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

Optimal. Leaf size=795 \[ -\frac {(4 e-3 f x) \left (c x^2+a\right )^{3/2}}{12 f^2}-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {c x^2+a}}{8 f^4}+\frac {\left (3 a^2 f^4+12 a c \left (e^2-d f\right ) f^2+8 c^2 \left (e^4-3 d f e^2+d^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {c x^2+a}}\right )}{8 \sqrt {c} f^5}-\frac {\left (a^2 \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right ) f^4+2 a c \left (e^4-\sqrt {e^2-4 d f} e^3-4 d f e^2+2 d f \sqrt {e^2-4 d f} e+2 d^2 f^2\right ) f^2+c^2 \left (e^6-\sqrt {e^2-4 d f} e^5-6 d f e^4+4 d f \sqrt {e^2-4 d f} e^3+9 d^2 f^2 e^2-3 d^2 f^2 \sqrt {e^2-4 d f} e-2 d^3 f^3\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right )}{\sqrt {2} f^5 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )}}+\frac {\left (a^2 \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right ) f^4+2 a c \left (e^4+\sqrt {e^2-4 d f} e^3-4 d f e^2-2 d f \sqrt {e^2-4 d f} e+2 d^2 f^2\right ) f^2+c^2 \left (e^6+\sqrt {e^2-4 d f} e^5-6 d f e^4-4 d f \sqrt {e^2-4 d f} e^3+9 d^2 f^2 e^2+3 d^2 f^2 \sqrt {e^2-4 d f} e-2 d^3 f^3\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right )}{\sqrt {2} f^5 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )}} \]

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Rubi [A]  time = 4.26, antiderivative size = 795, 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 = {1069, 1068, 1080, 217, 206, 1034, 725} \begin {gather*} -\frac {(4 e-3 f x) \left (c x^2+a\right )^{3/2}}{12 f^2}-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {c x^2+a}}{8 f^4}+\frac {\left (3 a^2 f^4+12 a c \left (e^2-d f\right ) f^2+8 c^2 \left (e^4-3 d f e^2+d^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {c x^2+a}}\right )}{8 \sqrt {c} f^5}-\frac {\left (a^2 \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right ) f^4+2 a c \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 ) f^2+c^2 \left (e^6-\sqrt {e^2-4 d f} e^5-6 d f e^4+4 d f \sqrt {e^2-4 d f} e^3+9 d^2 f^2 e^2-3 d^2 f^2 \sqrt {e^2-4 d f} e-2 d^3 f^3\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right )}{\sqrt {2} f^5 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )}}+\frac {\left (a^2 \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right ) f^4+2 a c \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 ) f^2+c^2 \left (e^6+\sqrt {e^2-4 d f} e^5-6 d f e^4-4 d f \sqrt {e^2-4 d f} e^3+9 d^2 f^2 e^2+3 d^2 f^2 \sqrt {e^2-4 d f} e-2 d^3 f^3\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right )}{\sqrt {2} f^5 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*e*(a*f^2 + c*(e^2 - 2*d*f)) - f*(3*a*f^2 + 4*c*(e^2 - d*f))*x)*Sqrt[a + c*x^2])/(8*f^4) - ((4*e - 3*f*x)*
(a + c*x^2)^(3/2))/(12*f^2) + ((3*a^2*f^4 + 12*a*c*f^2*(e^2 - d*f) + 8*c^2*(e^4 - 3*d*e^2*f + d^2*f^2))*ArcTan
h[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*Sqrt[c]*f^5) - ((a^2*f^4*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) + 2*a*c*f^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]) + c^2*(e^6 - 6*d*e^4*f + 9*d^2
*e^2*f^2 - 2*d^3*f^3 - e^5*Sqrt[e^2 - 4*d*f] + 4*d*e^3*f*Sqrt[e^2 - 4*d*f] - 3*d^2*e*f^2*Sqrt[e^2 - 4*d*f]))*A
rcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sq
rt[a + c*x^2])])/(Sqrt[2]*f^5*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + ((a^2
*f^4*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) + 2*a*c*f^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]) + c^2*(e^6 - 6*d*e^4*f + 9*d^2*e^2*f^2 - 2*d^3*f^3 + e^5*Sqrt[e^2 - 4*d*f] - 4*d*e^3*
f*Sqrt[e^2 - 4*d*f] + 3*d^2*e*f^2*Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*S
qrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*f^5*Sqrt[e^2 - 4*d*f]*Sqrt[2*
a*f^2 + c*(e^2 - 2*d*f + e*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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> 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 + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 1068

