3.101 \(\int \frac{1}{x^3 \sqrt{a+b x+c x^2} (d-f x^2)} \, dx\)

Optimal. Leaf size=376 \[ -\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 a^{5/2} d}+\frac{3 b \sqrt{a+b x+c x^2}}{4 a^2 d x}-\frac{f^{3/2} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 d^2 \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}+\frac{f^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 d^2 \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}-\frac{f \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+b x+c x^2}}{2 a d x^2} \]

[Out]

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

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Rubi [A]  time = 0.734105, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6725, 744, 806, 724, 206, 1033} \[ -\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 a^{5/2} d}+\frac{3 b \sqrt{a+b x+c x^2}}{4 a^2 d x}-\frac{f^{3/2} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+x \left (2 c \sqrt{d}-b \sqrt{f}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{2 d^2 \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}+\frac{f^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+x \left (b \sqrt{f}+2 c \sqrt{d}\right )+b \sqrt{d}}{2 \sqrt{a+b x+c x^2} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{2 d^2 \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}-\frac{f \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+b x+c x^2}}{2 a d x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b*x + c*x^2]*(d - f*x^2)),x]

[Out]

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

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 744

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))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

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]

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 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)]

Rubi steps

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

Mathematica [A]  time = 2.15525, size = 314, normalized size = 0.84 \[ \frac{2 \sqrt{a} \left (\frac{2 a^2 f^{3/2} \tanh ^{-1}\left (\frac{2 a \sqrt{f}+b \sqrt{d}+b \sqrt{f} x+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}\right )}{\sqrt{a f+b \sqrt{d} \sqrt{f}+c d}}-\frac{2 a^2 f^{3/2} \tanh ^{-1}\left (\frac{-2 a \sqrt{f}+b \left (\sqrt{d}-\sqrt{f} x\right )+2 c \sqrt{d} x}{2 \sqrt{a+x (b+c x)} \sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}\right )}{\sqrt{a f+b \left (-\sqrt{d}\right ) \sqrt{f}+c d}}-\frac{d (2 a-3 b x) \sqrt{a+x (b+c x)}}{x^2}\right )+\left (4 a (c d-2 a f)-3 b^2 d\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{8 a^{5/2} d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b*x + c*x^2]*(d - f*x^2)),x]

[Out]

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

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Maple [A]  time = 0.279, size = 519, normalized size = 1.4 \begin{align*} -{\frac{f}{{d}^{2}}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{f}{2\,{d}^{2}}\ln \left ({ \left ( 2\,{\frac{-b\sqrt{df}+af+cd}{f}}+{\frac{1}{f} \left ( -2\,c\sqrt{df}+bf \right ) \left ( x+{\frac{1}{f}\sqrt{df}} \right ) }+2\,\sqrt{{\frac{-b\sqrt{df}+af+cd}{f}}}\sqrt{ \left ( x+{\frac{\sqrt{df}}{f}} \right ) ^{2}c+{\frac{-2\,c\sqrt{df}+bf}{f} \left ( x+{\frac{\sqrt{df}}{f}} \right ) }+{\frac{-b\sqrt{df}+af+cd}{f}}} \right ) \left ( x+{\frac{1}{f}\sqrt{df}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{1}{f} \left ( -b\sqrt{df}+af+cd \right ) }}}}}-{\frac{1}{2\,ad{x}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,b}{4\,{a}^{2}dx}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{2}}{8\,d}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{c}{2\,d}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}+{\frac{f}{2\,{d}^{2}}\ln \left ({ \left ( 2\,{\frac{b\sqrt{df}+af+cd}{f}}+{\frac{1}{f} \left ( 2\,c\sqrt{df}+bf \right ) \left ( x-{\frac{1}{f}\sqrt{df}} \right ) }+2\,\sqrt{{\frac{b\sqrt{df}+af+cd}{f}}}\sqrt{ \left ( x-{\frac{\sqrt{df}}{f}} \right ) ^{2}c+{\frac{2\,c\sqrt{df}+bf}{f} \left ( x-{\frac{\sqrt{df}}{f}} \right ) }+{\frac{b\sqrt{df}+af+cd}{f}}} \right ) \left ( x-{\frac{1}{f}\sqrt{df}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{1}{f} \left ( b\sqrt{df}+af+cd \right ) }}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{\sqrt{c x^{2} + b x + a}{\left (f x^{2} - d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="maxima")

[Out]

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

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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(1/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{1}{- d x^{3} \sqrt{a + b x + c x^{2}} + f x^{5} \sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(-d*x**3*sqrt(a + b*x + c*x**2) + f*x**5*sqrt(a + b*x + c*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(1/x^3/(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x, algorithm="giac")

[Out]

Exception raised: TypeError