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

Optimal. Leaf size=394 \[ -\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{3/2} d}-\frac {2 f \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {f^{3/2} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {f^{3/2} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \]

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Rubi [A]  time = 1.18, antiderivative size = 394, 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, 12, 724, 206, 1018, 1033} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{3/2} d}-\frac {2 f \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\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 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\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 \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\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 \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(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*Sqrt[a + b*x + c*x^2]) - (2*f*(a*(2*c^2*d - b^2*f + 2*a*c*f) + b*
c*(c*d - a*f)*x))/((b^2 - 4*a*c)*d*(b^2*d*f - (c*d + a*f)^2)*Sqrt[a + b*x + c*x^2]) - ArcTanh[(2*a + b*x)/(2*S
qrt[a]*Sqrt[a + b*x + c*x^2])]/(a^(3/2)*d) - (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*(c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(
3/2)) + (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*(c*d + b*Sqrt[d]*Sqrt[f] + a*f)^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1018

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^(q + 1)*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*(
2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(2*a*f)) - h*(b*c*d + a*b*f))*x))/((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[(b*h - 2*g*c)*((c*d - a*f)^2 - (b*d)*(-(b*f)))*(p + 1) + (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*
(c*d - a*f)))*(a*f*(p + 1) - c*d*(p + 2)) - (2*f*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*
(2*a*f)))*(p + q + 2) - (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(b*f*(p + 1)))*x - c*f*(b^2*(g*f
) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, 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
])

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

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Mathematica [A]  time = 0.87, size = 436, normalized size = 1.11 \begin {gather*} \frac {-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{a^{3/2}}-\frac {2 f \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a 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^{3/2} \left (\left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{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 )+\left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{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 \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )\right )}{2 \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d} \left ((a f+c d)^2-b^2 d f\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)}}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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IntegrateAlgebraic [C]  time = 1.59, size = 498, normalized size = 1.26 \begin {gather*} \frac {f^2 \text {RootSum}\left [\text {$\#$1}^4 (-f)+2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d-4 \text {$\#$1} b \sqrt {c} d-a^2 f+b^2 d\&,\frac {-\text {$\#$1}^2 c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 a f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a c d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b \sqrt {c} d \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{\text {$\#$1}^3 f-\text {$\#$1} a f-2 \text {$\#$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 {2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (2 a^2 c^2 f-4 a b^2 c f-3 a b c^2 f x+2 a c^3 d+b^4 f+b^3 c f x-b^2 c^2 d-b c^3 d x\right )}{a \left (b^2-4 a c\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*(a + b*x + c*x^2)^(3/2)*(d - f*x^2)),x]

[Out]

(2*(-(b^2*c^2*d) + 2*a*c^3*d + b^4*f - 4*a*b^2*c*f + 2*a^2*c^2*f - b*c^3*d*x + b^3*c*f*x - 3*a*b*c^2*f*x))/(a*
(b^2 - 4*a*c)*(-(c^2*d^2) + b^2*d*f - 2*a*c*d*f - a^2*f^2)*Sqrt[a + b*x + c*x^2]) + (2*ArcTanh[(Sqrt[c]*x)/Sqr
t[a] - Sqrt[a + b*x + c*x^2]/Sqrt[a]])/(a^(3/2)*d) + (f^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^2*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a*c*d*Log[-(Sqrt[c]*x) + S
qrt[a + b*x + c*x^2] - #1] + a^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*Sqrt[c]*d*Log[-(Sqrt[c
]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - a*f*Log[-(Sq
rt[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/(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2

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maple [B]  time = 0.02, size = 1518, normalized size = 3.85

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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

[Out]

int(1/(x*(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 \sqrt {a + b x + c x^{2}} + a f x^{3} \sqrt {a + b x + c x^{2}} - b d x^{2} \sqrt {a + b x + c x^{2}} + b f x^{4} \sqrt {a + b x + c x^{2}} - c d x^{3} \sqrt {a + b x + c x^{2}} + c f x^{5} \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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