3.111 \(\int \frac{d+e x+f x^2}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=116 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-4 c (a f+b e)+3 b^2 f+8 c^2 d\right )}{8 c^{5/2}}+\frac{\sqrt{a+b x+c x^2} (4 c e-3 b f)}{4 c^2}+\frac{f x \sqrt{a+b x+c x^2}}{2 c} \]

[Out]

((4*c*e - 3*b*f)*Sqrt[a + b*x + c*x^2])/(4*c^2) + (f*x*Sqrt[a + b*x + c*x^2])/(2*c) + ((8*c^2*d + 3*b^2*f - 4*
c*(b*e + a*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2))

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Rubi [A]  time = 0.110769, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1661, 640, 621, 206} \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-4 c (a f+b e)+3 b^2 f+8 c^2 d\right )}{8 c^{5/2}}+\frac{\sqrt{a+b x+c x^2} (4 c e-3 b f)}{4 c^2}+\frac{f x \sqrt{a+b x+c x^2}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

((4*c*e - 3*b*f)*Sqrt[a + b*x + c*x^2])/(4*c^2) + (f*x*Sqrt[a + b*x + c*x^2])/(2*c) + ((8*c^2*d + 3*b^2*f - 4*
c*(b*e + a*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2))

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

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]

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 \frac{d+e x+f x^2}{\sqrt{a+b x+c x^2}} \, dx &=\frac{f x \sqrt{a+b x+c x^2}}{2 c}+\frac{\int \frac{2 c d-a f+\frac{1}{2} (4 c e-3 b f) x}{\sqrt{a+b x+c x^2}} \, dx}{2 c}\\ &=\frac{(4 c e-3 b f) \sqrt{a+b x+c x^2}}{4 c^2}+\frac{f x \sqrt{a+b x+c x^2}}{2 c}+\frac{\left (2 c (2 c d-a f)-\frac{1}{2} b (4 c e-3 b f)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{4 c^2}\\ &=\frac{(4 c e-3 b f) \sqrt{a+b x+c x^2}}{4 c^2}+\frac{f x \sqrt{a+b x+c x^2}}{2 c}+\frac{\left (2 c (2 c d-a f)-\frac{1}{2} b (4 c e-3 b f)\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 )}{2 c^2}\\ &=\frac{(4 c e-3 b f) \sqrt{a+b x+c x^2}}{4 c^2}+\frac{f x \sqrt{a+b x+c x^2}}{2 c}+\frac{\left (8 c^2 d+3 b^2 f-4 c (b e+a f)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.150978, size = 96, normalized size = 0.83 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (-4 c (a f+b e)+3 b^2 f+8 c^2 d\right )}{8 c^{5/2}}+\frac{\sqrt{a+x (b+c x)} (-3 b f+4 c e+2 c f x)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

((4*c*e - 3*b*f + 2*c*f*x)*Sqrt[a + x*(b + c*x)])/(4*c^2) + ((8*c^2*d + 3*b^2*f - 4*c*(b*e + a*f))*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8*c^(5/2))

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Maple [A]  time = 0.062, size = 185, normalized size = 1.6 \begin{align*}{\frac{fx}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,bf}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}f}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{af}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{e}{c}\sqrt{c{x}^{2}+bx+a}}-{\frac{be}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{d\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*f*x*(c*x^2+b*x+a)^(1/2)/c-3/4*f*b/c^2*(c*x^2+b*x+a)^(1/2)+3/8*f*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+
b*x+a)^(1/2))-1/2*f*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+e/c*(c*x^2+b*x+a)^(1/2)-1/2*e*b/c^(3
/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+d*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92911, size = 549, normalized size = 4.73 \begin{align*} \left [-\frac{{\left (8 \, c^{2} d - 4 \, b c e +{\left (3 \, b^{2} - 4 \, a c\right )} f\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} f x + 4 \, c^{2} e - 3 \, b c f\right )} \sqrt{c x^{2} + b x + a}}{16 \, c^{3}}, -\frac{{\left (8 \, c^{2} d - 4 \, b c e +{\left (3 \, b^{2} - 4 \, a c\right )} f\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} f x + 4 \, c^{2} e - 3 \, b c f\right )} \sqrt{c x^{2} + b x + a}}{8 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((8*c^2*d - 4*b*c*e + (3*b^2 - 4*a*c)*f)*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*f*x + 4*c^2*e - 3*b*c*f)*sqrt(c*x^2 + b*x + a))/c^3, -1/8*((8*c^2*d
- 4*b*c*e + (3*b^2 - 4*a*c)*f)*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*f*x + 4*c^2*e - 3*b*c*f)*sqrt(c*x^2 + b*x + a))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x + f*x**2)/sqrt(a + b*x + c*x**2), x)

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Giac [A]  time = 1.29089, size = 132, normalized size = 1.14 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x + a}{\left (\frac{2 \, f x}{c} - \frac{3 \, b f - 4 \, c e}{c^{2}}\right )} - \frac{{\left (8 \, c^{2} d + 3 \, b^{2} f - 4 \, a c f - 4 \, b c e\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{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x + a)*(2*f*x/c - (3*b*f - 4*c*e)/c^2) - 1/8*(8*c^2*d + 3*b^2*f - 4*a*c*f - 4*b*c*e)*log(ab
s(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)