3.77 \(\int \frac{x^3 \sqrt{a+b x+c x^2}}{d-f x^2} \, dx\)

Optimal. Leaf size=392 \[ -\frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{5/2} f}+\frac{b (b+2 c x) \sqrt{a+b x+c x^2}}{8 c^2 f}-\frac{d \sqrt{a+b x+c x^2}}{f^2}-\frac{b d \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}-\frac{d \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 f^{5/2}}+\frac{d \sqrt{a f+b \sqrt{d} \sqrt{f}+c d} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 f^{5/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f} \]

[Out]

-((d*Sqrt[a + b*x + c*x^2])/f^2) + (b*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(8*c^2*f) - (a + b*x + c*x^2)^(3/2)/(
3*c*f) - (b*d*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c]*f^2) - (b*(b^2 - 4*a*c)*ArcTa
nh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(5/2)*f) - (d*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*Arc
Tanh[(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^(5/2)) + (d*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sq
rt[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^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.952051, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {6725, 640, 612, 621, 206, 1021, 1078, 1033, 724} \[ -\frac{b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{5/2} f}+\frac{b (b+2 c x) \sqrt{a+b x+c x^2}}{8 c^2 f}-\frac{d \sqrt{a+b x+c x^2}}{f^2}-\frac{b d \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}-\frac{d \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 f^{5/2}}+\frac{d \sqrt{a f+b \sqrt{d} \sqrt{f}+c d} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 f^{5/2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Sqrt[a + b*x + c*x^2])/f^2) + (b*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(8*c^2*f) - (a + b*x + c*x^2)^(3/2)/(
3*c*f) - (b*d*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c]*f^2) - (b*(b^2 - 4*a*c)*ArcTa
nh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(5/2)*f) - (d*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*Arc
Tanh[(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^(5/2)) + (d*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sq
rt[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^(5/2))

Rule 6725

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]

Rule 640

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

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 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 1021

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(h*(a + b*x + c*x^2)^p*(d + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] - Dist[1/(2*f*(p + q + 1)), Int[(a + b*x +
c*x^2)^(p - 1)*(d + f*x^2)^q*Simp[h*p*(b*d) + a*(-2*g*f)*(p + q + 1) + (2*h*p*(c*d - a*f) + b*(-2*g*f)*(p + q
+ 1))*x + (h*p*(-(b*f)) + c*(-2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}, x] && NeQ
[b^2 - 4*a*c, 0] && GtQ[p, 0] && NeQ[p + q + 1, 0]

Rule 1078

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + e*x + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + B*c*x)/((a + c*x^2)*Sqrt[d
+ e*x + f*x^2]), x], x] /; FreeQ[{a, c, d, e, f, A, B, C}, x] && NeQ[e^2 - 4*d*f, 0]

Rule 1033

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 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]

