3.389 \(\int \frac{1}{x^{9/2} \sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=326 \[ \frac{7 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{14 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{14 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}} \]

[Out]

(14*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^
4]) - (2*Sqrt[b*x^2 + c*x^4])/(9*b*x^(11/2)) + (14*c*Sqrt[b*x^2 + c*x^4])/(45*b^
2*x^(7/2)) - (14*c^2*Sqrt[b*x^2 + c*x^4])/(15*b^3*x^(3/2)) - (14*c^(9/4)*x*(Sqrt
[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c
^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4]) + (7*c^(9/4)*x
*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*Arc
Tan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.746663, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286 \[ \frac{7 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{14 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{14 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{14 c^2 \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{14 c \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{2 \sqrt{b x^2+c x^4}}{9 b x^{11/2}} \]

Antiderivative was successfully verified.

[In]  Int[1/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(14*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^
4]) - (2*Sqrt[b*x^2 + c*x^4])/(9*b*x^(11/2)) + (14*c*Sqrt[b*x^2 + c*x^4])/(45*b^
2*x^(7/2)) - (14*c^2*Sqrt[b*x^2 + c*x^4])/(15*b^3*x^(3/2)) - (14*c^(9/4)*x*(Sqrt
[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c
^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4]) + (7*c^(9/4)*x
*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*Arc
Tan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4])

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Rubi in Sympy [A]  time = 76.4918, size = 308, normalized size = 0.94 \[ - \frac{2 \sqrt{b x^{2} + c x^{4}}}{9 b x^{\frac{11}{2}}} + \frac{14 c \sqrt{b x^{2} + c x^{4}}}{45 b^{2} x^{\frac{7}{2}}} + \frac{14 c^{\frac{5}{2}} \sqrt{b x^{2} + c x^{4}}}{15 b^{3} \sqrt{x} \left (\sqrt{b} + \sqrt{c} x\right )} - \frac{14 c^{2} \sqrt{b x^{2} + c x^{4}}}{15 b^{3} x^{\frac{3}{2}}} - \frac{14 c^{\frac{9}{4}} \sqrt{\frac{b + c x^{2}}{\left (\sqrt{b} + \sqrt{c} x\right )^{2}}} \left (\sqrt{b} + \sqrt{c} x\right ) \sqrt{b x^{2} + c x^{4}} E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{11}{4}} x \left (b + c x^{2}\right )} + \frac{7 c^{\frac{9}{4}} \sqrt{\frac{b + c x^{2}}{\left (\sqrt{b} + \sqrt{c} x\right )^{2}}} \left (\sqrt{b} + \sqrt{c} x\right ) \sqrt{b x^{2} + c x^{4}} F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{11}{4}} x \left (b + c x^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/x**(9/2)/(c*x**4+b*x**2)**(1/2),x)

[Out]

-2*sqrt(b*x**2 + c*x**4)/(9*b*x**(11/2)) + 14*c*sqrt(b*x**2 + c*x**4)/(45*b**2*x
**(7/2)) + 14*c**(5/2)*sqrt(b*x**2 + c*x**4)/(15*b**3*sqrt(x)*(sqrt(b) + sqrt(c)
*x)) - 14*c**2*sqrt(b*x**2 + c*x**4)/(15*b**3*x**(3/2)) - 14*c**(9/4)*sqrt((b +
c*x**2)/(sqrt(b) + sqrt(c)*x)**2)*(sqrt(b) + sqrt(c)*x)*sqrt(b*x**2 + c*x**4)*el
liptic_e(2*atan(c**(1/4)*sqrt(x)/b**(1/4)), 1/2)/(15*b**(11/4)*x*(b + c*x**2)) +
7*c**(9/4)*sqrt((b + c*x**2)/(sqrt(b) + sqrt(c)*x)**2)*(sqrt(b) + sqrt(c)*x)*sq
rt(b*x**2 + c*x**4)*elliptic_f(2*atan(c**(1/4)*sqrt(x)/b**(1/4)), 1/2)/(15*b**(1
1/4)*x*(b + c*x**2))

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Mathematica [C]  time = 0.260137, size = 210, normalized size = 0.64 \[ \frac{-2 \sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}} \left (5 b^3-2 b^2 c x^2+14 b c^2 x^4+21 c^3 x^6\right )-42 \sqrt{b} c^{5/2} x^5 \sqrt{\frac{c x^2}{b}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}}\right )\right |-1\right )+42 \sqrt{b} c^{5/2} x^5 \sqrt{\frac{c x^2}{b}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}}\right )\right |-1\right )}{45 b^3 x^{7/2} \sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]  Integrate[1/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(-2*Sqrt[(I*Sqrt[c]*x)/Sqrt[b]]*(5*b^3 - 2*b^2*c*x^2 + 14*b*c^2*x^4 + 21*c^3*x^6
) + 42*Sqrt[b]*c^(5/2)*x^5*Sqrt[1 + (c*x^2)/b]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[
c]*x)/Sqrt[b]]], -1] - 42*Sqrt[b]*c^(5/2)*x^5*Sqrt[1 + (c*x^2)/b]*EllipticF[I*Ar
cSinh[Sqrt[(I*Sqrt[c]*x)/Sqrt[b]]], -1])/(45*b^3*x^(7/2)*Sqrt[(I*Sqrt[c]*x)/Sqrt
[b]]*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.025, size = 230, normalized size = 0.7 \[ -{\frac{1}{45\,{b}^{3}} \left ( 21\,\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{2}{x}^{4}b{c}^{2}-42\,\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{2}{x}^{4}b{c}^{2}+42\,{c}^{3}{x}^{6}+28\,b{c}^{2}{x}^{4}-4\,{b}^{2}c{x}^{2}+10\,{b}^{3} \right ){x}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/x^(9/2)/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/45/(c*x^4+b*x^2)^(1/2)/x^(7/2)*(21*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*((
-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x
+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x^4*b*c^2-42*((c*x+(-b*c
)^(1/2))/(-b*c)^(1/2))^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*
c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*2
^(1/2)*x^4*b*c^2+42*c^3*x^6+28*b*c^2*x^4-4*b^2*c*x^2+10*b^3)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{1}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)),x, algorithm="fricas")

[Out]

integral(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)

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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(1/x**(9/2)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)),x, algorithm="giac")

[Out]

integrate(1/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)