\(\int \sqrt {c+b x+a x^2} \, dx\) [955]

Optimal result

Integrand size = 14, antiderivative size = 73 \[ \int \sqrt {c+b x+a x^2} \, dx=\frac {(b+2 a x) \sqrt {c+b x+a x^2}}{4 a}+\frac {\left (b^2-4 a c\right ) \log \left (b+2 a x-2 \sqrt {a} \sqrt {c+b x+a x^2}\right )}{8 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {626, 635, 212} \[ \int \sqrt {c+b x+a x^2} \, dx=\frac {(2 a x+b) \sqrt {a x^2+b x+c}}{4 a}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a x+b}{2 \sqrt {a} \sqrt {a x^2+b x+c}}\right )}{8 a^{3/2}} \]

[In]

Int[Sqrt[c + b*x + a*x^2],x]

[Out]

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

Rule 212

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

Rule 626

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 635

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]

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \sqrt {c+b x+a x^2} \, dx=\frac {\sqrt {a} (b+2 a x) \sqrt {c+x (b+a x)}+\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {c}-\sqrt {c+x (b+a x)}}\right )}{4 a^{3/2}} \]

[In]

Integrate[Sqrt[c + b*x + a*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\left (2 a x +b \right ) \sqrt {a \,x^{2}+b x +c}}{4 a}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right )}{8 a^{\frac {3}{2}}}\)	\(65\)
risch	\(\frac {\left (2 a x +b \right ) \sqrt {a \,x^{2}+b x +c}}{4 a}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right )}{8 a^{\frac {3}{2}}}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.42 \[ \int \sqrt {c+b x+a x^2} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {a} \log \left (-8 \, a^{2} x^{2} - 8 \, a b x - 4 \, \sqrt {a x^{2} + b x + c} {\left (2 \, a x + b\right )} \sqrt {a} - b^{2} - 4 \, a c\right ) - 4 \, {\left (2 \, a^{2} x + a b\right )} \sqrt {a x^{2} + b x + c}}{16 \, a^{2}}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {a x^{2} + b x + c} {\left (2 \, a x + b\right )} \sqrt {-a}}{2 \, {\left (a^{2} x^{2} + a b x + a c\right )}}\right ) + 2 \, {\left (2 \, a^{2} x + a b\right )} \sqrt {a x^{2} + b x + c}}{8 \, a^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.63 \[ \int \sqrt {c+b x+a x^2} \, dx=\begin {cases} \left (\frac {c}{2} - \frac {b^{2}}{8 a}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {a} \sqrt {a x^{2} + b x + c} + 2 a x + b \right )}}{\sqrt {a}} & \text {for}\: c - \frac {b^{2}}{4 a} \neq 0 \\\frac {\left (x + \frac {b}{2 a}\right ) \log {\left (x + \frac {b}{2 a} \right )}}{\sqrt {a \left (x + \frac {b}{2 a}\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \left (\frac {x}{2} + \frac {b}{4 a}\right ) \sqrt {a x^{2} + b x + c} & \text {for}\: a \neq 0 \\\frac {2 \left (b x + c\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((c/2 - b**2/(8*a))*Piecewise((log(2*sqrt(a)*sqrt(a*x**2 + b*x + c) + 2*a*x + b)/sqrt(a), Ne(c - b**
2/(4*a), 0)), ((x + b/(2*a))*log(x + b/(2*a))/sqrt(a*(x + b/(2*a))**2), True)) + (x/2 + b/(4*a))*sqrt(a*x**2 +
b*x + c), Ne(a, 0)), (2*(b*x + c)**(3/2)/(3*b), Ne(b, 0)), (sqrt(c)*x, True))

Maxima [F(-2)]

Exception generated. \[ \int \sqrt {c+b x+a x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a*x^2+b*x+c)^(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 deta

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \sqrt {c+b x+a x^2} \, dx=\frac {1}{4} \, \sqrt {a x^{2} + b x + c} {\left (2 \, x + \frac {b}{a}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x + c}\right )} \sqrt {a} + b \right |}\right )}{8 \, a^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \sqrt {c+b x+a x^2} \, dx=\left (\frac {x}{2}+\frac {b}{4\,a}\right )\,\sqrt {a\,x^2+b\,x+c}+\frac {\ln \left (\frac {\frac {b}{2}+a\,x}{\sqrt {a}}+\sqrt {a\,x^2+b\,x+c}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,a^{3/2}} \]

[In]

int((c + b*x + a*x^2)^(1/2),x)

[Out]

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