Rubi steps

\begin{align*} \int \frac{x^3 \sqrt{a+b x+c x^2}}{d-f x^2} \, dx &=\int \left (-\frac{x \sqrt{a+b x+c x^2}}{f}+\frac{d x \sqrt{a+b x+c x^2}}{f \left (d-f x^2\right )}\right ) \, dx\\ &=-\frac{\int x \sqrt{a+b x+c x^2} \, dx}{f}+\frac{d \int \frac{x \sqrt{a+b x+c x^2}}{d-f x^2} \, dx}{f}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{f^2}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f}+\frac{d \int \frac{\frac{b d}{2}+(c d+a f) x+\frac{1}{2} b f x^2}{\sqrt{a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f^2}+\frac{b \int \sqrt{a+b x+c x^2} \, dx}{2 c f}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{f^2}+\frac{b (b+2 c x) \sqrt{a+b x+c x^2}}{8 c^2 f}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac{d \int \frac{-b d f-f (c d+a f) x}{\sqrt{a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f^3}-\frac{(b d) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2 f^2}-\frac{\left (b \left (b^2-4 a c\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c^2 f}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{f^2}+\frac{b (b+2 c x) \sqrt{a+b x+c x^2}}{8 c^2 f}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac{(b d) \operatorname{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{\left (b \left (b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{8 c^2 f}+\frac{\left (d \left (c d-b \sqrt{d} \sqrt{f}+a f\right )\right ) \int \frac{1}{\left (-\sqrt{d} \sqrt{f}-f x\right ) \sqrt{a+b x+c x^2}} \, dx}{2 f^2}+\frac{\left (d \left (c d+b \sqrt{d} \sqrt{f}+a f\right )\right ) \int \frac{1}{\left (\sqrt{d} \sqrt{f}-f x\right ) \sqrt{a+b x+c x^2}} \, dx}{2 f^2}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{f^2}+\frac{b (b+2 c x) \sqrt{a+b x+c x^2}}{8 c^2 f}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac{b d \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}-\frac{b \left (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 )}{16 c^{5/2} f}-\frac{\left (d \left (c d-b \sqrt{d} \sqrt{f}+a f\right )\right ) \operatorname{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 )}{f^2}-\frac{\left (d \left (c d+b \sqrt{d} \sqrt{f}+a f\right )\right ) \operatorname{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 )}{f^2}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{f^2}+\frac{b (b+2 c x) \sqrt{a+b x+c x^2}}{8 c^2 f}-\frac{\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac{b d \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c} f^2}-\frac{b \left (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 )}{16 c^{5/2} f}-\frac{d \sqrt{c d-b \sqrt{d} \sqrt{f}+a f} \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^{5/2}}+\frac{d \sqrt{c d+b \sqrt{d} \sqrt{f}+a f} \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^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.02982, size = 327, normalized size = 0.83 \[ \frac{-\frac{2 \sqrt{f} \sqrt{a+x (b+c x)} \left (2 c f (4 a+b x)-3 b^2 f+8 c^2 \left (3 d+f x^2\right )\right )}{c^2}-\frac{3 b \sqrt{f} \left (-4 a c f+b^2 f+8 c^2 d\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c^{5/2}}+24 d \sqrt{a f+b \sqrt{d} \sqrt{f}+c d} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+b \sqrt{d}+b \sqrt{f} x+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )-24 d \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+b \left (\sqrt{d}-\sqrt{f} x\right )+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{48 f^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[f]*Sqrt[a + x*(b + c*x)]*(-3*b^2*f + 2*c*f*(4*a + b*x) + 8*c^2*(3*d + f*x^2)))/c^2 - (3*b*Sqrt[f]*(8
*c^2*d + b^2*f - 4*a*c*f)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(5/2) + 24*d*Sqrt[c*d + b*
Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + 2*c*Sqrt[d]*x + b*Sqrt[f]*x)/(2*Sqrt[c*d + b*Sqrt[d]
*Sqrt[f] + a*f]*Sqrt[a + x*(b + c*x)])] - 24*d*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(-2*a*Sqrt[f] + 2*c
*Sqrt[d]*x + b*(Sqrt[d] - Sqrt[f]*x))/(2*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + x*(b + c*x)])])/(48*f^(5
/2))

________________________________________________________________________________________

Maple [B]  time = 0.266, size = 1817, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

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+d),x)

[Out]

-1/3*(c*x^2+b*x+a)^(3/2)/c/f+1/4/f*b/c*x*(c*x^2+b*x+a)^(1/2)+1/8/f*b^2/c^2*(c*x^2+b*x+a)^(1/2)+1/4/f*b/c^(3/2)
*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/16/f*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-
1/2/f^2*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)+a*f+c*d))^(1
/2)+1/2/f^3*d*ln((1/2/f*(-2*c*(d*f)^(1/2)+b*f)+(x+(d*f)^(1/2)/f)*c)/c^(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)+a*f+c*d))^(1/2))*c^(1/2)*(d*f)^(1/2)-1/4/f^2*d*ln((1/2/f
*(-2*c*(d*f)^(1/2)+b*f)+(x+(d*f)^(1/2)/f)*c)/c^(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)+a*f+c*d))^(1/2))/c^(1/2)*b-1/2/f^3*d/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2
/f*(-b*(d*f)^(1/2)+a*f+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)+a*f+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)+a*f+c*d))^(1/2))/(x+
(d*f)^(1/2)/f))*b*(d*f)^(1/2)+1/2/f^2*d/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+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)+a*f+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)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))*a+1/2/f^3*d^
2/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+a*f+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)+a*f+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)+a*f+c*d))^(1/2))/(x+(d*f)^(1/2)/f))*c-1/2/f^2*d*((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)+a*f+c*d)/f)^(1/2)-1/2/f^3*d*ln((1/2*(2*c*(d*f)^(1/2)+b*f)/f+(x-(d*f)
^(1/2)/f)*c)/c^(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)+a*f+c*d)/
f)^(1/2))*c^(1/2)*(d*f)^(1/2)-1/4/f^2*d*ln((1/2*(2*c*(d*f)^(1/2)+b*f)/f+(x-(d*f)^(1/2)/f)*c)/c^(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)+a*f+c*d)/f)^(1/2))/c^(1/2)*b+1/2/f^3*d/
((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*
((b*(d*f)^(1/2)+a*f+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)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*b*(d*f)^(1/2)+1/2/f^2*d/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(
d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+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)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*a+
1/2/f^3*d^2/((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)
^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+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)+a*f+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{3} \sqrt{a + b x + c x^{2}}}{- d + f x^{2}}\, dx \end{align*}

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+d),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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+d),x, algorithm="giac")

[Out]

Exception raised: TypeError