### 3.2334 $$\int (d+e x)^2 \sqrt{a+b x+c x^2} \, dx$$

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

[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))

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Rubi [A]  time = 0.234412, antiderivative size = 191, normalized size of antiderivative = 1., 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} $\frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right )}{64 c^3}-\frac{\left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{7/2}}+\frac{5 e \left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac{e (d+e x) \left (a+b x+c x^2\right )^{3/2}}{4 c}$

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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.179127, size = 162, normalized size = 0.85 $\frac{\frac{\left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \left (2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )}{32 c^{5/2}}+\frac{5 e (a+x (b+c x))^{3/2} (2 c d-b e)}{6 c}+e (d+e x) (a+x (b+c x))^{3/2}}{4 c}$

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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Maple [B]  time = 0.05, size = 484, normalized size = 2.5 \begin{align*}{\frac{{e}^{2}x}{4\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,b{e}^{2}}{24\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}{e}^{2}x}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{3}{e}^{2}}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}{e}^{2}a}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{5\,{e}^{2}{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}-{\frac{a{e}^{2}x}{8\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{a{e}^{2}b}{16\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{{a}^{2}{e}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{2\,de}{3\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{bdex}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{{b}^{2}de}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{abde}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{de{b}^{3}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{{d}^{2}x}{2}\sqrt{c{x}^{2}+bx+a}}+{\frac{{d}^{2}b}{4\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{a{d}^{2}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{2}{d}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

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

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

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

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Fricas [A]  time = 2.61507, size = 1137, normalized size = 5.95 \begin{align*} \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{align*}

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

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 = 1.68407, size = 317, normalized size = 1.66 \begin{align*} \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{align*}

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)