3.2.7 \(\int \frac {(a+b x+c x^2)^{3/2}}{(d+e x+f x^2)^2} \, dx\) [107]

Optimal. Leaf size=704 \[ -\frac {(c e-2 b f-2 c f x) \sqrt {a+b x+c x^2}}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\left ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e-\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a f)\right )\right )\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {2} f^2 \left (e^2-4 d f\right )^{3/2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}+\frac {\left ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e+\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a f)\right )\right )\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {2} f^2 \left (e^2-4 d f\right )^{3/2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 7.55, antiderivative size = 704, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {985, 1080, 1090, 635, 212, 1046, 738} \begin {gather*} -\frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right )-2 f \left (f \left (-b e (3 a f+c d)+4 a f (a f+c d)+2 b^2 d f\right )+2 c^2 d \left (e^2-4 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 )}{2 \sqrt {2} f^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 (\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right )-2 f \left (f \left (-b e (3 a f+c d)+4 a f (a f+c d)+2 b^2 d f\right )+2 c^2 d \left (e^2-4 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 )}{2 \sqrt {2} f^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 {c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\sqrt {a+b x+c x^2} (-2 b f+c e-2 c f x)}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((c*e - 2*b*f - 2*c*f*x)*Sqrt[a + b*x + c*x^2])/(f*(e^2 - 4*d*f))) - ((e + 2*f*x)*(a + b*x + c*x^2)^(3/2))/(
(e^2 - 4*d*f)*(d + e*x + f*x^2)) + (c^(3/2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/f^2 - (((c
*e - b*f)*(f*(b*e - 2*a*f) + 2*c*(e^2 - 5*d*f))*(e - Sqrt[e^2 - 4*d*f]) - 2*f*(2*c^2*d*(e^2 - 4*d*f) + f*(2*b^
2*d*f + 4*a*f*(c*d + a*f) - b*e*(c*d + 3*a*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])])/(2*Sqrt[2]*f^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)*S
qrt[e^2 - 4*d*f]]) + (((c*e - b*f)*(f*(b*e - 2*a*f) + 2*c*(e^2 - 5*d*f))*(e + Sqrt[e^2 - 4*d*f]) - 2*f*(2*c^2*
d*(e^2 - 4*d*f) + f*(2*b^2*d*f + 4*a*f*(c*d + a*f) - b*e*(c*d + 3*a*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])])/(2*Sqrt[2]*f^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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 738

