3.106 \(\int \frac {(a+b x+c x^2)^{3/2}}{d+e x+f x^2} \, dx\)

Optimal. Leaf size=679 \[ \frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (-f^2 \left (b^2 d-a^2 f\right )+2 c d f (b e-a f)+c^2 (-d) \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\left (\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (-f^2 \left (b^2 d-a^2 f\right )+2 c d f (b e-a f)+c^2 (-d) \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right )}{8 \sqrt {c} f^3}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2} \]

[Out]

1/8*(3*b^2*f^2-12*c*f*(-a*f+b*e)+8*c^2*(-d*f+e^2))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/f^3/c^(1
/2)-1/4*(-2*c*f*x-5*b*f+4*c*e)*(c*x^2+b*x+a)^(1/2)/f^2+1/2*arctanh(1/4*(4*a*f+2*x*(b*f-c*(e-(-4*d*f+e^2)^(1/2)
))-b*(e-(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+b*x+a)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4*d*f+e^2)^
(1/2))^(1/2))*(-2*f*(2*c*d*f*(-a*f+b*e)-f^2*(-a^2*f+b^2*d)-c^2*d*(-d*f+e^2))+(-b*f+c*e)*(f*(-2*a*f+b*e)-c*(-2*
d*f+e^2))*(e-(-4*d*f+e^2)^(1/2)))/f^3*2^(1/2)/(-4*d*f+e^2)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4*d
*f+e^2)^(1/2))^(1/2)-1/2*arctanh(1/4*(4*a*f-b*(e+(-4*d*f+e^2)^(1/2))+2*x*(b*f-c*(e+(-4*d*f+e^2)^(1/2))))*2^(1/
2)/(c*x^2+b*x+a)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+e^2)^(1/2))^(1/2))*(-2*f*(2*c*d*f*(-a*f
+b*e)-f^2*(-a^2*f+b^2*d)-c^2*d*(-d*f+e^2))+(-b*f+c*e)*(f*(-2*a*f+b*e)-c*(-2*d*f+e^2))*(e+(-4*d*f+e^2)^(1/2)))/
f^3*2^(1/2)/(-4*d*f+e^2)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+e^2)^(1/2))^(1/2)

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Rubi [A]  time = 11.03, antiderivative size = 678, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {977, 1076, 621, 206, 1032, 724} \[ \frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (-f^2 \left (b^2 d-a^2 f\right )+2 c d f (b e-a f)+c^2 (-d) \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {\left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right )}{8 \sqrt {c} f^3}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(d + e*x + f*x^2),x]

[Out]

-((4*c*e - 5*b*f - 2*c*f*x)*Sqrt[a + b*x + c*x^2])/(4*f^2) + ((3*b^2*f^2 - 12*c*f*(b*e - a*f) + 8*c^2*(e^2 - d
*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c]*f^3) + (((c*e - b*f)*(e - Sqrt[e^2 - 4
*d*f])*(f*(b*e - 2*a*f) - c*(e^2 - 2*d*f)) - 2*f*(2*c*d*f*(b*e - a*f) - f^2*(b^2*d - a^2*f) - c^2*d*(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^3*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]]) + ((4*c*d*f^2*(b*e - a*f) - 2*
f^3*(b^2*d - a^2*f) - 2*c^2*d*f*(e^2 - d*f) - (c*e - b*f)*(e + Sqrt[e^2 - 4*d*f])*(f*(b*e - 2*a*f) - c*(e^2 -
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^3*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 977

