3.72 \(\int \frac {x^2}{(a+c x^2)^{3/2} (d+e x+f x^2)} \, dx\)
Optimal. Leaf size=410 \[ -\frac {f \left (2 d (c d-a f)+a e \left (e-\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (2 d (c d-a f)+a e \left (\sqrt {e^2-4 d f}+e\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {x (c d-a f)+a e}{\sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )} \]
[Out]
(-a*e-(-a*f+c*d)*x)/(a*c*e^2+(-a*f+c*d)^2)/(c*x^2+a)^(1/2)-1/2*f*arctanh(1/2*(2*a*f-c*x*(e-(-4*d*f+e^2)^(1/2))
)*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+(-(-4*d*f+e^2)^(1/2)*e-2*d*f+e^2)*c)^(1/2))*(2*d*(-a*f+c*d)+a*e*(e-(-4*d*f+
e^2)^(1/2)))/(a*c*e^2+(-a*f+c*d)^2)*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+(-(-4*d*f+e^2)^(1/2)*e-2*d*f+e^2)*c)^(
1/2)+1/2*f*arctanh(1/2*(2*a*f-c*x*(e+(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+((-4*d*f+e^2)^(1/2)
*e-2*d*f+e^2)*c)^(1/2))*(2*d*(-a*f+c*d)+a*e*(e+(-4*d*f+e^2)^(1/2)))/(a*c*e^2+(-a*f+c*d)^2)*2^(1/2)/(-4*d*f+e^2
)^(1/2)/(2*a*f^2+((-4*d*f+e^2)^(1/2)*e-2*d*f+e^2)*c)^(1/2)
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Rubi [A] time = 0.71, antiderivative size = 410, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used =
{1063, 1034, 725, 206} \[ -\frac {f \left (2 d (c d-a f)+a e \left (e-\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (2 d (c d-a f)+a e \left (\sqrt {e^2-4 d f}+e\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {x (c d-a f)+a e}{\sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[x^2/((a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]
[Out]
-((a*e + (c*d - a*f)*x)/((a*c*e^2 + (c*d - a*f)^2)*Sqrt[a + c*x^2])) - (f*(2*d*(c*d - a*f) + a*e*(e - Sqrt[e^2
- 4*d*f]))*ArcTanh[(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])]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*Sqrt[2*a*f^2 + c*(e^2 - 2*d
*f - e*Sqrt[e^2 - 4*d*f])]) + (f*(2*d*(c*d - a*f) + a*e*(e + Sqrt[e^2 - 4*d*f]))*ArcTanh[(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])]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sq
rt[e^2 - 4*d*f]*(a*c*e^2 + (c*d - a*f)^2)*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 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 1063
Int[((a_) + (c_.)*(x_)^2)^(p_)*((A_.) + (C_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Si
mp[((a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*((A*c - a*C)*(2*a*c*e) + c*(A*(2*c^2*d - c*(2*a*f)) + C*(-2*
a*(c*d - a*f)))*x))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), x] + Dist[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)
^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(-2*A*c - 2*a*C)*((c*d - a*f)^2 - (-(a*e))*(c*e
))*(p + 1) + (2*(A*c*(c*d - a*f) - a*(c*C*d - a*C*f)))*(a*f*(p + 1) - c*d*(p + 2)) - e*((A*c - a*C)*(2*a*c*e))
*(p + q + 2) - (2*f*((A*c - a*C)*(2*a*c*e))*(p + q + 2) - (2*(A*c*(c*d - a*f) - a*(c*C*d - a*C*f)))*(-(c*e*(2*
p + q + 4))))*x - c*f*(2*(A*c*(c*d - a*f) - a*(c*C*d - a*C*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c
, d, e, f, A, C, q}, x] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] && !( !Intege
rQ[p] && ILtQ[q, -1]) && !IGtQ[q, 0]
Rubi steps
\begin {align*} \int \frac {x^2}{\left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=-\frac {a e+(c d-a f) x}{\left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {\int \frac {2 a c d (c d-a f)-2 a^2 c e f x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 a c \left (a c e^2+(c d-a f)^2\right )}\\ &=-\frac {a e+(c d-a f) x}{\left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {\left (f \left (2 d (c d-a f)+a e \left (e-\sqrt {e^2-4 d f}\right )\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}-\frac {\left (f \left (2 d (c d-a f)+a e \left (e+\sqrt {e^2-4 d f}\right )\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=-\frac {a e+(c d-a f) x}{\left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}-\frac {\left (f \left (2 d (c d-a f)+a e \left (e-\sqrt {e^2-4 d f}\right )\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 )}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}+\frac {\left (f \left (2 d (c d-a f)+a e \left (e+\sqrt {e^2-4 d f}\right )\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 )}{\sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=-\frac {a e+(c d-a f) x}{\left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}-\frac {f \left (2 d (c d-a f)+a e \left (e-\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} \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {f \left (2 d (c d-a f)+a e \left (e+\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} \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \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 = 2.07, size = 509, normalized size = 1.24 \[ \frac {-\frac {\left (\frac {2 d f-e^2}{\sqrt {e^2-4 d f}}+e\right ) \left (2 a f+c x \left (e-\sqrt {e^2-4 d f}\right )\right )}{a \sqrt {a+c x^2} \left (4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )}-\frac {\left (\frac {e^2-2 d f}{\sqrt {e^2-4 d f}}+e\right ) \left (2 a f+c x \left (\sqrt {e^2-4 d f}+e\right )\right )}{a \sqrt {a+c x^2} \left (4 a f^2+c \left (\sqrt {e^2-4 d f}+e\right )^2\right )}+\frac {\sqrt {2} f^2 \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right ) \tanh ^{-1}\left (\frac {2 a f+c x \left (\sqrt {e^2-4 d f}-e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2-2 c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt {e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac {\sqrt {2} f^2 \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f} \left (2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac {x}{a \sqrt {a+c x^2}}}{f} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2/((a + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]
[Out]
(x/(a*Sqrt[a + c*x^2]) - ((e + (-e^2 + 2*d*f)/Sqrt[e^2 - 4*d*f])*(2*a*f + c*(e - Sqrt[e^2 - 4*d*f])*x))/(a*(4*
a*f^2 + c*(e - Sqrt[e^2 - 4*d*f])^2)*Sqrt[a + c*x^2]) - ((e + (e^2 - 2*d*f)/Sqrt[e^2 - 4*d*f])*(2*a*f + c*(e +
Sqrt[e^2 - 4*d*f])*x))/(a*(4*a*f^2 + c*(e + Sqrt[e^2 - 4*d*f])^2)*Sqrt[a + c*x^2]) + (Sqrt[2]*f^2*(-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])])/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]))^(
3/2)) + (Sqrt[2]*f^2*(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])])/(Sqrt[e^2 - 4*d*f]*(2*a*f^2 + c*(e^2 - 2*
d*f + e*Sqrt[e^2 - 4*d*f]))^(3/2)))/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(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 \[ \mathit {sage}_{2} \]
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 = 0.02, size = 4752, normalized size = 11.59 \[ \text {output too large to display} \]
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]
1/f*x/a/(c*x^2+a)^(1/2)-1/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c
-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e
)/f^2)^(1/2)*e+2/(-4*d*f+e^2)^(1/2)*f/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/((x+1/2*(e+(-4*d*f+e^2)^(
1/2))/f)^2*c-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^
2)^(1/2)*c*e)/f^2)^(1/2)*d-1/(-4*d*f+e^2)^(1/2)/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/((x+1/2*(e+(-4*
d*f+e^2)^(1/2))/f)^2*c-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2+
(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*e^2-2*(-4*d*f+e^2)^(1/2)/f*c^2/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*
e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e+(-4*d*f+e^2)^(1
/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*x*e+4*
c^2/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x+1/2*
(e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f
+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*x*d-4/f*c^2/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*
c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*
(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*x*e^2+4/(-4*d*f+e^
2)^(1/2)*c^2/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))
/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^
2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*e*x*d-2/(-4*d*f+e^2)^(1/2)/f*c^2/(2*a*f^2-2*c*d*f+c*e^2+(-4
*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-
(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)
/f^2)^(1/2)*e^3*x+1/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)*2^(1/2)/((2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2
)^(1/2)*c*e)/f^2)^(1/2)*ln((-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+(2*a*f^2-2*c*d*f+c*e^
2+(-4*d*f+e^2)^(1/2)*c*e)/f^2+1/2*2^(1/2)*((2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*(4*(x+1/2
*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+2*(2*a*f^2-2*c*d*
f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))*e-2/(-4*d*f+e^2)^(1/2)*f/(2*a*f^
2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)*2^(1/2)/((2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*ln(
(-(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f
^2+1/2*2^(1/2)*((2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^
2*c-4*(e+(-4*d*f+e^2)^(1/2))*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*
