3.859 \(\int \frac{(e x)^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=248 \[ \frac{e \sqrt{e x} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right )}{6 c d^3 \sqrt{c+d x^2}}-\frac{e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{5/4} d^{13/4} \sqrt{c+d x^2}}+\frac{(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}} \]

[Out]

((b*c - a*d)^2*(e*x)^(5/2))/(3*c*d^2*e*(c + d*x^2)^(3/2)) + ((15*b^2*c^2 - 10*a*
b*c*d - a^2*d^2)*e*Sqrt[e*x])/(6*c*d^3*Sqrt[c + d*x^2]) + (2*b^2*(e*x)^(5/2))/(3
*d^2*e*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*e^(3/2)*(Sqrt[c]
+ Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/
4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(5/4)*d^(13/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.506208, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179 \[ \frac{e \sqrt{e x} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right )}{6 c d^3 \sqrt{c+d x^2}}-\frac{e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (-a^2 d^2-10 a b c d+15 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{12 c^{5/4} d^{13/4} \sqrt{c+d x^2}}+\frac{(e x)^{5/2} (b c-a d)^2}{3 c d^2 e \left (c+d x^2\right )^{3/2}}+\frac{2 b^2 (e x)^{5/2}}{3 d^2 e \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

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

[Out]

((b*c - a*d)^2*(e*x)^(5/2))/(3*c*d^2*e*(c + d*x^2)^(3/2)) + ((15*b^2*c^2 - 10*a*
b*c*d - a^2*d^2)*e*Sqrt[e*x])/(6*c*d^3*Sqrt[c + d*x^2]) + (2*b^2*(e*x)^(5/2))/(3
*d^2*e*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*e^(3/2)*(Sqrt[c]
+ Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/
4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(12*c^(5/4)*d^(13/4)*Sqrt[c + d*x^2])

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Rubi in Sympy [A]  time = 61.5478, size = 226, normalized size = 0.91 \[ \frac{2 b^{2} \left (e x\right )^{\frac{5}{2}}}{3 d^{2} e \sqrt{c + d x^{2}}} + \frac{\left (e x\right )^{\frac{5}{2}} \left (a d - b c\right )^{2}}{3 c d^{2} e \left (c + d x^{2}\right )^{\frac{3}{2}}} - \frac{e \sqrt{e x} \left (a^{2} d^{2} + 10 a b c d - 15 b^{2} c^{2}\right )}{6 c d^{3} \sqrt{c + d x^{2}}} + \frac{e^{\frac{3}{2}} \sqrt{\frac{c + d x^{2}}{\left (\sqrt{c} + \sqrt{d} x\right )^{2}}} \left (\sqrt{c} + \sqrt{d} x\right ) \left (a^{2} d^{2} + 10 a b c d - 15 b^{2} c^{2}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}} \right )}\middle | \frac{1}{2}\right )}{12 c^{\frac{5}{4}} d^{\frac{13}{4}} \sqrt{c + d x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2*b**2*(e*x)**(5/2)/(3*d**2*e*sqrt(c + d*x**2)) + (e*x)**(5/2)*(a*d - b*c)**2/(3
*c*d**2*e*(c + d*x**2)**(3/2)) - e*sqrt(e*x)*(a**2*d**2 + 10*a*b*c*d - 15*b**2*c
**2)/(6*c*d**3*sqrt(c + d*x**2)) + e**(3/2)*sqrt((c + d*x**2)/(sqrt(c) + sqrt(d)
*x)**2)*(sqrt(c) + sqrt(d)*x)*(a**2*d**2 + 10*a*b*c*d - 15*b**2*c**2)*elliptic_f
(2*atan(d**(1/4)*sqrt(e*x)/(c**(1/4)*sqrt(e))), 1/2)/(12*c**(5/4)*d**(13/4)*sqrt
(c + d*x**2))