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 985

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 1046

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 1080

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

Rule 1090

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}}{\left (d+e x+f x^2\right )^2} \, dx &=-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}-\frac {\int \frac {\sqrt {a+b x+c x^2} \left (\frac {1}{2} (3 b e-4 a f)+(3 c e+b f) x+4 c f x^2\right )}{d+e x+f x^2} \, dx}{-e^2+4 d f}\\ &=-\frac {(c e-2 b f-2 c f x) \sqrt {a+b x+c x^2}}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}-\frac {\int \frac {c f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a f)\right )-c f \left (2 c^2 d e+2 a c e f+b f (b e-2 a f)+b c \left (e^2-10 d f\right )\right ) x-2 c^3 f \left (e^2-4 d f\right ) x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 c f^2 \left (e^2-4 d f\right )}\\ &=-\frac {(c e-2 b f-2 c f x) \sqrt {a+b x+c x^2}}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {c^2 \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{f^2}-\frac {\int \frac {2 c^3 d f \left (e^2-4 d f\right )+c f^2 \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a f)\right )+\left (2 c^3 e f \left (e^2-4 d f\right )-c f^2 \left (2 c^2 d e+2 a c e f+b f (b e-2 a f)+b c \left (e^2-10 d f\right )\right )\right ) x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 c f^3 \left (e^2-4 d f\right )}\\ &=-\frac {(c e-2 b f-2 c f x) \sqrt {a+b x+c x^2}}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {\left (2 c^2\right ) \text {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 ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e-\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a 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}{2 f^2 \left (e^2-4 d f\right )^{3/2}}-\frac {\left ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e+\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a 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}{2 f^2 \left (e^2-4 d f\right )^{3/2}}\\ &=-\frac {(c e-2 b f-2 c f x) \sqrt {a+b x+c x^2}}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\left ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e-\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a f)\right )\right )\right ) \text {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 \left (e^2-4 d f\right )^{3/2}}+\frac {\left ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e+\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a f)\right )\right )\right ) \text {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 \left (e^2-4 d f\right )^{3/2}}\\ &=-\frac {(c e-2 b f-2 c f x) \sqrt {a+b x+c x^2}}{f \left (e^2-4 d f\right )}-\frac {(e+2 f x) \left (a+b x+c x^2\right )^{3/2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\left ((c e-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e-\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a 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 )}{2 \sqrt {2} f^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-b f) \left (f (b e-2 a f)+2 c \left (e^2-5 d f\right )\right ) \left (e+\sqrt {e^2-4 d f}\right )-2 f \left (2 c^2 d \left (e^2-4 d f\right )+f \left (2 b^2 d f+4 a f (c d+a f)-b e (c d+3 a 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 )}{2 \sqrt {2} f^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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.79, size = 2416, normalized size = 3.43 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*f^3*Sqrt[a + x*(b + c*x)]*(c*e^2*x - b*f*(2*d + e*x) + c*d*(e - 2*f*x) + a*f*(e + 2*f*x)))/((e^2 - 4*d*f)
*(d + x*(e + f*x))) - 2*c^(3/2)*f^2*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]] - 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 & , (4*c^3*e^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 8*c^3*d*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x
+ c*x^2] - #1] - 9*b*c^2*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 10*b*c^2*d*f^2*Log[-(Sqrt[c]*x
) + Sqrt[a + b*x + c*x^2] - #1] + 6*b^2*c*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 6*a*c^2*e*f^2
*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - b^3*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 8*a
*b*c*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*c^(5/2)*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*
x^2] - #1]*#1 - 4*c^(5/2)*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 4*b*c^(3/2)*e*f^2*Log[-(Sq
rt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*b^2*Sqrt[c]*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#
1 + 4*a*c^(3/2)*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*c^2*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a
+ b*x + c*x^2] - #1]*#1^2 - 2*b*c*f^3*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) & ] + 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 & , (8*c
^3*e^5*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 48*c^3*d*e^3*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2
] - #1] - 18*b*c^2*e^4*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 64*c^3*d^2*e*f^2*Log[-(Sqrt[c]*x) +
Sqrt[a + b*x + c*x^2] - #1] + 90*b*c^2*d*e^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 12*b^2*c*e^3
*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 14*a*c^2*e^3*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2
] - #1] - 72*b*c^2*d^2*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 47*b^2*c*d*e*f^3*Log[-(Sqrt[c]*x)
+ Sqrt[a + b*x + c*x^2] - #1] - 58*a*c^2*d*e*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b^3*e^2*f^
3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 17*a*b*c*e^2*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] -
#1] + 6*b^3*d*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 70*a*b*c*d*f^4*Log[-(Sqrt[c]*x) + Sqrt[a +
b*x + c*x^2] - #1] + 2*a*b^2*e*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*a^2*c*e*f^4*Log[-(Sqrt[
c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*a^2*b*f^5*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 4*c^(5/2)*e
^4*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 20*c^(5/2)*d*e^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x
+ c*x^2] - #1]*#1 - 8*b*c^(3/2)*e^3*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 16*c^(5/2)*d^2*f^3
*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 30*b*c^(3/2)*d*e*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*
x^2] - #1]*#1 + 4*b^2*Sqrt[c]*e^2*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 8*a*c^(3/2)*e^2*f^3*
Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 12*b^2*Sqrt[c]*d*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x
^2] - #1]*#1 - 24*a*c^(3/2)*d*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 6*a*b*Sqrt[c]*e*f^4*Log[
-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 8*a^2*Sqrt[c]*f^5*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #
1]*#1 + 2*c^2*e^3*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - 6*c^2*d*e*f^3*Log[-(Sqrt[c]*x) + S
qrt[a + b*x + c*x^2] - #1]*#1^2 - 3*b*c*e^2*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + 6*b*c*d*
f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + b^2*e*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] -
#1]*#1^2 + 2*a*c*e*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - 2*a*b*f^5*Log[-(Sqrt[c]*x) + Sqr
t[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) & ]/(e^2 - 4*d*f))/(2*f^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(7855\) vs. \(2(640)=1280\).
time = 0.16, size = 7856, normalized size = 11.16

method	result	size
default	\(\text {Expression too large to display}\)	\(7856\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
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((c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
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((c*x**2+b*x+a)**(3/2)/(f*x**2+e*x+d)**2,x)

[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((c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{[-1,0]:[1,0,%%%{-1,[1]%%%}]%%},[8,4,8,0,0,0]%%%}+%%%{
%%{[%%%{16,

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (f\,x^2+e\,x+d\right )}^2} \,d x \end {gather*}

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)^2,x)

[Out]

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

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