### 3.2335 $$\int (d+e x) \sqrt{a+b x+c x^2} \, dx$$

Optimal. Leaf size=115 $-\frac{\left (b^2-4 a c\right ) (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{5/2}}+\frac{(b+2 c x) \sqrt{a+b x+c x^2} (2 c d-b e)}{8 c^2}+\frac{e \left (a+b x+c x^2\right )^{3/2}}{3 c}$

[Out]

((2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(8*c^2) + (e*(a + b*x + c*x^2)^(3/2))/(3*c) - ((b^2 - 4*a*c)
*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(5/2))

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Rubi [A]  time = 0.0428137, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {640, 612, 621, 206} $-\frac{\left (b^2-4 a c\right ) (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{5/2}}+\frac{(b+2 c x) \sqrt{a+b x+c x^2} (2 c d-b e)}{8 c^2}+\frac{e \left (a+b x+c x^2\right )^{3/2}}{3 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*c*d - b*e)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(8*c^2) + (e*(a + b*x + c*x^2)^(3/2))/(3*c) - ((b^2 - 4*a*c)
*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(5/2))

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

Mathematica [A]  time = 0.105502, size = 114, normalized size = 0.99 $\frac{2 \sqrt{c} \sqrt{a+x (b+c x)} \left (4 c (2 a e+c x (3 d+2 e x))-3 b^2 e+2 b c (3 d+e x)\right )+3 \left (b^2-4 a c\right ) (b e-2 c d) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{48 c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-3*b^2*e + 2*b*c*(3*d + e*x) + 4*c*(2*a*e + c*x*(3*d + 2*e*x))) + 3*(b^2 - 4
*a*c)*(-2*c*d + b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(48*c^(5/2))

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Maple [B]  time = 0.046, size = 229, normalized size = 2. \begin{align*}{\frac{e}{3\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{bxe}{4\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{{b}^{2}e}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{aeb}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{e{b}^{3}}{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{dx}{2}\sqrt{c{x}^{2}+bx+a}}+{\frac{bd}{4\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{ad}{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}{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)*(c*x^2+b*x+a)^(1/2),x)

[Out]

1/3*e*(c*x^2+b*x+a)^(3/2)/c-1/4*e*b/c*x*(c*x^2+b*x+a)^(1/2)-1/8*e*b^2/c^2*(c*x^2+b*x+a)^(1/2)-1/4*e*b/c^(3/2)*
ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+1/16*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*x*(c*x^2+b*x+a)^(1/2)+1/4*d/c*(c*x^2+b*x+a)^(1/2)*b+1/2*d/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1
/2))*a-1/8*d/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)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.42144, size = 683, normalized size = 5.94 \begin{align*} \left [\frac{3 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d -{\left (b^{3} - 4 \, a b c\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 (8 \, c^{3} e x^{2} + 6 \, b c^{2} d -{\left (3 \, b^{2} c - 8 \, a c^{2}\right )} e + 2 \,{\left (6 \, c^{3} d + b c^{2} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{96 \, c^{3}}, \frac{3 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d -{\left (b^{3} - 4 \, a b c\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 (8 \, c^{3} e x^{2} + 6 \, b c^{2} d -{\left (3 \, b^{2} c - 8 \, a c^{2}\right )} e + 2 \,{\left (6 \, c^{3} d + b c^{2} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{48 \, c^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(2*(b^2*c - 4*a*c^2)*d - (b^3 - 4*a*b*c)*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*(8*c^3*e*x^2 + 6*b*c^2*d - (3*b^2*c - 8*a*c^2)*e + 2*(6*c^3*d + b*c^2
*e)*x)*sqrt(c*x^2 + b*x + a))/c^3, 1/48*(3*(2*(b^2*c - 4*a*c^2)*d - (b^3 - 4*a*b*c)*e)*sqrt(-c)*arctan(1/2*sqr
t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(8*c^3*e*x^2 + 6*b*c^2*d - (3*b^2*c - 8*a
*c^2)*e + 2*(6*c^3*d + b*c^2*e)*x)*sqrt(c*x^2 + b*x + a))/c^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right ) \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)*(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.17386, size = 174, normalized size = 1.51 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \, x e + \frac{6 \, c^{2} d + b c e}{c^{2}}\right )} x + \frac{6 \, b c d - 3 \, b^{2} e + 8 \, a c e}{c^{2}}\right )} + \frac{{\left (2 \, b^{2} c d - 8 \, a c^{2} d - b^{3} e + 4 \, a b c e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*x*e + (6*c^2*d + b*c*e)/c^2)*x + (6*b*c*d - 3*b^2*e + 8*a*c*e)/c^2) + 1/16*(2
*b^2*c*d - 8*a*c^2*d - b^3*e + 4*a*b*c*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)