3.1650 \(\int \frac{(b+2 c x) (d+e x)^{5/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=494 \[ -\frac{20 \sqrt{2} e \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{10 e \sqrt{d+e x} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{5 \sqrt{2} e \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 (d+e x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

(-2*(d + e*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (10*e*Sqrt[d + e*x]*(b*d - 2*
a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (5*Sqrt[2]*e*(
2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*Elliptic
E[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*S
qrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*Sqrt[b^2 - 4*a*c]
*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2])
- (20*Sqrt[2]*e*(c*d^2 - b*d*e + a*e^2)*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^
2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sq
rt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4
*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*Sqrt[b^2 - 4*a*c]*Sqrt[d + e
*x]*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 1.10298, antiderivative size = 494, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2 \[ -\frac{20 \sqrt{2} e \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (a e^2-b d e+c d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{10 e \sqrt{d+e x} (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{5 \sqrt{2} e \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{3 c \sqrt{b^2-4 a c} \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{2 (d+e x)^{5/2}}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*(d + e*x)^(5/2))/(3*(a + b*x + c*x^2)^(3/2)) - (10*e*Sqrt[d + e*x]*(b*d - 2*
a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (5*Sqrt[2]*e*(
2*c*d - b*e)*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*Elliptic
E[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*S
qrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*Sqrt[b^2 - 4*a*c]
*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[a + b*x + c*x^2])
- (20*Sqrt[2]*e*(c*d^2 - b*d*e + a*e^2)*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^
2 - 4*a*c])*e)]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sq
rt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4
*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(3*c*Sqrt[b^2 - 4*a*c]*Sqrt[d + e
*x]*Sqrt[a + b*x + c*x^2])

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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Mathematica [C]  time = 12.6985, size = 1973, normalized size = 3.99 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[d + e*x]*(a + b*x + c*x^2)^3*((-2*(c*d^2 - a*e^2 + 2*c*d*e*x - b*e^2*x))/(
3*c*(a + b*x + c*x^2)^2) + (2*(5*b*c*d*e + b^2*e^2 - 14*a*c*e^2 + 10*c^2*d*e*x -
5*b*c*e^2*x))/(3*c*(-b^2 + 4*a*c)*(a + b*x + c*x^2))))/(a + x*(b + c*x))^(5/2)
+ (5*(a + b*x + c*x^2)^(5/2)*((2*(2*c*d - b*e)*(d + e*x)^(3/2)*(c + (c*d^2)/(d +
e*x)^2 - (b*d*e)/(d + e*x)^2 + (a*e^2)/(d + e*x)^2 - (2*c*d)/(d + e*x) + (b*e)/
(d + e*x)))/(c*Sqrt[((d + e*x)^2*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d)/(d + e*
x) + (a*e)/(d + e*x)))/(d + e*x)))/e^2]) - (2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)*
Sqrt[c + (c*d^2)/(d + e*x)^2 - (b*d*e)/(d + e*x)^2 + (a*e^2)/(d + e*x)^2 - (2*c*
d)/(d + e*x) + (b*e)/(d + e*x)]*((I*c*d*(2*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2]
)*Sqrt[1 - (2*(c*d^2 - b*d*e + a*e^2))/((2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2]
)*(d + e*x))]*Sqrt[1 - (2*(c*d^2 - b*d*e + a*e^2))/((2*c*d - b*e + Sqrt[b^2*e^2
- 4*a*c*e^2])*(d + e*x))]*(EllipticE[I*ArcSinh[(Sqrt[2]*Sqrt[-((c*d^2 - b*d*e +
a*e^2)/(2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2]))])/Sqrt[d + e*x]], (2*c*d - b*e
- Sqrt[b^2*e^2 - 4*a*c*e^2])/(2*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2])] - Ellip
ticF[I*ArcSinh[(Sqrt[2]*Sqrt[-((c*d^2 - b*d*e + a*e^2)/(2*c*d - b*e - Sqrt[b^2*e
^2 - 4*a*c*e^2]))])/Sqrt[d + e*x]], (2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2])/(2
*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2])]))/(Sqrt[2]*(c*d^2 - b*d*e + a*e^2)*Sqrt
[-((c*d^2 - b*d*e + a*e^2)/(2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2]))]*Sqrt[c +
(c*d^2 - b*d*e + a*e^2)/(d + e*x)^2 + (-2*c*d + b*e)/(d + e*x)]) - ((I/2)*b*e*(2
*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2])*Sqrt[1 - (2*(c*d^2 - b*d*e + a*e^2))/((2
*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2])*(d + e*x))]*Sqrt[1 - (2*(c*d^2 - b*d*e +
a*e^2))/((2*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2])*(d + e*x))]*(EllipticE[I*Arc
Sinh[(Sqrt[2]*Sqrt[-((c*d^2 - b*d*e + a*e^2)/(2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c
*e^2]))])/Sqrt[d + e*x]], (2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2])/(2*c*d - b*e
+ Sqrt[b^2*e^2 - 4*a*c*e^2])] - EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[-((c*d^2 - b*
d*e + a*e^2)/(2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2]))])/Sqrt[d + e*x]], (2*c*d
- b*e - Sqrt[b^2*e^2 - 4*a*c*e^2])/(2*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2])]))
/(Sqrt[2]*(c*d^2 - b*d*e + a*e^2)*Sqrt[-((c*d^2 - b*d*e + a*e^2)/(2*c*d - b*e -
Sqrt[b^2*e^2 - 4*a*c*e^2]))]*Sqrt[c + (c*d^2 - b*d*e + a*e^2)/(d + e*x)^2 + (-2*
c*d + b*e)/(d + e*x)]) + (I*Sqrt[2]*c*Sqrt[1 - (2*(c*d^2 - b*d*e + a*e^2))/((2*c
*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2])*(d + e*x))]*Sqrt[1 - (2*(c*d^2 - b*d*e + a
*e^2))/((2*c*d - b*e + Sqrt[b^2*e^2 - 4*a*c*e^2])*(d + e*x))]*EllipticF[I*ArcSin
h[(Sqrt[2]*Sqrt[-((c*d^2 - b*d*e + a*e^2)/(2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^
2]))])/Sqrt[d + e*x]], (2*c*d - b*e - Sqrt[b^2*e^2 - 4*a*c*e^2])/(2*c*d - b*e +
Sqrt[b^2*e^2 - 4*a*c*e^2])])/(Sqrt[-((c*d^2 - b*d*e + a*e^2)/(2*c*d - b*e - Sqrt
[b^2*e^2 - 4*a*c*e^2]))]*Sqrt[c + (c*d^2 - b*d*e + a*e^2)/(d + e*x)^2 + (-2*c*d
+ b*e)/(d + e*x)])))/(c*Sqrt[((d + e*x)^2*(c*(-1 + d/(d + e*x))^2 + (e*(b - (b*d
)/(d + e*x) + (a*e)/(d + e*x)))/(d + e*x)))/e^2])))/(3*(b^2 - 4*a*c)*(a + x*(b +
c*x))^(5/2))

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Maple [B]  time = 0.152, size = 5517, normalized size = 11.2 \[ \text{output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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GIAC/XCAS [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out