Int[((a_) + (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*(-(c*e*(2*p + q + 2))) + 2*c*C*f*(p + q + 1)*x)*(a + 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 + c*x^2)^(p - 1)*(d + e*x + f*x^2)^q*Simp[p*(-(a*e))*(C*(c*e)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3
)) + (p + q + 1)*(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)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)*(C*e*f*p*(-4*a*c)))*x + (p*(c*e)*(C*(c*e)*(q + 1
) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)*(C*f^2*p*(-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] /; FreeQ[{a, c, d, e, f, A, B, C, q}, x] && 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 1069

Int[((a_) + (c_.)*(x_)^2)^(p_)*((A_.) + (C_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Si
mp[((C*(-(c*e*(2*p + q + 2))) + 2*c*C*f*(p + q + 1)*x)*(a + 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 + c*x^2)^(p - 1)*(d + e*x +
f*x^2)^q*Simp[p*(-(a*e))*(C*(c*e)*(q + 1) - c*(C*e)*(2*p + 2*q + 3)) + (p + q + 1)*(a*c*(C*(2*d*f - e^2*(2*p
+ q + 2)) + f*(-2*A*f)*(2*p + 2*q + 3))) + (2*p*(c*d - a*f)*(C*(c*e)*(q + 1) - c*(C*e)*(2*p + 2*q + 3)) + (p +
q + 1)*(C*e*f*p*(-4*a*c)))*x + (p*(c*e)*(C*(c*e)*(q + 1) - c*(C*e)*(2*p + 2*q + 3)) + (p + q + 1)*(C*f^2*p*(-
4*a*c) - c^2*(C*(e^2 - 4*d*f)*(2*p + q + 2) + f*(2*C*d + 2*A*f)*(2*p + 2*q + 3))))*x^2, x], x], x] /; FreeQ[{a
, c, d, e, f, A, C, q}, x] && 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 1080

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + 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 + f*x^2]), x], x] /; FreeQ[{a, b, c, d, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx &=-\frac {(4 e-3 f x) \left (a+c x^2\right )^{3/2}}{12 f^2}-\frac {\int \frac {\sqrt {a+c x^2} \left (3 a c d f-3 c e (4 c d-a f) x-3 c \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x^2\right )}{d+e x+f x^2} \, dx}{12 c f^2}\\ &=-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {a+c x^2}}{8 f^4}-\frac {(4 e-3 f x) \left (a+c x^2\right )^{3/2}}{12 f^2}+\frac {\int \frac {-3 a c^2 d f \left (5 a f^2+4 c \left (e^2-d f\right )\right )-3 c^2 e \left (5 a^2 f^3+4 a c f \left (e^2-5 d f\right )-8 c^2 d \left (e^2-2 d f\right )\right ) x+3 c^2 \left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right ) x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{24 c^2 f^4}\\ &=-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {a+c x^2}}{8 f^4}-\frac {(4 e-3 f x) \left (a+c x^2\right )^{3/2}}{12 f^2}+\frac {\int \frac {-3 a c^2 d f^2 \left (5 a f^2+4 c \left (e^2-d f\right )\right )-3 c^2 d \left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right )+\left (-3 c^2 e f \left (5 a^2 f^3+4 a c f \left (e^2-5 d f\right )-8 c^2 d \left (e^2-2 d f\right )\right )-3 c^2 e \left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{24 c^2 f^5}+\frac {\left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 f^5}\\ &=-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {a+c x^2}}{8 f^4}-\frac {(4 e-3 f x) \left (a+c x^2\right )^{3/2}}{12 f^2}+\frac {\left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 f^5}+\frac {\left (a^2 f^4 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+2 a c f^2 \left (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}\right )+c^2 \left (e^6-6 d e^4 f+9 d^2 e^2 f^2-2 d^3 f^3-e^5 \sqrt {e^2-4 d f}+4 d e^3 f \sqrt {e^2-4 d f}-3 d^2 e f^2 \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+c x^2}} \, dx}{f^5 \sqrt {e^2-4 d f}}-\frac {\left (a^2 f^4 \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )+2 a c f^2 \left (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}\right )+c^2 \left (e^6-6 d e^4 f+9 d^2 e^2 f^2-2 d^3 f^3+e^5 \sqrt {e^2-4 d f}-4 d e^3 f \sqrt {e^2-4 d f}+3 d^2 e f^2 \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+c x^2}} \, dx}{f^5 \sqrt {e^2-4 d f}}\\ &=-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {a+c x^2}}{8 f^4}-\frac {(4 e-3 f x) \left (a+c x^2\right )^{3/2}}{12 f^2}+\frac {\left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c} f^5}-\frac {\left (a^2 f^4 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+2 a c f^2 \left (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}\right )+c^2 \left (e^6-6 d e^4 f+9 d^2 e^2 f^2-2 d^3 f^3-e^5 \sqrt {e^2-4 d f}+4 d e^3 f \sqrt {e^2-4 d f}-3 d^2 e f^2 \sqrt {e^2-4 d f}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {a+c x^2}}\right )}{f^5 \sqrt {e^2-4 d f}}+\frac {\left (a^2 f^4 \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )+2 a c f^2 \left (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}\right )+c^2 \left (e^6-6 d e^4 f+9 d^2 e^2 f^2-2 d^3 f^3+e^5 \sqrt {e^2-4 d f}-4 d e^3 f \sqrt {e^2-4 d f}+3 d^2 e f^2 \sqrt {e^2-4 d f}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a f^2+c \left (e+\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {a+c x^2}}\right )}{f^5 \sqrt {e^2-4 d f}}\\ &=-\frac {\left (8 e \left (a f^2+c \left (e^2-2 d f\right )\right )-f \left (3 a f^2+4 c \left (e^2-d f\right )\right ) x\right ) \sqrt {a+c x^2}}{8 f^4}-\frac {(4 e-3 f x) \left (a+c x^2\right )^{3/2}}{12 f^2}+\frac {\left (3 a^2 f^4+12 a c f^2 \left (e^2-d f\right )+8 c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 \sqrt {c} f^5}-\frac {\left (a^2 f^4 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+2 a c f^2 \left (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}\right )+c^2 \left (e^6-6 d e^4 f+9 d^2 e^2 f^2-2 d^3 f^3-e^5 \sqrt {e^2-4 d f}+4 d e^3 f \sqrt {e^2-4 d f}-3 d^2 e f^2 \sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^5 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\left (a^2 f^4 \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )+2 a c f^2 \left (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}\right )+c^2 \left (e^6-6 d e^4 f+9 d^2 e^2 f^2-2 d^3 f^3+e^5 \sqrt {e^2-4 d f}-4 d e^3 f \sqrt {e^2-4 d f}+3 d^2 e f^2 \sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^5 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}\\ \end {align*}

