\(\int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx\) [225]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 29, antiderivative size = 63 \[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=-\frac {\sqrt {x} \sqrt {1-a x}}{4 a}+\frac {1}{2} x^{3/2} \sqrt {1-a x}+\frac {\arcsin \left (\sqrt {a} \sqrt {x}\right )}{4 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {862, 52, 56, 222} \[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\frac {\arcsin \left (\sqrt {a} \sqrt {x}\right )}{4 a^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {1-a x}-\frac {\sqrt {x} \sqrt {1-a x}}{4 a} \]

[In]

Int[(Sqrt[x]*Sqrt[1 - a^2*x^2])/Sqrt[1 + a*x],x]

[Out]

-1/4*(Sqrt[x]*Sqrt[1 - a*x])/a + (x^(3/2)*Sqrt[1 - a*x])/2 + ArcSin[Sqrt[a]*Sqrt[x]]/(4*a^(3/2))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
 a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

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

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\frac {\sqrt {a} \sqrt {x} \sqrt {1-a x} (-1+2 a x)+\arcsin \left (\sqrt {a} \sqrt {x}\right )}{4 a^{3/2}} \]

[In]

Integrate[(Sqrt[x]*Sqrt[1 - a^2*x^2])/Sqrt[1 + a*x],x]

[Out]

(Sqrt[a]*Sqrt[x]*Sqrt[1 - a*x]*(-1 + 2*a*x) + ArcSin[Sqrt[a]*Sqrt[x]])/(4*a^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(43)=86\).

Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.46

method result size
default \(\frac {\sqrt {x}\, \sqrt {-a^{2} x^{2}+1}\, \left (4 a^{\frac {3}{2}} x \sqrt {-x \left (a x -1\right )}-2 \sqrt {a}\, \sqrt {-x \left (a x -1\right )}+\arctan \left (\frac {2 a x -1}{2 \sqrt {a}\, \sqrt {-x \left (a x -1\right )}}\right )\right )}{8 a^{\frac {3}{2}} \sqrt {a x +1}\, \sqrt {-x \left (a x -1\right )}}\) \(92\)
risch \(-\frac {\left (2 a x -1\right ) \sqrt {x}\, \left (a x -1\right ) \sqrt {\frac {x \left (-a^{2} x^{2}+1\right )}{a x +1}}\, \sqrt {a x +1}}{4 a \sqrt {-x \left (a x -1\right )}\, \sqrt {-a^{2} x^{2}+1}}+\frac {\arctan \left (\frac {\sqrt {a}\, \left (x -\frac {1}{2 a}\right )}{\sqrt {-a \,x^{2}+x}}\right ) \sqrt {\frac {x \left (-a^{2} x^{2}+1\right )}{a x +1}}\, \sqrt {a x +1}}{8 a^{\frac {3}{2}} \sqrt {x}\, \sqrt {-a^{2} x^{2}+1}}\) \(141\)

[In]

int(x^(1/2)*(-a^2*x^2+1)^(1/2)/(a*x+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (43) = 86\).

Time = 0.33 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.51 \[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\left [\frac {4 \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a^{2} x - a\right )} \sqrt {a x + 1} \sqrt {x} - {\left (a x + 1\right )} \sqrt {-a} \log \left (-\frac {8 \, a^{3} x^{3} - 4 \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x - 1\right )} \sqrt {a x + 1} \sqrt {-a} \sqrt {x} - 7 \, a x + 1}{a x + 1}\right )}{16 \, {\left (a^{3} x + a^{2}\right )}}, \frac {2 \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a^{2} x - a\right )} \sqrt {a x + 1} \sqrt {x} - {\left (a x + 1\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {a x + 1} \sqrt {a} \sqrt {x}}{2 \, a^{2} x^{2} + a x - 1}\right )}{8 \, {\left (a^{3} x + a^{2}\right )}}\right ] \]

[In]

integrate(x^(1/2)*(-a^2*x^2+1)^(1/2)/(a*x+1)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(4*sqrt(-a^2*x^2 + 1)*(2*a^2*x - a)*sqrt(a*x + 1)*sqrt(x) - (a*x + 1)*sqrt(-a)*log(-(8*a^3*x^3 - 4*sqrt(
-a^2*x^2 + 1)*(2*a*x - 1)*sqrt(a*x + 1)*sqrt(-a)*sqrt(x) - 7*a*x + 1)/(a*x + 1)))/(a^3*x + a^2), 1/8*(2*sqrt(-
a^2*x^2 + 1)*(2*a^2*x - a)*sqrt(a*x + 1)*sqrt(x) - (a*x + 1)*sqrt(a)*arctan(2*sqrt(-a^2*x^2 + 1)*sqrt(a*x + 1)
*sqrt(a)*sqrt(x)/(2*a^2*x^2 + a*x - 1)))/(a^3*x + a^2)]

Sympy [F]

\[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\int \frac {\sqrt {x} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\sqrt {a x + 1}}\, dx \]

[In]

integrate(x**(1/2)*(-a**2*x**2+1)**(1/2)/(a*x+1)**(1/2),x)

[Out]

Integral(sqrt(x)*sqrt(-(a*x - 1)*(a*x + 1))/sqrt(a*x + 1), x)

Maxima [F]

\[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {x}}{\sqrt {a x + 1}} \,d x } \]

[In]

integrate(x^(1/2)*(-a^2*x^2+1)^(1/2)/(a*x+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*x^2 + 1)*sqrt(x)/sqrt(a*x + 1), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^(1/2)*(-a^2*x^2+1)^(1/2)/(a*x+1)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x} \sqrt {1-a^2 x^2}}{\sqrt {1+a x}} \, dx=\int \frac {\sqrt {x}\,\sqrt {1-a^2\,x^2}}{\sqrt {a\,x+1}} \,d x \]

[In]

int((x^(1/2)*(1 - a^2*x^2)^(1/2))/(a*x + 1)^(1/2),x)

[Out]

int((x^(1/2)*(1 - a^2*x^2)^(1/2))/(a*x + 1)^(1/2), x)