3.2.25 \(\int \frac {1}{\sqrt {\frac {b^2+4 c}{4 c}+b x+c x^2}} \, dx\) [125]

Optimal. Leaf size=22 \[ \frac {\sinh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{\sqrt {c}} \]

[Out]

arcsinh(1/2*(2*c*x+b)/c^(1/2))/c^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {633, 221} \begin {gather*} \frac {\sinh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c}}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(b^2 + 4*c)/(4*c) + b*x + c*x^2],x]

[Out]

ArcSinh[(b + 2*c*x)/(2*Sqrt[c])]/Sqrt[c]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 633

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

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 44, normalized size = 2.00 \begin {gather*} -\frac {\log \left (b+2 c x-\sqrt {c} \sqrt {4+\frac {b^2}{c}+4 b x+4 c x^2}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(b^2 + 4*c)/(4*c) + b*x + c*x^2],x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(50\) vs. \(2(16)=32\).
time = 0.68, size = 51, normalized size = 2.32

method	result	size
default	\(\frac {\ln \left (\frac {\left (4 c x +2 b \right ) \sqrt {4}}{4 \sqrt {c}}+\sqrt {\frac {b^{2}+4 c}{c}+4 b x +4 c \,x^{2}}\right ) \sqrt {4}}{2 \sqrt {c}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/((b^2+4*c)/c+4*b*x+4*c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 16, normalized size = 0.73 \begin {gather*} \frac {\operatorname {arsinh}\left (\frac {2 \, c x + b}{2 \, \sqrt {c}}\right )}{\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((b^2+4*c)/c+4*b*x+4*c*x^2)^(1/2),x, algorithm="maxima")

[Out]

arcsinh(1/2*(2*c*x + b)/sqrt(c))/sqrt(c)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (16) = 32\).
time = 2.12, size = 137, normalized size = 6.23 \begin {gather*} \left [\frac {\log \left (-4 \, c^{2} x^{2} - 4 \, b c x - b^{2} - {\left (2 \, c x + b\right )} \sqrt {c} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2} + 4 \, c}{c}} - 2 \, c\right )}{2 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2} + 4 \, c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2} + 4 \, c}\right )}{c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((b^2+4*c)/c+4*b*x+4*c*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*log(-4*c^2*x^2 - 4*b*c*x - b^2 - (2*c*x + b)*sqrt(c)*sqrt((4*c^2*x^2 + 4*b*c*x + b^2 + 4*c)/c) - 2*c)/sqr
t(c), -sqrt(-c)*arctan((2*c*x + b)*sqrt(-c)*sqrt((4*c^2*x^2 + 4*b*c*x + b^2 + 4*c)/c)/(4*c^2*x^2 + 4*b*c*x + b
^2 + 4*c))/c]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \int \frac {1}{\sqrt {\frac {b^{2}}{c} + 4 b x + 4 c x^{2} + 4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((b**2+4*c)/c+4*b*x+4*c*x**2)**(1/2),x)

[Out]

2*Integral(1/sqrt(b**2/c + 4*b*x + 4*c*x**2 + 4), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (16) = 32\).
time = 0.75, size = 60, normalized size = 2.73 \begin {gather*} -\frac {\log \left ({\left | -b c^{2} - {\left (2 \, \sqrt {c^{3}} x - \sqrt {4 \, c^{3} x^{2} + 4 \, b c^{2} x + b^{2} c + 4 \, c^{2}}\right )} \sqrt {c} {\left | c \right |} \right |}\right )}{\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/((b^2+4*c)/c+4*b*x+4*c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.42, size = 40, normalized size = 1.82 \begin {gather*} \frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+\sqrt {\frac {b^2+4\,c}{c}+4\,b\,x+4\,c\,x^2}\right )}{\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/((4*c + b^2)/c + 4*b*x + 4*c*x^2)^(1/2),x)

[Out]

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

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