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

Optimal. Leaf size=299 \[ -\frac {2 \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}-\frac {\sqrt {f} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {\sqrt {f} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \]

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Rubi [A]  time = 0.40, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1018, 1033, 724, 206} \begin {gather*} -\frac {2 \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+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}-\frac {\sqrt {f} \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 \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {\sqrt {f} \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 \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 356, normalized size = 1.19 \begin {gather*} \frac {2 \left (\frac {2 a^2 c f+a \left (b^2 (-f)-b c f x+2 c^2 d\right )+b c^2 d x}{\sqrt {a+x (b+c x)}}-\frac {\sqrt {f} \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 )}{4 \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {\sqrt {f} \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 )}{4 \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{\left (b^2-4 a c\right ) \left ((a f+c d)^2-b^2 d f\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 0.91, size = 419, normalized size = 1.40 \begin {gather*} \frac {f \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 \left (a^2 f^2+2 a c d f+b^2 (-d) f+c^2 d^2\right )}-\frac {2 \left (2 a^2 c f-a b^2 f-a b c f x+2 a c^2 d+b c^2 d x\right )}{\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[x/((a + b*x + c*x^2)^(3/2)*(d - f*x^2)),x]

[Out]

(-2*(2*a*c^2*d - a*b^2*f + 2*a^2*c*f + b*c^2*d*x - a*b*c*f*x))/((-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]) + (f*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) + Sqrt[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[-(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*(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(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(-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(x/(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 Valueint()  Bad Argument Typ
e

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maple [B]  time = 0.02, size = 1360, normalized size = 4.55

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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/(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/(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-1/f/(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/(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/(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))-1/2/(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/(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/(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+1/f/(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/(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/(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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((c*sqrt(4*d*f))/(2*f^2)>0)',
see `assume?` for more details)Is ((c*sqrt(4*d*f))/(2*f^2)    +b/(2*f))    ^2    -(c*((b*sqrt(4*d*f))
/(2*f)                  +(c*d)/f+a))     /f^2 zero or nonzero?

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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