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

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

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Rubi [A]  time = 9.00, antiderivative size = 615, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.09, size = 517, normalized size = 0.84 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [C]  time = 1.21, size = 893, normalized size = 1.45 \begin {gather*} \frac {\sqrt {c x^2+b x+a} B}{f}+\frac {(2 B c e-b B f-2 A c f) \log \left (b+2 c x-2 \sqrt {c} \sqrt {c x^2+b x+a}\right )}{2 \sqrt {c} f^2}+\frac {\text {RootSum}\left [f \text {$\#$1}^4-2 \sqrt {c} e \text {$\#$1}^3+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+b^2 d-a b e+a^2 f\&,\frac {B d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b^2-A f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2 b+B e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2 b-B c d e \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b+A c d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b-a B e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b-2 B \sqrt {c} d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1} b-B c e^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2-a B f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2+B c d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2+A c e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2+a B c e^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+a^2 B f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )-a B c d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )-a A c e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+2 a A \sqrt {c} f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}+2 B c^{3/2} d e \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}-2 A c^{3/2} d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}}{-2 f \text {$\#$1}^3+3 \sqrt {c} e \text {$\#$1}^2-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+2 b \sqrt {c} d-a \sqrt {c} e}\&\right ]}{f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((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)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(
2*Sqrt[c]*f^2) + RootSum[b^2*d - a*b*e + a^2*f - 4*b*Sqrt[c]*d*#1 + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 + b*e*#1^2 -
2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (-(b*B*c*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]) + a*B
*c*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + b^2*B*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #
1] + A*b*c*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*B*c*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*
x^2] - #1] - a*b*B*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*A*c*e*f*Log[-(Sqrt[c]*x) + Sqrt[a +
b*x + c*x^2] - #1] + a^2*B*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*B*c^(3/2)*d*e*Log[-(Sqrt[c]*
x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*b*B*Sqrt[c]*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*
A*c^(3/2)*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*a*A*Sqrt[c]*f^2*Log[-(Sqrt[c]*x) + Sqrt[a
+ b*x + c*x^2] - #1]*#1 - B*c*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + B*c*d*f*Log[-(Sqrt[c]*
x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + b*B*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + A*c*e*f*
Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - A*b*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#
1^2 - a*B*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(2*b*Sqrt[c]*d - a*Sqrt[c]*e - 4*c*d*#1 - b
*e*#1 + 2*a*f*#1 + 3*Sqrt[c]*e*#1^2 - 2*f*#1^3) & ]/f^2

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.05, size = 16209, normalized size = 26.27 \begin {gather*} \text {output too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` f
or more details)Is 4*d*f-e^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{f\,x^2+e\,x+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \sqrt {a + b x + c x^{2}}}{d + e x + f x^{2}}\, dx \end {gather*}

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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