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

Optimal. Leaf size=444 \[ \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (-2 c x \left (-c^2 \left (16 a^2 f^2+12 a b e f-\left (b^2 \left (2 d f+e^2\right )\right )\right )+b^2 c f (14 a f+b e)-c^3 \left (8 b d e-4 a \left (2 d f+e^2\right )\right )-2 b^4 f^2+8 c^4 d^2\right )-4 b c^2 \left (8 a^2 f^2+a c \left (2 d f+e^2\right )+2 c^2 d^2\right )+48 a^2 c^3 e f+b^3 c \left (10 a f^2-c \left (2 d f+e^2\right )\right )+4 b^2 c^2 e (2 c d-3 a f)+b^5 \left (-f^2\right )+2 b^4 c e f\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {f^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

2/3*(2*a*b^2*c*e*f-a*b^3*f^2+4*a*c^2*e*(-a*f+c*d)-b*c*(c^2*d^2-3*a^2*f^2+a*c*(2*d*f+e^2))-(2*c^4*d^2+b^4*f^2-2
*b^2*c*f*(2*a*f+b*e)-2*c^3*(b*d*e+a*(2*d*f+e^2))+c^2*(6*a*b*e*f+2*a^2*f^2+b^2*(2*d*f+e^2)))*x)/c^3/(-4*a*c+b^2
)/(c*x^2+b*x+a)^(3/2)+f^2*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)-2/3*(2*b^4*c*e*f+48*a^2*c
^3*e*f-b^5*f^2+4*b^2*c^2*e*(-3*a*f+2*c*d)+b^3*c*(10*a*f^2-c*(2*d*f+e^2))-4*b*c^2*(2*c^2*d^2+8*a^2*f^2+a*c*(2*d
*f+e^2))-2*c*(8*c^4*d^2-2*b^4*f^2+b^2*c*f*(14*a*f+b*e)-c^3*(8*b*d*e-4*a*(2*d*f+e^2))-c^2*(12*a*b*e*f+16*a^2*f^
2-b^2*(2*d*f+e^2)))*x)/c^3/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.45, antiderivative size = 444, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1660, 12, 621, 206} \[ -\frac {2 \left (-2 c x \left (-c^2 \left (16 a^2 f^2+12 a b e f+b^2 \left (-\left (2 d f+e^2\right )\right )\right )+b^2 c f (14 a f+b e)-c^3 \left (8 b d e-4 a \left (2 d f+e^2\right )\right )-2 b^4 f^2+8 c^4 d^2\right )-4 b c^2 \left (8 a^2 f^2+a c \left (2 d f+e^2\right )+2 c^2 d^2\right )+48 a^2 c^3 e f+4 b^2 c^2 e (2 c d-3 a f)+b^3 c \left (10 a f^2-c \left (2 d f+e^2\right )\right )+2 b^4 c e f+b^5 \left (-f^2\right )\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )+2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {f^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*a*b^2*c*e*f - a*b^3*f^2 + 4*a*c^2*e*(c*d - a*f) - b*c*(c^2*d^2 - 3*a^2*f^2 + a*c*(e^2 + 2*d*f)) - (2*c^4
*d^2 + b^4*f^2 - 2*b^2*c*f*(b*e + 2*a*f) - 2*c^3*(b*d*e + a*(e^2 + 2*d*f)) + c^2*(6*a*b*e*f + 2*a^2*f^2 + b^2*
(e^2 + 2*d*f)))*x))/(3*c^3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (2*(2*b^4*c*e*f + 48*a^2*c^3*e*f - b^5*f^2
+ 4*b^2*c^2*e*(2*c*d - 3*a*f) + b^3*c*(10*a*f^2 - c*(e^2 + 2*d*f)) - 4*b*c^2*(2*c^2*d^2 + 8*a^2*f^2 + a*c*(e^
2 + 2*d*f)) - 2*c*(8*c^4*d^2 - 2*b^4*f^2 + b^2*c*f*(b*e + 14*a*f) - c^3*(8*b*d*e - 4*a*(e^2 + 2*d*f)) - c^2*(1
2*a*b*e*f + 16*a^2*f^2 - b^2*(e^2 + 2*d*f)))*x))/(3*c^3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]) + (f^2*ArcTanh[
(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(5/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=\frac {2 \left (2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)-b c \left (c^2 d^2-3 a^2 f^2+a c \left (e^2+2 d f\right )\right )-\left (2 c^4 d^2+b^4 f^2-2 b^2 c f (b e+2 a f)-2 c^3 \left (b d e+a \left (e^2+2 d f\right )\right )+c^2 \left (6 a b e f+2 a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {\frac {8 c^4 d^2+b^4 f^2-b^2 c f (2 b e+a f)-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (4 