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

Optimal. Leaf size=609 \[ \frac {2 \left (a \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )+c x (a b f-2 a c e+b c d)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac {f \left (2 d (c e-b f)-\left (e-\sqrt {e^2-4 d f}\right ) (c d-a f)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {f \left (2 d (c e-b f)-\left (\sqrt {e^2-4 d f}+e\right ) (c d-a f)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}} \]

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Rubi [A]  time = 5.64, antiderivative size = 609, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1016, 1032, 724, 206} \begin {gather*} \frac {2 \left (a \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )+c x (a b f-2 a c e+b c d)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac {f \left (2 d (c e-b f)-\left (e-\sqrt {e^2-4 d f}\right ) (c d-a f)\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} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \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 {f \left (2 d (c e-b f)-\left (\sqrt {e^2-4 d f}+e\right ) (c d-a f)\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} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*(2*c^2*d - b*c*e + b^2*f - 2*a*c*f) + c*(b*c*d - 2*a*c*e + a*b*f)*x))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b
*d - a*e)*(c*e - b*f))*Sqrt[a + b*x + c*x^2]) + (f*(2*d*(c*e - b*f) - (c*d - a*f)*(e - Sqrt[e^2 - 4*d*f]))*Arc
Tanh[(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]*Sqrt[e^2 - 4*d*f]*((c*
d - a*f)^2 - (b*d - a*e)*(c*e - b*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]])
- (f*(2*d*(c*e - b*f) - (c*d - a*f)*(e + Sqrt[e^2 - 4*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]*Sqrt[e^2 - 4*d*f]*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*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 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 1016

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Sy
mbol] :> Simp[((a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1)*(g*c*(2*a*c*e - b*(c*d + a*f)) + (g*b - a*h
)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) - h*(b*c*d - 2*a*c*e + a*b*f)
)*x))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), x] + Dist[1/((b^2 - 4*a*c)*((c*d - a*
f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Simp[(b*h - 2*g*c)
*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1) + (b^2*(g*f) - b*(h*c*d + g*c*e + a*h*f) + 2*(g*c*(c*d - a*
f) - a*(-(h*c*e))))*(a*f*(p + 1) - c*d*(p + 2)) - e*((g*c)*(2*a*c*e - b*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d +
b^2*f - c*(b*e + 2*a*f)))*(p + q + 2) - (2*f*((g*c)*(2*a*c*e - b*(c*d + a*f)) + (g*b - a*h)*(2*c^2*d + b^2*f -
c*(b*e + 2*a*f)))*(p + q + 2) - (b^2*g*f - b*(h*c*d + g*c*e + a*h*f) + 2*(g*c*(c*d - a*f) - a*(-(h*c*e))))*(b
*f*(p + 1) - c*e*(2*p + q + 4)))*x - c*f*(b^2*(g*f) - b*(h*c*d + g*c*e + a*h*f) + 2*(g*c*(c*d - a*f) + a*h*c*e
))*(2*p + 2*q + 5)*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] && LtQ[p, -1] && NeQ[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1
])

