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

Optimal. Leaf size=214 \[ \frac{4 a^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (11 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{77 b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (11 A b-a B)}{77 b e}+\frac{4 a \sqrt{e x} \sqrt{a+b x^2} (11 A b-a B)}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e} \]

[Out]

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

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Rubi [A]  time = 0.358399, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154 \[ \frac{4 a^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (11 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{77 b^{5/4} \sqrt{e} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (11 A b-a B)}{77 b e}+\frac{4 a \sqrt{e x} \sqrt{a+b x^2} (11 A b-a B)}{77 b e}+\frac{2 B \sqrt{e x} \left (a+b x^2\right )^{5/2}}{11 b e} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 34.239, size = 192, normalized size = 0.9 \[ \frac{2 B \sqrt{e x} \left (a + b x^{2}\right )^{\frac{5}{2}}}{11 b e} + \frac{4 a^{\frac{7}{4}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (11 A b - B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{77 b^{\frac{5}{4}} \sqrt{e} \sqrt{a + b x^{2}}} + \frac{4 a \sqrt{e x} \sqrt{a + b x^{2}} \left (11 A b - B a\right )}{77 b e} + \frac{2 \sqrt{e x} \left (a + b x^{2}\right )^{\frac{3}{2}} \left (11 A b - B a\right )}{77 b e} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [C]  time = 0.408146, size = 155, normalized size = 0.72 \[ \frac{2 x \left (\left (a+b x^2\right ) \left (4 a^2 B+a b \left (33 A+13 B x^2\right )+b^2 x^2 \left (11 A+7 B x^2\right )\right )-\frac{4 i a^2 \sqrt{x} \sqrt{\frac{a}{b x^2}+1} (a B-11 A b) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}\right )}{77 b \sqrt{e x} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

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

[Out]

(2*x*((a + b*x^2)*(4*a^2*B + b^2*x^2*(11*A + 7*B*x^2) + a*b*(33*A + 13*B*x^2)) -
((4*I)*a^2*(-11*A*b + a*B)*Sqrt[1 + a/(b*x^2)]*Sqrt[x]*EllipticF[I*ArcSinh[Sqrt
[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1])/Sqrt[(I*Sqrt[a])/Sqrt[b]]))/(77*b*Sqrt[e*x]
*Sqrt[a + b*x^2])

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Maple [A]  time = 0.022, size = 272, normalized size = 1.3 \[{\frac{2}{77\,{b}^{2}} \left ( 7\,B{x}^{7}{b}^{4}+22\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{2}\sqrt{-ab}{a}^{2}b+11\,A{x}^{5}{b}^{4}-2\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{2}\sqrt{-ab}{a}^{3}+20\,B{x}^{5}a{b}^{3}+44\,A{x}^{3}a{b}^{3}+17\,B{x}^{3}{a}^{2}{b}^{2}+33\,Ax{a}^{2}{b}^{2}+4\,Bx{a}^{3}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2/77/(b*x^2+a)^(1/2)*(7*B*x^7*b^4+22*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+11*A
*x^5*b^4-2*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+20*B*x^5*a*b^3+44*A*x^3*a*b^3+17
*B*x^3*a^2*b^2+33*A*x*a^2*b^2+4*B*x*a^3*b)/b^2/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*b*x^4 + (B*a + A*b)*x^2 + A*a)*sqrt(b*x^2 + a)/sqrt(e*x), x)

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Sympy [A]  time = 51.0508, size = 199, normalized size = 0.93 \[ \frac{A a^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{5}{4}\right )} + \frac{A \sqrt{a} b x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{9}{4}\right )} + \frac{B a^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{9}{4}\right )} + \frac{B \sqrt{a} b x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{e} \Gamma \left (\frac{13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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