3.110 \(\int \frac {(d+e x+f x^2)^2}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=316 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (48 c^2 \left (a^2 f^2+4 a b e f+b^2 \left (2 d f+e^2\right )\right )-40 b^2 c f (3 a f+2 b e)-64 c^3 \left (a \left (2 d f+e^2\right )+2 b d e\right )+35 b^4 f^2+128 c^4 d^2\right )}{128 c^{9/2}}+\frac {\sqrt {a+b x+c x^2} \left (-16 c^2 \left (16 a e f+9 b \left (2 d f+e^2\right )\right )+20 b c f (11 a f+12 b e)-105 b^3 f^2+384 c^3 d e\right )}{192 c^4}+\frac {x \sqrt {a+b x+c x^2} \left (-4 c f (9 a f+20 b e)+35 b^2 f^2+48 c^2 \left (2 d f+e^2\right )\right )}{96 c^3}+\frac {f x^2 \sqrt {a+b x+c x^2} (16 c e-7 b f)}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c} \]

[Out]

1/128*(128*c^4*d^2+35*b^4*f^2-40*b^2*c*f*(3*a*f+2*b*e)-64*c^3*(2*b*d*e+a*(2*d*f+e^2))+48*c^2*(4*a*b*e*f+a^2*f^
2+b^2*(2*d*f+e^2)))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(9/2)+1/192*(384*c^3*d*e-105*b^3*f^2+
20*b*c*f*(11*a*f+12*b*e)-16*c^2*(16*a*e*f+9*b*(2*d*f+e^2)))*(c*x^2+b*x+a)^(1/2)/c^4+1/96*(35*b^2*f^2-4*c*f*(9*
a*f+20*b*e)+48*c^2*(2*d*f+e^2))*x*(c*x^2+b*x+a)^(1/2)/c^3+1/24*f*(-7*b*f+16*c*e)*x^2*(c*x^2+b*x+a)^(1/2)/c^2+1
/4*f^2*x^3*(c*x^2+b*x+a)^(1/2)/c

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Rubi [A]  time = 0.63, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1661, 640, 621, 206} \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (48 c^2 \left (a^2 f^2+4 a b e f+b^2 \left (2 d f+e^2\right )\right )-40 b^2 c f (3 a f+2 b e)-64 c^3 \left (a \left (2 d f+e^2\right )+2 b d e\right )+35 b^4 f^2+128 c^4 d^2\right )}{128 c^{9/2}}+\frac {x \sqrt {a+b x+c x^2} \left (-4 c f (9 a f+20 b e)+35 b^2 f^2+48 c^2 \left (2 d f+e^2\right )\right )}{96 c^3}+\frac {\sqrt {a+b x+c x^2} \left (-16 c^2 \left (16 a e f+9 b \left (2 d f+e^2\right )\right )+20 b c f (11 a f+12 b e)-105 b^3 f^2+384 c^3 d e\right )}{192 c^4}+\frac {f x^2 \sqrt {a+b x+c x^2} (16 c e-7 b f)}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((384*c^3*d*e - 105*b^3*f^2 + 20*b*c*f*(12*b*e + 11*a*f) - 16*c^2*(16*a*e*f + 9*b*(e^2 + 2*d*f)))*Sqrt[a + b*x
+ c*x^2])/(192*c^4) + ((35*b^2*f^2 - 4*c*f*(20*b*e + 9*a*f) + 48*c^2*(e^2 + 2*d*f))*x*Sqrt[a + b*x + c*x^2])/
(96*c^3) + (f*(16*c*e - 7*b*f)*x^2*Sqrt[a + b*x + c*x^2])/(24*c^2) + (f^2*x^3*Sqrt[a + b*x + c*x^2])/(4*c) + (
(128*c^4*d^2 + 35*b^4*f^2 - 40*b^2*c*f*(2*b*e + 3*a*f) - 64*c^3*(2*b*d*e + a*(e^2 + 2*d*f)) + 48*c^2*(4*a*b*e*
f + a^2*f^2 + b^2*(e^2 + 2*d*f)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(9/2))

