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

Optimal. Leaf size=213 \[ -\frac{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{13/4} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{e x} (3 A b-a B)}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{\sqrt{e x} (3 A b-a B)}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

[Out]

(-2*A)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(3/2)) - ((3*A*b - a*B)*Sqrt[e*x])/(3*a^2*
e^3*(a + b*x^2)^(3/2)) - (5*(3*A*b - a*B)*Sqrt[e*x])/(6*a^3*e^3*Sqrt[a + b*x^2])
- (5*(3*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])/(12*a^(13/4
)*b^(1/4)*e^(5/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.368943, antiderivative size = 213, 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 \[ -\frac{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{13/4} \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}-\frac{5 \sqrt{e x} (3 A b-a B)}{6 a^3 e^3 \sqrt{a+b x^2}}-\frac{\sqrt{e x} (3 A b-a B)}{3 a^2 e^3 \left (a+b x^2\right )^{3/2}}-\frac{2 A}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*A)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(3/2)) - ((3*A*b - a*B)*Sqrt[e*x])/(3*a^2*
e^3*(a + b*x^2)^(3/2)) - (5*(3*A*b - a*B)*Sqrt[e*x])/(6*a^3*e^3*Sqrt[a + b*x^2])
- (5*(3*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])/(12*a^(13/4
)*b^(1/4)*e^(5/2)*Sqrt[a + b*x^2])

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Rubi in Sympy [A]  time = 37.8319, size = 197, normalized size = 0.92 \[ - \frac{2 A}{3 a e \left (e x\right )^{\frac{3}{2}} \left (a + b x^{2}\right )^{\frac{3}{2}}} - \frac{\sqrt{e x} \left (3 A b - B a\right )}{3 a^{2} e^{3} \left (a + b x^{2}\right )^{\frac{3}{2}}} - \frac{5 \sqrt{e x} \left (3 A b - B a\right )}{6 a^{3} e^{3} \sqrt{a + b x^{2}}} - \frac{5 \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (3 A b - B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{12 a^{\frac{13}{4}} \sqrt [4]{b} e^{\frac{5}{2}} \sqrt{a + b x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*A/(3*a*e*(e*x)**(3/2)*(a + b*x**2)**(3/2)) - sqrt(e*x)*(3*A*b - B*a)/(3*a**2*
e**3*(a + b*x**2)**(3/2)) - 5*sqrt(e*x)*(3*A*b - B*a)/(6*a**3*e**3*sqrt(a + b*x*
*2)) - 5*sqrt((a + b*x**2)/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(3*A*
b - B*a)*elliptic_f(2*atan(b**(1/4)*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(12*a**(
13/4)*b**(1/4)*e**(5/2)*sqrt(a + b*x**2))

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Mathematica [C]  time = 0.446226, size = 166, normalized size = 0.78 \[ \frac{x^{5/2} \left (\frac{a^2 \left (7 B x^2-4 A\right )+a \left (5 b B x^4-21 A b x^2\right )-15 A b^2 x^4}{a^3 x^{3/2} \left (a+b x^2\right )}+\frac{5 i x \sqrt{\frac{a}{b x^2}+1} (a B-3 A b) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )}{a^3 \sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}\right )}{6 (e x)^{5/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

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

[Out]

(x^(5/2)*((-15*A*b^2*x^4 + a^2*(-4*A + 7*B*x^2) + a*(-21*A*b*x^2 + 5*b*B*x^4))/(
a^3*x^(3/2)*(a + b*x^2)) + ((5*I)*(-3*A*b + a*B)*Sqrt[1 + a/(b*x^2)]*x*EllipticF
[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1])/(a^3*Sqrt[(I*Sqrt[a])/Sqrt[b
]])))/(6*(e*x)^(5/2)*Sqrt[a + b*x^2])

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Maple [B]  time = 0.035, size = 446, normalized size = 2.1 \[ -{\frac{1}{12\,x{e}^{2}{a}^{3}b} \left ( 15\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{x}^{3}{b}^{2}-5\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{x}^{3}ab+15\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}xab-5\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}x{a}^{2}+30\,A{x}^{4}{b}^{3}-10\,B{x}^{4}a{b}^{2}+42\,A{x}^{2}a{b}^{2}-14\,B{x}^{2}{a}^{2}b+8\,A{a}^{2}b \right ){\frac{1}{\sqrt{ex}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/12*(15*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^3*b^2-5*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^3*a*b+15*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-5*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+30*A*x^4*b^3-10*B*x^4*a*b^2+42*A*x^2*a*b^2-14*B*x^2*a^2*b+8*A*a
^2*b)/x/e^2/(e*x)^(1/2)/a^3/b/(b*x^2+a)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{B x^{2} + A}{{\left (b^{2} e^{2} x^{6} + 2 \, a b e^{2} x^{4} + a^{2} e^{2} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{e x}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*x^2 + A)/((b^2*e^2*x^6 + 2*a*b*e^2*x^4 + a^2*e^2*x^2)*sqrt(b*x^2 + a
)*sqrt(e*x)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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