3.606 \(\int \frac{\left (a+b x^2\right )^{3/2}}{(c x)^{11/2}} \, dx\)

Optimal. Leaf size=331 \[ \frac{4 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}-\frac{8 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt{a+b x^2}}{15 a c^6 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}} \]

[Out]

(-4*b*Sqrt[a + b*x^2])/(15*c^3*(c*x)^(5/2)) - (8*b^2*Sqrt[a + b*x^2])/(15*a*c^5*
Sqrt[c*x]) + (8*b^(5/2)*Sqrt[c*x]*Sqrt[a + b*x^2])/(15*a*c^6*(Sqrt[a] + Sqrt[b]*
x)) - (2*(a + b*x^2)^(3/2))/(9*c*(c*x)^(9/2)) - (8*b^(9/4)*(Sqrt[a] + Sqrt[b]*x)
*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x]
)/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*x^2]) + (4*b^(9/4)*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTa
n[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*
x^2])

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Rubi [A]  time = 0.686009, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316 \[ \frac{4 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}-\frac{8 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt{a+b x^2}}{15 a c^6 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x^2)^(3/2)/(c*x)^(11/2),x]

[Out]

(-4*b*Sqrt[a + b*x^2])/(15*c^3*(c*x)^(5/2)) - (8*b^2*Sqrt[a + b*x^2])/(15*a*c^5*
Sqrt[c*x]) + (8*b^(5/2)*Sqrt[c*x]*Sqrt[a + b*x^2])/(15*a*c^6*(Sqrt[a] + Sqrt[b]*
x)) - (2*(a + b*x^2)^(3/2))/(9*c*(c*x)^(9/2)) - (8*b^(9/4)*(Sqrt[a] + Sqrt[b]*x)
*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x]
)/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*x^2]) + (4*b^(9/4)*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTa
n[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*
x^2])

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Rubi in Sympy [A]  time = 70.9209, size = 304, normalized size = 0.92 \[ - \frac{4 b \sqrt{a + b x^{2}}}{15 c^{3} \left (c x\right )^{\frac{5}{2}}} - \frac{2 \left (a + b x^{2}\right )^{\frac{3}{2}}}{9 c \left (c x\right )^{\frac{9}{2}}} + \frac{8 b^{\frac{5}{2}} \sqrt{c x} \sqrt{a + b x^{2}}}{15 a c^{6} \left (\sqrt{a} + \sqrt{b} x\right )} - \frac{8 b^{2} \sqrt{a + b x^{2}}}{15 a c^{5} \sqrt{c x}} - \frac{8 b^{\frac{9}{4}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}} \right )}\middle | \frac{1}{2}\right )}{15 a^{\frac{3}{4}} c^{\frac{11}{2}} \sqrt{a + b x^{2}}} + \frac{4 b^{\frac{9}{4}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}} \right )}\middle | \frac{1}{2}\right )}{15 a^{\frac{3}{4}} c^{\frac{11}{2}} \sqrt{a + b x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x**2+a)**(3/2)/(c*x)**(11/2),x)

[Out]

-4*b*sqrt(a + b*x**2)/(15*c**3*(c*x)**(5/2)) - 2*(a + b*x**2)**(3/2)/(9*c*(c*x)*
*(9/2)) + 8*b**(5/2)*sqrt(c*x)*sqrt(a + b*x**2)/(15*a*c**6*(sqrt(a) + sqrt(b)*x)
) - 8*b**2*sqrt(a + b*x**2)/(15*a*c**5*sqrt(c*x)) - 8*b**(9/4)*sqrt((a + b*x**2)
/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*elliptic_e(2*atan(b**(1/4)*sqrt
(c*x)/(a**(1/4)*sqrt(c))), 1/2)/(15*a**(3/4)*c**(11/2)*sqrt(a + b*x**2)) + 4*b**
(9/4)*sqrt((a + b*x**2)/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*elliptic
_f(2*atan(b**(1/4)*sqrt(c*x)/(a**(1/4)*sqrt(c))), 1/2)/(15*a**(3/4)*c**(11/2)*sq
rt(a + b*x**2))

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Mathematica [C]  time = 0.424198, size = 213, normalized size = 0.64 \[ -\frac{2 \sqrt{c x} \left (\sqrt{\frac{i \sqrt{b} x}{\sqrt{a}}} \left (5 a^3+16 a^2 b x^2+23 a b^2 x^4+12 b^3 x^6\right )+12 \sqrt{a} b^{5/2} x^5 \sqrt{\frac{b x^2}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b} x}{\sqrt{a}}}\right )\right |-1\right )-12 \sqrt{a} b^{5/2} x^5 \sqrt{\frac{b x^2}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b} x}{\sqrt{a}}}\right )\right |-1\right )\right )}{45 a c^6 x^5 \sqrt{\frac{i \sqrt{b} x}{\sqrt{a}}} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(a + b*x^2)^(3/2)/(c*x)^(11/2),x]

[Out]

(-2*Sqrt[c*x]*(Sqrt[(I*Sqrt[b]*x)/Sqrt[a]]*(5*a^3 + 16*a^2*b*x^2 + 23*a*b^2*x^4
+ 12*b^3*x^6) - 12*Sqrt[a]*b^(5/2)*x^5*Sqrt[1 + (b*x^2)/a]*EllipticE[I*ArcSinh[S
qrt[(I*Sqrt[b]*x)/Sqrt[a]]], -1] + 12*Sqrt[a]*b^(5/2)*x^5*Sqrt[1 + (b*x^2)/a]*El
lipticF[I*ArcSinh[Sqrt[(I*Sqrt[b]*x)/Sqrt[a]]], -1]))/(45*a*c^6*x^5*Sqrt[(I*Sqrt
[b]*x)/Sqrt[a]]*Sqrt[a + b*x^2])

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Maple [A]  time = 0.046, size = 234, normalized size = 0.7 \[{\frac{2}{45\,a{x}^{4}{c}^{5}} \left ( 12\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{2}{x}^{4}a{b}^{2}-6\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{2}{x}^{4}a{b}^{2}-12\,{b}^{3}{x}^{6}-23\,a{b}^{2}{x}^{4}-16\,{a}^{2}b{x}^{2}-5\,{a}^{3} \right ){\frac{1}{\sqrt{cx}}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a)^(3/2)/(c*x)^(11/2),x)

[Out]

2/45/x^4*(12*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)
^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/
2))^(1/2),1/2*2^(1/2))*2^(1/2)*x^4*a*b^2-6*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/
2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticF(
((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x^4*a*b^2-12*b^3*x^
6-23*a*b^2*x^4-16*a^2*b*x^2-5*a^3)/(b*x^2+a)^(1/2)/a/c^5/(c*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (c x\right )^{\frac{11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\sqrt{c x} c^{5} x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/2)/(sqrt(c*x)*c^5*x^5), 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((b*x**2+a)**(3/2)/(c*x)**(11/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (c x\right )^{\frac{11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2), x)