3.362 \(\int \frac{\left (d+e x^2\right )^{3/2}}{x^3 \left (a+b x^2+c x^4\right )} \, dx\)

Optimal. Leaf size=417 \[ -\frac{\sqrt{c} \left (-2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{c} \left (-2 a \left (c d^2-e \left (d \sqrt{b^2-4 a c}+a e\right )\right )-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 7.54212, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241 \[ -\frac{\sqrt{c} \left (b d \left (d \sqrt{b^2-4 a c}-2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}-a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{c} \left (-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+2 a e \left (d \sqrt{b^2-4 a c}+a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]  Int[(d + e*x^2)^(3/2)/(x^3*(a + b*x^2 + c*x^4)),x]

[Out]

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

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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((e*x**2+d)**(3/2)/x**3/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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Mathematica [A]  time = 2.01722, size = 382, normalized size = 0.92 \[ \frac{\frac{\sqrt{2} \left (\frac{c \left (2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (2 a e-d \sqrt{b^2-4 a c}\right )-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{c \left (b d \left (d \sqrt{b^2-4 a c}+2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}+a e\right )+2 a c d^2-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{c} \sqrt{b^2-4 a c}}+\sqrt{d} (2 b d-3 a e) \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-\sqrt{d} \log (x) (2 b d-3 a e)-\frac{a d \sqrt{d+e x^2}}{x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]  Integrate[(d + e*x^2)^(3/2)/(x^3*(a + b*x^2 + c*x^4)),x]

[Out]

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

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Maple [C]  time = 0.046, size = 555, normalized size = 1.3 \[ -{\frac{1}{2\,ad{x}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{e}{2\,ad} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{3\,e}{2\,a}\sqrt{d}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) }+{\frac{e}{a}\sqrt{e{x}^{2}+d}}-{\frac{b{x}^{3}}{6\,{a}^{2}}{e}^{{\frac{3}{2}}}}+{\frac{{x}^{2}be}{8\,{a}^{2}}\sqrt{e{x}^{2}+d}}-{\frac{3\,bdx}{4\,{a}^{2}}\sqrt{e}}-{\frac{7\,b}{24\,{a}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{x}{2\,a}{e}^{{\frac{3}{2}}}}-{\frac{3\,bd}{8\,{a}^{2}}\sqrt{e{x}^{2}+d}}+{\frac{1}{4\,{a}^{2}}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{8}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{6}+ \left ( 16\,a{e}^{2}-8\,bde+6\,c{d}^{2} \right ){{\it \_Z}}^{4}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){{\it \_Z}}^{2}+c{d}^{4} \right ) }{\frac{cd \left ( -2\,ae+bd \right ){{\it \_R}}^{6}+ \left ( 4\,{a}^{2}{e}^{3}-8\,abd{e}^{2}+2\,ac{d}^{2}e+4\,{b}^{2}{d}^{2}e-3\,bc{d}^{3} \right ){{\it \_R}}^{4}+d \left ( -4\,{a}^{2}{e}^{3}+8\,abd{e}^{2}-2\,ac{d}^{2}e-4\,{b}^{2}{d}^{2}e+3\,bc{d}^{3} \right ){{\it \_R}}^{2}+2\,ac{d}^{4}e-bc{d}^{5}}{{{\it \_R}}^{7}c+3\,{{\it \_R}}^{5}be-3\,{{\it \_R}}^{5}cd+8\,{{\it \_R}}^{3}a{e}^{2}-4\,{{\it \_R}}^{3}bde+3\,{{\it \_R}}^{3}c{d}^{2}+{\it \_R}\,b{d}^{2}e-{\it \_R}\,c{d}^{3}}\ln \left ( \sqrt{e{x}^{2}+d}-x\sqrt{e}-{\it \_R} \right ) }}-{\frac{de}{2\,a} \left ( \sqrt{e{x}^{2}+d}-x\sqrt{e} \right ) ^{-1}}+{\frac{5\,b{d}^{2}}{8\,{a}^{2}} \left ( \sqrt{e{x}^{2}+d}-x\sqrt{e} \right ) ^{-1}}+{\frac{b{d}^{3}}{24\,{a}^{2}} \left ( \sqrt{e{x}^{2}+d}-x\sqrt{e} \right ) ^{-3}}+{\frac{b}{{a}^{2}}{d}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x^2+d)^(3/2)/x^3/(c*x^4+b*x^2+a),x)

[Out]

-1/2/a/d/x^2*(e*x^2+d)^(5/2)+1/2/a/d*e*(e*x^2+d)^(3/2)-3/2/a*d^(1/2)*e*ln((2*d+2
*d^(1/2)*(e*x^2+d)^(1/2))/x)+1/a*e*(e*x^2+d)^(1/2)-1/6/a^2*e^(3/2)*x^3*b+1/8/a^2
*e*(e*x^2+d)^(1/2)*x^2*b-3/4/a^2*e^(1/2)*x*b*d-7/24/a^2*(e*x^2+d)^(3/2)*b+1/2/a*
e^(3/2)*x-3/8/a^2*(e*x^2+d)^(1/2)*b*d+1/4/a^2*sum((c*d*(-2*a*e+b*d)*_R^6+(4*a^2*
e^3-8*a*b*d*e^2+2*a*c*d^2*e+4*b^2*d^2*e-3*b*c*d^3)*_R^4+d*(-4*a^2*e^3+8*a*b*d*e^
2-2*a*c*d^2*e-4*b^2*d^2*e+3*b*c*d^3)*_R^2+2*a*c*d^4*e-b*c*d^5)/(_R^7*c+3*_R^5*b*
e-3*_R^5*c*d+8*_R^3*a*e^2-4*_R^3*b*d*e+3*_R^3*c*d^2+_R*b*d^2*e-_R*c*d^3)*ln((e*x
^2+d)^(1/2)-x*e^(1/2)-_R),_R=RootOf(c*_Z^8+(4*b*e-4*c*d)*_Z^6+(16*a*e^2-8*b*d*e+
6*c*d^2)*_Z^4+(4*b*d^2*e-4*c*d^3)*_Z^2+c*d^4))-1/2/a*d/((e*x^2+d)^(1/2)-x*e^(1/2
))*e+5/8/a^2*d^2/((e*x^2+d)^(1/2)-x*e^(1/2))*b+1/24/a^2*b*d^3/((e*x^2+d)^(1/2)-x
*e^(1/2))^3+b/a^2*d^(3/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}{{\left (c x^{4} + b x^{2} + a\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^2 + d)^(3/2)/((c*x^4 + b*x^2 + a)*x^3),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^(3/2)/((c*x^4 + b*x^2 + a)*x^3), 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((e*x^2 + d)^(3/2)/((c*x^4 + b*x^2 + a)*x^3),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x**2+d)**(3/2)/x**3/(c*x**4+b*x**2+a),x)

[Out]

Timed out

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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((e*x^2 + d)^(3/2)/((c*x^4 + b*x^2 + a)*x^3),x, algorithm="giac")

[Out]

Timed out