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

Optimal. Leaf size=185 \[ \frac{e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{5/4} b^{9/4} \sqrt{a+b x^2}}-\frac{e \sqrt{e x} (5 a B+A b)}{6 a b^2 \sqrt{a+b x^2}}+\frac{(e x)^{5/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

[Out]

((A*b - a*B)*(e*x)^(5/2))/(3*a*b*e*(a + b*x^2)^(3/2)) - ((A*b + 5*a*B)*e*Sqrt[e*
x])/(6*a*b^2*Sqrt[a + b*x^2]) + ((A*b + 5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqr
t[(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^(5/4)*b^(9/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.306688, antiderivative size = 185, 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 \[ \frac{e^{3/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 a^{5/4} b^{9/4} \sqrt{a+b x^2}}-\frac{e \sqrt{e x} (5 a B+A b)}{6 a b^2 \sqrt{a+b x^2}}+\frac{(e x)^{5/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

((A*b - a*B)*(e*x)^(5/2))/(3*a*b*e*(a + b*x^2)^(3/2)) - ((A*b + 5*a*B)*e*Sqrt[e*
x])/(6*a*b^2*Sqrt[a + b*x^2]) + ((A*b + 5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqr
t[(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^(5/4)*b^(9/4)*Sqrt[a + b*x^2])

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Rubi in Sympy [A]  time = 32.0885, size = 163, normalized size = 0.88 \[ \frac{\left (e x\right )^{\frac{5}{2}} \left (A b - B a\right )}{3 a b e \left (a + b x^{2}\right )^{\frac{3}{2}}} - \frac{e \sqrt{e x} \left (A b + 5 B a\right )}{6 a b^{2} \sqrt{a + b x^{2}}} + \frac{e^{\frac{3}{2}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (A b + 5 B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{12 a^{\frac{5}{4}} b^{\frac{9}{4}} \sqrt{a + b x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

(e*x)**(5/2)*(A*b - B*a)/(3*a*b*e*(a + b*x**2)**(3/2)) - e*sqrt(e*x)*(A*b + 5*B*
a)/(6*a*b**2*sqrt(a + b*x**2)) + e**(3/2)*sqrt((a + b*x**2)/(sqrt(a) + sqrt(b)*x
)**2)*(sqrt(a) + sqrt(b)*x)*(A*b + 5*B*a)*elliptic_f(2*atan(b**(1/4)*sqrt(e*x)/(
a**(1/4)*sqrt(e))), 1/2)/(12*a**(5/4)*b**(9/4)*sqrt(a + b*x**2))

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Mathematica [C]  time = 0.319532, size = 163, normalized size = 0.88 \[ \frac{e \sqrt{e x} \left (\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}} \left (-5 a^2 B-a b \left (A+7 B x^2\right )+A b^2 x^2\right )+i \sqrt{x} \sqrt{\frac{a}{b x^2}+1} \left (a+b x^2\right ) (5 a B+A b) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )\right )}{6 a b^2 \sqrt{\frac{i \sqrt{a}}{\sqrt{b}}} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [B]  time = 0.028, size = 429, normalized size = 2.3 \[{\frac{e}{12\,ax{b}^{3}} \left ( A\sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{1 \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}{x}^{2}{b}^{2}+5\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{x}^{2}ab+A\sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{1 \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{1 \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}ab+5\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{a}^{2}+2\,A{x}^{3}{b}^{3}-14\,B{x}^{3}a{b}^{2}-2\,Axa{b}^{2}-10\,Bx{a}^{2}b \right ) \sqrt{ex} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/12*(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^2*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^2*a*b+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+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
)*a^2+2*A*x^3*b^3-14*B*x^3*a*b^2-2*A*x*a*b^2-10*B*x*a^2*b)*e/x*(e*x)^(1/2)/a/b^3
/(b*x^2+a)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B e x^{3} + A e x\right )} \sqrt{e x}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*e*x^3 + A*e*x)*sqrt(e*x)/((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(b*x^2 + a
)), 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)**(3/2)*(B*x**2+A)/(b*x**2+a)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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