3.24.45 \(\int (d+e x) (a+b x+c x^2)^{3/2} \, dx\) [2345]

Optimal. Leaf size=161 \[ -\frac {3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}} \]

[Out]

1/16*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c^2+1/5*e*(c*x^2+b*x+a)^(5/2)/c+3/256*(-4*a*c+b^2)^2*(-b*e+2*c
*d)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)-3/128*(-4*a*c+b^2)*(-b*e+2*c*d)*(2*c*x+b)*(c*x^
2+b*x+a)^(1/2)/c^3

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Rubi [A]
time = 0.04, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 626, 635, 212} \begin {gather*} \frac {3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e)}{128 c^3}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2 - 4*a*c)*(2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^3) + ((2*c*d - b*e)*(b + 2*c*x)*(a +
b*x + c*x^2)^(3/2))/(16*c^2) + (e*(a + b*x + c*x^2)^(5/2))/(5*c) + (3*(b^2 - 4*a*c)^2*(2*c*d - b*e)*ArcTanh[(
b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*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]

Rubi steps

\begin {align*} \int (d+e x) \left (a+b x+c x^2\right )^{3/2} \, dx &=\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {(2 c d-b e) \int \left (a+b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 \left (b^2-4 a c\right ) (2 c d-b e)\right ) \int \sqrt {a+b x+c x^2} \, dx}{32 c^2}\\ &=-\frac {3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 \left (b^2-4 a c\right )^2 (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^3}\\ &=-\frac {3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 \left (b^2-4 a c\right )^2 (2 c d-b e)\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 )}{128 c^3}\\ &=-\frac {3 \left (b^2-4 a c\right ) (2 c d-b e) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}+\frac {(2 c d-b e) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {e \left (a+b x+c x^2\right )^{5/2}}{5 c}+\frac {3 \left (b^2-4 a c\right )^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.68, size = 187, normalized size = 1.16 \begin {gather*} \frac {\sqrt {a+x (b+c x)} \left (15 b^4 e-10 b^3 c (3 d+e x)+8 b c^2 \left (25 a d+7 a e x+30 c d x^2+22 c e x^3\right )+4 b^2 c (-25 a e+c x (5 d+2 e x))+16 c^2 \left (8 a^2 e+2 c^2 x^3 (5 d+4 e x)+a c x (25 d+16 e x)\right )\right )}{640 c^3}+\frac {3 \left (b^2-4 a c\right )^2 (-2 c d+b e) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{256 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*(15*b^4*e - 10*b^3*c*(3*d + e*x) + 8*b*c^2*(25*a*d + 7*a*e*x + 30*c*d*x^2 + 22*c*e*x^3)
+ 4*b^2*c*(-25*a*e + c*x*(5*d + 2*e*x)) + 16*c^2*(8*a^2*e + 2*c^2*x^3*(5*d + 4*e*x) + a*c*x*(25*d + 16*e*x)))
)/(640*c^3) + (3*(b^2 - 4*a*c)^2*(-2*c*d + b*e)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(256*c^(7/2)
)

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Maple [A]
time = 0.75, size = 236, normalized size = 1.47

method	result	size
default	\(e \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )+d \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )\)	\(236\)
risch	\(\frac {\left (128 c^{4} e \,x^{4}+176 b \,c^{3} e \,x^{3}+160 c^{4} d \,x^{3}+256 a \,c^{3} e \,x^{2}+8 b^{2} c^{2} e \,x^{2}+240 b \,c^{3} d \,x^{2}+56 a b \,c^{2} e x +400 a \,c^{3} d x -10 b^{3} c e x +20 x \,b^{2} c^{2} d +128 a^{2} c^{2} e -100 a \,b^{2} c e +200 a b \,c^{2} d +15 b^{4} e -30 b^{3} d c \right ) \sqrt {c \,x^{2}+b x +a}}{640 c^{3}}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{2} b e}{16 c^{\frac {3}{2}}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{2} d}{8 \sqrt {c}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,b^{3} e}{32 c^{\frac {5}{2}}}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,b^{2} d}{16 c^{\frac {3}{2}}}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{5} e}{256 c^{\frac {7}{2}}}+\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{4} d}{128 c^{\frac {5}{2}}}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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)*(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 deta