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )}{2 c^3}-\frac {3 \left (b^2-4 a c\right ) f (2 c e-b f) x}{2 c^2}+\frac {3}{2} \left (4 a-\frac {b^2}{c}\right ) f^2 x^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=\frac {2 \left (2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)-b c \left (c^2 d^2-3 a^2 f^2+a c \left (e^2+2 d f\right )\right )-\left (2 c^4 d^2+b^4 f^2-2 b^2 c f (b e+2 a f)-2 c^3 \left (b d e+a \left (e^2+2 d f\right )\right )+c^2 \left (6 a b e f+2 a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 b^4 c e f+48 a^2 c^3 e f-b^5 f^2+4 b^2 c^2 e (2 c d-3 a f)+b^3 c \left (10 a f^2-c \left (e^2+2 d f\right )\right )-4 b c^2 \left (2 c^2 d^2+8 a^2 f^2+a c \left (e^2+2 d f\right )\right )-2 c \left (8 c^4 d^2-2 b^4 f^2+b^2 c f (b e+14 a f)-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (12 a b e f+16 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {4 \int \frac {3 \left (b^2-4 a c\right )^2 f^2}{4 c^2 \sqrt {a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )^2}\\ &=\frac {2 \left (2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)-b c \left (c^2 d^2-3 a^2 f^2+a c \left (e^2+2 d f\right )\right )-\left (2 c^4 d^2+b^4 f^2-2 b^2 c f (b e+2 a f)-2 c^3 \left (b d e+a \left (e^2+2 d f\right )\right )+c^2 \left (6 a b e f+2 a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 b^4 c e f+48 a^2 c^3 e f-b^5 f^2+4 b^2 c^2 e (2 c d-3 a f)+b^3 c \left (10 a f^2-c \left (e^2+2 d f\right )\right )-4 b c^2 \left (2 c^2 d^2+8 a^2 f^2+a c \left (e^2+2 d f\right )\right )-2 c \left (8 c^4 d^2-2 b^4 f^2+b^2 c f (b e+14 a f)-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (12 a b e f+16 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {f^2 \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c^2}\\ &=\frac {2 \left (2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)-b c \left (c^2 d^2-3 a^2 f^2+a c \left (e^2+2 d f\right )\right )-\left (2 c^4 d^2+b^4 f^2-2 b^2 c f (b e+2 a f)-2 c^3 \left (b d e+a \left (e^2+2 d f\right )\right )+c^2 \left (6 a b e f+2 a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 b^4 c e f+48 a^2 c^3 e f-b^5 f^2+4 b^2 c^2 e (2 c d-3 a f)+b^3 c \left (10 a f^2-c \left (e^2+2 d f\right )\right )-4 b c^2 \left (2 c^2 d^2+8 a^2 f^2+a c \left (e^2+2 d f\right )\right )-2 c \left (8 c^4 d^2-2 b^4 f^2+b^2 c f (b e+14 a f)-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (12 a b e f+16 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{c^2}\\ &=\frac {2 \left (2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)-b c \left (c^2 d^2-3 a^2 f^2+a c \left (e^2+2 d f\right )\right )-\left (2 c^4 d^2+b^4 f^2-2 b^2 c f (b e+2 a f)-2 c^3 \left (b d e+a \left (e^2+2 d f\right )\right )+c^2 \left (6 a b e f+2 a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 b^4 c e f+48 a^2 c^3 e f-b^5 f^2+4 b^2 c^2 e (2 c d-3 a f)+b^3 c \left (10 a f^2-c \left (e^2+2 d f\right )\right )-4 b c^2 \left (2 c^2 d^2+8 a^2 f^2+a c \left (e^2+2 d f\right )\right )-2 c \left (8 c^4 d^2-2 b^4 f^2+b^2 c f (b e+14 a f)-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (12 a b e f+16 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {f^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 