3.460 \(\int \frac{(e x)^{3/2} (A+B x)}{\sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=326 \[ -\frac{a^{3/4} e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{6 a^{5/4} B e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}-\frac{6 a B e^2 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c} \]

[Out]

(2*A*e*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c) + (2*B*(e*x)^(3/2)*Sqrt[a + c*x^2])/(5*c
) - (6*a*B*e^2*x*Sqrt[a + c*x^2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) +
(6*a^(5/4)*B*e^2*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[
c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*c^(7/4)*Sqrt[e*
x]*Sqrt[a + c*x^2]) - (a^(3/4)*(9*Sqrt[a]*B + 5*A*Sqrt[c])*e^2*Sqrt[x]*(Sqrt[a]
+ Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/
4)*Sqrt[x])/a^(1/4)], 1/2])/(15*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.796803, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25 \[ -\frac{a^{3/4} e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{6 a^{5/4} B e^2 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{2 A e \sqrt{e x} \sqrt{a+c x^2}}{3 c}-\frac{6 a B e^2 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 B (e x)^{3/2} \sqrt{a+c x^2}}{5 c} \]

Antiderivative was successfully verified.

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

[Out]

(2*A*e*Sqrt[e*x]*Sqrt[a + c*x^2])/(3*c) + (2*B*(e*x)^(3/2)*Sqrt[a + c*x^2])/(5*c
) - (6*a*B*e^2*x*Sqrt[a + c*x^2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) +
(6*a^(5/4)*B*e^2*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[
c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*c^(7/4)*Sqrt[e*
x]*Sqrt[a + c*x^2]) - (a^(3/4)*(9*Sqrt[a]*B + 5*A*Sqrt[c])*e^2*Sqrt[x]*(Sqrt[a]
+ Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/
4)*Sqrt[x])/a^(1/4)], 1/2])/(15*c^(7/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi in Sympy [A]  time = 72.5054, size = 304, normalized size = 0.93 \[ \frac{2 A e \sqrt{e x} \sqrt{a + c x^{2}}}{3 c} + \frac{6 B a^{\frac{5}{4}} e^{2} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 c^{\frac{7}{4}} \sqrt{e x} \sqrt{a + c x^{2}}} - \frac{6 B a e^{2} x \sqrt{a + c x^{2}}}{5 c^{\frac{3}{2}} \sqrt{e x} \left (\sqrt{a} + \sqrt{c} x\right )} + \frac{2 B \left (e x\right )^{\frac{3}{2}} \sqrt{a + c x^{2}}}{5 c} - \frac{a^{\frac{3}{4}} e^{2} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) \left (5 A \sqrt{c} + 9 B \sqrt{a}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 c^{\frac{7}{4}} \sqrt{e x} \sqrt{a + c x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**(3/2)*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

2*A*e*sqrt(e*x)*sqrt(a + c*x**2)/(3*c) + 6*B*a**(5/4)*e**2*sqrt(x)*sqrt((a + c*x
**2)/(sqrt(a) + sqrt(c)*x)**2)*(sqrt(a) + sqrt(c)*x)*elliptic_e(2*atan(c**(1/4)*
sqrt(x)/a**(1/4)), 1/2)/(5*c**(7/4)*sqrt(e*x)*sqrt(a + c*x**2)) - 6*B*a*e**2*x*s
qrt(a + c*x**2)/(5*c**(3/2)*sqrt(e*x)*(sqrt(a) + sqrt(c)*x)) + 2*B*(e*x)**(3/2)*
sqrt(a + c*x**2)/(5*c) - a**(3/4)*e**2*sqrt(x)*sqrt((a + c*x**2)/(sqrt(a) + sqrt
(c)*x)**2)*(sqrt(a) + sqrt(c)*x)*(5*A*sqrt(c) + 9*B*sqrt(a))*elliptic_f(2*atan(c
**(1/4)*sqrt(x)/a**(1/4)), 1/2)/(15*c**(7/4)*sqrt(e*x)*sqrt(a + c*x**2))

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Mathematica [C]  time = 1.22426, size = 229, normalized size = 0.7 \[ -\frac{2 e^2 \left (-9 a^{3/2} B \sqrt{c} x^{3/2} \sqrt{\frac{a}{c x^2}+1} E\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )+\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \left (a+c x^2\right ) (9 a B-c x (5 A+3 B x))+a \sqrt{c} x^{3/2} \sqrt{\frac{a}{c x^2}+1} \left (9 \sqrt{a} B+5 i A \sqrt{c}\right ) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )\right )}{15 c^2 \sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*e^2*(Sqrt[(I*Sqrt[a])/Sqrt[c]]*(a + c*x^2)*(9*a*B - c*x*(5*A + 3*B*x)) - 9*a
^(3/2)*B*Sqrt[c]*Sqrt[1 + a/(c*x^2)]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[a]
)/Sqrt[c]]/Sqrt[x]], -1] + a*(9*Sqrt[a]*B + (5*I)*A*Sqrt[c])*Sqrt[c]*Sqrt[1 + a/
(c*x^2)]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[c]]/Sqrt[x]], -1]))/(
15*Sqrt[(I*Sqrt[a])/Sqrt[c]]*c^2*Sqrt[e*x]*Sqrt[a + c*x^2])

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Maple [A]  time = 0.02, size = 316, normalized size = 1. \[ -{\frac{e}{15\,{c}^{2}x}\sqrt{ex} \left ( 5\,A{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}\sqrt{-ac}a+18\,B{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{a}^{2}-9\,B{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{a}^{2}-6\,B{c}^{2}{x}^{4}-10\,A{c}^{2}{x}^{3}-6\,aBc{x}^{2}-10\,aAcx \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^(3/2)*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

-1/15*e/x*(e*x)^(1/2)/(c*x^2+a)^(1/2)/c^2*(5*A*EllipticF(((c*x+(-a*c)^(1/2))/(-a
*c)^(1/2))^(1/2),1/2*2^(1/2))*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((
-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*(-a*c)^(1/2)*a+
18*B*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*((c*x+(-a*c)
^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x
*c/(-a*c)^(1/2))^(1/2)*a^2-9*B*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)
,1/2*2^(1/2))*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2
))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*a^2-6*B*c^2*x^4-10*A*c^2*x^3-6*
a*B*c*x^2-10*a*A*c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*e*x^2 + A*e*x)*sqrt(e*x)/sqrt(c*x^2 + a), x)

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Sympy [A]  time = 37.2202, size = 94, normalized size = 0.29 \[ \frac{A e^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{B e^{\frac{3}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x)**(3/2)*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/
a)/(2*sqrt(a)*gamma(9/4)) + B*e**(3/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11
/4,), c*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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