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Mathematica [C]  time = 0.423975, size = 204, normalized size = 0.82 \[ \frac{(e x)^{3/2} \left (\frac{\sqrt{x} \left (a^2 d^2 \left (d x^2-c\right )-2 a b c d \left (5 c+7 d x^2\right )+b^2 c \left (15 c^2+21 c d x^2+4 d^2 x^4\right )\right )}{c d^3 \left (c+d x^2\right )}+\frac{i x \sqrt{\frac{c}{d x^2}+1} \left (a^2 d^2+10 a b c d-15 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right )\right |-1\right )}{c d^3 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{6 x^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

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

[Out]

((e*x)^(3/2)*((Sqrt[x]*(a^2*d^2*(-c + d*x^2) - 2*a*b*c*d*(5*c + 7*d*x^2) + b^2*c
*(15*c^2 + 21*c*d*x^2 + 4*d^2*x^4)))/(c*d^3*(c + d*x^2)) + (I*(-15*b^2*c^2 + 10*
a*b*c*d + a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sq
rt[d]]/Sqrt[x]], -1])/(c*Sqrt[(I*Sqrt[c])/Sqrt[d]]*d^3)))/(6*x^(3/2)*Sqrt[c + d*
x^2])

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Maple [B]  time = 0.032, size = 674, normalized size = 2.7 \[{\frac{e}{12\,cx{d}^{4}} \left ({\it EllipticF} \left ( \sqrt{{1 \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{1 \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{2}\sqrt{{1 \left ( -dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{-{dx{\frac{1}{\sqrt{-cd}}}}}\sqrt{-cd}{x}^{2}{a}^{2}{d}^{3}+10\,{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}\sqrt{-cd}{x}^{2}abc{d}^{2}-15\,{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}\sqrt{-cd}{x}^{2}{b}^{2}{c}^{2}d+\sqrt{-cd}\sqrt{{1 \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{2}\sqrt{{1 \left ( -dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}}\sqrt{-{dx{\frac{1}{\sqrt{-cd}}}}}{\it EllipticF} \left ( \sqrt{{1 \left ( dx+\sqrt{-cd} \right ){\frac{1}{\sqrt{-cd}}}}},{\frac{\sqrt{2}}{2}} \right ){a}^{2}c{d}^{2}+10\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) ab{c}^{2}d-15\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){b}^{2}{c}^{3}+8\,{x}^{5}{b}^{2}c{d}^{3}+2\,{x}^{3}{a}^{2}{d}^{4}-28\,{x}^{3}abc{d}^{3}+42\,{x}^{3}{b}^{2}{c}^{2}{d}^{2}-2\,x{a}^{2}c{d}^{3}-20\,xab{c}^{2}{d}^{2}+30\,x{b}^{2}{c}^{3}d \right ) \sqrt{ex} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/12*(EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*((d*x+(-c*d
)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-
x/(-c*d)^(1/2)*d)^(1/2)*(-c*d)^(1/2)*x^2*a^2*d^3+10*EllipticF(((d*x+(-c*d)^(1/2)
)/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/
2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*(-c*d)^(1/
2)*x^2*a*b*c*d^2-15*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2
))*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(
1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*(-c*d)^(1/2)*x^2*b^2*c^2*d+(-c*d)^(1/2)*((
d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))
^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/
2),1/2*2^(1/2))*a^2*c*d^2+10*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2
)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*Ell
ipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^2*d-15*(-c*d)^
(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d
)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1
/2))^(1/2),1/2*2^(1/2))*b^2*c^3+8*x^5*b^2*c*d^3+2*x^3*a^2*d^4-28*x^3*a*b*c*d^3+4
2*x^3*b^2*c^2*d^2-2*x*a^2*c*d^3-20*x*a*b*c^2*d^2+30*x*b^2*c^3*d)*e/x*(e*x)^(1/2)
/c/d^4/(d*x^2+c)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((b*x^2 + a)^2*(e*x)^(3/2)/(d*x^2 + c)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (b^{2} e x^{5} + 2 \, a b e x^{3} + a^{2} e x\right )} \sqrt{e x}}{{\left (d^{2} x^{4} + 2 \, c d x^{2} + c^{2}\right )} \sqrt{d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((b*x^2 + a)^2*(e*x)^(3/2)/(d*x^2 + c)^(5/2), x)