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((b*f
*(3*p + 2*q) - c*e*(2*p + q) + 2*c*f*(p + q)*x)*(a + b*x + c*x^2)^(p - 1)*(d + e*x + f*x^2)^(q + 1))/(2*f^2*(p
+ q)*(2*p + 2*q + 1)), x] - Dist[1/(2*f^2*(p + q)*(2*p + 2*q + 1)), Int[(a + b*x + c*x^2)^(p - 2)*(d + e*x +
f*x^2)^q*Simp[(b*d - a*e)*(c*e - b*f)*(1 - p)*(2*p + q) - (p + q)*(b^2*d*f*(1 - p) - a*(f*(b*e - 2*a*f)*(2*p +
2*q + 1) + c*(2*d*f - e^2*(2*p + q)))) + (2*(c*d - a*f)*(c*e - b*f)*(1 - p)*(2*p + q) - (p + q)*((b^2 - 4*a*c
)*e*f*(1 - p) + b*(c*(e^2 - 4*d*f)*(2*p + q) + f*(2*c*d - b*e + 2*a*f)*(2*p + 2*q + 1))))*x + ((c*e - b*f)^2*(
1 - p)*p + c*(p + q)*(f*(b*e - 2*a*f)*(4*p + 2*q - 1) - c*(2*d*f*(1 - 2*p) + e^2*(3*p + q - 1))))*x^2, x], x],
x] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && GtQ[p, 1] && NeQ[p + q
, 0] && NeQ[2*p + 2*q + 1, 0] &&  !IGtQ[p, 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 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 {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx &=-\frac {(4 c e-5 b f-2 c f x) \sqrt {a+b x+c x^2}}{4 f^2}-\frac {\int \frac {\frac {1}{4} \left (-4 b c d e+5 b^2 d f+4 a f (c d-2 a f)\right )-\frac {1}{4} \left (8 c^2 d e-4 a c e f-b f (5 b e-16 a f)+4 b c \left (e^2-4 d f\right )\right ) x-\frac {1}{4} \left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right ) x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 f^2}\\ &=-\frac {(4 c e-5 b f-2 c f x) \sqrt {a+b x+c x^2}}{4 f^2}-\frac {\int \frac {\frac {1}{4} f \left (-4 b c d e+5 b^2 d f+4 a f (c d-2 a f)\right )-\frac {1}{4} d \left (-3 b^2 f^2+12 c f (b e-a f)-8 c^2 \left (e^2-d f\right )\right )+\left (\frac {1}{4} f \left (-8 c^2 d e+4 a c e f+b f (5 b e-16 a f)-4 b c \left (e^2-4 d f\right )\right )-\frac {1}{4} e \left (-3 b^2 f^2+12 c f (b e-a f)-8 c^2 \left (e^2-d f\right )\right )\right ) x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 f^3}+\frac {\left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 f^3}\\ &=-\frac {(4 c e-5 b f-2 c f x) \sqrt {a+b x+c x^2}}{4 f^2}+\frac {\left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right ) \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 )}{4 f^3}-\frac {\left ((c e-b f) \left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \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^3 \sqrt {e^2-4 d f}}+\frac {\left (2 f \left (\frac {1}{4} f \left (-4 b c d e+5 b^2 d f+4 a f (c d-2 a f)\right )-\frac {1}{4} d \left (-3 b^2 f^2+12 c f (b e-a f)-8 c^2 \left (e^2-d f\right )\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (\frac {1}{4} f \left (-8 c^2 d e+4 a c e f+b f (5 b e-16 a f)-4 b c \left (e^2-4 d f\right )\right )-\frac {1}{4} e \left (-3 b^2 f^2+12 c f (b e-a f)-8 c^2 \left (e^2-d f\right )\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}{2 f^3 \sqrt {e^2-4 d f}}\\ &=-\frac {(4 c e-5 b f-2 c f x) \sqrt {a+b x+c x^2}}{4 f^2}+\frac {\left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} f^3}+\frac {\left (2 \left ((c e-b f) \left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \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^3 \sqrt {e^2-4 d f}}-\frac {\left (2 f \left (\frac {1}{4} f \left (-4 b c d e+5 b^2 d f+4 a f (c d-2 a f)\right )-\frac {1}{4} d \left (-3 b^2 f^2+12 c f (b e-a f)-8 c^2 \left (e^2-d f\right )\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (\frac {1}{4} f \left (-8 c^2 d e+4 a c e f+b f (5 b e-16 a f)-4 b c \left (e^2-4 d f\right )\right )-\frac {1}{4} e \left (-3 b^2 f^2+12 c f (b e-a f)-8 c^2 \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^3 \sqrt {e^2-4 d f}}\\ &=-\frac {(4 c e-5 b f-2 c f x) \sqrt {a+b x+c x^2}}{4 f^2}+\frac {\left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} f^3}+\frac {\left ((c e-b f) \left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \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^3 \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 ((c e-b f) \left (e+\sqrt {e^2-4 d f}\right ) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \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^3 \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 = 4.58, size = 1232, normalized size = 1.81 \[ \frac {\sqrt {a+x (b+c x)} \left (-4 \left (e+\sqrt {e^2-4 d f}\right )^2 c^2+4 f \left (e+\sqrt {e^2-4 d f}\right ) x c^2-16 a f^2 c+10 b f \left (e+\sqrt {e^2-4 d f}\right ) c-4 b f^2 x c-2 b^2 f^2\right )+2 \sqrt {a+x (b+c x)} \left (-2 \left (-2 e^2+f x e+2 \sqrt {e^2-4 d f} e+4 d f-f \sqrt {e^2-4 d f} x\right ) c^2+f \left (8 a f+b \left (-5 e+2 f x+5 \sqrt {e^2-4 d f}\right )\right ) c+b^2 f^2\right )-\frac {\left (b f+c \left (\sqrt {e^2-4 d f}-e\right )\right ) \left (4 \left (-e^2+\sqrt {e^2-4 d f} e+2 d f\right ) c^2-4 f \left (3 a f+b \left (\sqrt {e^2-4 d f}-e\right )\right ) c+b^2 f^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c} f}+\frac {\left (c \left (e+\sqrt {e^2-4 d f}\right )-b f\right ) \left (4 \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right ) c^2-4 f \left (b \left (e+\sqrt {e^2-4 d f}\right )-3 a f\right ) c-b^2 f^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c} f}+\frac {8 \sqrt {2} c \left (\left (e^4+\sqrt {e^2-4 d f} e^3-4 d f e^2-2 d f \sqrt {e^2-4 d f} e+2 d^2 f^2\right ) c^2+2 f \left (a f \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )-b \left (e^3+\sqrt {e^2-4 d f} e^2-3 d f e-d f \sqrt {e^2-4 d f}\right )\right ) c+f^2 \left (\left (e^2+\sqrt {e^2-4 d f} e-2 d f\right ) b^2-2 a f \left (e+\sqrt {e^2-4 d f}\right ) b+2 a^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {4 a f-2 c \left (e+\sqrt {e^2-4 d f}\right ) x-b \left (e-2 f x+\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e+\sqrt {e^2-4 d f}\right )\right )} \sqrt {a+x (b+c x)}}\right )}{f \sqrt {c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e+\sqrt {e^2-4 d f}\right )\right )}}+\frac {8 \sqrt {2} c \left (\left (-e^4+\sqrt {e^2-4 d f} e^3+4 d f e^2-2 d f \sqrt {e^2-4 d f} e-2 d^2 f^2\right ) c^2+2 f \left (a f \left (-e^2+\sqrt {e^2-4 d f} e+2 d f\right )+b \left (e^3-\sqrt {e^2-4 d f} e^2-3 d f e+d f \sqrt {e^2-4 d f}\right )\right ) c+f^2 \left (\left (-e^2+\sqrt {e^2-4 d f} e+2 d f\right ) b^2+2 a f \left (e-\sqrt {e^2-4 d f}\right ) b-2 a^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 c \left (\sqrt {e^2-4 d f}-e\right ) x+b \left (-e+2 f x+\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )+f \left (2 a f+b \left (\sqrt {e^2-4 d f}-e\right )\right )} \sqrt {a+x (b+c x)}}\right )}{f \sqrt {c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )+f \left (2 a f+b \left (\sqrt {e^2-4 d f}-e\right )\right )}}}{16 c f^2 \sqrt {e^2-4 d f}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(d + e*x + f*x^2),x]

