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

Optimal. Leaf size=431 \[ -\frac{\sqrt{f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}+\frac{\sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 1.05133, antiderivative size = 431, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {989, 621, 206, 1032, 724} \[ -\frac{\sqrt{f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}+\frac{\sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 989

Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (e_.)*(x_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sq
rt[a + b*x + c*x^2], x], x] - Dist[1/f, Int[(c*d - a*f + (c*e - b*f)*x)/(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^
2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 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 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 1032

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbo
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + e*x + f*x^2])
, x], x] - Dist[(2*c*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b,
c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && PosQ[b^2 - 4*a*c]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x+c x^2}}{d+e x+f x^2} \, dx &=-\frac{\int \frac{c d-a f+(c e-b f) x}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}+\frac{c \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{f}\\ &=\frac{(2 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 )}{f}-\frac{\left (2 f (c d-a f)-(c e-b f) \left (e-\sqrt{e^2-4 d f}\right )\right ) \int \frac{1}{\left (e-\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+b x+c x^2}} \, dx}{f \sqrt{e^2-4 d f}}+\frac{\left (2 f (c d-a f)-(c e-b f) \left (e+\sqrt{e^2-4 d f}\right )\right ) \int \frac{1}{\left (e+\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+b x+c x^2}} \, dx}{f \sqrt{e^2-4 d f}}\\ &=\frac{\sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{f}+\frac{\left (2 \left (2 f (c d-a f)-(c e-b f) \left (e-\sqrt{e^2-4 d f}\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e-\sqrt{e^2-4 d f}\right )+4 c \left (e-\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{f \sqrt{e^2-4 d f}}-\frac{\left (2 \left (2 f (c d-a f)-(c e-b f) \left (e+\sqrt{e^2-4 d f}\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e+\sqrt{e^2-4 d f}\right )+4 c \left (e+\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{f \sqrt{e^2-4 d f}}\\ &=\frac{\sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{f}-\frac{\sqrt{c \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right )+f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )} \tanh ^{-1}\left (\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}+\frac{\sqrt{c \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )+f \left (2 a f-b \left (e+\sqrt{e^2-4 d f}\right )\right )} \tanh ^{-1}\left (\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}\\ \end{align*}

Mathematica [A]  time = 1.25489, size = 417, normalized size = 0.97 \[ \frac{\sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f-b \left (\sqrt{e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )-\sqrt{f \left (2 a f+b \sqrt{e^2-4 d f}+b (-e)\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+b \left (\sqrt{e^2-4 d f}-e+2 f x\right )+2 c x \left (\sqrt{e^2-4 d f}-e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f+b \sqrt{e^2-4 d f}+b (-e)\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f \sqrt{e^2-4 d f}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.433, size = 6019, normalized size = 14. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError