3.785 \(\int (e x)^{3/2} \sqrt{a+b x^2} \left (A+B x^2\right ) \, dx\)

Optimal. Leaf size=212 \[ -\frac{2 a^{7/4} e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (11 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a+b x^2}}+\frac{4 a e \sqrt{e x} \sqrt{a+b x^2} (11 A b-5 a B)}{231 b^2}+\frac{2 (e x)^{5/2} \sqrt{a+b x^2} (11 A b-5 a B)}{77 b e}+\frac{2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e} \]

[Out]

(4*a*(11*A*b - 5*a*B)*e*Sqrt[e*x]*Sqrt[a + b*x^2])/(231*b^2) + (2*(11*A*b - 5*a*
B)*(e*x)^(5/2)*Sqrt[a + b*x^2])/(77*b*e) + (2*B*(e*x)^(5/2)*(a + b*x^2)^(3/2))/(
11*b*e) - (2*a^(7/4)*(11*A*b - 5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*
x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sq
rt[e])], 1/2])/(231*b^(9/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.378458, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192 \[ -\frac{2 a^{7/4} e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (11 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a+b x^2}}+\frac{4 a e \sqrt{e x} \sqrt{a+b x^2} (11 A b-5 a B)}{231 b^2}+\frac{2 (e x)^{5/2} \sqrt{a+b x^2} (11 A b-5 a B)}{77 b e}+\frac{2 B (e x)^{5/2} \left (a+b x^2\right )^{3/2}}{11 b e} \]

Antiderivative was successfully verified.

[In]  Int[(e*x)^(3/2)*Sqrt[a + b*x^2]*(A + B*x^2),x]

[Out]

(4*a*(11*A*b - 5*a*B)*e*Sqrt[e*x]*Sqrt[a + b*x^2])/(231*b^2) + (2*(11*A*b - 5*a*
B)*(e*x)^(5/2)*Sqrt[a + b*x^2])/(77*b*e) + (2*B*(e*x)^(5/2)*(a + b*x^2)^(3/2))/(
11*b*e) - (2*a^(7/4)*(11*A*b - 5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*
x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sq
rt[e])], 1/2])/(231*b^(9/4)*Sqrt[a + b*x^2])

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Rubi in Sympy [A]  time = 34.7358, size = 199, normalized size = 0.94 \[ \frac{2 B \left (e x\right )^{\frac{5}{2}} \left (a + b x^{2}\right )^{\frac{3}{2}}}{11 b e} - \frac{2 a^{\frac{7}{4}} e^{\frac{3}{2}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (11 A b - 5 B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{231 b^{\frac{9}{4}} \sqrt{a + b x^{2}}} + \frac{4 a e \sqrt{e x} \sqrt{a + b x^{2}} \left (11 A b - 5 B a\right )}{231 b^{2}} + \frac{2 \left (e x\right )^{\frac{5}{2}} \sqrt{a + b x^{2}} \left (11 A b - 5 B a\right )}{77 b e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**(3/2)*(B*x**2+A)*(b*x**2+a)**(1/2),x)

[Out]

2*B*(e*x)**(5/2)*(a + b*x**2)**(3/2)/(11*b*e) - 2*a**(7/4)*e**(3/2)*sqrt((a + b*
x**2)/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(11*A*b - 5*B*a)*elliptic_
f(2*atan(b**(1/4)*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(231*b**(9/4)*sqrt(a + b*x
**2)) + 4*a*e*sqrt(e*x)*sqrt(a + b*x**2)*(11*A*b - 5*B*a)/(231*b**2) + 2*(e*x)**
(5/2)*sqrt(a + b*x**2)*(11*A*b - 5*B*a)/(77*b*e)

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Mathematica [C]  time = 0.498514, size = 159, normalized size = 0.75 \[ \frac{2 e \sqrt{e x} \left (-\left (a+b x^2\right ) \left (10 a^2 B-2 a b \left (11 A+3 B x^2\right )-3 b^2 x^2 \left (11 A+7 B x^2\right )\right )+\frac{2 i a^2 \sqrt{x} \sqrt{\frac{a}{b x^2}+1} (5 a B-11 A b) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}\right )}{231 b^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(e*x)^(3/2)*Sqrt[a + b*x^2]*(A + B*x^2),x]

[Out]

(2*e*Sqrt[e*x]*(-((a + b*x^2)*(10*a^2*B - 2*a*b*(11*A + 3*B*x^2) - 3*b^2*x^2*(11
*A + 7*B*x^2))) + ((2*I)*a^2*(-11*A*b + 5*a*B)*Sqrt[1 + a/(b*x^2)]*Sqrt[x]*Ellip
ticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1])/Sqrt[(I*Sqrt[a])/Sqrt[b]
]))/(231*b^2*Sqrt[a + b*x^2])

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Maple [A]  time = 0.059, size = 276, normalized size = 1.3 \[ -{\frac{2\,e}{231\,x{b}^{3}}\sqrt{ex} \left ( -21\,B{x}^{7}{b}^{4}+11\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{2}\sqrt{-ab}{a}^{2}b-33\,A{x}^{5}{b}^{4}-5\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{2}\sqrt{-ab}{a}^{3}-27\,B{x}^{5}a{b}^{3}-55\,A{x}^{3}a{b}^{3}+4\,B{x}^{3}{a}^{2}{b}^{2}-22\,Ax{a}^{2}{b}^{2}+10\,Bx{a}^{3}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^(3/2)*(B*x^2+A)*(b*x^2+a)^(1/2),x)

[Out]

-2/231*e/x*(e*x)^(1/2)/(b*x^2+a)^(1/2)*(-21*B*x^7*b^4+11*A*((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)*(-a
*b)^(1/2)*a^2*b-33*A*x^5*b^4-5*B*((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)*(-a*b)^(1/2)*a^3-27*B*x^5*a*b
^3-55*A*x^3*a*b^3+4*B*x^3*a^2*b^2-22*A*x*a^2*b^2+10*B*x*a^3*b)/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2),x, algorithm="fricas")

[Out]

integral((B*e*x^3 + A*e*x)*sqrt(b*x^2 + a)*sqrt(e*x), x)

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Sympy [A]  time = 134.715, size = 97, normalized size = 0.46 \[ \frac{A \sqrt{a} e^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{9}{4}\right )} + \frac{B \sqrt{a} e^{\frac{3}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x)**(3/2)*(B*x**2+A)*(b*x**2+a)**(1/2),x)

[Out]

A*sqrt(a)*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**2*exp_pol
ar(I*pi)/a)/(2*gamma(9/4)) + B*sqrt(a)*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((-1/2,
9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(13/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)*(e*x)^(3/2), x)