3.807 \(\int \frac{(e x)^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx\)

Optimal. Leaf size=211 \[ -\frac{5 a^{3/4} e^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-9 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{42 b^{13/4} \sqrt{a+b x^2}}+\frac{5 e^3 \sqrt{e x} \sqrt{a+b x^2} (7 A b-9 a B)}{21 b^3}-\frac{e (e x)^{5/2} (7 A b-9 a B)}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}} \]

[Out]

-((7*A*b - 9*a*B)*e*(e*x)^(5/2))/(7*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(9/2))/(7*
b*e*Sqrt[a + b*x^2]) + (5*(7*A*b - 9*a*B)*e^3*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*b^3
) - (5*a^(3/4)*(7*A*b - 9*a*B)*e^(7/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])]
, 1/2])/(42*b^(13/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.363669, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192 \[ -\frac{5 a^{3/4} e^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (7 A b-9 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{42 b^{13/4} \sqrt{a+b x^2}}+\frac{5 e^3 \sqrt{e x} \sqrt{a+b x^2} (7 A b-9 a B)}{21 b^3}-\frac{e (e x)^{5/2} (7 A b-9 a B)}{7 b^2 \sqrt{a+b x^2}}+\frac{2 B (e x)^{9/2}}{7 b e \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]  Int[((e*x)^(7/2)*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

-((7*A*b - 9*a*B)*e*(e*x)^(5/2))/(7*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(9/2))/(7*
b*e*Sqrt[a + b*x^2]) + (5*(7*A*b - 9*a*B)*e^3*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*b^3
) - (5*a^(3/4)*(7*A*b - 9*a*B)*e^(7/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])]
, 1/2])/(42*b^(13/4)*Sqrt[a + b*x^2])

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Rubi in Sympy [A]  time = 36.77, size = 199, normalized size = 0.94 \[ \frac{2 B \left (e x\right )^{\frac{9}{2}}}{7 b e \sqrt{a + b x^{2}}} - \frac{5 a^{\frac{3}{4}} e^{\frac{7}{2}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (7 A b - 9 B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{42 b^{\frac{13}{4}} \sqrt{a + b x^{2}}} - \frac{e \left (e x\right )^{\frac{5}{2}} \left (7 A b - 9 B a\right )}{7 b^{2} \sqrt{a + b x^{2}}} + \frac{5 e^{3} \sqrt{e x} \sqrt{a + b x^{2}} \left (7 A b - 9 B a\right )}{21 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**(7/2)*(B*x**2+A)/(b*x**2+a)**(3/2),x)

[Out]

2*B*(e*x)**(9/2)/(7*b*e*sqrt(a + b*x**2)) - 5*a**(3/4)*e**(7/2)*sqrt((a + b*x**2
)/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(7*A*b - 9*B*a)*elliptic_f(2*a
tan(b**(1/4)*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(42*b**(13/4)*sqrt(a + b*x**2))
- e*(e*x)**(5/2)*(7*A*b - 9*B*a)/(7*b**2*sqrt(a + b*x**2)) + 5*e**3*sqrt(e*x)*s
qrt(a + b*x**2)*(7*A*b - 9*B*a)/(21*b**3)

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Mathematica [C]  time = 0.29038, size = 168, normalized size = 0.8 \[ \frac{e^3 \sqrt{e x} \left (\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}} \left (-45 a^2 B+a b \left (35 A-18 B x^2\right )+2 b^2 x^2 \left (7 A+3 B x^2\right )\right )-5 i a \sqrt{x} \sqrt{\frac{a}{b x^2}+1} (7 A b-9 a B) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )\right )}{21 b^3 \sqrt{\frac{i \sqrt{a}}{\sqrt{b}}} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]  Integrate[((e*x)^(7/2)*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

(e^3*Sqrt[e*x]*(Sqrt[(I*Sqrt[a])/Sqrt[b]]*(-45*a^2*B + a*b*(35*A - 18*B*x^2) + 2
*b^2*x^2*(7*A + 3*B*x^2)) - (5*I)*a*(7*A*b - 9*a*B)*Sqrt[1 + a/(b*x^2)]*Sqrt[x]*
EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1]))/(21*Sqrt[(I*Sqrt[a
])/Sqrt[b]]*b^3*Sqrt[a + b*x^2])

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Maple [A]  time = 0.053, size = 252, normalized size = 1.2 \[ -{\frac{{e}^{3}}{42\,x{b}^{4}}\sqrt{ex} \left ( 35\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}ab-45\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{a}^{2}-12\,B{x}^{5}{b}^{3}-28\,A{x}^{3}{b}^{3}+36\,B{x}^{3}a{b}^{2}-70\,Axa{b}^{2}+90\,Bx{a}^{2}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^(7/2)*(B*x^2+A)/(b*x^2+a)^(3/2),x)

[Out]

-1/42*e^3/x*(e*x)^(1/2)*(35*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)*a*b-45*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)*a^2-12*B*x^5*b^3-28*A*x^3*b^3+36*B*x^3*a*b^2-70*A*x*a*b^2+9
0*B*x*a^2*b)/(b*x^2+a)^(1/2)/b^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x^{2} + A\right )} \left (e x\right )^{\frac{7}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(e*x)^(7/2)/(b*x^2 + a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(e*x)^(7/2)/(b*x^2 + a)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(e*x)^(7/2)/(b*x^2 + a)^(3/2),x, algorithm="fricas")

[Out]

integral((B*e^3*x^5 + A*e^3*x^3)*sqrt(e*x)/(b*x^2 + a)^(3/2), 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((e*x)**(7/2)*(B*x**2+A)/(b*x**2+a)**(3/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x^{2} + A\right )} \left (e x\right )^{\frac{7}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(e*x)^(7/2)/(b*x^2 + a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(e*x)^(7/2)/(b*x^2 + a)^(3/2), x)