3.374 \(\int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{17/2}} \, dx\)

Optimal. Leaf size=320 \[ \frac{4 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{b x^2+c x^4}}-\frac{8 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{b x^2+c x^4}}+\frac{8 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{8 c^2 \sqrt{b x^2+c x^4}}{15 b x^{3/2}}-\frac{4 c \sqrt{b x^2+c x^4}}{15 x^{7/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}} \]

[Out]

(8*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4])
- (4*c*Sqrt[b*x^2 + c*x^4])/(15*x^(7/2)) - (8*c^2*Sqrt[b*x^2 + c*x^4])/(15*b*x^
(3/2)) - (2*(b*x^2 + c*x^4)^(3/2))/(9*x^(15/2)) - (8*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^(3/4)*Sqrt[b*x^2 + c*x^4]) + (4*c^(9/4)*x*(Sqrt[b] + S
qrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*
Sqrt[x])/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.730754, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333 \[ \frac{4 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{b x^2+c x^4}}-\frac{8 c^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{b x^2+c x^4}}+\frac{8 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{8 c^2 \sqrt{b x^2+c x^4}}{15 b x^{3/2}}-\frac{4 c \sqrt{b x^2+c x^4}}{15 x^{7/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{15/2}} \]

Antiderivative was successfully verified.

[In]  Int[(b*x^2 + c*x^4)^(3/2)/x^(17/2),x]

[Out]

(8*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4])
- (4*c*Sqrt[b*x^2 + c*x^4])/(15*x^(7/2)) - (8*c^2*Sqrt[b*x^2 + c*x^4])/(15*b*x^
(3/2)) - (2*(b*x^2 + c*x^4)^(3/2))/(9*x^(15/2)) - (8*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^(3/4)*Sqrt[b*x^2 + c*x^4]) + (4*c^(9/4)*x*(Sqrt[b] + S
qrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*
Sqrt[x])/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[b*x^2 + c*x^4])

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Rubi in Sympy [A]  time = 65.8946, size = 299, normalized size = 0.93 \[ - \frac{4 c \sqrt{b x^{2} + c x^{4}}}{15 x^{\frac{7}{2}}} - \frac{2 \left (b x^{2} + c x^{4}\right )^{\frac{3}{2}}}{9 x^{\frac{15}{2}}} + \frac{8 c^{\frac{5}{2}} \sqrt{b x^{2} + c x^{4}}}{15 b \sqrt{x} \left (\sqrt{b} + \sqrt{c} x\right )} - \frac{8 c^{2} \sqrt{b x^{2} + c x^{4}}}{15 b x^{\frac{3}{2}}} - \frac{8 c^{\frac{9}{4}} \sqrt{\frac{b + c x^{2}}{\left (\sqrt{b} + \sqrt{c} x\right )^{2}}} \left (\sqrt{b} + \sqrt{c} x\right ) \sqrt{b x^{2} + c x^{4}} E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} x \left (b + c x^{2}\right )} + \frac{4 c^{\frac{9}{4}} \sqrt{\frac{b + c x^{2}}{\left (\sqrt{b} + \sqrt{c} x\right )^{2}}} \left (\sqrt{b} + \sqrt{c} x\right ) \sqrt{b x^{2} + c x^{4}} F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} x \left (b + c x^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((c*x**4+b*x**2)**(3/2)/x**(17/2),x)

[Out]

-4*c*sqrt(b*x**2 + c*x**4)/(15*x**(7/2)) - 2*(b*x**2 + c*x**4)**(3/2)/(9*x**(15/
2)) + 8*c**(5/2)*sqrt(b*x**2 + c*x**4)/(15*b*sqrt(x)*(sqrt(b) + sqrt(c)*x)) - 8*
c**2*sqrt(b*x**2 + c*x**4)/(15*b*x**(3/2)) - 8*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)*elliptic_e(2*ata
n(c**(1/4)*sqrt(x)/b**(1/4)), 1/2)/(15*b**(3/4)*x*(b + c*x**2)) + 4*c**(9/4)*sqr
t((b + c*x**2)/(sqrt(b) + sqrt(c)*x)**2)*(sqrt(b) + sqrt(c)*x)*sqrt(b*x**2 + c*x
**4)*elliptic_f(2*atan(c**(1/4)*sqrt(x)/b**(1/4)), 1/2)/(15*b**(3/4)*x*(b + c*x*
*2))

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Mathematica [C]  time = 0.476275, size = 209, normalized size = 0.65 \[ -\frac{2 \left (\sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}} \left (5 b^3+16 b^2 c x^2+23 b c^2 x^4+12 c^3 x^6\right )+12 \sqrt{b} c^{5/2} x^5 \sqrt{\frac{c x^2}{b}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}}\right )\right |-1\right )-12 \sqrt{b} c^{5/2} x^5 \sqrt{\frac{c x^2}{b}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}}\right )\right |-1\right )\right )}{45 b x^{7/2} \sqrt{\frac{i \sqrt{c} x}{\sqrt{b}}} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]  Integrate[(b*x^2 + c*x^4)^(3/2)/x^(17/2),x]

[Out]

(-2*(Sqrt[(I*Sqrt[c]*x)/Sqrt[b]]*(5*b^3 + 16*b^2*c*x^2 + 23*b*c^2*x^4 + 12*c^3*x
^6) - 12*Sqrt[b]*c^(5/2)*x^5*Sqrt[1 + (c*x^2)/b]*EllipticE[I*ArcSinh[Sqrt[(I*Sqr
t[c]*x)/Sqrt[b]]], -1] + 12*Sqrt[b]*c^(5/2)*x^5*Sqrt[1 + (c*x^2)/b]*EllipticF[I*
ArcSinh[Sqrt[(I*Sqrt[c]*x)/Sqrt[b]]], -1]))/(45*b*x^(7/2)*Sqrt[(I*Sqrt[c]*x)/Sqr
t[b]]*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.024, size = 239, normalized size = 0.8 \[{\frac{2}{45\, \left ( c{x}^{2}+b \right ) ^{2}b} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 12\,\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{2}{x}^{4}b{c}^{2}-6\,\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{2}{x}^{4}b{c}^{2}-12\,{c}^{3}{x}^{6}-23\,b{c}^{2}{x}^{4}-16\,{b}^{2}c{x}^{2}-5\,{b}^{3} \right ){x}^{-{\frac{15}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((c*x^4+b*x^2)^(3/2)/x^(17/2),x)

[Out]

2/45*(c*x^4+b*x^2)^(3/2)/x^(15/2)/(c*x^2+b)^2*(12*((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)*Ell
ipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x^4*b*c^2-6*
((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^2-12*c^3*x^6-23*b*c^2*x^4-16*b^2*c*x^2-5*b^3)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}{x^{\frac{17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2}}{\left (c x^{2} + b\right )}}{x^{\frac{13}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*(c*x^2 + b)/x^(13/2), 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((c*x**4+b*x**2)**(3/2)/x**(17/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}{x^{\frac{17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(17/2), x)