3.798 \(\int \frac{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=210 \[ \frac{4 a^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 a B+7 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{21 \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (3 a B+7 A b)}{21 a e^3}+\frac{4 \sqrt{e x} \sqrt{a+b x^2} (3 a B+7 A b)}{21 e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}} \]

[Out]

(4*(7*A*b + 3*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*e^3) + (2*(7*A*b + 3*a*B)*Sqrt
[e*x]*(a + b*x^2)^(3/2))/(21*a*e^3) - (2*A*(a + b*x^2)^(5/2))/(3*a*e*(e*x)^(3/2)
) + (4*a^(3/4)*(7*A*b + 3*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])/
(21*b^(1/4)*e^(5/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.361701, antiderivative size = 210, 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 \[ \frac{4 a^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 a B+7 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{21 \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (3 a B+7 A b)}{21 a e^3}+\frac{4 \sqrt{e x} \sqrt{a+b x^2} (3 a B+7 A b)}{21 e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]  Int[((a + b*x^2)^(3/2)*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

(4*(7*A*b + 3*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*e^3) + (2*(7*A*b + 3*a*B)*Sqrt
[e*x]*(a + b*x^2)^(3/2))/(21*a*e^3) - (2*A*(a + b*x^2)^(5/2))/(3*a*e*(e*x)^(3/2)
) + (4*a^(3/4)*(7*A*b + 3*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])/
(21*b^(1/4)*e^(5/2)*Sqrt[a + b*x^2])

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Rubi in Sympy [A]  time = 35.2711, size = 197, normalized size = 0.94 \[ - \frac{2 A \left (a + b x^{2}\right )^{\frac{5}{2}}}{3 a e \left (e x\right )^{\frac{3}{2}}} + \frac{4 a^{\frac{3}{4}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (7 A b + 3 B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{21 \sqrt [4]{b} e^{\frac{5}{2}} \sqrt{a + b x^{2}}} + \frac{4 \sqrt{e x} \sqrt{a + b x^{2}} \left (7 A b + 3 B a\right )}{21 e^{3}} + \frac{2 \sqrt{e x} \left (a + b x^{2}\right )^{\frac{3}{2}} \left (7 A b + 3 B a\right )}{21 a e^{3}} \]

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)**(5/2),x)

[Out]

-2*A*(a + b*x**2)**(5/2)/(3*a*e*(e*x)**(3/2)) + 4*a**(3/4)*sqrt((a + b*x**2)/(sq
rt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(7*A*b + 3*B*a)*elliptic_f(2*atan(b
**(1/4)*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(21*b**(1/4)*e**(5/2)*sqrt(a + b*x**
2)) + 4*sqrt(e*x)*sqrt(a + b*x**2)*(7*A*b + 3*B*a)/(21*e**3) + 2*sqrt(e*x)*(a +
b*x**2)**(3/2)*(7*A*b + 3*B*a)/(21*a*e**3)

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Mathematica [C]  time = 0.390099, size = 140, normalized size = 0.67 \[ \frac{2 x \left (\left (a+b x^2\right ) \left (-7 a A+9 a B x^2+7 A b x^2+3 b B x^4\right )+\frac{4 i a x^{5/2} \sqrt{\frac{a}{b x^2}+1} (3 a B+7 A b) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}\right )}{21 (e x)^{5/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

(2*x*((a + b*x^2)*(-7*a*A + 7*A*b*x^2 + 9*a*B*x^2 + 3*b*B*x^4) + ((4*I)*a*(7*A*b
+ 3*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])/Sqrt[b]]))/(21*(e*x)^(5/2)*Sqrt[a + b*x^2])

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Maple [A]  time = 0.025, size = 255, normalized size = 1.2 \[{\frac{2}{21\,bx{e}^{2}} \left ( 14\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}xab+6\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}x{a}^{2}+3\,B{x}^{6}{b}^{3}+7\,A{x}^{4}{b}^{3}+12\,B{x}^{4}a{b}^{2}+9\,B{x}^{2}{a}^{2}b-7\,A{a}^{2}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x)

[Out]

2/21/(b*x^2+a)^(1/2)/x*(14*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*a*b+6*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^2+3*B*x^6*b^3+7*A*x^4*b^3+12*B*x^4*a*b^2+9*B*x^2*a^2*b-
7*A*a^2*b)/b/e^2/(e*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x^{2} + A\right )}{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2),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)*e^2*x^2),
x)

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Sympy [A]  time = 127.336, size = 202, normalized size = 0.96 \[ \frac{A a^{\frac{3}{2}} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{A \sqrt{a} b \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{B a^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{B \sqrt{a} b x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x**2+a)**(3/2)*(B*x**2+A)/(e*x)**(5/2),x)

[Out]

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

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x^{2} + A\right )}{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2), x)