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

Optimal. Leaf size=341 \[ \frac{4 a^{3/4} \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{24 a^{5/4} B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 \sqrt{a+c x^2} (9 a B-5 A c x)}{15 e^2 \sqrt{e x}}-\frac{2 \left (a+c x^2\right )^{3/2} (5 A-3 B x)}{15 e (e x)^{3/2}}+\frac{24 a B \sqrt{c} x \sqrt{a+c x^2}}{5 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

[Out]

(24*a*B*Sqrt[c]*x*Sqrt[a + c*x^2])/(5*e^2*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (4*
(9*a*B - 5*A*c*x)*Sqrt[a + c*x^2])/(15*e^2*Sqrt[e*x]) - (2*(5*A - 3*B*x)*(a + c*
x^2)^(3/2))/(15*e*(e*x)^(3/2)) - (24*a^(5/4)*B*c^(1/4)*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*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]) + (4*a^(3/4)*(9*Sqrt[a]*B
+ 5*A*Sqrt[c])*c^(1/4)*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*e^2*Sqrt
[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.774437, antiderivative size = 341, 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 \[ \frac{4 a^{3/4} \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (9 \sqrt{a} B+5 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{24 a^{5/4} B \sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 e^2 \sqrt{e x} \sqrt{a+c x^2}}-\frac{4 \sqrt{a+c x^2} (9 a B-5 A c x)}{15 e^2 \sqrt{e x}}-\frac{2 \left (a+c x^2\right )^{3/2} (5 A-3 B x)}{15 e (e x)^{3/2}}+\frac{24 a B \sqrt{c} x \sqrt{a+c x^2}}{5 e^2 \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )} \]

Antiderivative was successfully verified.

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

[Out]

(24*a*B*Sqrt[c]*x*Sqrt[a + c*x^2])/(5*e^2*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (4*
(9*a*B - 5*A*c*x)*Sqrt[a + c*x^2])/(15*e^2*Sqrt[e*x]) - (2*(5*A - 3*B*x)*(a + c*
x^2)^(3/2))/(15*e*(e*x)^(3/2)) - (24*a^(5/4)*B*c^(1/4)*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*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]) + (4*a^(3/4)*(9*Sqrt[a]*B
+ 5*A*Sqrt[c])*c^(1/4)*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*e^2*Sqrt
[e*x]*Sqrt[a + c*x^2])

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Rubi in Sympy [A]  time = 92.0238, size = 330, normalized size = 0.97 \[ - \frac{24 B a^{\frac{5}{4}} \sqrt [4]{c} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 e^{2} \sqrt{e x} \sqrt{a + c x^{2}}} + \frac{24 B a \sqrt{c} x \sqrt{a + c x^{2}}}{5 e^{2} \sqrt{e x} \left (\sqrt{a} + \sqrt{c} x\right )} + \frac{4 a^{\frac{3}{4}} \sqrt [4]{c} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) \left (5 A \sqrt{c} + 9 B \sqrt{a}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{15 e^{2} \sqrt{e x} \sqrt{a + c x^{2}}} - \frac{4 \left (\frac{5 A}{2} - \frac{3 B x}{2}\right ) \left (a + c x^{2}\right )^{\frac{3}{2}}}{15 e \left (e x\right )^{\frac{3}{2}}} - \frac{8 \sqrt{a + c x^{2}} \left (- \frac{5 A c x}{2} + \frac{9 B a}{2}\right )}{15 e^{2} \sqrt{e x}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-24*B*a**(5/4)*c**(1/4)*sqrt(x)*sqrt((a + c*x**2)/(sqrt(a) + sqrt(c)*x)**2)*(sqr
t(a) + sqrt(c)*x)*elliptic_e(2*atan(c**(1/4)*sqrt(x)/a**(1/4)), 1/2)/(5*e**2*sqr
t(e*x)*sqrt(a + c*x**2)) + 24*B*a*sqrt(c)*x*sqrt(a + c*x**2)/(5*e**2*sqrt(e*x)*(
sqrt(a) + sqrt(c)*x)) + 4*a**(3/4)*c**(1/4)*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*a
tan(c**(1/4)*sqrt(x)/a**(1/4)), 1/2)/(15*e**2*sqrt(e*x)*sqrt(a + c*x**2)) - 4*(5
*A/2 - 3*B*x/2)*(a + c*x**2)**(3/2)/(15*e*(e*x)**(3/2)) - 8*sqrt(a + c*x**2)*(-5
*A*c*x/2 + 9*B*a/2)/(15*e**2*sqrt(e*x))

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Mathematica [C]  time = 0.812857, size = 233, normalized size = 0.68 \[ \frac{x \left (-72 a^{3/2} B \sqrt{c} x^{5/2} \sqrt{\frac{a}{c x^2}+1} E\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )+8 a \sqrt{c} x^{5/2} \sqrt{\frac{a}{c x^2}+1} \left (9 \sqrt{a} B+5 i A \sqrt{c}\right ) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )+2 \sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \left (a+c x^2\right ) \left (-5 a A+21 a B x+5 A c x^2+3 B c x^3\right )\right )}{15 \sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} (e x)^{5/2} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

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

[Out]

(x*(2*Sqrt[(I*Sqrt[a])/Sqrt[c]]*(a + c*x^2)*(-5*a*A + 21*a*B*x + 5*A*c*x^2 + 3*B
*c*x^3) - 72*a^(3/2)*B*Sqrt[c]*Sqrt[1 + a/(c*x^2)]*x^(5/2)*EllipticE[I*ArcSinh[S
qrt[(I*Sqrt[a])/Sqrt[c]]/Sqrt[x]], -1] + 8*a*(9*Sqrt[a]*B + (5*I)*A*Sqrt[c])*Sqr
t[c]*Sqrt[1 + a/(c*x^2)]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[c]]/S
qrt[x]], -1]))/(15*Sqrt[(I*Sqrt[a])/Sqrt[c]]*(e*x)^(5/2)*Sqrt[a + c*x^2])

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Maple [A]  time = 0.026, size = 325, normalized size = 1. \[{\frac{2}{15\,x{e}^{2}} \left ( 10\,A\sqrt{-ac}\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) xa+36\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) x{a}^{2}-18\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) x{a}^{2}+3\,B{c}^{2}{x}^{5}+5\,A{c}^{2}{x}^{4}-12\,aBc{x}^{3}-15\,{a}^{2}Bx-5\,A{a}^{2} \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2/15/x*(10*A*(-a*c)^(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)*EllipticF(((c*x+(-a
*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a+36*B*((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)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a^2-18*
B*((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)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))
^(1/2),1/2*2^(1/2))*x*a^2+3*B*c^2*x^5+5*A*c^2*x^4-12*a*B*c*x^3-15*a^2*B*x-5*A*a^
2)/(c*x^2+a)^(1/2)/e^2/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 92.4588, size = 206, normalized size = 0.6 \[ \frac{A a^{\frac{3}{2}} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{A \sqrt{a} c \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{B a^{\frac{3}{2}} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{B \sqrt{a} c x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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