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

Optimal. Leaf size=73 \[ \frac {\left (b^2-4 a c\right ) \log \left (-2 \sqrt {a} \sqrt {a x^2+b x+c}+2 a x+b\right )}{8 a^{3/2}}+\frac {(2 a x+b) \sqrt {a x^2+b x+c}}{4 a} \]

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.03, 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} \begin {gather*} \frac {(2 a x+b) \sqrt {a x^2+b x+c}}{4 a}-\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a x+b}{2 \sqrt {a} \sqrt {a x^2+b x+c}}\right )}{8 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[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 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])

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]

Rubi steps

\begin {align*} \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 ) \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 ) \operatorname {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 ) \tanh ^{-1}\left (\frac {b+2 a x}{2 \sqrt {a} \sqrt {c+b x+a x^2}}\right )}{8 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 71, normalized size = 0.97 \begin {gather*} \frac {(2 a x+b) \sqrt {x (a x+b)+c}}{4 a}-\frac {\left (b^2-4 a c\right ) \log \left (2 \sqrt {a} \sqrt {x (a x+b)+c}+2 a x+b\right )}{8 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + b*x + a*x^2],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 77, normalized size = 1.05 \begin {gather*} \frac {(b+2 a x) \sqrt {c+b x+a x^2}}{4 a}+\frac {\left (b^2-4 a c\right ) \log \left (a b+2 a^2 x-2 a^{3/2} \sqrt {c+b x+a x^2}\right )}{8 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.64, size = 177, normalized size = 2.42 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.38, size = 68, normalized size = 0.93 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.19, size = 65, normalized size = 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	\(\frac {\left (2 a x +b \right ) \sqrt {a \,x^{2}+b x +c}}{4 a}+\frac {\ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right ) c}{2 \sqrt {a}}-\frac {\ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right ) b^{2}}{8 a^{\frac {3}{2}}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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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((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 details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 0.12, size = 63, normalized size = 0.86 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x**2 + b*x + c), x)

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