3.53 \(\int \frac{x \sqrt{a+c x^2}}{d+e x+f x^2} \, dx\)

Optimal. Leaf size=395 \[ -\frac{\left (2 c d e f-\left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 c d e f-\left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\sqrt{c} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{\sqrt{a+c x^2}}{f} \]

[Out]

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

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Rubi [A]  time = 3.70065, antiderivative size = 395, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28 \[ -\frac{\left (2 c d e f-\left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 c d e f-\left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\sqrt{c} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{\sqrt{a+c x^2}}{f} \]

Antiderivative was successfully verified.

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

[Out]

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

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

[Out]

Timed out

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Mathematica [A]  time = 1.68396, size = 785, normalized size = 1.99 \[ \frac{-\frac{\sqrt{2} \left (a f^2 \left (\sqrt{e^2-4 d f}-e\right )+c \left (e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}+3 d e f-e^3\right )\right ) \log \left (\sqrt{2} \sqrt{a+c x^2} \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}+2 a f \sqrt{e^2-4 d f}+c x \left (-e \sqrt{e^2-4 d f}-4 d f+e^2\right )\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\sqrt{2} \left (a f^2 \left (\sqrt{e^2-4 d f}+e\right )+c \left (e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}-3 d e f+e^3\right )\right ) \log \left (\sqrt{2} \sqrt{a+c x^2} \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}+2 a f \sqrt{e^2-4 d f}-c x \left (e \sqrt{e^2-4 d f}-4 d f+e^2\right )\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} \left (a f^2 \left (\sqrt{e^2-4 d f}-e\right )+c \left (e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}+3 d e f-e^3\right )\right ) \log \left (\sqrt{e^2-4 d f}-e-2 f x\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} \left (a f^2 \left (\sqrt{e^2-4 d f}+e\right )+c \left (e^2 \sqrt{e^2-4 d f}-d f \sqrt{e^2-4 d f}-3 d e f+e^3\right )\right ) \log \left (\sqrt{e^2-4 d f}+e+2 f x\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-2 \sqrt{c} e \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )+2 f \sqrt{a+c x^2}}{2 f^2} \]

Antiderivative was successfully verified.

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

[Out]

(2*f*Sqrt[a + c*x^2] + (Sqrt[2]*(a*f^2*(-e + Sqrt[e^2 - 4*d*f]) + c*(-e^3 + 3*d*
e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f]))*Log[-e + Sqrt[e^2 - 4*d*f]
- 2*f*x])/(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]*(a*f^2*(e + Sqrt[e^2 - 4*d*f]) + c*(e^3 - 3*d*e*f + e^2*Sqrt[e^2
- 4*d*f] - d*f*Sqrt[e^2 - 4*d*f]))*Log[e + Sqrt[e^2 - 4*d*f] + 2*f*x])/(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*Sqrt[c]*e*
Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]] - (Sqrt[2]*(a*f^2*(-e + Sqrt[e^2 - 4*d*f]) +
c*(-e^3 + 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f]))*Log[2*a*f*Sq
rt[e^2 - 4*d*f] + c*(e^2 - 4*d*f - e*Sqrt[e^2 - 4*d*f])*x + Sqrt[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[a + c*x^2]])/(S
qrt[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]*(a*f^2*(e + Sqrt[e^2 - 4*d*f]) + c*(e^3 - 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d
*f*Sqrt[e^2 - 4*d*f]))*Log[2*a*f*Sqrt[e^2 - 4*d*f] - c*(e^2 - 4*d*f + e*Sqrt[e^2
- 4*d*f])*x + Sqrt[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[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*f^2)

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Maple [B]  time = 0.02, size = 5581, normalized size = 14.1 \[ \text{output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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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(sqrt(c*x^2 + a)*x/(f*x^2 + e*x + d),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\sqrt{c x^{2} + a} x}{f x^{2} + e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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