1.26, size = 387, normalized size = 0.87 \[ \frac {2 \left (b^3 \left (-3 a^2 f^2+18 a c f^2 x^2+c^2 \left (-d^2+6 d x (f x-e)+e x^2 (3 e+2 f x)\right )\right )+2 b^2 c \left (21 a^2 f^2 x-2 a c \left (d (e-6 f x)+x \left (-3 e^2+3 e f x-7 f^2 x^2\right )\right )+c^2 x \left (3 d^2+2 d x (f x-6 e)+e^2 x^2\right )\right )+4 b c \left (5 a^3 f^2+2 a^2 c \left (2 d f+e^2-6 e f x\right )+3 a c^2 (d-e x) (d+x (2 f x-e))+2 c^3 d x^2 (3 d-2 e x)\right )+8 c^2 \left (a^3 (-f) (4 e+3 f x)-2 a^2 c \left (d e+f x^2 (3 e+2 f x)\right )+a c^2 x \left (3 d^2+2 d f x^2+e^2 x^2\right )+2 c^3 d^2 x^3\right )-2 b^4 f^2 x \left (3 a+2 c x^2\right )-3 b^5 f^2 x^2\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}+\frac {f^2 \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-3*b^5*f^2*x^2 - 2*b^4*f^2*x*(3*a + 2*c*x^2) + 4*b*c*(5*a^3*f^2 + 2*c^3*d*x^2*(3*d - 2*e*x) + 2*a^2*c*(e^2
+ 2*d*f - 6*e*f*x) + 3*a*c^2*(d - e*x)*(d + x*(-e + 2*f*x))) + b^3*(-3*a^2*f^2 + 18*a*c*f^2*x^2 + c^2*(-d^2 +
6*d*x*(-e + f*x) + e*x^2*(3*e + 2*f*x))) + 8*c^2*(2*c^3*d^2*x^3 - a^3*f*(4*e + 3*f*x) + a*c^2*x*(3*d^2 + e^2*
x^2 + 2*d*f*x^2) - 2*a^2*c*(d*e + f*x^2*(3*e + 2*f*x))) + 2*b^2*c*(21*a^2*f^2*x + c^2*x*(3*d^2 + e^2*x^2 + 2*d
*x*(-6*e + f*x)) - 2*a*c*(d*(e - 6*f*x) + x*(-3*e^2 + 3*e*f*x - 7*f^2*x^2)))))/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(
b + c*x))^(3/2)) + (f^2*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/c^(5/2)

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fricas [A]  time = 6.52, size = 1581, normalized size = 3.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*f^2*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*f^2*x^3 + (b^6
- 6*a*b^4*c + 32*a^3*c^3)*f^2*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*f^2*x + (a^2*b^4 - 8*a^3*b^2*c + 16
*a^4*c^2)*f^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) +
4*(8*a^2*b*c^3*e^2 + 2*(8*c^6*d^2 - 8*b*c^5*d*e + (b^2*c^4 + 4*a*c^5)*e^2 - 2*(b^4*c^2 - 7*a*b^2*c^3 + 8*a^2*
c^4)*f^2 + (2*(b^2*c^4 + 4*a*c^5)*d + (b^3*c^3 - 12*a*b*c^4)*e)*f)*x^3 - (b^3*c^3 - 12*a*b*c^4)*d^2 - 4*(a*b^2
*c^3 + 4*a^2*c^4)*d*e - (3*a^2*b^3*c - 20*a^3*b*c^2)*f^2 + 3*(8*b*c^5*d^2 - 8*b^2*c^4*d*e + (b^3*c^3 + 4*a*b*c
^4)*e^2 - (b^5*c - 6*a*b^3*c^2)*f^2 + 2*((b^3*c^3 + 4*a*b*c^4)*d - 2*(a*b^2*c^3 + 4*a^2*c^4)*e)*f)*x^2 + 16*(a
^2*b*c^3*d - 2*a^3*c^3*e)*f + 6*(2*a*b^2*c^3*e^2 + (b^2*c^4 + 4*a*c^5)*d^2 - (b^3*c^3 + 4*a*b*c^4)*d*e - (a*b^
4*c - 7*a^2*b^2*c^2 + 4*a^3*c^3)*f^2 + 4*(a*b^2*c^3*d - 2*a^2*b*c^3*e)*f)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c
^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2
*b*c^6)*x^3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^4 + 16*a^3*b*c^5)*x), -1/3
*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*f^2*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*f^2*x^3 + (b^6 - 6*
a*b^4*c + 32*a^3*c^3)*f^2*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*f^2*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4
*c^2)*f^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(8*a^2*
b*c^3*e^2 + 2*(8*c^6*d^2 - 8*b*c^5*d*e + (b^2*c^4 + 4*a*c^5)*e^2 - 2*(b^4*c^2 - 7*a*b^2*c^3 + 8*a^2*c^4)*f^2 +
(2*(b^2*c^4 + 4*a*c^5)*d + (b^3*c^3 - 12*a*b*c^4)*e)*f)*x^3 - (b^3*c^3 - 12*a*b*c^4)*d^2 - 4*(a*b^2*c^3 + 4*a
