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

Optimal. Leaf size=454 \[ \frac {3 b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{5/2} d}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 d x \left (b^2-4 a c\right )}-\frac {2 f \left (b \left (b^2 f-c (3 a f+c d)\right )-c x \left (2 a c f+b^2 (-f)+2 c^2 d\right )\right )}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a d x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {f^2 \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^{3/2} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {f^2 \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^{3/2} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \]

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Rubi [A]  time = 1.19, antiderivative size = 454, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6725, 740, 806, 724, 206, 975, 1033} \begin {gather*} -\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x+c x^2}}{a^2 d x \left (b^2-4 a c\right )}+\frac {3 b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{5/2} d}-\frac {2 f \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 )}{d \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 \left (-2 a c+b^2+b c x\right )}{a d x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {f^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^{3/2} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {f^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^{3/2} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 740

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 1.19, size = 488, normalized size = 1.07 \begin {gather*} \frac {\frac {3 b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{a^{5/2}}+\frac {2 \left (8 a c-3 b^2\right ) \sqrt {a+x (b+c x)}}{a^2 x}-\frac {f^2 \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 \sqrt {d}+b \sqrt {f} x-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 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}}\right )}{\sqrt {d} \left ((a f+c d)^2-b^2 d f\right )}+\frac {4 \left (-2 a c+b^2+b c x\right )}{a x \sqrt {a+x (b+c x)}}-\frac {4 f \left (-b c (3 a f+c d)-2 c^2 x (a f+c d)+b^3 f+b^2 c f x\right )}{\sqrt {a+x (b+c x)} \left (b^2 d f-(a f+c d)^2\right )}}{2 d \left (b^2-4 a c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 2.92, size = 649, normalized size = 1.43 \begin {gather*} -\frac {f^2 \text {RootSum}\left [\text {$\#$1}^4 (-f)+2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d-4 \text {$\#$1} b \sqrt {c} d-a^2 f+b^2 d\&,\frac {\text {$\#$1}^2 (-b) f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} c^{3/2} d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a \sqrt {c} f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 a b f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{\text {$\#$1}^3 f-\text {$\#$1} a f-2 \text {$\#$1} c d+b \sqrt {c} d}\&\right ]}{2 d \left (a^2 f^2+2 a c d f+b^2 (-d) f+c^2 d^2\right )}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {-4 a^4 c f^2+a^3 b^2 f^2-4 a^3 b c f^2 x-8 a^3 c^2 d f-4 a^3 c^2 f^2 x^2+a^2 b^3 f^2 x+6 a^2 b^2 c d f+a^2 b^2 c f^2 x^2-18 a^2 b c^2 d f x-4 a^2 c^3 d^2-12 a^2 c^3 d f x^2-a b^4 d f+16 a b^3 c d f x+a b^2 c^2 d^2+14 a b^2 c^2 d f x^2-10 a b c^3 d^2 x-8 a c^4 d^2 x^2-3 b^5 d f x-3 b^4 c d f x^2+3 b^3 c^2 d^2 x+3 b^2 c^3 d^2 x^2}{a^2 d x \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2} \left (a^2 f^2+2 a c d f+b^2 (-d) f+c^2 d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2*c^2*d^2 - 4*a^2*c^3*d^2 - a*b^4*d*f + 6*a^2*b^2*c*d*f - 8*a^3*c^2*d*f + a^3*b^2*f^2 - 4*a^4*c*f^2 + 3*b
^3*c^2*d^2*x - 10*a*b*c^3*d^2*x - 3*b^5*d*f*x + 16*a*b^3*c*d*f*x - 18*a^2*b*c^2*d*f*x + a^2*b^3*f^2*x - 4*a^3*
b*c*f^2*x + 3*b^2*c^3*d^2*x^2 - 8*a*c^4*d^2*x^2 - 3*b^4*c*d*f*x^2 + 14*a*b^2*c^2*d*f*x^2 - 12*a^2*c^3*d*f*x^2
+ a^2*b^2*c*f^2*x^2 - 4*a^3*c^2*f^2*x^2)/(a^2*(-b^2 + 4*a*c)*d*(c^2*d^2 - b^2*d*f + 2*a*c*d*f + a^2*f^2)*x*Sqr
t[a + b*x + c*x^2]) - (3*b*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + b*x + c*x^2]/Sqrt[a]])/(a^(5/2)*d) - (f^2*Ro
otSum[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*d*(c^2*d^2
- b^2*d*f + 2*a*c*d*f + a^2*f^2))

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fricas [F(-1)]  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(1/x^2/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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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(1/x^2/(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueEvaluation time: 1.84in
t()  Bad Argument Type

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maple [B]  time = 0.02, size = 1656, normalized size = 3.65

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (f x^{2} - d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^2\,\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(1/(x^2*(d - f*x^2)*(a + b*x + c*x^2)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{- a d x^{2} \sqrt {a + b x + c x^{2}} + a f x^{4} \sqrt {a + b x + c x^{2}} - b d x^{3} \sqrt {a + b x + c x^{2}} + b f x^{5} \sqrt {a + b x + c x^{2}} - c d x^{4} \sqrt {a + b x + c x^{2}} + c f x^{6} \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(-a*d*x**2*sqrt(a + b*x + c*x**2) + a*f*x**4*sqrt(a + b*x + c*x**2) - b*d*x**3*sqrt(a + b*x + c*x*
*2) + b*f*x**5*sqrt(a + b*x + c*x**2) - c*d*x**4*sqrt(a + b*x + c*x**2) + c*f*x**6*sqrt(a + b*x + c*x**2)), x)

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