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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 1.34656, antiderivative size = 466, normalized size of antiderivative = 1., 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 = {6725, 613, 738, 640, 621, 206, 975, 1033, 724} \[ -\frac{2 d^2 \left (b \left (b^2 f-c (3 a f+c d)\right )-c x \left (2 a c f+b^2 (-f)+2 c^2 d\right )\right )}{f^2 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac{2 d (b+2 c x)}{f^2 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 b \sqrt{a+b x+c x^2}}{c f \left (b^2-4 a c\right )}-\frac{2 x (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2} f}+\frac{d^{3/2} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 f \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right )^{3/2}}+\frac{d^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 f \left (a f+b \sqrt{d} \sqrt{f}+c d\right )^{3/2}} \]

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 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 613

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 738

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 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 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 975

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

Mathematica [A]  time = 1.64, size = 562, normalized size = 1.21 \[ \frac{-\frac{2 d^2 \left (-b c (3 a f+c d)-2 c^2 x (a f+c d)+b^2 c f x+b^3 f\right )}{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \left (b^2 d f-(a f+c d)^2\right )}+\frac{f \left (a \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{c} \left (a (b-2 c x)+b c x^2\right ) \sqrt{a+x (b+c x)}\right )}{a c^{3/2} \left (4 a c-b^2\right )}+\frac{d^{3/2} f \left (\frac{\left (b^2-4 a c\right ) \left (a f+b \sqrt{d} \sqrt{f}+c d\right ) \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+b \left (\sqrt{d}-\sqrt{f} x\right )+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{\sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}+\frac{\left (4 a c-b^2\right ) \left (a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d\right ) \tanh ^{-1}\left (\frac{-2 \left (a \sqrt{f}+c \sqrt{d} x\right )-b \left (\sqrt{d}+\sqrt{f} x\right )}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{\sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 \left (b^2-4 a c\right ) \left ((a f+c d)^2-b^2 d f\right )}+\frac{2 d (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)}}-\frac{2 f x^3 \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)}}}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.283, size = 1648, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

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

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

[Out]

Timed out

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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^4/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x, algorithm="giac")

[Out]

Exception raised: TypeError