c*e)/f^2)^(1/2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))*d+1/(-4*d*f+e^2)^(1/2)/(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^
(1/2)*c*e)*2^(1/2)/((2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*ln((-(e+(-4*d*f+e^2)^(1/2))*(x+1
/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2+1/2*2^(1/2)*((2*a*f^2-2*c*
d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*(e+(-4*d*f+e^2)^(1/2))*
(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)*c/f+2*(2*a*f^2-2*c*d*f+c*e^2+(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2))/(x+1/2*(e+(-
4*d*f+e^2)^(1/2))/f))*e^2-1/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^
2*c-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)
*c*e)/f^2)^(1/2)*e-2/(-4*d*f+e^2)^(1/2)*f/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/((x-1/2*(-e+(-4*d*f+e
^2)^(1/2))/f)^2*c-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2-(-4*
d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*d+1/(-4*d*f+e^2)^(1/2)/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/((x-1/2*(
-e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*
f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*e^2+2*(-4*d*f+e^2)^(1/2)/f*c^2/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^
(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e-(-4*d*
f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1
/2)*x*e+4*c^2/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2)
)/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a
*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*x*d-4/f*c^2/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*
e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e-(-4*d*f+e^2)^(
1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*x*e^
2-4/(-4*d*f+e^2)^(1/2)*c^2/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2
*(-4*d*f+e^2))/((x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)
*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*e*x*d+2/(-4*d*f+e^2)^(1/2)/f*c^2/(2*a*f^2-2
*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/(4*a*c-4*c^2/f*d+c^2/f^2*e^2-c^2/f^2*(-4*d*f+e^2))/((x-1/2*(-e+(-4*d*f+e^
2)^(1/2))/f)^2*c-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+1/2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d
*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*e^3*x+1/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)*2^(1/2)/((2*a*f^2-2*c*d*f
+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*ln((-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+(2
*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2+1/2*2^(1/2)*((2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f
^2)^(1/2)*(4*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*
c/f+2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2))/(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f))*e+2/(-4*d*
f+e^2)^(1/2)*f/(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)*2^(1/2)/((2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/
2)*c*e)/f^2)^(1/2)*ln((-(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+(2*a*f^2-2*c*d*f+c*e^2-(-
4*d*f+e^2)^(1/2)*c*e)/f^2+1/2*2^(1/2)*((2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*(4*(x-1/2*(-e
+(-4*d*f+e^2)^(1/2))/f)^2*c-4*(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+2*(2*a*f^2-2*c*d*f+
c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2))/(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f))*d-1/(-4*d*f+e^2)^(1/2)/(2*a*f^2-2
*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)*2^(1/2)/((2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*ln((-(
e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2
+1/2*2^(1/2)*((2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*c*e)/f^2)^(1/2)*(4*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)^2
*c-4*(e-(-4*d*f+e^2)^(1/2))*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)*c/f+2*(2*a*f^2-2*c*d*f+c*e^2-(-4*d*f+e^2)^(1/2)*
c*e)/f^2)^(1/2))/(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f))*e^2
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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(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 \[ \int \frac {x^2}{{\left (c\,x^2+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right )} \,d x \]
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] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x + f x^{2}\right )}\, dx \]
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]
Integral(x**2/((a + c*x**2)**(3/2)*(d + e*x + f*x**2)), x)
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