3.2.14 \(\int \frac {x^3}{\sqrt {a+b x+c x^2} (d+e x+f x^2)} \, dx\) [114]

Optimal. Leaf size=545 \[ \frac {\sqrt {a+b x+c x^2}}{c f}-\frac {e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}-\frac {b \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} f}-\frac {\left (2 d e f-\left (e^2-d f\right ) \left (e-\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}+\frac {\left (2 d e f-\left (e^2-d f\right ) \left (e+\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 2.30, antiderivative size = 545, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6860, 635, 212, 654, 1046, 738} \begin {gather*} -\frac {b \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} f}-\frac {\left (2 d e f-\left (e^2-d f\right ) \left (e-\sqrt {e^2-4 d f}\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^2 \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 d e f-\left (e^2-d f\right ) \left (\sqrt {e^2-4 d f}+e\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^2 \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 {e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}+\frac {\sqrt {a+b x+c x^2}}{c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a + b*x + c*x^2]/(c*f) - (e*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(Sqrt[c]*f^2) - (b*Ar
cTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c^(3/2)*f) - ((2*d*e*f - (e^2 - d*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]*f^2*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]]) + ((2*d*e*f - (e^2 - d*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])])/(S
qrt[2]*f^2*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 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 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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 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 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.66, size = 423, normalized size = 0.78 \begin {gather*} \frac {\frac {2 f \sqrt {a+x (b+c x)}}{c}+\frac {(2 c e+b f) \log \left (c \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{c^{3/2}}-2 \text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 \sqrt {c} d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{2 b \sqrt {c} d-a \sqrt {c} e-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{2 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*f*Sqrt[a + x*(b + c*x)])/c + ((2*c*e + b*f)*Log[c*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(3/2)
- 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*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*e^2*Log[-(Sqrt[c]*x)
+ Sqrt[a + b*x + c*x^2] - #1] + a*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*Sqrt[c]*d*e*Log[-(Sqr
t[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - d*f*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) & ])/(2*f^2)

________________________________________________________________________________________

Maple [A]
time = 0.19, size = 930, normalized size = 1.71

method	result	size
default	\(\frac {\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}}{f}-\frac {e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{f^{2} \sqrt {c}}-\frac {\left (e^{3}-3 d e f +e^{2} \sqrt {-4 d f +e^{2}}-d f \sqrt {-4 d f +e^{2}}\right ) \sqrt {2}\, \ln \left (\frac {\frac {-b f \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}+\frac {\left (-c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-b f \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c +\frac {4 \left (-c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 b f \sqrt {-4 d f +e^{2}}+2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{3} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-b f \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}}-\frac {\left (-d f \sqrt {-4 d f +e^{2}}+e^{2} \sqrt {-4 d f +e^{2}}+3 d e f -e^{3}\right ) \sqrt {2}\, \ln \left (\frac {\frac {b f \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}+\frac {\left (c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {b f \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c +\frac {4 \left (c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 b f \sqrt {-4 d f +e^{2}}-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{3} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {b f \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}}\)	\(930\)
risch	\(\text {Expression too large to display}\)	\(3131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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^3/(c*x^2+b*x+a)^(1/2)/(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?`
for more det

________________________________________________________________________________________

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(x^3/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {a + b x + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(c*x**2+b*x+a)**(1/2)/(f*x**2+e*x+d),x)

[Out]

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

________________________________________________________________________________________

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(x^3/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

________________________________________________________________________________________

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

[Out]

int(x^3/((a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)), x)

________________________________________________________________________________________