3.1.94 \(\int \frac {x^4}{\sqrt {a+b x+c x^2} (d-f x^2)} \, dx\) [94]

Optimal. Leaf size=369 \[ \frac {3 b \sqrt {a+b x+c x^2}}{4 c^2 f}-\frac {x \sqrt {a+b x+c x^2}}{2 c f}-\frac {d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}-\frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} f}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f}} \]

[Out]

-1/8*(-4*a*c+3*b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)/f-d*arctanh(1/2*(2*c*x+b)/c^(1/
2)/(c*x^2+b*x+a)^(1/2))/f^2/c^(1/2)+3/4*b*(c*x^2+b*x+a)^(1/2)/c^2/f-1/2*x*(c*x^2+b*x+a)^(1/2)/c/f+1/2*d^(3/2)*
arctanh(1/2*(b*d^(1/2)-2*a*f^(1/2)+x*(2*c*d^(1/2)-b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1/2))^
(1/2))/f^2/(c*d+a*f-b*d^(1/2)*f^(1/2))^(1/2)+1/2*d^(3/2)*arctanh(1/2*(b*d^(1/2)+2*a*f^(1/2)+x*(2*c*d^(1/2)+b*f
^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2))/f^2/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2)

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Rubi [A]
time = 0.51, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6857, 635, 212, 756, 654, 998, 738} \begin {gather*} -\frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} f}+\frac {3 b \sqrt {a+b x+c x^2}}{4 c^2 f}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 f^2 \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 f^2 \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}-\frac {d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}-\frac {x \sqrt {a+b x+c x^2}}{2 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[a + b*x + c*x^2]*(d - f*x^2)),x]

[Out]

(3*b*Sqrt[a + b*x + c*x^2])/(4*c^2*f) - (x*Sqrt[a + b*x + c*x^2])/(2*c*f) - (d*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*
Sqrt[a + b*x + c*x^2])])/(Sqrt[c]*f^2) - ((3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2]
)])/(8*c^(5/2)*f) + (d^(3/2)*ArcTanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[c*d - b*S
qrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])])/(2*f^2*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]) + (d^(3/2)*ArcTanh[
(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sqrt[d] + b*Sqrt[f])*x)/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x +
c*x^2])])/(2*f^2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*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 756

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 998

Int[1/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[1/2, Int[1/((a - Rt[(
-a)*c, 2]*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[1/2, Int[1/((a + Rt[(-a)*c, 2]*x)*Sqrt[d + e*x + f*x^2]), x
], x] /; FreeQ[{a, c, d, e, f}, x] && NeQ[e^2 - 4*d*f, 0] && PosQ[(-a)*c]

