3.11.79 \(\int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx\)

Optimal. Leaf size=38 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{b \sqrt {c}} \]

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {63, 217, 203} \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{b \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

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

Rule 217

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 48, normalized size = 1.26 \begin {gather*} -\frac {2 \sqrt {a-b x} \tan ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {a+b x}}\right )}{b \sqrt {c (a-b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 39, normalized size = 1.03 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {a c-b c x}}{\sqrt {c} \sqrt {a+b x}}\right )}{b \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 108, normalized size = 2.84 \begin {gather*} \left [-\frac {\sqrt {-c} \log \left (2 \, b^{2} c x^{2} - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right )}{2 \, b c}, -\frac {\arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right )}{b \sqrt {c}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-c)*log(2*b^2*c*x^2 - 2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(-c)*x - a^2*c)/(b*c), -arctan(sqrt(
-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - a^2*c))/(b*sqrt(c))]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.00, size = 71, normalized size = 1.87 \begin {gather*} \frac {\sqrt {\left (b x +a \right ) \left (-b c x +a c \right )}\, \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right )}{\sqrt {b x +a}\, \sqrt {-b c x +a c}\, \sqrt {b^{2} c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.93, size = 14, normalized size = 0.37 \begin {gather*} \frac {\arcsin \left (\frac {b x}{a}\right )}{b \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(b*x/a)/(b*sqrt(c))

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mupad [B]  time = 0.18, size = 53, normalized size = 1.39 \begin {gather*} -\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {b^2\,c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {b^2\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*atan((b*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/((b^2*c)^(1/2)*((a + b*x)^(1/2) - a^(1/2)))))/(b^2*c)^(1/2)

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sympy [C]  time = 4.69, size = 90, normalized size = 2.37 \begin {gather*} - \frac {i {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b \sqrt {c}} + \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), a**2/(b**2*x**2))/(4*pi**(3/2)*b*sq
rt(c)) + meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), a**2*exp_polar(-2*I*pi)/(
b**2*x**2))/(4*pi**(3/2)*b*sqrt(c))

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