3.11.50 \(\int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=55 \[ \frac {\tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {c} d \sqrt {b^2-4 a c}} \]

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {688, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {c} d \sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]]/(Sqrt[c]*Sqrt[b^2 - 4*a*c]*d)

Rule 205

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

Rule 688

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
- 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx &=(4 c) \operatorname {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )\\ &=\frac {\tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {c} \sqrt {b^2-4 a c} d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.98 \begin {gather*} \frac {\tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {c} d \sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

ArcTan[(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])/Sqrt[b^2 - 4*a*c]]/(Sqrt[c]*Sqrt[b^2 - 4*a*c]*d)

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IntegrateAlgebraic [A]  time = 0.35, size = 87, normalized size = 1.58 \begin {gather*} -\frac {2 \tan ^{-1}\left (-\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {c} d \sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

(-2*ArcTan[b/Sqrt[b^2 - 4*a*c] + (2*c*x)/Sqrt[b^2 - 4*a*c] - (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*
c]])/(Sqrt[c]*Sqrt[b^2 - 4*a*c]*d)

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fricas [A]  time = 0.43, size = 174, normalized size = 3.16 \begin {gather*} \left [-\frac {\sqrt {-b^{2} c + 4 \, a c^{2}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {-b^{2} c + 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d}, -\frac {\arctan \left (\frac {\sqrt {b^{2} c - 4 \, a c^{2}} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right )}{\sqrt {b^{2} c - 4 \, a c^{2}} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-b^2*c + 4*a*c^2)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(-b^2*c + 4*a*c^2)*sqrt(c*x^2 + b
*x + a))/(4*c^2*x^2 + 4*b*c*x + b^2))/((b^2*c - 4*a*c^2)*d), -arctan(1/2*sqrt(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*
x + a)/(c^2*x^2 + b*c*x + a*c))/(sqrt(b^2*c - 4*a*c^2)*d)]

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giac [A]  time = 0.24, size = 65, normalized size = 1.18 \begin {gather*} \frac {2 \, \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{\sqrt {b^{2} c - 4 \, a c^{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^2))/(sqrt(b^2*c - 4*a*c^2)*
d)

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maple [B]  time = 0.05, size = 101, normalized size = 1.84 \begin {gather*} -\frac {\ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\sqrt {\frac {4 a c -b^{2}}{c}}\, c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\left (b\,d+2\,c\,d\,x\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{b \sqrt {a + b x + c x^{2}} + 2 c x \sqrt {a + b x + c x^{2}}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(b*sqrt(a + b*x + c*x**2) + 2*c*x*sqrt(a + b*x + c*x**2)), x)/d

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