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

Optimal. Leaf size=24 \[ \frac {2 \tan ^{-1}\left (\sqrt {\frac {a+b x}{c-b x}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1961, 12, 203} \[ \frac {2 \tan ^{-1}\left (\sqrt {\frac {a+b x}{c-b x}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1961

Int[(u_)^(r_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[SimplifyIntegrand[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(1/n -
1)*(u /. x -> (-(a*e) + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r)/(b*e - d*x^q)^(1/n + 1), x], x], x, ((e*(a + b*x
^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && PolynomialQ[u, x] && FractionQ[p] && IntegerQ[1/
n] && IntegerQ[r]

Rubi steps

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

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Mathematica [B]  time = 0.23, size = 93, normalized size = 3.88 \[ \frac {2 b \sqrt {c-b x} \sqrt {\frac {a+b x}{c-b x}} \sin ^{-1}\left (\frac {b \sqrt {c-b x}}{\sqrt {-b} \sqrt {-b (a+c)}}\right )}{(-b)^{3/2} \sqrt {-b (a+c)} \sqrt {\frac {a+b x}{a+c}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 24, normalized size = 1.00 \[ \frac {2 \, \arctan \left (\sqrt {-\frac {b x + a}{b x - c}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.50, size = 41, normalized size = 1.71 \[ -\frac {\arcsin \left (-\frac {2 \, b x + a - c}{a + c}\right ) \mathrm {sgn}\left (-a b - b c\right ) \mathrm {sgn}\left (b x - c\right )}{{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(-(2*b*x + a - c)/(a + c))*sgn(-a*b - b*c)*sgn(b*x - c)/abs(b)

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maple [B]  time = 0.03, size = 85, normalized size = 3.54 \[ -\frac {\left (b x -c \right ) \sqrt {-\frac {b x +a}{b x -c}}\, \arctan \left (\frac {\sqrt {b^{2}}\, \left (2 b x +a -c \right )}{2 \sqrt {-\left (b x +a \right ) \left (b x -c \right )}\, b}\right )}{\sqrt {b^{2}}\, \sqrt {-\left (b x +a \right ) \left (b x -c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.94, size = 24, normalized size = 1.00 \[ \frac {2 \, \arctan \left (\sqrt {-\frac {b x + a}{b x - c}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 36, normalized size = 1.50 \[ -\frac {2\,\sqrt {-b}\,\mathrm {atanh}\left (\frac {\sqrt {-b}\,\sqrt {\frac {a+b\,x}{c-b\,x}}}{\sqrt {b}}\right )}{b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a + b x}{- b x + c}}}{a + b x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((a + b*x)/(-b*x + c))/(a + b*x), x)

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