3.1.24 \(\int \frac {x (1+a x)}{\sqrt {a x} \sqrt {1-a x}} \, dx\)

Optimal. Leaf size=63 \[ -\frac {\sqrt {1-a x} (a x)^{3/2}}{2 a^2}-\frac {7 \sqrt {1-a x} \sqrt {a x}}{4 a^2}-\frac {7 \sin ^{-1}(1-2 a x)}{8 a^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {16, 80, 50, 53, 619, 216} \begin {gather*} -\frac {\sqrt {1-a x} (a x)^{3/2}}{2 a^2}-\frac {7 \sqrt {1-a x} \sqrt {a x}}{4 a^2}-\frac {7 \sin ^{-1}(1-2 a x)}{8 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*Sqrt[a*x]*Sqrt[1 - a*x])/(4*a^2) - ((a*x)^(3/2)*Sqrt[1 - a*x])/(2*a^2) - (7*ArcSin[1 - 2*a*x])/(8*a^2)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 50

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 53

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

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 216

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 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps

\begin {align*} \int \frac {x (1+a x)}{\sqrt {a x} \sqrt {1-a x}} \, dx &=\frac {\int \frac {\sqrt {a x} (1+a x)}{\sqrt {1-a x}} \, dx}{a}\\ &=-\frac {(a x)^{3/2} \sqrt {1-a x}}{2 a^2}+\frac {7 \int \frac {\sqrt {a x}}{\sqrt {1-a x}} \, dx}{4 a}\\ &=-\frac {7 \sqrt {a x} \sqrt {1-a x}}{4 a^2}-\frac {(a x)^{3/2} \sqrt {1-a x}}{2 a^2}+\frac {7 \int \frac {1}{\sqrt {a x} \sqrt {1-a x}} \, dx}{8 a}\\ &=-\frac {7 \sqrt {a x} \sqrt {1-a x}}{4 a^2}-\frac {(a x)^{3/2} \sqrt {1-a x}}{2 a^2}+\frac {7 \int \frac {1}{\sqrt {a x-a^2 x^2}} \, dx}{8 a}\\ &=-\frac {7 \sqrt {a x} \sqrt {1-a x}}{4 a^2}-\frac {(a x)^{3/2} \sqrt {1-a x}}{2 a^2}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,a-2 a^2 x\right )}{8 a^3}\\ &=-\frac {7 \sqrt {a x} \sqrt {1-a x}}{4 a^2}-\frac {(a x)^{3/2} \sqrt {1-a x}}{2 a^2}-\frac {7 \sin ^{-1}(1-2 a x)}{8 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 73, normalized size = 1.16 \begin {gather*} \frac {\sqrt {a} x \left (2 a^2 x^2+5 a x-7\right )+7 \sqrt {x} \sqrt {1-a x} \sin ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{4 a^{3/2} \sqrt {-a x (a x-1)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*x*(-7 + 5*a*x + 2*a^2*x^2) + 7*Sqrt[x]*Sqrt[1 - a*x]*ArcSin[Sqrt[a]*Sqrt[x]])/(4*a^(3/2)*Sqrt[-(a*x*(
-1 + a*x))])

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IntegrateAlgebraic [A]  time = 0.08, size = 84, normalized size = 1.33 \begin {gather*} -\frac {\sqrt {1-a x} \left (\frac {7 (1-a x)}{a x}+9\right )}{4 a^2 \sqrt {a x} \left (\frac {1-a x}{a x}+1\right )^2}-\frac {7 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x}}\right )}{4 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(1 + a*x))/(Sqrt[a*x]*Sqrt[1 - a*x]),x]

[Out]

-1/4*(Sqrt[1 - a*x]*(9 + (7*(1 - a*x))/(a*x)))/(a^2*Sqrt[a*x]*(1 + (1 - a*x)/(a*x))^2) - (7*ArcTan[Sqrt[1 - a*
x]/Sqrt[a*x]])/(4*a^2)

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fricas [A]  time = 1.31, size = 49, normalized size = 0.78 \begin {gather*} -\frac {{\left (2 \, a x + 7\right )} \sqrt {a x} \sqrt {-a x + 1} + 7 \, \arctan \left (\frac {\sqrt {a x} \sqrt {-a x + 1}}{a x}\right )}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((2*a*x + 7)*sqrt(a*x)*sqrt(-a*x + 1) + 7*arctan(sqrt(a*x)*sqrt(-a*x + 1)/(a*x)))/a^2

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giac [A]  time = 1.28, size = 40, normalized size = 0.63 \begin {gather*} -\frac {\sqrt {a x} \sqrt {-a x + 1} {\left (2 \, x + \frac {7}{a}\right )} - \frac {7 \, \arcsin \left (\sqrt {a x}\right )}{a}}{4 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(a*x)*sqrt(-a*x + 1)*(2*x + 7/a) - 7*arcsin(sqrt(a*x))/a)/a

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maple [C]  time = 0.02, size = 90, normalized size = 1.43 \begin {gather*} -\frac {\sqrt {-a x +1}\, \left (4 \sqrt {-\left (a x -1\right ) a x}\, a x \,\mathrm {csgn}\relax (a )-7 \arctan \left (\frac {\left (2 a x -1\right ) \mathrm {csgn}\relax (a )}{2 \sqrt {-\left (a x -1\right ) a x}}\right )+14 \sqrt {-\left (a x -1\right ) a x}\, \mathrm {csgn}\relax (a )\right ) x \,\mathrm {csgn}\relax (a )}{8 \sqrt {a x}\, \sqrt {-\left (a x -1\right ) a x}\, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-a*x+1)^(1/2)*x/a*(4*(-(a*x-1)*a*x)^(1/2)*a*x*csgn(a)+14*(-(a*x-1)*a*x)^(1/2)*csgn(a)-7*arctan(1/2*(2*a*
x-1)/(-(a*x-1)*a*x)^(1/2)*csgn(a)))*csgn(a)/(a*x)^(1/2)/(-(a*x-1)*a*x)^(1/2)

