3.459 \(\int \frac{(e x)^{5/2} (A+B x)}{\sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=356 \[ \frac{a^{5/4} e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (25 \sqrt{a} B-63 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{6 a^{5/4} A e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{6 a A e^3 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 A e (e x)^{3/2} \sqrt{a+c x^2}}{5 c}-\frac{10 a B e^2 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}+\frac{2 B (e x)^{5/2} \sqrt{a+c x^2}}{7 c} \]

[Out]

(-10*a*B*e^2*Sqrt[e*x]*Sqrt[a + c*x^2])/(21*c^2) + (2*A*e*(e*x)^(3/2)*Sqrt[a + c
*x^2])/(5*c) + (2*B*(e*x)^(5/2)*Sqrt[a + c*x^2])/(7*c) - (6*a*A*e^3*x*Sqrt[a + c
*x^2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (6*a^(5/4)*A*e^3*Sqrt[x]*(S
qrt[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^(5
/4)*(25*Sqrt[a]*B - 63*A*Sqrt[c])*e^3*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
])/(105*c^(9/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.958589, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25 \[ \frac{a^{5/4} e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (25 \sqrt{a} B-63 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 c^{9/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{6 a^{5/4} A e^3 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 c^{7/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{6 a A e^3 x \sqrt{a+c x^2}}{5 c^{3/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{2 A e (e x)^{3/2} \sqrt{a+c x^2}}{5 c}-\frac{10 a B e^2 \sqrt{e x} \sqrt{a+c x^2}}{21 c^2}+\frac{2 B (e x)^{5/2} \sqrt{a+c x^2}}{7 c} \]

Antiderivative was successfully verified.

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

[Out]

(-10*a*B*e^2*Sqrt[e*x]*Sqrt[a + c*x^2])/(21*c^2) + (2*A*e*(e*x)^(3/2)*Sqrt[a + c
*x^2])/(5*c) + (2*B*(e*x)^(5/2)*Sqrt[a + c*x^2])/(7*c) - (6*a*A*e^3*x*Sqrt[a + c
*x^2])/(5*c^(3/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (6*a^(5/4)*A*e^3*Sqrt[x]*(S
qrt[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^(5
/4)*(25*Sqrt[a]*B - 63*A*Sqrt[c])*e^3*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
])/(105*c^(9/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi in Sympy [A]  time = 126.308, size = 335, normalized size = 0.94 \[ \frac{6 A a^{\frac{5}{4}} e^{3} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{5 c^{\frac{7}{4}} \sqrt{e x} \sqrt{a + c x^{2}}} - \frac{6 A a e^{3} x \sqrt{a + c x^{2}}}{5 c^{\frac{3}{2}} \sqrt{e x} \left (\sqrt{a} + \sqrt{c} x\right )} + \frac{2 A e \left (e x\right )^{\frac{3}{2}} \sqrt{a + c x^{2}}}{5 c} - \frac{10 B a e^{2} \sqrt{e x} \sqrt{a + c x^{2}}}{21 c^{2}} + \frac{2 B \left (e x\right )^{\frac{5}{2}} \sqrt{a + c x^{2}}}{7 c} - \frac{a^{\frac{5}{4}} e^{3} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) \left (63 A \sqrt{c} - 25 B \sqrt{a}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{105 c^{\frac{9}{4}} \sqrt{e x} \sqrt{a + c x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**(5/2)*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

6*A*a**(5/4)*e**3*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*A*a*e**3*x*sqrt(a + c*x**2)/(5*c**(3/2)*sqrt(e*x)*(sq
rt(a) + sqrt(c)*x)) + 2*A*e*(e*x)**(3/2)*sqrt(a + c*x**2)/(5*c) - 10*B*a*e**2*sq
rt(e*x)*sqrt(a + c*x**2)/(21*c**2) + 2*B*(e*x)**(5/2)*sqrt(a + c*x**2)/(7*c) - a
**(5/4)*e**3*sqrt(x)*sqrt((a + c*x**2)/(sqrt(a) + sqrt(c)*x)**2)*(sqrt(a) + sqrt
(c)*x)*(63*A*sqrt(c) - 25*B*sqrt(a))*elliptic_f(2*atan(c**(1/4)*sqrt(x)/a**(1/4)
), 1/2)/(105*c**(9/4)*sqrt(e*x)*sqrt(a + c*x**2))

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Mathematica [C]  time = 1.14155, size = 236, normalized size = 0.66 \[ -\frac{2 e^3 \left (a^{3/2} x^{3/2} \sqrt{\frac{a}{c x^2}+1} \left (63 A \sqrt{c}-25 i \sqrt{a} B\right ) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )-63 a^{3/2} A \sqrt{c} x^{3/2} \sqrt{\frac{a}{c x^2}+1} E\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )+\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \left (a+c x^2\right ) \left (a (63 A+25 B x)-3 c x^2 (7 A+5 B x)\right )\right )}{105 c^2 \sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \sqrt{e x} \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*e^3*(Sqrt[(I*Sqrt[a])/Sqrt[c]]*(a + c*x^2)*(-3*c*x^2*(7*A + 5*B*x) + a*(63*A
+ 25*B*x)) - 63*a^(3/2)*A*Sqrt[c]*Sqrt[1 + a/(c*x^2)]*x^(3/2)*EllipticE[I*ArcSi
nh[Sqrt[(I*Sqrt[a])/Sqrt[c]]/Sqrt[x]], -1] + a^(3/2)*((-25*I)*Sqrt[a]*B + 63*A*S
qrt[c])*Sqrt[1 + a/(c*x^2)]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[c]
]/Sqrt[x]], -1]))/(105*Sqrt[(I*Sqrt[a])/Sqrt[c]]*c^2*Sqrt[e*x]*Sqrt[a + c*x^2])

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Maple [A]  time = 0.039, size = 336, normalized size = 0.9 \[ -{\frac{{e}^{2}}{105\,x{c}^{3}}\sqrt{ex} \left ( 126\,A\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{2}c-63\,A\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{2}c-25\,B\sqrt{-ac}\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{2}-30\,B{c}^{3}{x}^{5}-42\,A{c}^{3}{x}^{4}+20\,aB{c}^{2}{x}^{3}-42\,aA{c}^{2}{x}^{2}+50\,{a}^{2}Bcx \right ){\frac{1}{\sqrt{c{x}^{2}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^(5/2)*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

-1/105/x*e^2*(e*x)^(1/2)/(c*x^2+a)^(1/2)*(126*A*((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))*a^2*c-63*A*((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))*a^2*c-25*B*(-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)*Ellipt
icF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^2-30*B*c^3*x^5-42*A*c
^3*x^4+20*a*B*c^2*x^3-42*a*A*c^2*x^2+50*a^2*B*c*x)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*e^2*x^3 + A*e^2*x^2)*sqrt(e*x)/sqrt(c*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)**(5/2)*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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