3.789 \(\int \frac{\sqrt{a+b x^2} \left (A+B x^2\right )}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=172 \[ \frac{2 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \sqrt{a+b x^2} (a B+A b)}{3 a e^3}-\frac{2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.288693, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154 \[ \frac{2 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \sqrt{a+b x^2} (a B+A b)}{3 a e^3}-\frac{2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]  Int[(Sqrt[a + b*x^2]*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

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

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Rubi in Sympy [A]  time = 28.8323, size = 156, normalized size = 0.91 \[ - \frac{2 A \left (a + b x^{2}\right )^{\frac{3}{2}}}{3 a e \left (e x\right )^{\frac{3}{2}}} + \frac{2 \sqrt{e x} \sqrt{a + b x^{2}} \left (A b + B a\right )}{3 a e^{3}} + \frac{2 \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (A b + B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt [4]{b} e^{\frac{5}{2}} \sqrt{a + b x^{2}}} \]

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)**(5/2),x)

[Out]

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

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Mathematica [C]  time = 0.365106, size = 120, normalized size = 0.7 \[ \frac{2 x \left (\left (a+b x^2\right ) \left (B x^2-A\right )+\frac{2 i x^{5/2} \sqrt{\frac{a}{b x^2}+1} (a B+A b) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}\right )}{3 (e x)^{5/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(Sqrt[a + b*x^2]*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.041, size = 234, normalized size = 1.4 \[{\frac{2}{3\,x{e}^{2}b} \left ( A\sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{1 \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}xb+B\sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{1 \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}xa+{b}^{2}B{x}^{4}-A{x}^{2}{b}^{2}+B{x}^{2}ab-abA \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(5/2),x)

[Out]

2/3/(b*x^2+a)^(1/2)/x*(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)^(1/2)*x*b+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*b)^(1/2)*x*a+b^2*B*x^4-A*x^2*b^2+B*x^2*a*b-a*b*A)/e^2/(e*x)^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 65.5552, size = 100, normalized size = 0.58 \[ \frac{A \sqrt{a} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{B \sqrt{a} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x**2+A)*(b*x**2+a)**(1/2)/(e*x)**(5/2),x)

[Out]

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

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(5/2), x)