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

Optimal. Leaf size=333 \[ \frac{2 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+5 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{4 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+5 A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 (e x)^{3/2} \sqrt{a+b x^2} (a B+5 A b)}{5 a e^3}+\frac{4 \sqrt{e x} \sqrt{a+b x^2} (a B+5 A b)}{5 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt{e x}} \]

[Out]

(2*(5*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(5*a*e^3) + (4*(5*A*b + a*B)*Sqrt[
e*x]*Sqrt[a + b*x^2])/(5*Sqrt[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) - (2*A*(a + b*x^2)^(
3/2))/(a*e*Sqrt[e*x]) - (4*a^(1/4)*(5*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a +
b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)
*Sqrt[e])], 1/2])/(5*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (2*a^(1/4)*(5*A*b + a*B)
*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*Arc
Tan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(5*b^(3/4)*e^(3/2)*Sqrt[a + b*
x^2])

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Rubi [A]  time = 0.636198, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231 \[ \frac{2 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+5 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{4 \sqrt [4]{a} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+5 A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 (e x)^{3/2} \sqrt{a+b x^2} (a B+5 A b)}{5 a e^3}+\frac{4 \sqrt{e x} \sqrt{a+b x^2} (a B+5 A b)}{5 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt{e x}} \]

Antiderivative was successfully verified.

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

[Out]

(2*(5*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(5*a*e^3) + (4*(5*A*b + a*B)*Sqrt[
e*x]*Sqrt[a + b*x^2])/(5*Sqrt[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) - (2*A*(a + b*x^2)^(
3/2))/(a*e*Sqrt[e*x]) - (4*a^(1/4)*(5*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a +
b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)
*Sqrt[e])], 1/2])/(5*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (2*a^(1/4)*(5*A*b + a*B)
*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*Arc
Tan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(5*b^(3/4)*e^(3/2)*Sqrt[a + b*
x^2])

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Rubi in Sympy [A]  time = 63.9852, size = 309, normalized size = 0.93 \[ - \frac{2 A \left (a + b x^{2}\right )^{\frac{3}{2}}}{a e \sqrt{e x}} - \frac{4 \sqrt [4]{a} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (5 A b + B a\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{5 b^{\frac{3}{4}} e^{\frac{3}{2}} \sqrt{a + b x^{2}}} + \frac{2 \sqrt [4]{a} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (5 A b + B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{5 b^{\frac{3}{4}} e^{\frac{3}{2}} \sqrt{a + b x^{2}}} + \frac{4 \sqrt{e x} \sqrt{a + b x^{2}} \left (5 A b + B a\right )}{5 \sqrt{b} e^{2} \left (\sqrt{a} + \sqrt{b} x\right )} + \frac{2 \left (e x\right )^{\frac{3}{2}} \sqrt{a + b x^{2}} \left (5 A b + B a\right )}{5 a e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*A*(a + b*x**2)**(3/2)/(a*e*sqrt(e*x)) - 4*a**(1/4)*sqrt((a + b*x**2)/(sqrt(a)
+ sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(5*A*b + B*a)*elliptic_e(2*atan(b**(1/4)
*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(5*b**(3/4)*e**(3/2)*sqrt(a + b*x**2)) + 2*
a**(1/4)*sqrt((a + b*x**2)/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(5*A*
b + B*a)*elliptic_f(2*atan(b**(1/4)*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(5*b**(3
/4)*e**(3/2)*sqrt(a + b*x**2)) + 4*sqrt(e*x)*sqrt(a + b*x**2)*(5*A*b + B*a)/(5*s
qrt(b)*e**2*(sqrt(a) + sqrt(b)*x)) + 2*(e*x)**(3/2)*sqrt(a + b*x**2)*(5*A*b + B*
a)/(5*a*e**3)

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Mathematica [C]  time = 1.15575, size = 186, normalized size = 0.56 \[ \frac{x^{3/2} \left (\frac{2 \sqrt{a+b x^2} \left (B x^2-5 A\right )}{\sqrt{x}}-\frac{4 x (a B+5 A b) \left (-\sqrt{x} \left (\frac{a}{x^2}+b\right )+\frac{i a \sqrt{\frac{a}{b x^2}+1} \left (E\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )\right )}{\left (\frac{i \sqrt{a}}{\sqrt{b}}\right )^{3/2}}\right )}{b \sqrt{a+b x^2}}\right )}{5 (e x)^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(x^(3/2)*((2*Sqrt[a + b*x^2]*(-5*A + B*x^2))/Sqrt[x] - (4*(5*A*b + a*B)*x*(-((b
+ a/x^2)*Sqrt[x]) + (I*a*Sqrt[1 + a/(b*x^2)]*(EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[a
])/Sqrt[b]]/Sqrt[x]], -1] - EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x
]], -1]))/((I*Sqrt[a])/Sqrt[b])^(3/2)))/(b*Sqrt[a + b*x^2])))/(5*(e*x)^(3/2))

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Maple [A]  time = 0.049, size = 391, normalized size = 1.2 \[{\frac{2}{5\,be} \left ( 10\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) ab-5\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) ab+2\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{2}-B\sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{1 \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ){a}^{2}+{b}^{2}B{x}^{4}-5\,A{x}^{2}{b}^{2}+B{x}^{2}ab-5\,abA \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2/5*(10*A*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*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))*a*b-5*A*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*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))*a*b+2*B*((b*x+(-a*b)^(1/2)
)/(-a*b)^(1/2))^(1/2)*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))*
a^2-B*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*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))*a^2+b^2*B*x^4-5*A*x^2*b^2+B*x^2*a*b-5*a*b*A)/(b*x^2+a)^(
1/2)/b/e/(e*x)^(1/2)

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

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. \[{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{\sqrt{e x} e x}, x\right ) \]

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

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Sympy [A]  time = 12.7345, size = 100, normalized size = 0.3 \[ \frac{A \sqrt{a} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{B \sqrt{a} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*sqrt(a)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*e
**(3/2)*sqrt(x)*gamma(3/4)) + B*sqrt(a)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (
7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(7/4))

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

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)