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

Optimal. Leaf size=488 \[ -\frac{\left (f (b e-4 a f)-\left (e-\sqrt{e^2-4 d f}\right ) (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \left (e^2-4 d f\right )^{3/2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{\left (f (b e-4 a f)-\left (\sqrt{e^2-4 d f}+e\right ) (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \left (e^2-4 d f\right )^{3/2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{(e+2 f x) \sqrt{a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )} \]

[Out]

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

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Rubi [A]  time = 2.92941, antiderivative size = 488, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {971, 1032, 724, 206} \[ -\frac{\left (f (b e-4 a f)-\left (e-\sqrt{e^2-4 d f}\right ) (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \left (e^2-4 d f\right )^{3/2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{\left (f (b e-4 a f)-\left (\sqrt{e^2-4 d f}+e\right ) (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} \left (e^2-4 d f\right )^{3/2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{(e+2 f x) \sqrt{a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 971

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((b +
2*c*x)*(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q)/((b^2 - 4*a*c)*(p + 1)), x] - Dist[1/((b^2 - 4*a*c)*(p
+ 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q - 1)*Simp[2*c*d*(2*p + 3) + b*e*q + (2*b*f*q + 2*c*e
*(2*p + q + 3))*x + 2*c*f*(2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !IGtQ[q, 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]

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

Mathematica [A]  time = 5.4495, size = 555, normalized size = 1.14 \[ \frac{4 f (e+2 f x) \sqrt{a+x (b+c x)}}{\left (e^2-4 d f\right ) \left (\sqrt{e^2-4 d f}-e-2 f x\right ) \left (\sqrt{e^2-4 d f}+e+2 f x\right )}+\frac{\left (c e \left (\sqrt{e^2-4 d f}-e\right )-f \left (4 a f+b \left (\sqrt{e^2-4 d f}-2 e\right )\right )\right ) \tanh ^{-1}\left (\frac{-4 a f+b \left (-\sqrt{e^2-4 d f}+e-2 f x\right )+2 c x \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \left (e^2-4 d f\right )^{3/2} \sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\left (f \left (4 a f-b \left (\sqrt{e^2-4 d f}+2 e\right )\right )+c e \left (\sqrt{e^2-4 d f}+e\right )\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (\sqrt{e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} \left (e^2-4 d f\right )^{3/2} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.371, size = 22287, normalized size = 45.7 \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)^2,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x + a}}{{\left (f x^{2} + e x + d\right )}^{2}}\,{d x} \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)^2,x, algorithm="maxima")

[Out]

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

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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)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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)**2,x)

[Out]

Timed out

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Giac [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)^2,x, algorithm="giac")

[Out]

Timed out