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

Optimal. Leaf size=191 \[ -\frac {\left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right )}{64 c^3}+\frac {5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c} \]

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Rubi [A]  time = 0.23, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {742, 640, 612, 621, 206} \begin {gather*} \frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right )}{64 c^3}-\frac {\left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}+\frac {5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^3) + (5*e*(2*c*d - b*
e)*(a + b*x + c*x^2)^(3/2))/(24*c^2) + (e*(d + e*x)*(a + b*x + c*x^2)^(3/2))/(4*c) - ((b^2 - 4*a*c)*(16*c^2*d^
2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(7/2))

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 612

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

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 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rubi steps

\begin {align*} \int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx &=\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}+\frac {\int \left (\frac {1}{2} \left (8 c d^2-2 e \left (\frac {3 b d}{2}+a e\right )\right )+\frac {5}{2} e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2} \, dx}{4 c}\\ &=\frac {5 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}+\frac {\left (-\frac {5}{2} b e (2 c d-b e)+c \left (8 c d^2-2 e \left (\frac {3 b d}{2}+a e\right )\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{8 c^2}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {5 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (\left (b^2-4 a c\right ) \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {5 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (\left (b^2-4 a c\right ) \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\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^3}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {5 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2-4 a c\right ) \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 162, normalized size = 0.85 \begin {gather*} \frac {\frac {\left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{32 c^{5/2}}+\frac {5 e (a+x (b+c x))^{3/2} (2 c d-b e)}{6 c}+e (d+e x) (a+x (b+c x))^{3/2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*e*(2*c*d - b*e)*(a + x*(b + c*x))^(3/2))/(6*c) + e*(d + e*x)*(a + x*(b + c*x))^(3/2) + ((16*c^2*d^2 + 5*b^
2*e^2 - 4*c*e*(4*b*d + a*e))*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/
(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(32*c^(5/2)))/(4*c)

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IntegrateAlgebraic [A]  time = 0.80, size = 239, normalized size = 1.25 \begin {gather*} \frac {\left (16 a^2 c^2 e^2-24 a b^2 c e^2+64 a b c^2 d e-64 a c^3 d^2+5 b^4 e^2-16 b^3 c d e+16 b^2 c^2 d^2\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{128 c^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (-52 a b c e^2+128 a c^2 d e+24 a c^2 e^2 x+15 b^3 e^2-48 b^2 c d e-10 b^2 c e^2 x+48 b c^2 d^2+32 b c^2 d e x+8 b c^2 e^2 x^2+96 c^3 d^2 x+128 c^3 d e x^2+48 c^3 e^2 x^3\right )}{192 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(48*b*c^2*d^2 - 48*b^2*c*d*e + 128*a*c^2*d*e + 15*b^3*e^2 - 52*a*b*c*e^2 + 96*c^3*d^2*x
+ 32*b*c^2*d*e*x - 10*b^2*c*e^2*x + 24*a*c^2*e^2*x + 128*c^3*d*e*x^2 + 8*b*c^2*e^2*x^2 + 48*c^3*e^2*x^3))/(19
2*c^3) + ((16*b^2*c^2*d^2 - 64*a*c^3*d^2 - 16*b^3*c*d*e + 64*a*b*c^2*d*e + 5*b^4*e^2 - 24*a*b^2*c*e^2 + 16*a^2
*c^2*e^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(128*c^(7/2))