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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (d+e x+f x^2\right )^2}{\sqrt {a+b x+c x^2}} \, dx &=\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {4 c d^2+8 c d e x-\left (3 a f^2-4 c \left (e^2+2 d f\right )\right ) x^2+\frac {1}{2} f (16 c e-7 b f) x^3}{\sqrt {a+b x+c x^2}} \, dx}{4 c}\\ &=\frac {f (16 c e-7 b f) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {12 c^2 d^2+\left (24 c^2 d e-16 a c e f+7 a b f^2\right ) x+\frac {1}{4} \left (35 b^2 f^2-4 c f (20 b e+9 a f)+48 c^2 \left (e^2+2 d f\right )\right ) x^2}{\sqrt {a+b x+c x^2}} \, dx}{12 c^2}\\ &=\frac {\left (35 b^2 f^2-4 c f (20 b e+9 a f)+48 c^2 \left (e^2+2 d f\right )\right ) x \sqrt {a+b x+c x^2}}{96 c^3}+\frac {f (16 c e-7 b f) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\int \frac {\frac {1}{4} \left (96 c^3 d^2-35 a b^2 f^2+4 a c f (20 b e+9 a f)-48 a c^2 \left (e^2+2 d f\right )\right )+\frac {1}{8} \left (384 c^3 d e-105 b^3 f^2+20 b c f (12 b e+11 a f)-16 c^2 \left (16 a e f+9 b \left (e^2+2 d f\right )\right )\right ) x}{\sqrt {a+b x+c x^2}} \, dx}{24 c^3}\\ &=\frac {\left (384 c^3 d e-105 b^3 f^2+20 b c f (12 b e+11 a f)-16 c^2 \left (16 a e f+9 b \left (e^2+2 d f\right )\right )\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^2 f^2-4 c f (20 b e+9 a f)+48 c^2 \left (e^2+2 d f\right )\right ) x \sqrt {a+b x+c x^2}}{96 c^3}+\frac {f (16 c e-7 b f) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\left (128 c^4 d^2+35 b^4 f^2-40 b^2 c f (2 b e+3 a f)-64 c^3 \left (2 b d e+a \left (e^2+2 d f\right )\right )+48 c^2 \left (4 a b e f+a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^4}\\ &=\frac {\left (384 c^3 d e-105 b^3 f^2+20 b c f (12 b e+11 a f)-16 c^2 \left (16 a e f+9 b \left (e^2+2 d f\right )\right )\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^2 f^2-4 c f (20 b e+9 a f)+48 c^2 \left (e^2+2 d f\right )\right ) x \sqrt {a+b x+c x^2}}{96 c^3}+\frac {f (16 c e-7 b f) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\left (128 c^4 d^2+35 b^4 f^2-40 b^2 c f (2 b e+3 a f)-64 c^3 \left (2 b d e+a \left (e^2+2 d f\right )\right )+48 c^2 \left (4 a b e f+a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\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 )}{64 c^4}\\ &=\frac {\left (384 c^3 d e-105 b^3 f^2+20 b c f (12 b e+11 a f)-16 c^2 \left (16 a e f+9 b \left (e^2+2 d f\right )\right )\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^2 f^2-4 c f (20 b e+9 a f)+48 c^2 \left (e^2+2 d f\right )\right ) x \sqrt {a+b x+c x^2}}{96 c^3}+\frac {f (16 c e-7 b f) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {f^2 x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\left (128 c^4 d^2+35 b^4 f^2-40 b^2 c f (2 b e+3 a f)-64 c^3 \left (2 b d e+a \left (e^2+2 d f\right )\right )+48 c^2 \left (4 a b e f+a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 251, normalized size = 0.79 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) \left (48 c^2 \left (a^2 f^2+4 a b e f+b^2 \left (2 d f+e^2\right )\right )-40 b^2 c f (3 a f+2 b e)-64 c^3 \left (a \left (2 d f+e^2\right )+2 b d e\right )+35 b^4 f^2+128 c^4 d^2\right )}{128 c^{9/2}}+\frac {\sqrt {a+x (b+c x)} \left (-8 c^2 \left (a f (32 e+9 f x)+b \left (36 d f+18 e^2+20 e f x+7 f^2 x^2\right )\right )+10 b c f (22 a f+24 b e+7 b f x)-105 b^3 f^2+16 c^3 \left (12 d (2 e+f x)+x \left (6 e^2+8 e f x+3 f^2 x^2\right )\right )\right )}{192 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*(-105*b^3*f^2 + 10*b*c*f*(24*b*e + 22*a*f + 7*b*f*x) + 16*c^3*(12*d*(2*e + f*x) + x*(6*
e^2 + 8*e*f*x + 3*f^2*x^2)) - 8*c^2*(a*f*(32*e + 9*f*x) + b*(18*e^2 + 36*d*f + 20*e*f*x + 7*f^2*x^2))))/(192*c
^4) + ((128*c^4*d^2 + 35*b^4*f^2 - 40*b^2*c*f*(2*b*e + 3*a*f) - 64*c^3*(2*b*d*e + a*(e^2 + 2*d*f)) + 48*c^2*(4
*a*b*e*f + a^2*f^2 + b^2*(e^2 + 2*d*f)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(128*c^(9/2))

