} + 2\right )}{2 \, b} + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 4 \, a b x + 3 \, a^{2}} a x + \frac {\sqrt {b^{2} x^{2} + 4 \, a b x + 3 \, a^{2}} a^{2}}{b} + \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{3 \, b} \]

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a),x, algorithm="maxima")

[Out]

-1/2*I*a^3*arcsin(b*x/a + 2)/b + 1/2*sqrt(b^2*x^2 + 4*a*b*x + 3*a^2)*a*x + sqrt(b^2*x^2 + 4*a*b*x + 3*a^2)*a^2
/b + 1/3*(-b^2*x^2 + a^2)^(3/2)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{a+b x} \, dx=\frac {a^{3} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{2 \, {\left | b \right |}} - \frac {1}{6} \, \sqrt {-b^{2} x^{2} + a^{2}} {\left ({\left (2 \, b x - 3 \, a\right )} x - \frac {2 \, a^{2}}{b}\right )} \]

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a),x, algorithm="giac")

[Out]

1/2*a^3*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/6*sqrt(-b^2*x^2 + a^2)*((2*b*x - 3*a)*x - 2*a^2/b)

Mupad [F(-1)]

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

[In]

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

[Out]

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