[Out]

((-2*b^2*f^2 - 16*a*c*f^2 + 10*b*c*f*(e + Sqrt[e^2 - 4*d*f]) - 4*c^2*(e + Sqrt[e^2 - 4*d*f])^2 - 4*b*c*f^2*x +
4*c^2*f*(e + Sqrt[e^2 - 4*d*f])*x)*Sqrt[a + x*(b + c*x)] + 2*Sqrt[a + x*(b + c*x)]*(b^2*f^2 - 2*c^2*(-2*e^2 +
4*d*f + 2*e*Sqrt[e^2 - 4*d*f] + e*f*x - f*Sqrt[e^2 - 4*d*f]*x) + c*f*(8*a*f + b*(-5*e + 5*Sqrt[e^2 - 4*d*f] +
2*f*x))) - ((b*f + c*(-e + Sqrt[e^2 - 4*d*f]))*(b^2*f^2 + 4*c^2*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f]) - 4*c*f*
(3*a*f + b*(-e + Sqrt[e^2 - 4*d*f])))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(Sqrt[c]*f) + ((
-(b*f) + c*(e + Sqrt[e^2 - 4*d*f]))*(-(b^2*f^2) + 4*c^2*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) - 4*c*f*(-3*a*f +
b*(e + Sqrt[e^2 - 4*d*f])))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(Sqrt[c]*f) + (8*Sqrt[2]*c
*(c^2*(e^4 - 4*d*e^2*f + 2*d^2*f^2 + e^3*Sqrt[e^2 - 4*d*f] - 2*d*e*f*Sqrt[e^2 - 4*d*f]) + f^2*(2*a^2*f^2 - 2*a
*b*f*(e + Sqrt[e^2 - 4*d*f]) + b^2*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])) + 2*c*f*(a*f*(e^2 - 2*d*f + e*Sqrt[e^2
- 4*d*f]) - b*(e^3 - 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f])))*ArcTanh[(4*a*f - 2*c*(e + Sqr
t[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)])])/(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]))]) + (8*Sqrt[2]*c*(c^2*(-e^4 + 4*d*e^2*f - 2*d^2*f^2 + e^3*Sqrt[e^2 -
4*d*f] - 2*d*e*f*Sqrt[e^2 - 4*d*f]) + f^2*(-2*a^2*f^2 + 2*a*b*f*(e - Sqrt[e^2 - 4*d*f]) + b^2*(-e^2 + 2*d*f +
e*Sqrt[e^2 - 4*d*f])) + 2*c*f*(a*f*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f]) + b*(e^3 - 3*d*e*f - e^2*Sqrt[e^2 - 4*
d*f] + d*f*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)])])/(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]))]))/(
16*c*f^2*Sqrt[e^2 - 4*d*f])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(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 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(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 = 0.03, size = 22523, normalized size = 33.17 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(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 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{f\,x^2+e\,x+d} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/2)/(d + e*x + f*x^2),x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(d + e*x + f*x^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(f*x**2+e*x+d),x)

[Out]

Timed out

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