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fricas [A]  time = 1.44, size = 637, normalized size = 2.02 \[ \left [\frac {3 \, {\left (128 \, c^{4} d^{2} - 128 \, b c^{3} d e + 16 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} e^{2} + {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} f^{2} + 16 \, {\left (2 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d - {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} e\right )} f\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} f^{2} x^{3} + 384 \, c^{4} d e - 144 \, b c^{3} e^{2} - 5 \, {\left (21 \, b^{3} c - 44 \, a b c^{2}\right )} f^{2} + 8 \, {\left (16 \, c^{4} e f - 7 \, b c^{3} f^{2}\right )} x^{2} - 16 \, {\left (18 \, b c^{3} d - {\left (15 \, b^{2} c^{2} - 16 \, a c^{3}\right )} e\right )} f + 2 \, {\left (48 \, c^{4} e^{2} + {\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} f^{2} + 16 \, {\left (6 \, c^{4} d - 5 \, b c^{3} e\right )} f\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{5}}, -\frac {3 \, {\left (128 \, c^{4} d^{2} - 128 \, b c^{3} d e + 16 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} e^{2} + {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} f^{2} + 16 \, {\left (2 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d - {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} e\right )} f\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (48 \, c^{4} f^{2} x^{3} + 384 \, c^{4} d e - 144 \, b c^{3} e^{2} - 5 \, {\left (21 \, b^{3} c - 44 \, a b c^{2}\right )} f^{2} + 8 \, {\left (16 \, c^{4} e f - 7 \, b c^{3} f^{2}\right )} x^{2} - 16 \, {\left (18 \, b c^{3} d - {\left (15 \, b^{2} c^{2} - 16 \, a c^{3}\right )} e\right )} f + 2 \, {\left (48 \, c^{4} e^{2} + {\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} f^{2} + 16 \, {\left (6 \, c^{4} d - 5 \, b c^{3} e\right )} f\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{5}}\right ] \]

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)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(128*c^4*d^2 - 128*b*c^3*d*e + 16*(3*b^2*c^2 - 4*a*c^3)*e^2 + (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*f^
2 + 16*(2*(3*b^2*c^2 - 4*a*c^3)*d - (5*b^3*c - 12*a*b*c^2)*e)*f)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sq
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(48*c^4*f^2*x^3 + 384*c^4*d*e - 144*b*c^3*e^2 - 5*(21*b^3
*c - 44*a*b*c^2)*f^2 + 8*(16*c^4*e*f - 7*b*c^3*f^2)*x^2 - 16*(18*b*c^3*d - (15*b^2*c^2 - 16*a*c^3)*e)*f + 2*(4
8*c^4*e^2 + (35*b^2*c^2 - 36*a*c^3)*f^2 + 16*(6*c^4*d - 5*b*c^3*e)*f)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/384*(3
*(128*c^4*d^2 - 128*b*c^3*d*e + 16*(3*b^2*c^2 - 4*a*c^3)*e^2 + (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*f^2 + 16*(2
*(3*b^2*c^2 - 4*a*c^3)*d - (5*b^3*c - 12*a*b*c^2)*e)*f)*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*(48*c^4*f^2*x^3 + 384*c^4*d*e - 144*b*c^3*e^2 - 5*(21*b^3*c - 44*a*b*c^2
)*f^2 + 8*(16*c^4*e*f - 7*b*c^3*f^2)*x^2 - 16*(18*b*c^3*d - (15*b^2*c^2 - 16*a*c^3)*e)*f + 2*(48*c^4*e^2 + (35
*b^2*c^2 - 36*a*c^3)*f^2 + 16*(6*c^4*d - 5*b*c^3*e)*f)*x)*sqrt(c*x^2 + b*x + a))/c^5]

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giac [A]  time = 0.63, size = 304, normalized size = 0.96 \[ \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (\frac {6 \, f^{2} x}{c} - \frac {7 \, b c^{2} f^{2} - 16 \, c^{3} f e}{c^{4}}\right )} x + \frac {96 \, c^{3} d f + 35 \, b^{2} c f^{2} - 36 \, a c^{2} f^{2} - 80 \, b c^{2} f e + 48 \, c^{3} e^{2}}{c^{4}}\right )} x - \frac {288 \, b c^{2} d f + 105 \, b^{3} f^{2} - 220 \, a b c f^{2} - 384 \, c^{3} d e - 240 \, b^{2} c f e + 256 \, a c^{2} f e + 144 \, b c^{2} e^{2}}{c^{4}}\right )} - \frac {{\left (128 \, c^{4} d^{2} + 96 \, b^{2} c^{2} d f - 128 \, a c^{3} d f + 35 \, b^{4} f^{2} - 120 \, a b^{2} c f^{2} + 48 \, a^{2} c^{2} f^{2} - 128 \, b c^{3} d e - 80 \, b^{3} c f e + 192 \, a b c^{2} f e + 48 \, b^{2} c^{2} e^{2} - 64 \, a c^{3} e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {9}{2}}} \]