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Mathematica [A]  time = 3.45, size = 793, normalized size = 1.00 \begin {gather*} \frac {3 f \sqrt {a+c x^2} \left (\frac {3 a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {c} \sqrt {\frac {c x^2}{a}+1}}+5 a x+2 c x^3\right )-\frac {3 \left (\frac {2 d f-e^2}{\sqrt {e^2-4 d f}}+e\right ) \left (\frac {2 \left (2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \left (-\sqrt {4 a f^2-2 c e \sqrt {e^2-4 d f}-4 c d f+2 c e^2} \tanh ^{-1}\left (\frac {2 a f+c x \left (\sqrt {e^2-4 d f}-e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2-2 c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right )}}\right )+\sqrt {c} \left (\sqrt {e^2-4 d f}-e\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+2 f \sqrt {a+c x^2}\right )}{f^2}+\frac {2 \sqrt {c} \sqrt {a+c x^2} \left (\sqrt {e^2-4 d f}-e\right ) \left (\sqrt {c} x \sqrt {\frac {c x^2}{a}+1}+\sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )\right )}{\sqrt {\frac {c x^2}{a}+1}}\right )}{2 f}+\frac {3 \left (\frac {e^2-2 d f}{\sqrt {e^2-4 d f}}+e\right ) \left (\frac {2 \left (2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \left (\sqrt {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )+\sqrt {c} \left (\sqrt {e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-2 f \sqrt {a+c x^2}\right )}{f^2}+\frac {2 \sqrt {c} \sqrt {a+c x^2} \left (\sqrt {e^2-4 d f}+e\right ) \left (\sqrt {c} x \sqrt {\frac {c x^2}{a}+1}+\sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )\right )}{\sqrt {\frac {c x^2}{a}+1}}\right )}{2 f}-4 \left (a+c x^2\right )^{3/2} \left (\frac {e^2-2 d f}{\sqrt {e^2-4 d f}}+e\right )-4 \left (a+c x^2\right )^{3/2} \left (\frac {2 d f-e^2}{\sqrt {e^2-4 d f}}+e\right )}{24 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(e + (e^2 - 2*d*f)/Sqrt[e^2 - 4*d*f])*(a + c*x^2)^(3/2) - 4*(e + (-e^2 + 2*d*f)/Sqrt[e^2 - 4*d*f])*(a + c*
x^2)^(3/2) + 3*f*Sqrt[a + c*x^2]*(5*a*x + 2*c*x^3 + (3*a^(3/2)*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[c]*Sqrt[1 +
(c*x^2)/a])) - (3*(e + (-e^2 + 2*d*f)/Sqrt[e^2 - 4*d*f])*((2*Sqrt[c]*(-e + Sqrt[e^2 - 4*d*f])*Sqrt[a + c*x^2]
*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/a] + Sqrt[a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]]))/Sqrt[1 + (c*x^2)/a] + (2*(2*a*f^2 + c
*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]))*(2*f*Sqrt[a + c*x^2] + Sqrt[c]*(-e + Sqrt[e^2 - 4*d*f])*ArcTanh[(Sqrt[c]
*x)/Sqrt[a + c*x^2]] - Sqrt[2*c*e^2 - 4*c*d*f + 4*a*f^2 - 2*c*e*Sqrt[e^2 - 4*d*f]]*ArcTanh[(2*a*f + c*(-e + Sq
rt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])]))/f^2))/(2*f)
+ (3*(e + (e^2 - 2*d*f)/Sqrt[e^2 - 4*d*f])*((2*Sqrt[c]*(e + Sqrt[e^2 - 4*d*f])*Sqrt[a + c*x^2]*(Sqrt[c]*x*Sqrt
[1 + (c*x^2)/a] + Sqrt[a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]]))/Sqrt[1 + (c*x^2)/a] + (2*(2*a*f^2 + c*(e^2 - 2*d*f +
e*Sqrt[e^2 - 4*d*f]))*(-2*f*Sqrt[a + c*x^2] + Sqrt[c]*(e + Sqrt[e^2 - 4*d*f])*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x
^2]] + Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(
Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])]))/f^2))/(2*f))/(24*f^2)