Rule 6857

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

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx &=\int \left (-\frac {d}{f^2 \sqrt {a+b x+c x^2}}-\frac {x^2}{f \sqrt {a+b x+c x^2}}+\frac {d^2}{f^2 \sqrt {a+b x+c x^2} \left (d-f x^2\right )}\right ) \, dx\\ &=-\frac {d \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{f^2}+\frac {d^2 \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f^2}-\frac {\int \frac {x^2}{\sqrt {a+b x+c x^2}} \, dx}{f}\\ &=-\frac {x \sqrt {a+b x+c x^2}}{2 c f}-\frac {(2 d) \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 {d^2 \int \frac {1}{\left (d-\sqrt {d} \sqrt {f} x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f^2}+\frac {d^2 \int \frac {1}{\left (d+\sqrt {d} \sqrt {f} x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f^2}-\frac {\int \frac {-a-\frac {3 b x}{2}}{\sqrt {a+b x+c x^2}} \, dx}{2 c f}\\ &=\frac {3 b \sqrt {a+b x+c x^2}}{4 c^2 f}-\frac {x \sqrt {a+b x+c x^2}}{2 c f}-\frac {d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}-\frac {d^2 \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d^{3/2} \sqrt {f}+4 a d f-x^2} \, dx,x,\frac {-b d+2 a \sqrt {d} \sqrt {f}-\left (2 c d-b \sqrt {d} \sqrt {f}\right ) x}{\sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {d^2 \text {Subst}\left (\int \frac {1}{4 c d^2+4 b d^{3/2} \sqrt {f}+4 a d f-x^2} \, dx,x,\frac {-b d-2 a \sqrt {d} \sqrt {f}-\left (2 c d+b \sqrt {d} \sqrt {f}\right ) x}{\sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\left (3 b^2-4 a c\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c^2 f}\\ &=\frac {3 b \sqrt {a+b x+c x^2}}{4 c^2 f}-\frac {x \sqrt {a+b x+c x^2}}{2 c f}-\frac {d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f}}-\frac {\left (3 b^2-4 a c\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 )}{4 c^2 f}\\ &=\frac {3 b \sqrt {a+b x+c x^2}}{4 c^2 f}-\frac {x \sqrt {a+b x+c x^2}}{2 c f}-\frac {d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f^2}-\frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} f}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a 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 = 0.72, size = 252, normalized size = 0.68 \begin {gather*} \frac {2 \sqrt {c} f (3 b-2 c x) \sqrt {a+x (b+c x)}+\left (8 c^2 d+3 b^2 f-4 a c f\right ) \log \left (c^2 f^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )-4 c^{5/2} d^2 \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 \sqrt {c} \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{8 c^{5/2} f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[a + b*x + c*x^2]*(d - f*x^2)),x]

[Out]

(2*Sqrt[c]*f*(3*b - 2*c*x)*Sqrt[a + x*(b + c*x)] + (8*c^2*d + 3*b^2*f - 4*a*c*f)*Log[c^2*f^2*(b + 2*c*x - 2*Sq
rt[c]*Sqrt[a + x*(b + c*x)])] - 4*c^(5/2)*d^2*RootSum[b^2*d - a^2*f - 4*b*Sqrt[c]*d*#1 + 4*c*d*#1^2 + 2*a*f*#1
^2 - f*#1^4 & , (b*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*Sqrt[c]*Log[-(Sqrt[c]*x) + Sqrt[a + b*x
+ c*x^2] - #1]*#1)/(b*Sqrt[c]*d - 2*c*d*#1 - a*f*#1 + f*#1^3) & ])/(8*c^(5/2)*f^2)

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Maple [A]
time = 0.14, size = 513, normalized size = 1.39

method	result	size
risch	\(\frac {\left (-2 c x +3 b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{2} f}+\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}} f}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{2}}{8 c^{\frac {5}{2}} f}-\frac {d \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{f^{2} \sqrt {c}}+\frac {d^{2} \ln \left (\frac {\frac {2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {d f}\, \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}-\frac {d^{2} \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {d f}\, \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}\)	\(502\)
default	\(-\frac {\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}}{f}-\frac {d \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{f^{2} \sqrt {c}}-\frac {d^{2} \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {d f}\, \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}+\frac {d^{2} \ln \left (\frac {\frac {2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {d f}\, \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}\)	\(513\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x,method=_RETURNVERBOSE)

[Out]

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

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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^4/(c*x^2+b*x+a)^(1/2)/(-f*x^2+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(((c*sqrt(4*d*f))/(2*f^2)>0)',
see `assume?

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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(x^4/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x^{4}}{- d \sqrt {a + b x + c x^{2}} + f x^{2} \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(c*x**2+b*x+a)**(1/2)/(-f*x**2+d),x)

[Out]

-Integral(x**4/(-d*sqrt(a + b*x + c*x**2) + f*x**2*sqrt(a + b*x + c*x**2)), x)

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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(x^4/(c*x^2+b*x+a)^(1/2)/(-f*x^2+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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4}{\left (d-f\,x^2\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((d - f*x^2)*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int(x^4/((d - f*x^2)*(a + b*x + c*x^2)^(1/2)), x)

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