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)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*f^2*x/c - (7*b*c^2*f^2 - 16*c^3*f*e)/c^4)*x + (96*c^3*d*f + 35*b^2*c*f^2
- 36*a*c^2*f^2 - 80*b*c^2*f*e + 48*c^3*e^2)/c^4)*x - (288*b*c^2*d*f + 105*b^3*f^2 - 220*a*b*c*f^2 - 384*c^3*d*
e - 240*b^2*c*f*e + 256*a*c^2*f*e + 144*b*c^2*e^2)/c^4) - 1/128*(128*c^4*d^2 + 96*b^2*c^2*d*f - 128*a*c^3*d*f
+ 35*b^4*f^2 - 120*a*b^2*c*f^2 + 48*a^2*c^2*f^2 - 128*b*c^3*d*e - 80*b^3*c*f*e + 192*a*b*c^2*f*e + 48*b^2*c^2*
e^2 - 64*a*c^3*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.01, size = 706, normalized size = 2.23 \[ \frac {\sqrt {c \,x^{2}+b x +a}\, f^{2} x^{3}}{4 c}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, b \,f^{2} x^{2}}{24 c^{2}}+\frac {2 \sqrt {c \,x^{2}+b x +a}\, e f \,x^{2}}{3 c}+\frac {3 a^{2} f^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}-\frac {15 a \,b^{2} f^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {3 a b e f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}-\frac {a d f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}-\frac {a \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}+\frac {35 b^{4} f^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {9}{2}}}-\frac {5 b^{3} e f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {7}{2}}}+\frac {3 b^{2} d f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {5}{2}}}+\frac {3 b^{2} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}-\frac {b d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, a \,f^{2} x}{8 c^{2}}+\frac {35 \sqrt {c \,x^{2}+b x +a}\, b^{2} f^{2} x}{96 c^{3}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, b e f x}{6 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, d f x}{c}+\frac {\sqrt {c \,x^{2}+b x +a}\, e^{2} x}{2 c}+\frac {55 \sqrt {c \,x^{2}+b x +a}\, a b \,f^{2}}{48 c^{3}}-\frac {4 \sqrt {c \,x^{2}+b x +a}\, a e f}{3 c^{2}}-\frac {35 \sqrt {c \,x^{2}+b x +a}\, b^{3} f^{2}}{64 c^{4}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} e f}{4 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b d f}{2 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b \,e^{2}}{4 c^{2}}+\frac {2 \sqrt {c \,x^{2}+b x +a}\, d e}{c} \]

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)^(1/2),x)

[Out]

2*d*e/c*(c*x^2+b*x+a)^(1/2)+1/2*x/c*(c*x^2+b*x+a)^(1/2)*e^2-3/4*b/c^2*(c*x^2+b*x+a)^(1/2)*e^2+3/8*b^2/c^(5/2)*
ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e^2-1/2*a/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e^2-
35/64*f^2*b^3/c^4*(c*x^2+b*x+a)^(1/2)+35/128*f^2*b^4/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/8*f
^2*a^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-5/6*e*f*b/c^2*x*(c*x^2+b*x+a)^(1/2)+1/4*f^2*x^3*(c*
x^2+b*x+a)^(1/2)/c+3/2*e*f*b/c^(5/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+d^2*ln((c*x+1/2*b)/c^(1/2)+
(c*x^2+b*x+a)^(1/2))/c^(1/2)-5/8*e*f*b^3/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-4/3*e*f*a/c^2*(c*
x^2+b*x+a)^(1/2)+x/c*(c*x^2+b*x+a)^(1/2)*d*f-3/2*b/c^2*(c*x^2+b*x+a)^(1/2)*d*f+3/4*b^2/c^(5/2)*ln((c*x+1/2*b)/
c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*f-a/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*f-d*e*b/c^(3/2)*ln((c
*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/8*f^2*a/c^2*x*(c*x^2+b*x+a)^(1/2)+55/48*f^2*b/c^3*a*(c*x^2+b*x+a)^(1/
2)-7/24*f^2*b/c^2*x^2*(c*x^2+b*x+a)^(1/2)+35/96*f^2*b^2/c^3*x*(c*x^2+b*x+a)^(1/2)-15/16*f^2*b^2/c^(7/2)*a*ln((
c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2/3*e*f*x^2/c*(c*x^2+b*x+a)^(1/2)+5/4*e*f*b^2/c^3*(c*x^2+b*x+a)^(1/2)

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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)^(1/2),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 \[ \int \frac {{\left (f\,x^2+e\,x+d\right )}^2}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

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)^(1/2),x)

[Out]

int((d + e*x + f*x^2)^2/(a + b*x + c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x + f x^{2}\right )^{2}}{\sqrt {a + b x + c x^{2}}}\, dx \]

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)**(1/2),x)

[Out]

Integral((d + e*x + f*x**2)**2/sqrt(a + b*x + c*x**2), x)

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