3.69 \(\int \frac{1}{x^2 \sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx\)

Optimal. Leaf size=367 \[ -\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+c x^2}}{a d x} \]

[Out]

-(Sqrt[a + c*x^2]/(a*d*x)) - (f*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])*ArcTanh[(2*a
*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqr
t[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2
+ c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + (f*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f
])*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2
- 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f
]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]) + (e*ArcTanh[Sqrt[a + c
*x^2]/Sqrt[a]])/(Sqrt[a]*d^2)

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Rubi [A]  time = 2.67815, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296 \[ -\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+c x^2}}{a d x} \]

Antiderivative was successfully verified.

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

[Out]

-(Sqrt[a + c*x^2]/(a*d*x)) - (f*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])*ArcTanh[(2*a
*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqr
t[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2
+ c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + (f*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f
])*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2
- 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f
]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]) + (e*ArcTanh[Sqrt[a + c
*x^2]/Sqrt[a]])/(Sqrt[a]*d^2)

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/x**2/(f*x**2+e*x+d)/(c*x**2+a)**(1/2),x)

[Out]

Timed out

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Mathematica [A]  time = 1.14071, size = 567, normalized size = 1.54 \[ \frac{-\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \log \left (\sqrt{a+c x^2} \sqrt{4 a f^2-2 c e \sqrt{e^2-4 d f}-4 c d f+2 c e^2}+2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right ) \log \left (-\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}-2 a f+c x \sqrt{e^2-4 d f}+c e x\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \log \left (\sqrt{e^2-4 d f}-e-2 f x\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right ) \log \left (\sqrt{e^2-4 d f}+e+2 f x\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{2 d \sqrt{a+c x^2}}{a x}+\frac{2 e \log \left (\sqrt{a} \sqrt{a+c x^2}+a\right )}{\sqrt{a}}-\frac{2 e \log (x)}{\sqrt{a}}}{2 d^2} \]

Antiderivative was successfully verified.

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

[Out]

((-2*d*Sqrt[a + c*x^2])/(a*x) - (2*e*Log[x])/Sqrt[a] + (Sqrt[2]*f*(e^2 - 2*d*f +
 e*Sqrt[e^2 - 4*d*f])*Log[-e + Sqrt[e^2 - 4*d*f] - 2*f*x])/(Sqrt[e^2 - 4*d*f]*Sq
rt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + (Sqrt[2]*f*(-e^2 + 2*d*f
+ e*Sqrt[e^2 - 4*d*f])*Log[e + Sqrt[e^2 - 4*d*f] + 2*f*x])/(Sqrt[e^2 - 4*d*f]*Sq
rt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]) + (2*e*Log[a + Sqrt[a]*Sqrt
[a + c*x^2]])/Sqrt[a] - (Sqrt[2]*f*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])*Log[2*a*f
 + c*(-e + Sqrt[e^2 - 4*d*f])*x + Sqrt[2*c*e^2 - 4*c*d*f + 4*a*f^2 - 2*c*e*Sqrt[
e^2 - 4*d*f]]*Sqrt[a + c*x^2]])/(Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f
 - e*Sqrt[e^2 - 4*d*f])]) - (Sqrt[2]*f*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])*Log[
-2*a*f + c*e*x + c*Sqrt[e^2 - 4*d*f]*x - Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqr
t[e^2 - 4*d*f])]*Sqrt[a + c*x^2]])/(Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*
d*f + e*Sqrt[e^2 - 4*d*f])]))/(2*d^2)

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Maple [B]  time = 0.023, size = 736, normalized size = 2. \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/x^2/(f*x^2+e*x+d)/(c*x^2+a)^(1/2),x)

[Out]

4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))/a/x*(c*x^2+a)^(1/2)-4*f^2/(-e
+(-4*d*f+e^2)^(1/2))^2/(-4*d*f+e^2)^(1/2)*2^(1/2)/((-(-4*d*f+e^2)^(1/2)*c*e+2*a*
f^2-2*c*d*f+e^2*c)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+e^2*c
)/f^2-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*(
(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+e^2*c)/f^2)^(1/2)*(4*(x-1/2*(-e+(-4*d*f
+e^2)^(1/2))/f)^2*c-4*c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2*(-e+(-4*d*f+e^2)^(1/2))/
f)+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+e^2*c)/f^2)^(1/2))/(x-1/2*(-e+(-4*
d*f+e^2)^(1/2))/f))+4*f^2/(e+(-4*d*f+e^2)^(1/2))^2/(-4*d*f+e^2)^(1/2)*2^(1/2)/((
(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+e^2*c)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)
*c*e+2*a*f^2-2*c*d*f+e^2*c)/f^2-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2
)^(1/2))/f)+1/2*2^(1/2)*(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+e^2*c)/f^2)^(1/
2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(
e+(-4*d*f+e^2)^(1/2))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+e^2*c)/f^2)^(
1/2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))+16*f^2*e/(-e+(-4*d*f+e^2)^(1/2))^2/(e+(-
4*d*f+e^2)^(1/2))^2/a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{c x^{2} + a}{\left (f x^{2} + e x + d\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(c*x^2 + a)*(f*x^2 + e*x + d)*x^2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^2 + a)*(f*x^2 + e*x + d)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(c*x^2 + a)*(f*x^2 + e*x + d)*x^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{x^{2} \sqrt{a + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/x**2/(f*x**2+e*x+d)/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(x**2*sqrt(a + c*x**2)*(d + e*x + f*x**2)), x)

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GIAC/XCAS [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(c*x^2 + a)*(f*x^2 + e*x + d)*x^2),x, algorithm="giac")

[Out]

Timed out