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IntegrateAlgebraic [C]  time = 1.22, size = 960, normalized size = 1.21 \begin {gather*} \frac {\sqrt {c x^2+a} \left (-24 c e^3+12 c f x e^2-32 a f^2 e-8 c f^2 x^2 e+48 c d f e+6 c f^3 x^3+15 a f^3 x-12 c d f^2 x\right )}{24 f^4}+\frac {\left (-8 c^2 e^4-12 a c f^2 e^2+24 c^2 d f e^2-3 a^2 f^4+12 a c d f^3-8 c^2 d^2 f^2\right ) \log \left (\sqrt {c x^2+a}-\sqrt {c} x\right )}{8 \sqrt {c} f^5}+\frac {\text {RootSum}\left [f \text {$\#$1}^4-2 \sqrt {c} e \text {$\#$1}^3+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2+2 a \sqrt {c} e \text {$\#$1}+a^2 f\&,\frac {-c^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}^2 e^5+a c^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) e^5+2 c^{5/2} d \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1} e^4-2 a c f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}^2 e^3+4 c^2 d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}^2 e^3+2 a^2 c f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) e^3-4 a c^2 d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) e^3+4 a c^{3/2} d f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1} e^2-6 c^{5/2} d^2 f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1} e^2-a^2 f^4 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}^2 e+4 a c d f^3 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}^2 e-3 c^2 d^2 f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}^2 e+a^3 f^4 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) e-4 a^2 c d f^3 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) e+3 a c^2 d^2 f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) e+2 a^2 \sqrt {c} d f^4 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}-4 a c^{3/2} d^2 f^3 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}+2 c^{5/2} d^3 f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+a}\right ) \text {$\#$1}}{2 f \text {$\#$1}^3-3 \sqrt {c} e \text {$\#$1}^2+4 c d \text {$\#$1}-2 a f \text {$\#$1}+a \sqrt {c} e}\&\right ]}{f^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-24*c*e^3 + 48*c*d*e*f - 32*a*e*f^2 + 12*c*e^2*f*x - 12*c*d*f^2*x + 15*a*f^3*x - 8*c*e*f^2*x
^2 + 6*c*f^3*x^3))/(24*f^4) + ((-8*c^2*e^4 + 24*c^2*d*e^2*f - 8*c^2*d^2*f^2 - 12*a*c*e^2*f^2 + 12*a*c*d*f^3 -
3*a^2*f^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(8*Sqrt[c]*f^5) + RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^
2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (a*c^2*e^5*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - 4*a*c^2*d
*e^3*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 3*a*c^2*d^2*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] +
2*a^2*c*e^3*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - 4*a^2*c*d*e*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]
- #1] + a^3*e*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(5/2)*d*e^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^
2] - #1]*#1 - 6*c^(5/2)*d^2*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 2*c^(5/2)*d^3*f^2*Log[-(Sqrt[c
]*x) + Sqrt[a + c*x^2] - #1]*#1 + 4*a*c^(3/2)*d*e^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - 4*a*c^(3
/2)*d^2*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 2*a^2*Sqrt[c]*d*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^
2] - #1]*#1 - c^2*e^5*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 4*c^2*d*e^3*f*Log[-(Sqrt[c]*x) + Sqrt[a
+ c*x^2] - #1]*#1^2 - 3*c^2*d^2*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - 2*a*c*e^3*f^2*Log[-(Sqrt
[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 4*a*c*d*e*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - a^2*e*f^4*
Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^
3) & ]/f^5

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2

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maple [B]  time = 0.03, size = 19148, normalized size = 24.09 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^2+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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c*x^2+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 \begin {gather*} \int \frac {x^2\,{\left (c\,x^2+a\right )}^{3/2}}{f\,x^2+e\,x+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^2*(a + 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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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