^2*c^4)*d*e - (3*a^2*b^3*c - 20*a^3*b*c^2)*f^2 + 3*(8*b*c^5*d^2 - 8*b^2*c^4*d*e + (b^3*c^3 + 4*a*b*c^4)*e^2 -
(b^5*c - 6*a*b^3*c^2)*f^2 + 2*((b^3*c^3 + 4*a*b*c^4)*d - 2*(a*b^2*c^3 + 4*a^2*c^4)*e)*f)*x^2 + 16*(a^2*b*c^3*d
- 2*a^3*c^3*e)*f + 6*(2*a*b^2*c^3*e^2 + (b^2*c^4 + 4*a*c^5)*d^2 - (b^3*c^3 + 4*a*b*c^4)*d*e - (a*b^4*c - 7*a^
2*b^2*c^2 + 4*a^3*c^3)*f^2 + 4*(a*b^2*c^3*d - 2*a^2*b*c^3*e)*f)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3
*b^2*c^4 + 16*a^4*c^5 + (b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*x^
3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^4 + 16*a^3*b*c^5)*x)]

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giac [A]  time = 0.34, size = 587, normalized size = 1.32 \[ -\frac {f^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (8 \, c^{5} d^{2} + 2 \, b^{2} c^{3} d f + 8 \, a c^{4} d f - 2 \, b^{4} c f^{2} + 14 \, a b^{2} c^{2} f^{2} - 16 \, a^{2} c^{3} f^{2} - 8 \, b c^{4} d e + b^{3} c^{2} f e - 12 \, a b c^{3} f e + b^{2} c^{3} e^{2} + 4 \, a c^{4} e^{2}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac {3 \, {\left (8 \, b c^{4} d^{2} + 2 \, b^{3} c^{2} d f + 8 \, a b c^{3} d f - b^{5} f^{2} + 6 \, a b^{3} c f^{2} - 8 \, b^{2} c^{3} d e - 4 \, a b^{2} c^{2} f e - 16 \, a^{2} c^{3} f e + b^{3} c^{2} e^{2} + 4 \, a b c^{3} e^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {6 \, {\left (b^{2} c^{3} d^{2} + 4 \, a c^{4} d^{2} + 4 \, a b^{2} c^{2} d f - a b^{4} f^{2} + 7 \, a^{2} b^{2} c f^{2} - 4 \, a^{3} c^{2} f^{2} - b^{3} c^{2} d e - 4 \, a b c^{3} d e - 8 \, a^{2} b c^{2} f e + 2 \, a b^{2} c^{2} e^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {b^{3} c^{2} d^{2} - 12 \, a b c^{3} d^{2} - 16 \, a^{2} b c^{2} d f + 3 \, a^{2} b^{3} f^{2} - 20 \, a^{3} b c f^{2} + 4 \, a b^{2} c^{2} d e + 16 \, a^{2} c^{3} d e + 32 \, a^{3} c^{2} f e - 8 \, a^{2} b c^{2} e^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-f^2*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2) + 2/3*(((2*(8*c^5*d^2 + 2*b^2*c^3*d*
f + 8*a*c^4*d*f - 2*b^4*c*f^2 + 14*a*b^2*c^2*f^2 - 16*a^2*c^3*f^2 - 8*b*c^4*d*e + b^3*c^2*f*e - 12*a*b*c^3*f*e
+ b^2*c^3*e^2 + 4*a*c^4*e^2)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(8*b*c^4*d^2 + 2*b^3*c^2*d*f + 8*a*b*
c^3*d*f - b^5*f^2 + 6*a*b^3*c*f^2 - 8*b^2*c^3*d*e - 4*a*b^2*c^2*f*e - 16*a^2*c^3*f*e + b^3*c^2*e^2 + 4*a*b*c^3
*e^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 6*(b^2*c^3*d^2 + 4*a*c^4*d^2 + 4*a*b^2*c^2*d*f - a*b^4*f^2 + 7
*a^2*b^2*c*f^2 - 4*a^3*c^2*f^2 - b^3*c^2*d*e - 4*a*b*c^3*d*e - 8*a^2*b*c^2*f*e + 2*a*b^2*c^2*e^2)/(b^4*c^2 - 8
*a*b^2*c^3 + 16*a^2*c^4))*x - (b^3*c^2*d^2 - 12*a*b*c^3*d^2 - 16*a^2*b*c^2*d*f + 3*a^2*b^3*f^2 - 20*a^3*b*c*f^
2 + 4*a*b^2*c^2*d*e + 16*a^2*c^3*d*e + 32*a^3*c^2*f*e - 8*a^2*b*c^2*e^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))
/(c*x^2 + b*x + a)^(3/2)

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maple [B]  time = 0.01, size = 1786, normalized size = 4.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*d*e/c/(c*x^2+b*x+a)^(3/2)+2/3*d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*b-1/2*x/c/(c*x^2+b*x+a)^(3/2)*e^2+1/12*
