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

Optimal. Leaf size=367 \[ \frac{4 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{8 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 (e x)^{3/2} \left (a+b x^2\right )^{3/2} (a B+9 A b)}{9 a e^3}+\frac{4 (e x)^{3/2} \sqrt{a+b x^2} (a B+9 A b)}{15 e^3}+\frac{8 a \sqrt{e x} \sqrt{a+b x^2} (a B+9 A b)}{15 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}} \]

[Out]

(4*(9*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*e^3) + (8*a*(9*A*b + a*B)*Sqrt
[e*x]*Sqrt[a + b*x^2])/(15*Sqrt[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) + (2*(9*A*b + a*B)
*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*a*e^3) - (2*A*(a + b*x^2)^(5/2))/(a*e*Sqrt[e*
x]) - (8*a^(5/4)*(9*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] +
Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/
(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (4*a^(5/4)*(9*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])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.72001, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231 \[ \frac{4 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{8 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 (e x)^{3/2} \left (a+b x^2\right )^{3/2} (a B+9 A b)}{9 a e^3}+\frac{4 (e x)^{3/2} \sqrt{a+b x^2} (a B+9 A b)}{15 e^3}+\frac{8 a \sqrt{e x} \sqrt{a+b x^2} (a B+9 A b)}{15 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}} \]

Antiderivative was successfully verified.

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

[Out]

(4*(9*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*e^3) + (8*a*(9*A*b + a*B)*Sqrt
[e*x]*Sqrt[a + b*x^2])/(15*Sqrt[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) + (2*(9*A*b + a*B)
*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*a*e^3) - (2*A*(a + b*x^2)^(5/2))/(a*e*Sqrt[e*
x]) - (8*a^(5/4)*(9*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] +
Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/
(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (4*a^(5/4)*(9*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])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2])

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Rubi in Sympy [A]  time = 73.9658, size = 343, normalized size = 0.93 \[ - \frac{2 A \left (a + b x^{2}\right )^{\frac{5}{2}}}{a e \sqrt{e x}} - \frac{8 a^{\frac{5}{4}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (9 A b + B a\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} e^{\frac{3}{2}} \sqrt{a + b x^{2}}} + \frac{4 a^{\frac{5}{4}} \sqrt{\frac{a + b x^{2}}{\left (\sqrt{a} + \sqrt{b} x\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x\right ) \left (9 A b + B a\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{15 b^{\frac{3}{4}} e^{\frac{3}{2}} \sqrt{a + b x^{2}}} + \frac{8 a \sqrt{e x} \sqrt{a + b x^{2}} \left (9 A b + B a\right )}{15 \sqrt{b} e^{2} \left (\sqrt{a} + \sqrt{b} x\right )} + \frac{4 \left (e x\right )^{\frac{3}{2}} \sqrt{a + b x^{2}} \left (9 A b + B a\right )}{15 e^{3}} + \frac{2 \left (e x\right )^{\frac{3}{2}} \left (a + b x^{2}\right )^{\frac{3}{2}} \left (9 A b + B a\right )}{9 a e^{3}} \]

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)**(3/2),x)

[Out]

-2*A*(a + b*x**2)**(5/2)/(a*e*sqrt(e*x)) - 8*a**(5/4)*sqrt((a + b*x**2)/(sqrt(a)
+ sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(9*A*b + B*a)*elliptic_e(2*atan(b**(1/4)
*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(15*b**(3/4)*e**(3/2)*sqrt(a + b*x**2)) + 4
*a**(5/4)*sqrt((a + b*x**2)/(sqrt(a) + sqrt(b)*x)**2)*(sqrt(a) + sqrt(b)*x)*(9*A
*b + B*a)*elliptic_f(2*atan(b**(1/4)*sqrt(e*x)/(a**(1/4)*sqrt(e))), 1/2)/(15*b**
(3/4)*e**(3/2)*sqrt(a + b*x**2)) + 8*a*sqrt(e*x)*sqrt(a + b*x**2)*(9*A*b + B*a)/
(15*sqrt(b)*e**2*(sqrt(a) + sqrt(b)*x)) + 4*(e*x)**(3/2)*sqrt(a + b*x**2)*(9*A*b
+ B*a)/(15*e**3) + 2*(e*x)**(3/2)*(a + b*x**2)**(3/2)*(9*A*b + B*a)/(9*a*e**3)

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Mathematica [C]  time = 1.10047, size = 206, normalized size = 0.56 \[ \frac{x^{3/2} \left (\frac{2 \sqrt{a+b x^2} \left (-45 a A+11 a B x^2+9 A b x^2+5 b B x^4\right )}{3 \sqrt{x}}-\frac{8 a x (a B+9 A b) \left (-\sqrt{x} \left (\frac{a}{x^2}+b\right )+\frac{i a \sqrt{\frac{a}{b x^2}+1} \left (E\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{b}}}}{\sqrt{x}}\right )\right |-1\right )\right )}{\left (\frac{i \sqrt{a}}{\sqrt{b}}\right )^{3/2}}\right )}{b \sqrt{a+b x^2}}\right )}{15 (e x)^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(x^(3/2)*((2*Sqrt[a + b*x^2]*(-45*a*A + 9*A*b*x^2 + 11*a*B*x^2 + 5*b*B*x^4))/(3*
Sqrt[x]) - (8*a*(9*A*b + a*B)*x*(-((b + a/x^2)*Sqrt[x]) + (I*a*Sqrt[1 + a/(b*x^2
)]*(EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1] - EllipticF[I*Ar
cSinh[Sqrt[(I*Sqrt[a])/Sqrt[b]]/Sqrt[x]], -1]))/((I*Sqrt[a])/Sqrt[b])^(3/2)))/(b
*Sqrt[a + b*x^2])))/(15*(e*x)^(3/2))

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Maple [A]  time = 0.027, size = 421, normalized size = 1.2 \[{\frac{2}{45\,be} \left ( 5\,B{x}^{6}{b}^{3}+108\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{2}b-54\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{2}b+12\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}-6\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}+9\,A{x}^{4}{b}^{3}+16\,B{x}^{4}a{b}^{2}-36\,A{x}^{2}a{b}^{2}+11\,B{x}^{2}{a}^{2}b-45\,A{a}^{2}b \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2/45*(5*B*x^6*b^3+108*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)*EllipticE(((b*x+(-a*b
)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2*b-54*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^2*b+12*
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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))
^(1/2),1/2*2^(1/2))*a^3-6*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^3+9*A*x^4*b^3+16*B*x^4*a*b^2-36*
A*x^2*a*b^2+11*B*x^2*a^2*b-45*A*a^2*b)/(b*x^2+a)^(1/2)/b/e/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [A]  time = 68.5735, size = 202, normalized size = 0.55 \[ \frac{A a^{\frac{3}{2}} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{A \sqrt{a} b x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B a^{\frac{3}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B \sqrt{a} b x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*a**(3/2)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*
e**(3/2)*sqrt(x)*gamma(3/4)) + A*sqrt(a)*b*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4)
, (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(7/4)) + B*a**(3/2)*x**(3/2
)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*ga
mma(7/4)) + B*sqrt(a)*b*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**2*e
xp_polar(I*pi)/a)/(2*e**(3/2)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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