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fricas [A]  time = 0.45, size = 505, normalized size = 2.64 \begin {gather*} \left [\frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} - 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{2}\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} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 16 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} e^{2} + 8 \, {\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} - 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{2}\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} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 16 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} e^{2} + 8 \, {\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(16*(b^2*c^2 - 4*a*c^3)*d^2 - 16*(b^3*c - 4*a*b*c^2)*d*e + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*e^2)*sq
rt(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*(48*c^4*e^2*x^
3 + 48*b*c^3*d^2 - 16*(3*b^2*c^2 - 8*a*c^3)*d*e + (15*b^3*c - 52*a*b*c^2)*e^2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2
+ 2*(48*c^4*d^2 + 16*b*c^3*d*e - (5*b^2*c^2 - 12*a*c^3)*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/384*(3*(16*(b^2
*c^2 - 4*a*c^3)*d^2 - 16*(b^3*c - 4*a*b*c^2)*d*e + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*e^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*(48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 16*(3
*b^2*c^2 - 8*a*c^3)*d*e + (15*b^3*c - 52*a*b*c^2)*e^2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2 + 2*(48*c^4*d^2 + 16*b*
c^3*d*e - (5*b^2*c^2 - 12*a*c^3)*e^2)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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giac [A]  time = 0.29, size = 235, normalized size = 1.23 \begin {gather*} \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, x e^{2} + \frac {16 \, c^{3} d e + b c^{2} e^{2}}{c^{3}}\right )} x + \frac {48 \, c^{3} d^{2} + 16 \, b c^{2} d e - 5 \, b^{2} c e^{2} + 12 \, a c^{2} e^{2}}{c^{3}}\right )} x + \frac {48 \, b c^{2} d^{2} - 48 \, b^{2} c d e + 128 \, a c^{2} d e + 15 \, b^{3} e^{2} - 52 \, a b c e^{2}}{c^{3}}\right )} + \frac {{\left (16 \, b^{2} c^{2} d^{2} - 64 \, a c^{3} d^{2} - 16 \, b^{3} c d e + 64 \, a b c^{2} d e + 5 \, b^{4} e^{2} - 24 \, a b^{2} c e^{2} + 16 \, a^{2} c^{2} 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 {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*x*e^2 + (16*c^3*d*e + b*c^2*e^2)/c^3)*x + (48*c^3*d^2 + 16*b*c^2*d*e - 5*
b^2*c*e^2 + 12*a*c^2*e^2)/c^3)*x + (48*b*c^2*d^2 - 48*b^2*c*d*e + 128*a*c^2*d*e + 15*b^3*e^2 - 52*a*b*c*e^2)/c
^3) + 1/128*(16*b^2*c^2*d^2 - 64*a*c^3*d^2 - 16*b^3*c*d*e + 64*a*b*c^2*d*e + 5*b^4*e^2 - 24*a*b^2*c*e^2 + 16*a
^2*c^2*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.09, size = 484, normalized size = 2.53 \begin {gather*} -\frac {a^{2} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}+\frac {3 a \,b^{2} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {a b d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}+\frac {a \,d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\frac {5 b^{4} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}+\frac {b^{3} d e \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^{2} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a \,e^{2} x}{8 c}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} e^{2} x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, b d e x}{2 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, d^{2} x}{2}-\frac {\sqrt {c \,x^{2}+b x +a}\, a b \,e^{2}}{16 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{3} e^{2}}{64 c^{3}}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{2} d e}{4 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, b \,d^{2}}{4 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} e^{2} x}{4 c}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,e^{2}}{24 c^{2}}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d e}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^2*x*(c*x^2+b*x+a)^(3/2)/c-5/24*e^2*b/c^2*(c*x^2+b*x+a)^(3/2)+5/32*e^2*b^2/c^2*x*(c*x^2+b*x+a)^(1/2)+5/64
*e^2*b^3/c^3*(c*x^2+b*x+a)^(1/2)+3/16*e^2*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-5/128*e^2*
b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/8*e^2*a/c*x*(c*x^2+b*x+a)^(1/2)-1/16*e^2*a/c^2*(c*x^
2+b*x+a)^(1/2)*b-1/8*e^2*a^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2/3*d*e*(c*x^2+b*x+a)^(3/2)/c
-1/2*d*e*b/c*x*(c*x^2+b*x+a)^(1/2)-1/4*d*e*b^2/c^2*(c*x^2+b*x+a)^(1/2)-1/2*d*e*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2
)+(c*x^2+b*x+a)^(1/2))*a+1/8*d*e*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*d^2*x*(c*x^2+b*x+
a)^(1/2)+1/4*d^2/c*(c*x^2+b*x+a)^(1/2)*b+1/2*d^2/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/8*d^2
/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 positive, negative or zero?

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mupad [B]  time = 1.57, size = 332, normalized size = 1.74 \begin {gather*} d^2\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {e^2\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}-\frac {a\,e^2\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}+\frac {d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}-\frac {5\,b\,e^2\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {d\,e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{12\,c^2}+\frac {d\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{8\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (e^2*x*(a + b*x + c*x^2)^(3/2))/(4*c) - (a*e^2*((x/2 + b/(4*c))*
(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*
c) + (d^2*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)) - (5*b*e^2*((log((b +
2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*
x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) + (d*e*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2
))/(12*c^2) + (d*e*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(8*c^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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