3.24.34 \(\int (d+e x)^2 \sqrt {a+b x+c x^2} \, dx\) [2334]

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

[Out]

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

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Rubi [A]
time = 0.16, 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 = {756, 654, 626, 635, 212} \begin {gather*} -\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} \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 212

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

Rule 626

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 635

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 654

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 756

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 ) \text {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 = 0.64, size = 192, normalized size = 1.01 \begin {gather*} \frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^3 e^2-2 b^2 c e (24 d+5 e x)+4 b c \left (-13 a e^2+2 c \left (6 d^2+4 d e x+e^2 x^2\right )\right )+8 c^2 \left (a e (16 d+3 e x)+2 c x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )\right )+3 \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 ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{384 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^3*e^2 - 2*b^2*c*e*(24*d + 5*e*x) + 4*b*c*(-13*a*e^2 + 2*c*(6*d^2 + 4*d*
e*x + e^2*x^2)) + 8*c^2*(a*e*(16*d + 3*e*x) + 2*c*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2))) + 3*(b^2 - 4*a*c)*(16*c^2*
d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(384*c^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(169)=338\).
time = 0.91, size = 349, normalized size = 1.83

method	result	size
default	\(e^{2} \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+2 d e \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )+d^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )\)	\(349\)
risch	\(-\frac {\left (-48 e^{2} c^{3} x^{3}-8 e^{2} b \,c^{2} x^{2}-128 d e \,c^{3} x^{2}-24 a \,c^{2} e^{2} x +10 b^{2} e^{2} c x -32 d e b \,c^{2} x -96 c^{3} d^{2} x +52 a b c \,e^{2}-128 a \,c^{2} d e -15 b^{3} e^{2}+48 b^{2} d c e -48 d^{2} b \,c^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{3}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{2} e^{2}}{8 c^{\frac {3}{2}}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,b^{2} e^{2}}{16 c^{\frac {5}{2}}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a b d e}{2 c^{\frac {3}{2}}}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,d^{2}}{2 \sqrt {c}}-\frac {5 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{4} e^{2}}{128 c^{\frac {7}{2}}}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{3} d e}{8 c^{\frac {5}{2}}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{2} d^{2}}{8 c^{\frac {3}{2}}}\)	\(380\)

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,method=_RETURNVERBOSE)

[Out]

e^2*(1/4*x*(c*x^2+b*x+a)^(3/2)/c-5/8*b/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)
/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-1/4*a/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^
(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))+2*d*e*(1/3*(c*x^2+b*x+a)^(3/2)/c
-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/
2))))+d^2*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1
/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 deta

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Fricas [A]
time = 2.13, size = 489, normalized size = 2.56 \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 (96 \, c^{4} d^{2} x + 48 \, b c^{3} d^{2} + {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x\right )} e^{2} + 16 \, {\left (8 \, c^{4} d x^{2} + 2 \, b c^{3} d x - {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d\right )} e\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 (96 \, c^{4} d^{2} x + 48 \, b c^{3} d^{2} + {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x\right )} e^{2} + 16 \, {\left (8 \, c^{4} d x^{2} + 2 \, b c^{3} d x - {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} d\right )} e\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*(96*c^4*d^2*x
+ 48*b*c^3*d^2 + (48*c^4*x^3 + 8*b*c^3*x^2 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x)*e^2 + 16*(8*c
^4*d*x^2 + 2*b*c^3*d*x - (3*b^2*c^2 - 8*a*c^3)*d)*e)*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*(96*c^4*d^2*x + 48*b*c^3*d^2 + (48*c^4*x^3 + 8*b*c
^3*x^2 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x)*e^2 + 16*(8*c^4*d*x^2 + 2*b*c^3*d*x - (3*b^2*c^2
- 8*a*c^3)*d)*e)*sqrt(c*x^2 + b*x + a))/c^4]

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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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Giac [A]
time = 3.03, 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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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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