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maxima [A]  time = 0.96, size = 61, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + a x} x}{2 \, a} - \frac {7 \, \arcsin \left (-\frac {2 \, a^{2} x - a}{a}\right )}{8 \, a^{2}} - \frac {7 \, \sqrt {-a^{2} x^{2} + a x}}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-a^2*x^2 + a*x)*x/a - 7/8*arcsin(-(2*a^2*x - a)/a)/a^2 - 7/4*sqrt(-a^2*x^2 + a*x)/a^2

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mupad [B]  time = 4.53, size = 191, normalized size = 3.03 \begin {gather*} \frac {7\,\mathrm {atan}\left (\frac {\sqrt {a\,x}}{\sqrt {1-a\,x}-1}\right )}{2\,a^2}-\frac {\frac {2\,\sqrt {a\,x}}{\sqrt {1-a\,x}-1}-\frac {2\,{\left (a\,x\right )}^{3/2}}{{\left (\sqrt {1-a\,x}-1\right )}^3}}{a^2\,{\left (\frac {a\,x}{{\left (\sqrt {1-a\,x}-1\right )}^2}+1\right )}^2}-\frac {\frac {3\,\sqrt {a\,x}}{2\,\left (\sqrt {1-a\,x}-1\right )}+\frac {11\,{\left (a\,x\right )}^{3/2}}{2\,{\left (\sqrt {1-a\,x}-1\right )}^3}-\frac {11\,{\left (a\,x\right )}^{5/2}}{2\,{\left (\sqrt {1-a\,x}-1\right )}^5}-\frac {3\,{\left (a\,x\right )}^{7/2}}{2\,{\left (\sqrt {1-a\,x}-1\right )}^7}}{a^2\,{\left (\frac {a\,x}{{\left (\sqrt {1-a\,x}-1\right )}^2}+1\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*atan((a*x)^(1/2)/((1 - a*x)^(1/2) - 1)))/(2*a^2) - ((2*(a*x)^(1/2))/((1 - a*x)^(1/2) - 1) - (2*(a*x)^(3/2))
/((1 - a*x)^(1/2) - 1)^3)/(a^2*((a*x)/((1 - a*x)^(1/2) - 1)^2 + 1)^2) - ((3*(a*x)^(1/2))/(2*((1 - a*x)^(1/2) -
1)) + (11*(a*x)^(3/2))/(2*((1 - a*x)^(1/2) - 1)^3) - (11*(a*x)^(5/2))/(2*((1 - a*x)^(1/2) - 1)^5) - (3*(a*x)^
(7/2))/(2*((1 - a*x)^(1/2) - 1)^7))/(a^2*((a*x)/((1 - a*x)^(1/2) - 1)^2 + 1)^4)

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sympy [C]  time = 20.77, size = 269, normalized size = 4.27 \begin {gather*} a \left (\begin {cases} - \frac {3 i \operatorname {acosh}{\left (\sqrt {a} \sqrt {x} \right )}}{4 a^{3}} - \frac {i x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {a x - 1}} - \frac {i x^{\frac {3}{2}}}{4 a^{\frac {3}{2}} \sqrt {a x - 1}} + \frac {3 i \sqrt {x}}{4 a^{\frac {5}{2}} \sqrt {a x - 1}} & \text {for}\: \left |{a x}\right | > 1 \\\frac {3 \operatorname {asin}{\left (\sqrt {a} \sqrt {x} \right )}}{4 a^{3}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {- a x + 1}} + \frac {x^{\frac {3}{2}}}{4 a^{\frac {3}{2}} \sqrt {- a x + 1}} - \frac {3 \sqrt {x}}{4 a^{\frac {5}{2}} \sqrt {- a x + 1}} & \text {otherwise} \end {cases}\right ) + \begin {cases} - \frac {i \operatorname {acosh}{\left (\sqrt {a} \sqrt {x} \right )}}{a^{2}} - \frac {i \sqrt {x} \sqrt {a x - 1}}{a^{\frac {3}{2}}} & \text {for}\: \left |{a x}\right | > 1 \\\frac {\operatorname {asin}{\left (\sqrt {a} \sqrt {x} \right )}}{a^{2}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} \sqrt {- a x + 1}} - \frac {\sqrt {x}}{a^{\frac {3}{2}} \sqrt {- a x + 1}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((-3*I*acosh(sqrt(a)*sqrt(x))/(4*a**3) - I*x**(5/2)/(2*sqrt(a)*sqrt(a*x - 1)) - I*x**(3/2)/(4*a**(3
/2)*sqrt(a*x - 1)) + 3*I*sqrt(x)/(4*a**(5/2)*sqrt(a*x - 1)), Abs(a*x) > 1), (3*asin(sqrt(a)*sqrt(x))/(4*a**3)
+ x**(5/2)/(2*sqrt(a)*sqrt(-a*x + 1)) + x**(3/2)/(4*a**(3/2)*sqrt(-a*x + 1)) - 3*sqrt(x)/(4*a**(5/2)*sqrt(-a*x
+ 1)), True)) + Piecewise((-I*acosh(sqrt(a)*sqrt(x))/a**2 - I*sqrt(x)*sqrt(a*x - 1)/a**(3/2), Abs(a*x) > 1),
(asin(sqrt(a)*sqrt(x))/a**2 + x**(3/2)/(sqrt(a)*sqrt(-a*x + 1)) - sqrt(x)/(a**(3/2)*sqrt(-a*x + 1)), True))

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