3.2.2 \(\int \frac {x^4}{(a+b x+c x^2)^{3/2} (d-f x^2)} \, dx\) [102]

Optimal. Leaf size=466 \[ -\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt {a+b x+c x^2}}+\frac {2 d (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt {a+b x+c x^2}}-\frac {2 d^2 \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}+\frac {2 b \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right ) f}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} f}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2}}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.85, antiderivative size = 466, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {6857, 627, 752, 654, 635, 212, 989, 1047, 738} \begin {gather*} -\frac {2 d^2 \left (b \left (b^2 f-c (3 a f+c d)\right )-c x \left (2 a c f+b^2 (-f)+2 c^2 d\right )\right )}{f^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac {2 d (b+2 c x)}{f^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 b \sqrt {a+b x+c x^2}}{c f \left (b^2-4 a c\right )}-\frac {2 x (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} 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 \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\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 \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/((a + b*x + c*x^2)^(3/2)*(d - f*x^2)),x]

[Out]

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

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 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 752

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(d
*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 989

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b^3*f + b*c*(c*d
- 3*a*f) + c*(2*c^2*d + b^2*f - c*(2*a*f))*x)*(a + b*x + c*x^2)^(p + 1)*((d + f*x^2)^(q + 1)/((b^2 - 4*a*c)*(b
^2*d*f + (c*d - a*f)^2)*(p + 1))), x] - Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x
+ c*x^2)^(p + 1)*(d + f*x^2)^q*Simp[2*c*(b^2*d*f + (c*d - a*f)^2)*(p + 1) - (2*c^2*d + b^2*f - c*(2*a*f))*(a*
f*(p + 1) - c*d*(p + 2)) + (2*f*(b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(2*a*f))*(b*f*(
p + 1)))*x + c*f*(2*c^2*d + b^2*f - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &
&  !IGtQ[q, 0]

Rule 1047

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(-a)*c, 2]}, Dist[h/2 + c*(g/(2*q)), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - c*(g/
(2*q)), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, 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}{\left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx &=\int \left (-\frac {d}{f^2 \left (a+b x+c x^2\right )^{3/2}}-\frac {x^2}{f \left (a+b x+c x^2\right )^{3/2}}+\frac {d^2}{f^2 \left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )}\right ) \, dx\\ &=-\frac {d \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{f^2}+\frac {d^2 \int \frac {1}{\left (a+b x+c x^2\right )^{3/2} \left (d-f x^2\right )} \, dx}{f^2}-\frac {\int \frac {x^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{f}\\ &=-\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt {a+b x+c x^2}}+\frac {2 d (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt {a+b x+c x^2}}-\frac {2 d^2 \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}+\frac {2 \int \frac {2 a+b x}{\sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) f}-\frac {\left (2 d^2\right ) \int \frac {\frac {1}{2} \left (b^2-4 a c\right ) f (c d+a f)-\frac {1}{2} b \left (b^2-4 a c\right ) f^2 x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{\left (b^2-4 a c\right ) f^2 \left (b^2 d f-(c d+a f)^2\right )}\\ &=-\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt {a+b x+c x^2}}+\frac {2 d (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt {a+b x+c x^2}}-\frac {2 d^2 \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}+\frac {2 b \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right ) f}-\frac {\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c f}-\frac {d^{3/2} \int \frac {1}{\left (-\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 \sqrt {f} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )}+\frac {d^{3/2} \int \frac {1}{\left (\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 \sqrt {f} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )}\\ &=-\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt {a+b x+c x^2}}+\frac {2 d (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt {a+b x+c x^2}}-\frac {2 d^2 \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}+\frac {2 b \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right ) f}-\frac {2 \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 {d^{3/2} \text {Subst}\left (\int \frac {1}{4 c d f-4 b \sqrt {d} f^{3/2}+4 a f^2-x^2} \, dx,x,\frac {b \sqrt {d} \sqrt {f}-2 a f-\left (-2 c \sqrt {d} \sqrt {f}+b f\right ) x}{\sqrt {a+b x+c x^2}}\right )}{\sqrt {f} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )}-\frac {d^{3/2} \text {Subst}\left (\int \frac {1}{4 c d f+4 b \sqrt {d} f^{3/2}+4 a f^2-x^2} \, dx,x,\frac {-b \sqrt {d} \sqrt {f}-2 a f-\left (2 c \sqrt {d} \sqrt {f}+b f\right ) x}{\sqrt {a+b x+c x^2}}\right )}{\sqrt {f} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )}\\ &=-\frac {2 x (2 a+b x)}{\left (b^2-4 a c\right ) f \sqrt {a+b x+c x^2}}+\frac {2 d (b+2 c x)}{\left (b^2-4 a c\right ) f^2 \sqrt {a+b x+c x^2}}-\frac {2 d^2 \left (b \left (b^2 f-c (c d+3 a f)\right )-c \left (2 c^2 d-b^2 f+2 a c f\right ) x\right )}{\left (b^2-4 a c\right ) f^2 \left (b^2 d f-(c d+a f)^2\right ) \sqrt {a+b x+c x^2}}+\frac {2 b \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right ) f}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} 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 \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2}}+\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 \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.25, size = 453, normalized size = 0.97 \begin {gather*} \frac {2 \left (b^4 d x+a b^2 d (b-4 c x)-a^3 f (b-2 c x)-a^2 \left (3 b c d-2 c^2 d x+b^2 f x\right )\right )}{c \left (-b^2+4 a c\right ) \left (c^2 d^2+2 a c d f+f \left (-b^2 d+a^2 f\right )\right ) \sqrt {a+x (b+c x)}}+\frac {\log \left (c f \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{c^{3/2} f}-\frac {d^2 \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+2 a b f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 c^{3/2} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a \sqrt {c} f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-b f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{2 f \left (c^2 d^2+2 a c d f+f \left (-b^2 d+a^2 f\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/((a + b*x + c*x^2)^(3/2)*(d - f*x^2)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1063\) vs. \(2(394)=788\).
time = 0.13, size = 1064, normalized size = 2.28

