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

Optimal. Leaf size=617 \[ \frac{\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (e-\sqrt{e^2-4 d f}\right ) \left (B \left (f (b e-a f)-c \left (e^2-d f\right )\right )+A f (c e-b f)\right )\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (\sqrt{e^2-4 d f}+e\right ) \left (B \left (f (b e-a f)-c \left (e^2-d f\right )\right )+A f (c e-b f)\right )\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) (-2 A c f-b B f+2 B c e)}{2 \sqrt{c} f^2}+\frac{B \sqrt{a+b x+c x^2}}{f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 8.99756, antiderivative size = 615, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1019, 1076, 621, 206, 1032, 724} \[ \frac{\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (e-\sqrt{e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (\sqrt{e^2-4 d f}+e\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) (-2 A c f-b B f+2 B c e)}{2 \sqrt{c} f^2}+\frac{B \sqrt{a+b x+c x^2}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{(A+B x) \sqrt{a+b x+c x^2}}{d+e x+f x^2} \, dx &=\frac{B \sqrt{a+b x+c x^2}}{f}-\frac{\int \frac{\frac{1}{2} (b B d-2 a A f)-\frac{1}{2} (2 A b f-B (2 c d+b e-2 a f)) x+\frac{1}{2} (2 B c e-b B f-2 A c f) x^2}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}\\ &=\frac{B \sqrt{a+b x+c x^2}}{f}-\frac{\int \frac{\frac{1}{2} f (b B d-2 a A f)-\frac{1}{2} d (2 B c e-b B f-2 A c f)+\left (-\frac{1}{2} e (2 B c e-b B f-2 A c f)+\frac{1}{2} f (-2 A b f+B (2 c d+b e-2 a f))\right ) x}{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f^2}-\frac{(2 B c e-b B f-2 A c f) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2 f^2}\\ &=\frac{B \sqrt{a+b x+c x^2}}{f}-\frac{(2 B c e-b B f-2 A c 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 (A f (c d-a f)-B d (c e-b f))-\left (e+\sqrt{e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-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}{f^2 \sqrt{e^2-4 d f}}-\frac{\left (2 f \left (\frac{1}{2} f (b B d-2 a A f)-\frac{1}{2} d (2 B c e-b B f-2 A c f)\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (-\frac{1}{2} e (2 B c e-b B f-2 A c f)+\frac{1}{2} f (-2 A b f+B (2 c d+b e-2 a 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{B \sqrt{a+b x+c x^2}}{f}-\frac{(2 B c e-b B f-2 A c 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 (A f (c d-a f)-B d (c e-b f))-\left (e+\sqrt{e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-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 )}{f^2 \sqrt{e^2-4 d f}}+\frac{\left (2 \left (2 f \left (\frac{1}{2} f (b B d-2 a A f)-\frac{1}{2} d (2 B c e-b B f-2 A c f)\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (-\frac{1}{2} e (2 B c e-b B f-2 A c f)+\frac{1}{2} f (-2 A b f+B (2 c d+b e-2 a 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{B \sqrt{a+b x+c x^2}}{f}-\frac{(2 B c e-b B f-2 A c 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 (A f (c d-a f)-B d (c e-b f))-\left (e-\sqrt{e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B 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 (A f (c d-a f)-B d (c e-b f))-\left (e+\sqrt{e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B 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 = 2.15169, size = 517, normalized size = 0.84 \[ \frac{-\sqrt{2} \left (B \left (\sqrt{e^2-4 d f}+e\right )-2 A f\right ) \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f-b \left (\sqrt{e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )-\sqrt{2} \left (2 A f+B \left (\sqrt{e^2-4 d f}-e\right )\right ) \sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{4 a f+b \left (\sqrt{e^2-4 d f}-e+2 f x\right )+2 c x \left (\sqrt{e^2-4 d f}-e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )+4 B f \sqrt{e^2-4 d f} \sqrt{a+x (b+c x)}}{4 f^2 \sqrt{e^2-4 d f}}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) (2 A c f+b B f-2 B c e)}{2 \sqrt{c} f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.309, size = 16209, normalized size = 26.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError