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

Optimal. Leaf size=291 \[ \frac {b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2} d}+\frac {f \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^{3/2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {f \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^{3/2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}-\frac {\sqrt {a+b x+c x^2}}{a d x} \]

[Out]

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

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Rubi [A]  time = 0.66, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {6725, 730, 724, 206, 984} \[ \frac {b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2} d}+\frac {f \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^{3/2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\frac {f \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^{3/2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}-\frac {\sqrt {a+b x+c x^2}}{a d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x + c*x^2]/(a*d*x)) + (b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2*a^(3/2)*d) +
(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*d^(3/2)*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]) + (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*d^(3/2)*Sqrt[c*
d + b*Sqrt[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 730

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[(2*c*d - b*e)/(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, 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] && EqQ[m + 2*p + 3, 0]

Rule 984

Int[1/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[1/2, Int[1/((a - Rt[-
(a*c), 2]*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[1/2, Int[1/((a + Rt[-(a*c), 2]*x)*Sqrt[d + e*x + f*x^2]), x
], x] /; FreeQ[{a, c, d, e, f}, 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 \sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx &=\int \left (\frac {1}{d x^2 \sqrt {a+b x+c x^2}}+\frac {f}{d \sqrt {a+b x+c x^2} \left (d-f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {1}{x^2 \sqrt {a+b x+c x^2}} \, dx}{d}+\frac {f \int \frac {1}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{d}\\ &=-\frac {\sqrt {a+b x+c x^2}}{a d x}-\frac {b \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{2 a d}+\frac {f \int \frac {1}{\left (d-\sqrt {d} \sqrt {f} x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 d}+\frac {f \int \frac {1}{\left (d+\sqrt {d} \sqrt {f} x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 d}\\ &=-\frac {\sqrt {a+b x+c x^2}}{a d x}+\frac {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 d}-\frac {f \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d^{3/2} \sqrt {f}+4 a d f-x^2} \, dx,x,\frac {-b d+2 a \sqrt {d} \sqrt {f}-\left (2 c d-b \sqrt {d} \sqrt {f}\right ) x}{\sqrt {a+b x+c x^2}}\right )}{d}-\frac {f \operatorname {Subst}\left (\int \frac {1}{4 c d^2+4 b d^{3/2} \sqrt {f}+4 a d f-x^2} \, dx,x,\frac {-b d-2 a \sqrt {d} \sqrt {f}-\left (2 c d+b \sqrt {d} \sqrt {f}\right ) x}{\sqrt {a+b x+c x^2}}\right )}{d}\\ &=-\frac {\sqrt {a+b x+c x^2}}{a d x}+\frac {b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2} d}+\frac {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 d^{3/2} \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\frac {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 d^{3/2} \sqrt {c d+b \sqrt {d} \sqrt {f}+a f}}\\ \end {align*}

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Mathematica [A]  time = 1.03, size = 325, normalized size = 1.12 \[ \frac {\frac {b \sqrt {d} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{a^{3/2}}+\frac {f \tanh ^{-1}\left (\frac {2 a \sqrt {f}+b \sqrt {d}+b \sqrt {f} x+2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{\sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}+\frac {f \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+b \left (\sqrt {d}-\sqrt {f} x\right )+2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{\sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}-\frac {2 b \sqrt {d}}{a \sqrt {a+x (b+c x)}}-\frac {2 c \sqrt {d} x}{a \sqrt {a+x (b+c x)}}-\frac {2 \sqrt {d}}{x \sqrt {a+x (b+c x)}}}{2 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^2+b*x+a)^(1/2)/(-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 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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: 1.11sy
m2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.02, size = 427, normalized size = 1.47 \[ -\frac {f \ln \left (\frac {\frac {2 a f +2 c d -2 \sqrt {d f}\, b}{f}+\frac {\left (b f -2 \sqrt {d f}\, c \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {a f +c d -\sqrt {d f}\, b}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (b f -2 \sqrt {d f}\, c \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {a f +c d -\sqrt {d f}\, b}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 \sqrt {d f}\, \sqrt {\frac {a f +c d -\sqrt {d f}\, b}{f}}\, d}+\frac {f \ln \left (\frac {\frac {2 a f +2 c d +2 \sqrt {d f}\, b}{f}+\frac {\left (b f +2 \sqrt {d f}\, c \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {a f +c d +\sqrt {d f}\, b}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (b f +2 \sqrt {d f}\, c \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {a f +c d +\sqrt {d f}\, b}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 \sqrt {d f}\, \sqrt {\frac {a f +c d +\sqrt {d f}\, b}{f}}\, d}+\frac {b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}} d}-\frac {\sqrt {c \,x^{2}+b x +a}}{a d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*f/d/(d*f)^(1/2)/((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))-(c*x^2+b*x+a)^(1/2)/a/d/x+1/2/d*b/a^(3/2)*ln((b
*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+1/2*f/d/(d*f)^(1/2)/((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 \[ -\int \frac {1}{\sqrt {c x^{2} + b x + a} {\left (f x^{2} - d\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^2\,\left (d-f\,x^2\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(x^2*(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 \[ - \int \frac {1}{- d x^{2} \sqrt {a + b x + c x^{2}} + f x^{4} \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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