method	result	size
default	\(-\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}}{f}-\frac {2 d \left (2 c x +b \right )}{f^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {d^{2} \left (\frac {f}{\left (-b \sqrt {d f}+f a +c d \right ) \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}}}-\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (2 c \left (x +\frac {\sqrt {d f}}{f}\right )+\frac {-2 c \sqrt {d f}+b f}{f}\right )}{\left (-b \sqrt {d f}+f a +c d \right ) \left (\frac {4 c \left (-b \sqrt {d f}+f a +c d \right )}{f}-\frac {\left (-2 c \sqrt {d f}+b f \right )^{2}}{f^{2}}\right ) \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}}}-\frac {f \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 )}{\left (-b \sqrt {d f}+f a +c d \right ) \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}\right )}{2 f^{2} \sqrt {d f}}-\frac {d^{2} \left (\frac {f}{\left (b \sqrt {d f}+f a +c d \right ) \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}}}-\frac {\left (2 c \sqrt {d f}+b f \right ) \left (2 c \left (x -\frac {\sqrt {d f}}{f}\right )+\frac {2 c \sqrt {d f}+b f}{f}\right )}{\left (b \sqrt {d f}+f a +c d \right ) \left (\frac {4 c \left (b \sqrt {d f}+f a +c d \right )}{f}-\frac {\left (2 c \sqrt {d f}+b f \right )^{2}}{f^{2}}\right ) \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}}}-\frac {f \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 )}{\left (b \sqrt {d f}+f a +c d \right ) \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}\right )}{2 f^{2} \sqrt {d f}}\)	\(1064\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x,method=_RETURNVERBOSE)

[Out]

-1/f*(-x/c/(c*x^2+b*x+a)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
)+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))-2*d/f^2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+1/2
/f^2*d^2/(d*f)^(1/2)*(f/(-b*(d*f)^(1/2)+f*a+c*d)/((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)-(-2*c*(d*f)^(1/2)+b*f)/(-b*(d*f)^(1/2)+f*a+c*d)*(2*c*(x+(d*f)^(1/2)
/f)+1/f*(-2*c*(d*f)^(1/2)+b*f))/(4*c/f*(-b*(d*f)^(1/2)+f*a+c*d)-1/f^2*(-2*c*(d*f)^(1/2)+b*f)^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)-f/(-b*(d*f)^(1/2)+f*
a+c*d)/(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)*(1/(b*(d*f)^(1/2)+f*
a+c*d)*f/((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)-(2*
c*(d*f)^(1/2)+b*f)/(b*(d*f)^(1/2)+f*a+c*d)*(2*c*(x-(d*f)^(1/2)/f)+(2*c*(d*f)^(1/2)+b*f)/f)/(4*c*(b*(d*f)^(1/2)
+f*a+c*d)/f-(2*c*(d*f)^(1/2)+b*f)^2/f^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)-1/(b*(d*f)^(1/2)+f*a+c*d)*f/((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)^(3/2)/(-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)^(3/2)/(-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}}{- a d \sqrt {a + b x + c x^{2}} + a f x^{2} \sqrt {a + b x + c x^{2}} - b d x \sqrt {a + b x + c x^{2}} + b f x^{3} \sqrt {a + b x + c x^{2}} - c d x^{2} \sqrt {a + b x + c x^{2}} + c f x^{4} \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)**(3/2)/(-f*x**2+d),x)

[Out]

-Integral(x**4/(-a*d*sqrt(a + b*x + c*x**2) + a*f*x**2*sqrt(a + b*x + c*x**2) - b*d*x*sqrt(a + b*x + c*x**2) +
b*f*x**3*sqrt(a + b*x + c*x**2) - c*d*x**2*sqrt(a + b*x + c*x**2) + c*f*x**4*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)^(3/2)/(-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 )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,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)^(3/2)),x)

[Out]

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

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