b/c^2/(c*x^2+b*x+a)^(3/2)*e^2-1/3*f^2*x^3/c/(c*x^2+b*x+a)^(3/2)-1/48*f^2*b^3/c^4/(c*x^2+b*x+a)^(3/2)-f^2/c^2*x
/(c*x^2+b*x+a)^(1/2)+1/2*f^2/c^3*b/(c*x^2+b*x+a)^(1/2)-2*e*f*b/c*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+16/3*a/(4
*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*b*d*f-1/24*f^2*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-1/3*f^2*b^4/c^2/(4*a*
c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+1/4*f^2*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+2*f^2*b^3/c^2*a/(4*a*c-b^2)^2
/(c*x^2+b*x+a)^(1/2)+f^2/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-4/3*d*e*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x
-2/3*d*e*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-1/2*e*f*b/c^2*x/(c*x^2+b*x+a)^(3/2)+1/12*e*f*b^4/c^3/(4*a*c-b^2
)/(c*x^2+b*x+a)^(3/2)+2/3*e*f*b^4/c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/6*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(3
/2)*x*e^2+1/6*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*d*f+8/3*b^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*d*f+4/3*
b^3/c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*d*f+4/3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x*d*f+1/3*a/c/(4*a*c-b^2)/(c
*x^2+b*x+a)^(3/2)*b*e^2+16/3*a*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*e^2+f^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c
*x^2+b*x+a)^(1/2))-2*e*f*x^2/c/(c*x^2+b*x+a)^(3/2)+1/12*e*f*b^2/c^3/(c*x^2+b*x+a)^(3/2)-4/3*e*f*a/c^2/(c*x^2+b
*x+a)^(3/2)+1/2*f^2*b/c^2*x^2/(c*x^2+b*x+a)^(3/2)+1/8*f^2*b^2/c^3*x/(c*x^2+b*x+a)^(3/2)-1/48*f^2*b^5/c^4/(4*a*
c-b^2)/(c*x^2+b*x+a)^(3/2)-1/6*f^2*b^5/c^3/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/3*f^2*b/c^3*a/(c*x^2+b*x+a)^(3/
2)+1/2*f^2/c^3*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+2/3*b^3/c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*e^2+2/3*a/(4*a*
c-b^2)/(c*x^2+b*x+a)^(3/2)*x*e^2+8/3*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*b*e^2-16/3*d*e*b^2/(4*a*c-b^2)^2/(c*x
^2+b*x+a)^(1/2)+4/3*d^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x*c+32/3*d^2*c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x+1
6/3*d^2*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*b-x/c/(c*x^2+b*x+a)^(3/2)*d*f+1/6*b/c^2/(c*x^2+b*x+a)^(3/2)*d*f+32
/3*a*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*d*f-32/3*d*e*b*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-16*e*f*b*a/(4*
a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-8*e*f*b^2/c*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/6*e*f*b^3/c^2/(4*a*c-b^2)/(
c*x^2+b*x+a)^(3/2)*x+1/2*f^2*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+4*f^2*b^2/c*a/(4*a*c-b^2)^2/(c*x^2+b*
x+a)^(1/2)*x+1/3*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x*d*f+2/3*a/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*b*d*f+1/1
2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*e^2+4/3*b^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*e^2+4/3*e*f*b^3/c/(4
*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-e*f*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (f\,x^2+e\,x+d\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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