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

Optimal. Leaf size=549 \[ -\frac{\left (\left (e-\sqrt{e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{\left (\left (\sqrt{e^2-4 d f}+e\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{(2 c e-b f) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}+\frac{\sqrt{a+b x+c x^2}}{f} \]

[Out]

Sqrt[a + b*x + c*x^2]/f - ((2*c*e - b*f)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c]*f^
2) - ((2*d*f*(c*e - b*f) + (e - Sqrt[e^2 - 4*d*f])*(f*(b*e - a*f) - c*(e^2 - d*f)))*ArcTanh[(4*a*f - b*(e - Sq
rt[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^2*Sqrt[e^2 - 4*d*f]*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]]) + ((2*d*f*(c*e - b*f) + (e + Sqrt[e^2 - 4*d*f])*(f*(b*e - a*
f) - c*(e^2 - 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*Sq
rt[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^2*Sqrt[e^2 - 4*d*f]*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 = 7.02783, antiderivative size = 549, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1019, 1076, 621, 206, 1032, 724} \[ -\frac{\left (\left (e-\sqrt{e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{\left (\left (\sqrt{e^2-4 d f}+e\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{(2 c e-b f) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}+\frac{\sqrt{a+b x+c x^2}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a + b*x + c*x^2]/f - ((2*c*e - b*f)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c]*f^
2) - ((2*d*f*(c*e - b*f) + (e - Sqrt[e^2 - 4*d*f])*(f*(b*e - a*f) - c*(e^2 - d*f)))*ArcTanh[(4*a*f - b*(e - Sq
rt[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^2*Sqrt[e^2 - 4*d*f]*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]]) + ((2*d*f*(c*e - b*f) + (e + Sqrt[e^2 - 4*d*f])*(f*(b*e - a*
f) - c*(e^2 - 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*Sq
rt[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^2*Sqrt[e^2 - 4*d*f]*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 1019

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Sy
mbol] :> Simp[(h*(a + b*x + c*x^2)^p*(d + e*x + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] - Dist[1/(2*f*(p + q + 1
)), Int[(a + b*x + c*x^2)^(p - 1)*(d + e*x + f*x^2)^q*Simp[h*p*(b*d - a*e) + a*(h*e - 2*g*f)*(p + q + 1) + (2*
h*p*(c*d - a*f) + b*(h*e - 2*g*f)*(p + q + 1))*x + (h*p*(c*e - b*f) + c*(h*e - 2*g*f)*(p + q + 1))*x^2, x], x]
, x] /; FreeQ[{a, b, c, d, e, f, g, h, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && GtQ[p, 0] && Ne
Q[p + q + 1, 0]

Rule 1076

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x
_)^2]), x_Symbol] :> Dist[C/c, Int[1/Sqrt[d + e*x + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + (B*c - b*C)*x)
/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, 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{x \sqrt{a+b x+c x^2}}{d+e x+f x^2} \, dx &=\frac{\sqrt{a+b x+c x^2}}{f}-\frac{\int \frac{\frac{b d}{2}+\frac{1}{2} (2 c d+b e-2 a f) x+\frac{1}{2} (2 c e-b f) x^2}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}\\ &=\frac{\sqrt{a+b x+c x^2}}{f}-\frac{\int \frac{\frac{b d f}{2}-\frac{1}{2} d (2 c e-b f)+\left (\frac{1}{2} f (2 c d+b e-2 a f)-\frac{1}{2} e (2 c e-b f)\right ) x}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f^2}-\frac{(2 c e-b f) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2 f^2}\\ &=\frac{\sqrt{a+b x+c x^2}}{f}-\frac{(2 c e-b f) \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^2}-\frac{\left (2 f \left (\frac{b d f}{2}-\frac{1}{2} d (2 c e-b f)\right )-\left (\frac{1}{2} f (2 c d+b e-2 a f)-\frac{1}{2} e (2 c e-b f)\right ) \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^2 \sqrt{e^2-4 d f}}+\frac{\left (2 f \left (\frac{b d f}{2}-\frac{1}{2} d (2 c e-b f)\right )-\left (\frac{1}{2} f (2 c d+b e-2 a f)-\frac{1}{2} e (2 c e-b f)\right ) \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^2 \sqrt{e^2-4 d f}}\\ &=\frac{\sqrt{a+b x+c x^2}}{f}-\frac{(2 c e-b f) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}+\frac{\left (2 \left (2 f \left (\frac{b d f}{2}-\frac{1}{2} d (2 c e-b f)\right )-\left (\frac{1}{2} f (2 c d+b e-2 a f)-\frac{1}{2} e (2 c e-b f)\right ) \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^2 \sqrt{e^2-4 d f}}-\frac{\left (2 \left (2 f \left (\frac{b d f}{2}-\frac{1}{2} d (2 c e-b f)\right )-\left (\frac{1}{2} f (2 c d+b e-2 a f)-\frac{1}{2} e (2 c e-b f)\right ) \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^2 \sqrt{e^2-4 d f}}\\ &=\frac{\sqrt{a+b x+c x^2}}{f}-\frac{(2 c e-b f) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}-\frac{\left (2 f (c d e-b d f)+\left (e-\sqrt{e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-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} f^2 \sqrt{e^2-4 d f} \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 (2 f (c d e-b d f)+\left (e+\sqrt{e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-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} f^2 \sqrt{e^2-4 d f} \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 = 1.94204, size = 496, normalized size = 0.9 \[ \frac{4 f \sqrt{e^2-4 d f} \sqrt{a+x (b+c x)}-\sqrt{2} \left (\sqrt{e^2-4 d f}+e\right ) \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f-b \left (\sqrt{e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )-\sqrt{2} \left (\sqrt{e^2-4 d f}-e\right ) \sqrt{f \left (2 a f+b \sqrt{e^2-4 d f}+b (-e)\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+b \left (\sqrt{e^2-4 d f}-e+2 f x\right )+2 c x \left (\sqrt{e^2-4 d f}-e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f+b \sqrt{e^2-4 d f}+b (-e)\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{4 f^2 \sqrt{e^2-4 d f}}-\frac{(2 c e-b f) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{2 \sqrt{c} f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.321, size = 10138, normalized size = 18.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError