3.954 \(\int \frac {A+B x}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=67 \[ \frac {B \sqrt {a+b x+c x^2}}{c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}} \]

[Out]

-1/2*(-2*A*c+B*b)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)+B*(c*x^2+b*x+a)^(1/2)/c

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {640, 621, 206} \[ \frac {B \sqrt {a+b x+c x^2}}{c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sqrt[a + b*x + c*x^2])/c - ((b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c^(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 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 640

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 66, normalized size = 0.99 \[ \frac {(2 A c-b B) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 c^{3/2}}+\frac {B \sqrt {a+x (b+c x)}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 162, normalized size = 2.42 \[ \left [\frac {4 \, \sqrt {c x^{2} + b x + a} B c - {\left (B b - 2 \, A c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right )}{4 \, c^{2}}, \frac {2 \, \sqrt {c x^{2} + b x + a} B c + {\left (B b - 2 \, A c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right )}{2 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(c*x^2 + b*x + a)*B*c - (B*b - 2*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))/c^2, 1/2*(2*sqrt(c*x^2 + b*x + a)*B*c + (B*b - 2*A*c)*sqrt(-c)*arctan(1/2*s
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)))/c^2]

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giac [A]  time = 0.24, size = 62, normalized size = 0.93 \[ \frac {\sqrt {c x^{2} + b x + a} B}{c} + \frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 81, normalized size = 1.21 \[ \frac {A \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {B b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(c*x^2+b*x+a)^(1/2)/c-1/2*B*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+A*ln((c*x+1/2*b)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))/c^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(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 zero or nonzero?

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mupad [B]  time = 1.56, size = 80, normalized size = 1.19 \[ \frac {B\,\sqrt {c\,x^2+b\,x+a}}{c}+\frac {A\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{\sqrt {c}}-\frac {B\,b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{2\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{\sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/sqrt(a + b*x + c*x**2), x)

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