3.1341 \(\int \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{5/2}} \, dx\)

Optimal. Leaf size=219 \[ \frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{84 c^4 d^{5/2} \sqrt{a+b x+c x^2}}-\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{84 c^3 d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}} \]

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(84*c^3*d^3) + (5*S
qrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^(3/2))/(42*c^2*d^3) - (a + b*x + c*x^2)^(5/
2)/(3*c*d*(b*d + 2*c*d*x)^(3/2)) + (5*(b^2 - 4*a*c)^(9/4)*Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])], -1])/(84*c^4*d^(5/2)*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.518658, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179 \[ \frac{5 \left (b^2-4 a c\right )^{9/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{84 c^4 d^{5/2} \sqrt{a+b x+c x^2}}-\frac{5 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \sqrt{b d+2 c d x}}{84 c^3 d^3}+\frac{5 \left (a+b x+c x^2\right )^{3/2} \sqrt{b d+2 c d x}}{42 c^2 d^3}-\frac{\left (a+b x+c x^2\right )^{5/2}}{3 c d (b d+2 c d x)^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(-5*(b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/(84*c^3*d^3) + (5*S
qrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^(3/2))/(42*c^2*d^3) - (a + b*x + c*x^2)^(5/
2)/(3*c*d*(b*d + 2*c*d*x)^(3/2)) + (5*(b^2 - 4*a*c)^(9/4)*Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)
*Sqrt[d])], -1])/(84*c^4*d^(5/2)*Sqrt[a + b*x + c*x^2])

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Rubi in Sympy [A]  time = 111.101, size = 211, normalized size = 0.96 \[ - \frac{\left (a + b x + c x^{2}\right )^{\frac{5}{2}}}{3 c d \left (b d + 2 c d x\right )^{\frac{3}{2}}} + \frac{5 \sqrt{b d + 2 c d x} \left (a + b x + c x^{2}\right )^{\frac{3}{2}}}{42 c^{2} d^{3}} - \frac{5 \left (- 4 a c + b^{2}\right ) \sqrt{b d + 2 c d x} \sqrt{a + b x + c x^{2}}}{84 c^{3} d^{3}} + \frac{5 \sqrt{\frac{c \left (a + b x + c x^{2}\right )}{4 a c - b^{2}}} \left (- 4 a c + b^{2}\right )^{\frac{9}{4}} F\left (\operatorname{asin}{\left (\frac{\sqrt{b d + 2 c d x}}{\sqrt{d} \sqrt [4]{- 4 a c + b^{2}}} \right )}\middle | -1\right )}{84 c^{4} d^{\frac{5}{2}} \sqrt{a + b x + c x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-(a + b*x + c*x**2)**(5/2)/(3*c*d*(b*d + 2*c*d*x)**(3/2)) + 5*sqrt(b*d + 2*c*d*x
)*(a + b*x + c*x**2)**(3/2)/(42*c**2*d**3) - 5*(-4*a*c + b**2)*sqrt(b*d + 2*c*d*
x)*sqrt(a + b*x + c*x**2)/(84*c**3*d**3) + 5*sqrt(c*(a + b*x + c*x**2)/(4*a*c -
b**2))*(-4*a*c + b**2)**(9/4)*elliptic_f(asin(sqrt(b*d + 2*c*d*x)/(sqrt(d)*(-4*a
*c + b**2)**(1/4))), -1)/(84*c**4*d**(5/2)*sqrt(a + b*x + c*x**2))

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Mathematica [C]  time = 1.23309, size = 201, normalized size = 0.92 \[ \frac{\frac{(b+2 c x)^3 (a+x (b+c x)) \left (-\frac{7 \left (b^2-4 a c\right )^2}{(b+2 c x)^2}+64 a c-13 b^2+12 b c x+12 c^2 x^2\right )}{4 c^3}+\frac{5 i \left (b^2-4 a c\right )^2 (b+2 c x)^{7/2} \sqrt{\frac{c (a+x (b+c x))}{(b+2 c x)^2}} F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{-\sqrt{b^2-4 a c}}}{\sqrt{b+2 c x}}\right )\right |-1\right )}{c^4 \sqrt{-\sqrt{b^2-4 a c}}}}{84 \sqrt{a+x (b+c x)} (d (b+2 c x))^{5/2}} \]

Antiderivative was successfully verified.

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

[Out]

(((b + 2*c*x)^3*(a + x*(b + c*x))*(-13*b^2 + 64*a*c + 12*b*c*x + 12*c^2*x^2 - (7
*(b^2 - 4*a*c)^2)/(b + 2*c*x)^2))/(4*c^3) + ((5*I)*(b^2 - 4*a*c)^2*(b + 2*c*x)^(
7/2)*Sqrt[(c*(a + x*(b + c*x)))/(b + 2*c*x)^2]*EllipticF[I*ArcSinh[Sqrt[-Sqrt[b^
2 - 4*a*c]]/Sqrt[b + 2*c*x]], -1])/(c^4*Sqrt[-Sqrt[b^2 - 4*a*c]]))/(84*(d*(b + 2
*c*x))^(5/2)*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.042, size = 1000, normalized size = 4.6 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/168*(24*c^6*x^6+160*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-
(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^
(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/
2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*x*a^2*c^3-80*((b+2*c*x+(-4*a*c+b^2)^(1/2)
)/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4
*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^
(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*x*a*b^2*c^2
+10*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+
b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*Ellip
ticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2)
)*(-4*a*c+b^2)^(1/2)*x*b^4*c+72*b*c^5*x^5+80*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a
*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^
2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/
(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*a^2*b*c^2-40*((b+2
*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2
))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*
((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*(-4*a*c
+b^2)^(1/2)*a*b^3*c+5*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-
(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-b-2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^
(1/2))^(1/2)*EllipticF(1/2*((b+2*c*x+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/
2)*2^(1/2),2^(1/2))*(-4*a*c+b^2)^(1/2)*b^5+152*x^4*a*c^5+52*x^4*b^2*c^4+304*x^3*
a*b*c^4-16*b^3*c^3*x^3+72*x^2*a^2*c^4+192*x^2*a*b^2*c^3-30*x^2*b^4*c^2+72*a^2*b*
c^3*x+40*a*b^3*c^2*x-10*b^5*c*x-56*a^3*c^3+60*a^2*b^2*c^2-10*a*b^4*c)/d^3*(d*(2*
c*x+b))^(1/2)/(c*x^2+b*x+a)^(1/2)/(2*c*x+b)^2/c^4

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Integral((a + b*x + c*x**2)**(5/2)/(d*(b + 2*c*x))**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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