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Fricas [A]
time = 3.20, size = 523, normalized size = 3.25 \begin {gather*} \left [-\frac {15 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e\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 (160 \, c^{5} d x^{3} + 240 \, b c^{4} d x^{2} + 20 \, {\left (b^{2} c^{3} + 20 \, a c^{4}\right )} d x - 10 \, {\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d + {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x\right )} e\right )} \sqrt {c x^{2} + b x + a}}{2560 \, c^{4}}, -\frac {15 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e\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 (160 \, c^{5} d x^{3} + 240 \, b c^{4} d x^{2} + 20 \, {\left (b^{2} c^{3} + 20 \, a c^{4}\right )} d x - 10 \, {\left (3 \, b^{3} c^{2} - 20 \, a b c^{3}\right )} d + {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x\right )} e\right )} \sqrt {c x^{2} + b x + a}}{1280 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2560*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*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*(160*c^5*d*x^3 + 240*b*c^4*d*x^
2 + 20*(b^2*c^3 + 20*a*c^4)*d*x - 10*(3*b^3*c^2 - 20*a*b*c^3)*d + (128*c^5*x^4 + 176*b*c^4*x^3 + 15*b^4*c - 10
0*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^2 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x)*e)*sqrt(c*x^2 + b*x + a
))/c^4, -1/1280*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(-c)*arc
tan(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*(160*c^5*d*x^3 + 240*b*c^4*d*x
^2 + 20*(b^2*c^3 + 20*a*c^4)*d*x - 10*(3*b^3*c^2 - 20*a*b*c^3)*d + (128*c^5*x^4 + 176*b*c^4*x^3 + 15*b^4*c - 1
00*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^2 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x)*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 ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.39, size = 264, normalized size = 1.64 \begin {gather*} \frac {1}{640} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x e + \frac {10 \, c^{5} d + 11 \, b c^{4} e}{c^{4}}\right )} x + \frac {30 \, b c^{4} d + b^{2} c^{3} e + 32 \, a c^{4} e}{c^{4}}\right )} x + \frac {10 \, b^{2} c^{3} d + 200 \, a c^{4} d - 5 \, b^{3} c^{2} e + 28 \, a b c^{3} e}{c^{4}}\right )} x - \frac {30 \, b^{3} c^{2} d - 200 \, a b c^{3} d - 15 \, b^{4} c e + 100 \, a b^{2} c^{2} e - 128 \, a^{2} c^{3} e}{c^{4}}\right )} - \frac {3 \, {\left (2 \, b^{4} c d - 16 \, a b^{2} c^{2} d + 32 \, a^{2} c^{3} d - b^{5} e + 8 \, a b^{3} c e - 16 \, a^{2} b c^{2} e\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 {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*c*x*e + (10*c^5*d + 11*b*c^4*e)/c^4)*x + (30*b*c^4*d + b^2*c^3*e + 32*
a*c^4*e)/c^4)*x + (10*b^2*c^3*d + 200*a*c^4*d - 5*b^3*c^2*e + 28*a*b*c^3*e)/c^4)*x - (30*b^3*c^2*d - 200*a*b*c
^3*d - 15*b^4*c*e + 100*a*b^2*c^2*e - 128*a^2*c^3*e)/c^4) - 3/256*(2*b^4*c*d - 16*a*b^2*c^2*d + 32*a^2*c^3*d -
b^5*e + 8*a*b^3*c*e - 16*a^2*b*c^2*e)*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 = 1.51, size = 305, normalized size = 1.89 \begin {gather*} \frac {e\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5\,c}+\frac {d\,\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 )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}-\frac {b\,e\,\left (\frac {3\,a\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{4}+\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{16\,c}\right )}{2\,c}+\frac {d\,\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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