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

Optimal. Leaf size=220 \[ \frac {\tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \sqrt {f} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}-\frac {\tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \sqrt {f} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}} \]

[Out]

-1/2*arctanh(1/2*(b*d^(1/2)-2*a*f^(1/2)+x*(2*c*d^(1/2)-b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1
/2))^(1/2))/f^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1/2))^(1/2)+1/2*arctanh(1/2*(b*d^(1/2)+2*a*f^(1/2)+x*(2*c*d^(1/2)+b*
f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2))/f^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1033, 724, 206} \[ \frac {\tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \sqrt {f} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}-\frac {\tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \sqrt {f} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.17, size = 211, normalized size = 0.96 \[ -\frac {\frac {\tanh ^{-1}\left (\frac {-2 a \sqrt {f}+b \left (\sqrt {d}-\sqrt {f} x\right )+2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{\sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {\tanh ^{-1}\left (\frac {-2 \left (a \sqrt {f}+c \sqrt {d} x\right )-b \left (\sqrt {d}+\sqrt {f} x\right )}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{\sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}}{2 \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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] + 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])/Sqrt[f]

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fricas [B]  time = 2.40, size = 2753, normalized size = 12.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt((c*d + a*f + (c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2
- 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/(c^2*d^2*f + a^2*
f^3 - (b^2 - 2*a*c)*d*f^2))*log((2*b*c*d*x + b^2*d + 2*(b^2*d*f - (c^3*d^3*f + a^3*f^4 - (b^2*c - 3*a*c^2)*d^2
*f^2 - (a*b^2 - 3*a^2*c)*d*f^3)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b
^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))*sqrt(c*x^2 + b*x + a)*sqrt((c*d + a*f + (c^2*d^2*f
+ a^2*f^3 - (b^2 - 2*a*c)*d*f^2)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*
b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/(c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)) - (2*
a*c^2*d^2*f + 2*a^3*f^3 - 2*(a*b^2 - 2*a^2*c)*d*f^2 + (b*c^2*d^2*f + a^2*b*f^3 - (b^3 - 2*a*b*c)*d*f^2)*x)*sqr
t(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*
b^2 - 2*a^3*c)*d*f^4)))/x) - 1/4*sqrt((c*d + a*f + (c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)*sqrt(b^2*d/(c^4
*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*
c)*d*f^4)))/(c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2))*log((2*b*c*d*x + b^2*d - 2*(b^2*d*f - (c^3*d^3*f + a^
3*f^4 - (b^2*c - 3*a*c^2)*d^2*f^2 - (a*b^2 - 3*a^2*c)*d*f^3)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*
a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))*sqrt(c*x^2 + b*x + a)*
sqrt((c*d + a*f + (c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2
*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/(c^2*d^2*f + a^2*f^3
- (b^2 - 2*a*c)*d*f^2)) - (2*a*c^2*d^2*f + 2*a^3*f^3 - 2*(a*b^2 - 2*a^2*c)*d*f^2 + (b*c^2*d^2*f + a^2*b*f^3 -
(b^3 - 2*a*b*c)*d*f^2)*x)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c +
6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/x) + 1/4*sqrt((c*d + a*f - (c^2*d^2*f + a^2*f^3 - (b^2 -
2*a*c)*d*f^2)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*
d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/(c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2))*log((2*b*c*d*x + b^2*d +
2*(b^2*d*f + (c^3*d^3*f + a^3*f^4 - (b^2*c - 3*a*c^2)*d^2*f^2 - (a*b^2 - 3*a^2*c)*d*f^3)*sqrt(b^2*d/(c^4*d^4*
f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*
f^4)))*sqrt(c*x^2 + b*x + a)*sqrt((c*d + a*f - (c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)*sqrt(b^2*d/(c^4*d^4
*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d
*f^4)))/(c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)) + (2*a*c^2*d^2*f + 2*a^3*f^3 - 2*(a*b^2 - 2*a^2*c)*d*f^2
+ (b*c^2*d^2*f + a^2*b*f^3 - (b^3 - 2*a*b*c)*d*f^2)*x)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)
*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/x) - 1/4*sqrt((c*d + a*f - (
c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (
b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/(c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f
^2))*log((2*b*c*d*x + b^2*d - 2*(b^2*d*f + (c^3*d^3*f + a^3*f^4 - (b^2*c - 3*a*c^2)*d^2*f^2 - (a*b^2 - 3*a^2*c
)*d*f^3)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f
^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))*sqrt(c*x^2 + b*x + a)*sqrt((c*d + a*f - (c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*
c)*d*f^2)*sqrt(b^2*d/(c^4*d^4*f + a^4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*
f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))/(c^2*d^2*f + a^2*f^3 - (b^2 - 2*a*c)*d*f^2)) + (2*a*c^2*d^2*f + 2*a^3*f^3
- 2*(a*b^2 - 2*a^2*c)*d*f^2 + (b*c^2*d^2*f + a^2*b*f^3 - (b^3 - 2*a*b*c)*d*f^2)*x)*sqrt(b^2*d/(c^4*d^4*f + a^
4*f^5 - 2*(b^2*c^2 - 2*a*c^3)*d^3*f^2 + (b^4 - 4*a*b^2*c + 6*a^2*c^2)*d^2*f^3 - 2*(a^2*b^2 - 2*a^3*c)*d*f^4)))
/x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c*x^2+b*x+a)^(1/2)/(-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: 0.52sy
m2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.02, size = 354, normalized size = 1.61 \[ \frac {\ln \left (\frac {\frac {2 a f +2 c d -2 \sqrt {d f}\, b}{f}+\frac {\left (b f -2 \sqrt {d f}\, c \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {a f +c d -\sqrt {d f}\, b}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (b f -2 \sqrt {d f}\, c \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {a f +c d -\sqrt {d f}\, b}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 \sqrt {\frac {a f +c d -\sqrt {d f}\, b}{f}}\, f}+\frac {\ln \left (\frac {\frac {2 a f +2 c d +2 \sqrt {d f}\, b}{f}+\frac {\left (b f +2 \sqrt {d f}\, c \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {a f +c d +\sqrt {d f}\, b}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (b f +2 \sqrt {d f}\, c \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {a f +c d +\sqrt {d f}\, b}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 \sqrt {\frac {a f +c d +\sqrt {d f}\, b}{f}}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/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/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))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c*x^2+b*x+a)^(1/2)/(-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 \[ \int \frac {x}{\left (d-f\,x^2\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x}{- d \sqrt {a + b x + c x^{2}} + f x^{2} \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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