3.11.17 \(\int (b d+2 c d x)^2 (a+b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=165 \[ \frac {d^2 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}+\frac {d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {d^2 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c} \]

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Rubi [A]  time = 0.09, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {685, 692, 621, 206} \begin {gather*} \frac {d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {d^2 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2),x]

[Out]

((b^2 - 4*a*c)^2*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^2) - ((b^2 - 4*a*c)*d^2*(b + 2*c*x)^3*Sqrt[a +
b*x + c*x^2])/(64*c^2) + (d^2*(b + 2*c*x)^3*(a + b*x + c*x^2)^(3/2))/(12*c) + ((b^2 - 4*a*c)^3*d^2*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(5/2))

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 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps

\begin {align*} \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}-\frac {\left (b^2-4 a c\right ) \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx}{8 c}\\ &=-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (b^2-4 a c\right )^2 \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx}{128 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (\left (b^2-4 a c\right )^3 d^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (\left (b^2-4 a c\right )^3 d^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 )}{128 c^2}\\ &=\frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (b^2-4 a c\right )^3 d^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 211, normalized size = 1.28 \begin {gather*} \frac {1}{3} d^2 (b+2 c x) \sqrt {a+x (b+c x)} \left ((a+x (b+c x))^2-\frac {(a+x (b+c x)) \left (2 (b+2 c x) \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \left (4 c \left (5 a+2 c x^2\right )-3 b^2+8 b c x\right )+3 \sqrt {c} \sqrt {4 a-\frac {b^2}{c}} \left (4 a c-b^2\right ) \sinh ^{-1}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {4 a-\frac {b^2}{c}}}\right )\right )}{256 c (b+2 c x) \left (\frac {c (a+x (b+c x))}{4 a c-b^2}\right )^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2),x]

[Out]

(d^2*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*((a + x*(b + c*x))^2 - ((a + x*(b + c*x))*(2*(b + 2*c*x)*Sqrt[(c*(a + x
*(b + c*x)))/(-b^2 + 4*a*c)]*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2)) + 3*Sqrt[4*a - b^2/c]*Sqrt[c]*(-b^2 + 4*
a*c)*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])]))/(256*c*(b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*
c))^(3/2))))/3

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IntegrateAlgebraic [A]  time = 0.76, size = 245, normalized size = 1.48 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (48 a^2 b c^2 d^2+96 a^2 c^3 d^2 x+32 a b^3 c d^2+288 a b^2 c^2 d^2 x+672 a b c^3 d^2 x^2+448 a c^4 d^2 x^3-3 b^5 d^2+2 b^4 c d^2 x+152 b^3 c^2 d^2 x^2+528 b^2 c^3 d^2 x^3+640 b c^4 d^2 x^4+256 c^5 d^2 x^5\right )}{384 c^2}+\frac {\left (64 a^3 c^3 d^2-48 a^2 b^2 c^2 d^2+12 a b^4 c d^2+b^6 \left (-d^2\right )\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{256 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[a + b*x + c*x^2]*(-3*b^5*d^2 + 32*a*b^3*c*d^2 + 48*a^2*b*c^2*d^2 + 2*b^4*c*d^2*x + 288*a*b^2*c^2*d^2*x +
96*a^2*c^3*d^2*x + 152*b^3*c^2*d^2*x^2 + 672*a*b*c^3*d^2*x^2 + 528*b^2*c^3*d^2*x^3 + 448*a*c^4*d^2*x^3 + 640*
b*c^4*d^2*x^4 + 256*c^5*d^2*x^5))/(384*c^2) + ((-(b^6*d^2) + 12*a*b^4*c*d^2 - 48*a^2*b^2*c^2*d^2 + 64*a^3*c^3*
d^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(256*c^(5/2))

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fricas [A]  time = 0.44, size = 473, normalized size = 2.87 \begin {gather*} \left [-\frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} d^{2} \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 (256 \, c^{6} d^{2} x^{5} + 640 \, b c^{5} d^{2} x^{4} + 16 \, {\left (33 \, b^{2} c^{4} + 28 \, a c^{5}\right )} d^{2} x^{3} + 8 \, {\left (19 \, b^{3} c^{3} + 84 \, a b c^{4}\right )} d^{2} x^{2} + 2 \, {\left (b^{4} c^{2} + 144 \, a b^{2} c^{3} + 48 \, a^{2} c^{4}\right )} d^{2} x - {\left (3 \, b^{5} c - 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{1536 \, c^{3}}, -\frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} d^{2} \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 (256 \, c^{6} d^{2} x^{5} + 640 \, b c^{5} d^{2} x^{4} + 16 \, {\left (33 \, b^{2} c^{4} + 28 \, a c^{5}\right )} d^{2} x^{3} + 8 \, {\left (19 \, b^{3} c^{3} + 84 \, a b c^{4}\right )} d^{2} x^{2} + 2 \, {\left (b^{4} c^{2} + 144 \, a b^{2} c^{3} + 48 \, a^{2} c^{4}\right )} d^{2} x - {\left (3 \, b^{5} c - 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/1536*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*d^2*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*(256*c^6*d^2*x^5 + 640*b*c^5*d^2*x^4 + 16*(33*b^2*c^4 + 2
8*a*c^5)*d^2*x^3 + 8*(19*b^3*c^3 + 84*a*b*c^4)*d^2*x^2 + 2*(b^4*c^2 + 144*a*b^2*c^3 + 48*a^2*c^4)*d^2*x - (3*b
^5*c - 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^2)*sqrt(c*x^2 + b*x + a))/c^3, -1/768*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*
c^2 - 64*a^3*c^3)*sqrt(-c)*d^2*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*(256*c^6*d^2*x^5 + 640*b*c^5*d^2*x^4 + 16*(33*b^2*c^4 + 28*a*c^5)*d^2*x^3 + 8*(19*b^3*c^3 + 84*a*b*c^4)*d^
2*x^2 + 2*(b^4*c^2 + 144*a*b^2*c^3 + 48*a^2*c^4)*d^2*x - (3*b^5*c - 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^2)*sqrt(c*x
^2 + b*x + a))/c^3]

