\(\int (a+b x) (a^2-b^2 x^2)^{3/2} \, dx\) [790]

Optimal result

Integrand size = 22, antiderivative size = 100 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {3}{8} a^3 x \sqrt {a^2-b^2 x^2}+\frac {1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac {3 a^5 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b} \]

[Out]

1/4*a*x*(-b^2*x^2+a^2)^(3/2)-1/5*(-b^2*x^2+a^2)^(5/2)/b+3/8*a^5*arctan(b*x/(-b^2*x^2+a^2)^(1/2))/b+3/8*a^3*x*(
-b^2*x^2+a^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {655, 201, 223, 209} \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac {3 a^5 \arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b}+\frac {3}{8} a^3 x \sqrt {a^2-b^2 x^2} \]

[In]

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

[Out]

(3*a^3*x*Sqrt[a^2 - b^2*x^2])/8 + (a*x*(a^2 - b^2*x^2)^(3/2))/4 - (a^2 - b^2*x^2)^(5/2)/(5*b) + (3*a^5*ArcTan[
(b*x)/Sqrt[a^2 - b^2*x^2]])/(8*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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+a \int \left (a^2-b^2 x^2\right )^{3/2} \, dx \\ & = \frac {1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac {1}{4} \left (3 a^3\right ) \int \sqrt {a^2-b^2 x^2} \, dx \\ & = \frac {3}{8} a^3 x \sqrt {a^2-b^2 x^2}+\frac {1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac {1}{8} \left (3 a^5\right ) \int \frac {1}{\sqrt {a^2-b^2 x^2}} \, dx \\ & = \frac {3}{8} a^3 x \sqrt {a^2-b^2 x^2}+\frac {1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac {1}{8} \left (3 a^5\right ) \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}{8} a^3 x \sqrt {a^2-b^2 x^2}+\frac {1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac {\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac {3 a^5 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.11 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {\sqrt {a^2-b^2 x^2} \left (-8 a^4+25 a^3 b x+16 a^2 b^2 x^2-10 a b^3 x^3-8 b^4 x^4\right )}{40 b}-\frac {3 a^5 \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{8 \sqrt {-b^2}} \]

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^2]*(-8*a^4 + 25*a^3*b*x + 16*a^2*b^2*x^2 - 10*a*b^3*x^3 - 8*b^4*x^4))/(40*b) - (3*a^5*Log[-(
Sqrt[-b^2]*x) + Sqrt[a^2 - b^2*x^2]])/(8*Sqrt[-b^2])

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\left (8 b^{4} x^{4}+10 a \,b^{3} x^{3}-16 a^{2} b^{2} x^{2}-25 a^{3} b x +8 a^{4}\right ) \sqrt {-b^{2} x^{2}+a^{2}}}{40 b}+\frac {3 a^{5} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{8 \sqrt {b^{2}}}\)	\(94\)
default	\(a \left (\frac {x \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 a^{2} \left (\frac {x \sqrt {-b^{2} x^{2}+a^{2}}}{2}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{2 \sqrt {b^{2}}}\right )}{4}\right )-\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}}}{5 b}\)	\(96\)

[In]

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

[Out]

-1/40*(8*b^4*x^4+10*a*b^3*x^3-16*a^2*b^2*x^2-25*a^3*b*x+8*a^4)/b*(-b^2*x^2+a^2)^(1/2)+3/8*a^5/(b^2)^(1/2)*arct
an((b^2)^(1/2)*x/(-b^2*x^2+a^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=-\frac {30 \, a^{5} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + {\left (8 \, b^{4} x^{4} + 10 \, a b^{3} x^{3} - 16 \, a^{2} b^{2} x^{2} - 25 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{40 \, b} \]

[In]

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

[Out]

-1/40*(30*a^5*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + (8*b^4*x^4 + 10*a*b^3*x^3 - 16*a^2*b^2*x^2 - 25*a^3*
b*x + 8*a^4)*sqrt(-b^2*x^2 + a^2))/b

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.43 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 a^{5} \left (\begin {cases} \frac {\log {\left (- 2 b^{2} x + 2 \sqrt {- b^{2}} \sqrt {a^{2} - b^{2} x^{2}} \right )}}{\sqrt {- b^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- b^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {a^{2} - b^{2} x^{2}} \left (- \frac {a^{4}}{5 b} + \frac {5 a^{3} x}{8} + \frac {2 a^{2} b x^{2}}{5} - \frac {a b^{2} x^{3}}{4} - \frac {b^{3} x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\left (a x + \frac {b x^{2}}{2}\right ) \left (a^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*a**5*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))/8 + sqrt(a**2 - b**2*x**2)*(-a**4/(5*b) + 5*a**3*x/8 + 2*a**2*b*x**2/5 - a*
b**2*x**3/4 - b**3*x**4/5), Ne(b**2, 0)), ((a*x + b*x**2/2)*(a**2)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.73 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {3 \, a^{5} \arcsin \left (\frac {b x}{a}\right )}{8 \, b} + \frac {3}{8} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{3} x + \frac {1}{4} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a x - \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}}}{5 \, b} \]

[In]

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

[Out]

3/8*a^5*arcsin(b*x/a)/b + 3/8*sqrt(-b^2*x^2 + a^2)*a^3*x + 1/4*(-b^2*x^2 + a^2)^(3/2)*a*x - 1/5*(-b^2*x^2 + a^
2)^(5/2)/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.81 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {3 \, a^{5} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{8 \, {\left | b \right |}} - \frac {1}{40} \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (\frac {8 \, a^{4}}{b} - {\left (25 \, a^{3} + 2 \, {\left (8 \, a^{2} b - {\left (4 \, b^{3} x + 5 \, a b^{2}\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

3/8*a^5*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/40*sqrt(-b^2*x^2 + a^2)*(8*a^4/b - (25*a^3 + 2*(8*a^2*b - (4*b^
3*x + 5*a*b^2)*x)*x)*x)

Mupad [B] (verification not implemented)

Time = 9.91 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.67 \[ \int (a+b x) \left (a^2-b^2 x^2\right )^{3/2} \, dx=\frac {a\,x\,{\left (a^2-b^2\,x^2\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ \frac {b^2\,x^2}{a^2}\right )}{{\left (1-\frac {b^2\,x^2}{a^2}\right )}^{3/2}}-\frac {{\left (a^2-b^2\,x^2\right )}^{5/2}}{5\,b} \]

[In]

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

[Out]

(a*x*(a^2 - b^2*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, (b^2*x^2)/a^2))/(1 - (b^2*x^2)/a^2)^(3/2) - (a^2 - b^2*
x^2)^(5/2)/(5*b)