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=\frac {2 \left (a \left (2 c^2 d-b c e+b^2 f-2 a c f\right )+c (b c d-2 a c e+a b f) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} \left (b^2-4 a c\right ) d (c e-b f)+\frac {1}{2} \left (b^2-4 a c\right ) f (c d-a f) x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=\frac {2 \left (a \left (2 c^2 d-b c e+b^2 f-2 a c f\right )+c (b c d-2 a c e+a b f) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}-\frac {\left (f \left (2 d (c e-b f)-(c d-a f) \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+b x+c x^2}} \, dx}{\sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac {\left (f \left (2 d (c e-b f)-(c d-a f) \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+b x+c x^2}} \, dx}{\sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=\frac {2 \left (a \left (2 c^2 d-b c e+b^2 f-2 a c f\right )+c (b c d-2 a c e+a b f) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}+\frac {\left (2 f \left (2 d (c e-b f)-(c d-a f) \left (e-\sqrt {e^2-4 d 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 )}{\sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}-\frac {\left (2 f \left (2 d (c e-b f)-(c d-a f) \left (e+\sqrt {e^2-4 d 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 )}{\sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}\\ &=\frac {2 \left (a \left (2 c^2 d-b c e+b^2 f-2 a c f\right )+c (b c d-2 a c e+a b f) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}+\frac {f \left (2 d (c e-b f)-(c d-a f) \left (e-\sqrt {e^2-4 d f}\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} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \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 {f \left (2 d (c e-b f)-(c d-a f) \left (e+\sqrt {e^2-4 d f}\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} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \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 = 5.68, size = 770, normalized size = 1.26 \begin {gather*} \frac {2 \left (1-\frac {e}{\sqrt {e^2-4 d f}}\right ) \left (2 c \left (c x \left (\sqrt {e^2-4 d f}-e\right )-2 a f\right )+2 b^2 f+b c \left (\sqrt {e^2-4 d f}-e+2 f x\right )\right )}{\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \left (4 a f^2+2 b f \left (\sqrt {e^2-4 d f}-e\right )+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )}+\frac {2 \left (\frac {e}{\sqrt {e^2-4 d f}}+1\right ) \left (-2 c \left (2 a f+c x \left (\sqrt {e^2-4 d f}+e\right )\right )+2 b^2 f-b c \left (\sqrt {e^2-4 d f}+e-2 f x\right )\right )}{\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \left (4 a f^2-2 b f \left (\sqrt {e^2-4 d f}+e\right )+c \left (\sqrt {e^2-4 d f}+e\right )^2\right )}-\frac {\sqrt {2} f^2 \left (\sqrt {e^2-4 d f}+e\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 {e^2-4 d f} \left (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 )^{3/2}}-\frac {\sqrt {2} f^2 \left (\sqrt {e^2-4 d f}-e\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 {e^2-4 d f} \left (f \left (b \left (e-\sqrt {e^2-4 d f}\right )-2 a f\right )+c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(1 - e/Sqrt[e^2 - 4*d*f])*(2*b^2*f + b*c*(-e + Sqrt[e^2 - 4*d*f] + 2*f*x) + 2*c*(-2*a*f + c*(-e + Sqrt[e^2
- 4*d*f])*x)))/((b^2 - 4*a*c)*(4*a*f^2 + c*(e - Sqrt[e^2 - 4*d*f])^2 + 2*b*f*(-e + Sqrt[e^2 - 4*d*f]))*Sqrt[a
+ x*(b + c*x)]) + (2*(1 + e/Sqrt[e^2 - 4*d*f])*(2*b^2*f - b*c*(e + Sqrt[e^2 - 4*d*f] - 2*f*x) - 2*c*(2*a*f + c
*(e + Sqrt[e^2 - 4*d*f])*x)))/((b^2 - 4*a*c)*(4*a*f^2 - 2*b*f*(e + Sqrt[e^2 - 4*d*f]) + c*(e + Sqrt[e^2 - 4*d*
f])^2)*Sqrt[a + x*(b + c*x)]) - (Sqrt[2]*f^2*(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[e^2 - 4*d*f]*(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])))^(3/2)) - (Sqrt[2]*f^2*(-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[e^2 - 4*d*f]*(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])))^2)

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IntegrateAlgebraic [C]  time = 1.36, size = 608, normalized size = 1.00 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+\text {$\#$1}^2 b e+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e-4 \text {$\#$1} b \sqrt {c} d+a^2 f-a b e+b^2 d\&,\frac {-\text {$\#$1}^2 c d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 a f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a^2 f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} c^{3/2} d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-b c d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b \sqrt {c} d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a c d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{-2 \text {$\#$1}^3 f+3 \text {$\#$1}^2 \sqrt {c} e+2 \text {$\#$1} a f-\text {$\#$1} b e-4 \text {$\#$1} c d-a \sqrt {c} e+2 b \sqrt {c} d}\&\right ]}{-a^2 f^2+a b e f+2 a c d f-a c e^2+b^2 (-d) f+b c d e-c^2 d^2}+\frac {2 \left (2 a^2 c f-a b^2 f+a b c e-a b c f x-2 a c^2 d+2 a c^2 e x-b c^2 d x\right )}{\left (4 a c-b^2\right ) \sqrt {a+b x+c x^2} \left (a^2 f^2-a b e f-2 a c d f+a c e^2+b^2 d f-b c d e+c^2 d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2*a*c^2*d + a*b*c*e - a*b^2*f + 2*a^2*c*f - b*c^2*d*x + 2*a*c^2*e*x - a*b*c*f*x))/((-b^2 + 4*a*c)*(c^2*d^
2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2)*Sqrt[a + b*x + 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 & , (-(b*c*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]) + b^2*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x +
c*x^2] - #1] + a*c*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*
x + c*x^2] - #1] + 2*c^(3/2)*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*b*Sqrt[c]*d*f*Log[-(Sqr
t[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - c*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + a*f^2*L
og[-(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) & ]/(-(c^2*d^2) + b*c*d*e - a*c*e^2 - b^2*d*f + 2*a*c*d*f + a*b*e*f - a^2*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(x/(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]  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/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="giac")

[Out]

sage2

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maple [B]  time = 0.02, size = 7163, normalized size = 11.76 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/((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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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