3.61 \(\int \frac{\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx\)

Optimal. Leaf size=496 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d}-\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (\sqrt{e^2-4 d f}+e\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{c^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{\sqrt{a+c x^2} (c d-a f)}{d f}+\frac{a \sqrt{a+c x^2}}{d} \]

[Out]

(a*Sqrt[a + c*x^2])/d + ((c*d - a*f)*Sqrt[a + c*x^2])/(d*f) - (c^(3/2)*e*ArcTanh
[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/f^2 - ((2*e*f*(c^2*d^2 - a^2*f^2) - (c^2*d*e^2 -
f*(c*d - a*f)^2)*(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*f^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqr
t[e^2 - 4*d*f])]) + ((2*e*f*(c^2*d^2 - a^2*f^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*
(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*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])])
- (a^(3/2)*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/d

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Rubi [A]  time = 5.34175, antiderivative size = 496, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 11, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.407 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d}-\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (\sqrt{e^2-4 d f}+e\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{c^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{\sqrt{a+c x^2} (c d-a f)}{d f}+\frac{a \sqrt{a+c x^2}}{d} \]

Antiderivative was successfully verified.

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

[Out]

(a*Sqrt[a + c*x^2])/d + ((c*d - a*f)*Sqrt[a + c*x^2])/(d*f) - (c^(3/2)*e*ArcTanh
[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/f^2 - ((2*e*f*(c^2*d^2 - a^2*f^2) - (c^2*d*e^2 -
f*(c*d - a*f)^2)*(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*f^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqr
t[e^2 - 4*d*f])]) + ((2*e*f*(c^2*d^2 - a^2*f^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*
(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*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])])
- (a^(3/2)*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/d

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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((c*x**2+a)**(3/2)/x/(f*x**2+e*x+d),x)

[Out]

Timed out

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Mathematica [A]  time = 2.07136, size = 956, normalized size = 1.93 \[ \frac{\log (x) a^{3/2}}{d}-\frac{\log \left (a+\sqrt{c x^2+a} \sqrt{a}\right ) a^{3/2}}{d}-\frac{\left (a^2 \left (e+\sqrt{e^2-4 d f}\right ) f^3+2 a c d \left (e-\sqrt{e^2-4 d f}\right ) f^2+c^2 d \left (e^3-\sqrt{e^2-4 d f} e^2-3 d f e+d f \sqrt{e^2-4 d f}\right )\right ) \log \left (-e-2 f x+\sqrt{e^2-4 d f}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right )}}+\frac{\left (a^2 \left (e-\sqrt{e^2-4 d f}\right ) f^3+2 a c d \left (e+\sqrt{e^2-4 d f}\right ) f^2+c^2 d \left (e^3+\sqrt{e^2-4 d f} e^2-3 d f e-d f \sqrt{e^2-4 d f}\right )\right ) \log \left (e+2 f x+\sqrt{e^2-4 d f}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right )}}-\frac{c^{3/2} e \log \left (c x+\sqrt{c} \sqrt{c x^2+a}\right )}{f^2}+\frac{\left (a^2 \left (e+\sqrt{e^2-4 d f}\right ) f^3+2 a c d \left (e-\sqrt{e^2-4 d f}\right ) f^2+c^2 d \left (e^3-\sqrt{e^2-4 d f} e^2-3 d f e+d f \sqrt{e^2-4 d f}\right )\right ) \log \left (2 a \sqrt{e^2-4 d f} f+c \left (e^2-\sqrt{e^2-4 d f} e-4 d f\right ) x+\sqrt{2} \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right )} \sqrt{c x^2+a}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2-\sqrt{e^2-4 d f} e-2 d f\right )}}-\frac{\left (a^2 \left (e-\sqrt{e^2-4 d f}\right ) f^3+2 a c d \left (e+\sqrt{e^2-4 d f}\right ) f^2+c^2 d \left (e^3+\sqrt{e^2-4 d f} e^2-3 d f e-d f \sqrt{e^2-4 d f}\right )\right ) \log \left (2 a \sqrt{e^2-4 d f} f-c \left (e^2+\sqrt{e^2-4 d f} e-4 d f\right ) x+\sqrt{2} \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right )} \sqrt{c x^2+a}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2+\sqrt{e^2-4 d f} e-2 d f\right )}}+\frac{c \sqrt{c x^2+a}}{f} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [B]  time = 0.027, size = 9728, normalized size = 19.6 \[ \text{output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((c*x^2+a)^(3/2)/x/(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. \[ \int \frac{{\left (c x^{2} + a\right )}^{\frac{3}{2}}}{{\left (f x^{2} + e x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + a)^(3/2)/((f*x^2 + e*x + d)*x),x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^(3/2)/((f*x^2 + e*x + d)*x), 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((c*x^2 + a)^(3/2)/((f*x^2 + e*x + d)*x),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (a + c x^{2}\right )^{\frac{3}{2}}}{x \left (d + e x + f x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x**2+a)**(3/2)/x/(f*x**2+e*x+d),x)

[Out]

Integral((a + c*x**2)**(3/2)/(x*(d + e*x + f*x**2)), x)

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GIAC/XCAS [A]  time = 1.31481, size = 1, normalized size = 0. \[ \mathit{Done} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + a)^(3/2)/((f*x^2 + e*x + d)*x),x, algorithm="giac")

[Out]

Done