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giac [A]  time = 0.29, size = 259, normalized size = 1.57 \begin {gather*} \frac {1}{384} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, c^{3} d^{2} x + 5 \, b c^{2} d^{2}\right )} x + \frac {33 \, b^{2} c^{6} d^{2} + 28 \, a c^{7} d^{2}}{c^{5}}\right )} x + \frac {19 \, b^{3} c^{5} d^{2} + 84 \, a b c^{6} d^{2}}{c^{5}}\right )} x + \frac {b^{4} c^{4} d^{2} + 144 \, a b^{2} c^{5} d^{2} + 48 \, a^{2} c^{6} d^{2}}{c^{5}}\right )} x - \frac {3 \, b^{5} c^{3} d^{2} - 32 \, a b^{3} c^{4} d^{2} - 48 \, a^{2} b c^{5} d^{2}}{c^{5}}\right )} - \frac {{\left (b^{6} d^{2} - 12 \, a b^{4} c d^{2} + 48 \, a^{2} b^{2} c^{2} d^{2} - 64 \, a^{3} c^{3} d^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/384*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^3*d^2*x + 5*b*c^2*d^2)*x + (33*b^2*c^6*d^2 + 28*a*c^7*d^2)/c^5)*x
+ (19*b^3*c^5*d^2 + 84*a*b*c^6*d^2)/c^5)*x + (b^4*c^4*d^2 + 144*a*b^2*c^5*d^2 + 48*a^2*c^6*d^2)/c^5)*x - (3*b
^5*c^3*d^2 - 32*a*b^3*c^4*d^2 - 48*a^2*b*c^5*d^2)/c^5) - 1/256*(b^6*d^2 - 12*a*b^4*c*d^2 + 48*a^2*b^2*c^2*d^2
- 64*a^3*c^3*d^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.06, size = 406, normalized size = 2.46 \begin {gather*} -\frac {a^{3} \sqrt {c}\, d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4}+\frac {3 a^{2} b^{2} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 \sqrt {c}}-\frac {3 a \,b^{4} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {3}{2}}}+\frac {b^{6} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {5}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a^{2} c \,d^{2} x}{4}+\frac {\sqrt {c \,x^{2}+b x +a}\, a \,b^{2} d^{2} x}{8}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{4} d^{2} x}{64 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, a^{2} b \,d^{2}}{8}+\frac {\sqrt {c \,x^{2}+b x +a}\, a \,b^{3} d^{2}}{16 c}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a c \,d^{2} x}{6}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{5} d^{2}}{128 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} d^{2} x}{24}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b \,d^{2}}{12}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{3} d^{2}}{48 c}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} c \,d^{2} x}{3}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b \,d^{2}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(3/2),x)

[Out]

1/256*d^2*b^6/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/6*d^2*c*a*x*(c*x^2+b*x+a)^(3/2)-1/4*d^2*c*
a^2*(c*x^2+b*x+a)^(1/2)*x-1/4*d^2*c^(1/2)*a^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/8*d^2*b^2*(c*x^2+b
*x+a)^(1/2)*x*a-1/64*d^2/c*b^4*(c*x^2+b*x+a)^(1/2)*x+1/16*d^2/c*b^3*(c*x^2+b*x+a)^(1/2)*a-3/64*d^2*b^4/c^(3/2)
*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+3/16*d^2*b^2/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)*a^2+1/3*d^2*b*(c*x^2+b*x+a)^(5/2)-1/8*d^2*a^2*(c*x^2+b*x+a)^(1/2)*b+1/48*d^2/c*b^3*(c*x^2+b*x+a)^(3/2)-1/128
*d^2/c^2*b^5*(c*x^2+b*x+a)^(1/2)-1/12*d^2*a*(c*x^2+b*x+a)^(3/2)*b+2/3*d^2*c*x*(c*x^2+b*x+a)^(5/2)+1/24*d^2*b^2
*x*(c*x^2+b*x+a)^(3/2)

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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((2*c*d*x+b*d)^2*(c*x^2+b*x+a)^(3/2),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 [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,d+2\,c\,d\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2),x)

[Out]

int((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{2} \left (\int a b^{2} \sqrt {a + b x + c x^{2}}\, dx + \int b^{3} x \sqrt {a + b x + c x^{2}}\, dx + \int 4 c^{3} x^{4} \sqrt {a + b x + c x^{2}}\, dx + \int 4 a c^{2} x^{2} \sqrt {a + b x + c x^{2}}\, dx + \int 8 b c^{2} x^{3} \sqrt {a + b x + c x^{2}}\, dx + \int 5 b^{2} c x^{2} \sqrt {a + b x + c x^{2}}\, dx + \int 4 a b c x \sqrt {a + b x + c x^{2}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**2*(c*x**2+b*x+a)**(3/2),x)

[Out]

d**2*(Integral(a*b**2*sqrt(a + b*x + c*x**2), x) + Integral(b**3*x*sqrt(a + b*x + c*x**2), x) + Integral(4*c**
3*x**4*sqrt(a + b*x + c*x**2), x) + Integral(4*a*c**2*x**2*sqrt(a + b*x + c*x**2), x) + Integral(8*b*c**2*x**3
*sqrt(a + b*x + c*x**2), x) + Integral(5*b**2*c*x**2*sqrt(a + b*x + c*x**2), x) + Integral(4*a*b*c*x*sqrt(a +
b*x + c*x**2), x))

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