### 3.2336 $$\int \sqrt{a+b x+c x^2} \, dx$$

Optimal. Leaf size=75 $\frac{(b+2 c x) \sqrt{a+b x+c x^2}}{4 c}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{3/2}}$

[Out]

((b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2
])])/(8*c^(3/2))

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Rubi [A]  time = 0.0196978, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.214, Rules used = {612, 621, 206} $\frac{(b+2 c x) \sqrt{a+b x+c x^2}}{4 c}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[a + b*x + c*x^2],x]

[Out]

((b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2
])])/(8*c^(3/2))

Rule 612

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

Rule 621

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.064003, size = 71, normalized size = 0.95 $\frac{(b+2 c x) \sqrt{a+x (b+c x)}}{4 c}-\frac{\left (b^2-4 a c\right ) \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{8 c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[a + b*x + c*x^2],x]

[Out]

((b + 2*c*x)*Sqrt[a + x*(b + c*x)])/(4*c) - ((b^2 - 4*a*c)*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(
8*c^(3/2))

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Maple [A]  time = 0.042, size = 89, normalized size = 1.2 \begin{align*}{\frac{2\,cx+b}{4\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{a}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/2/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/8/c^(3/2)*ln((
1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.43896, size = 435, normalized size = 5.8 \begin{align*} \left [-\frac{{\left (b^{2} - 4 \, a c\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (2 \, c^{2} x + b c\right )} \sqrt{c x^{2} + b x + a}}{16 \, c^{2}}, \frac{{\left (b^{2} - 4 \, a c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (2 \, c^{2} x + b c\right )} \sqrt{c x^{2} + b x + a}}{8 \, c^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b x + c x^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*x + c*x**2), x)

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Giac [A]  time = 1.17612, size = 92, normalized size = 1.23 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x + a}{\left (2 \, x + \frac{b}{c}\right )} + \frac{{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x + a)*(2*x + b/c) + 1/8*(b^2 - 4*a*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(
c) - b))/c^(3/2)