### 3.2375 $$\int \frac{(d+e x)^2}{\sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=127 $\frac{\left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{5/2}}+\frac{3 e \sqrt{a+b x+c x^2} (2 c d-b e)}{4 c^2}+\frac{e (d+e x) \sqrt{a+b x+c x^2}}{2 c}$

[Out]

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

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Rubi [A]  time = 0.116312, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {742, 640, 621, 206} $\frac{\left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{5/2}}+\frac{3 e \sqrt{a+b x+c x^2} (2 c d-b e)}{4 c^2}+\frac{e (d+e x) \sqrt{a+b x+c x^2}}{2 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.120862, size = 103, normalized size = 0.81 $\frac{\left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{8 c^{5/2}}+\frac{e \sqrt{a+x (b+c x)} (-3 b e+8 c d+2 c e x)}{4 c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 198, normalized size = 1.6 \begin{align*}{\frac{{e}^{2}x}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,b{e}^{2}}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{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{5}{2}}}}-{\frac{a{e}^{2}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+2\,{\frac{de\sqrt{c{x}^{2}+bx+a}}{c}}-{bde\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{{d}^{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}} \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]

1/2*e^2*x/c*(c*x^2+b*x+a)^(1/2)-3/4*e^2*b/c^2*(c*x^2+b*x+a)^(1/2)+3/8*e^2*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))-1/2*e^2*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2*d*e/c*(c*x^2+b*x+a)^(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))+d^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1
/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 = 2.52757, size = 582, normalized size = 4.58 \begin{align*} \left [-\frac{{\left (8 \, c^{2} d^{2} - 8 \, b c d e +{\left (3 \, b^{2} - 4 \, a c\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 (2 \, c^{2} e^{2} x + 8 \, c^{2} d e - 3 \, b c e^{2}\right )} \sqrt{c x^{2} + b x + a}}{16 \, c^{3}}, -\frac{{\left (8 \, c^{2} d^{2} - 8 \, b c d e +{\left (3 \, b^{2} - 4 \, a c\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 (2 \, c^{2} e^{2} x + 8 \, c^{2} d e - 3 \, b c e^{2}\right )} \sqrt{c x^{2} + b x + a}}{8 \, c^{3}}\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]

[-1/16*((8*c^2*d^2 - 8*b*c*d*e + (3*b^2 - 4*a*c)*e^2)*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*(2*c^2*e^2*x + 8*c^2*d*e - 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/c^3, -1
/8*((8*c^2*d^2 - 8*b*c*d*e + (3*b^2 - 4*a*c)*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*(2*c^2*e^2*x + 8*c^2*d*e - 3*b*c*e^2)*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 \frac{\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 = 1.12988, size = 142, normalized size = 1.12 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x + a}{\left (\frac{2 \, x e^{2}}{c} + \frac{8 \, c d e - 3 \, b e^{2}}{c^{2}}\right )} - \frac{{\left (8 \, c^{2} d^{2} - 8 \, b c d e + 3 \, b^{2} e^{2} - 4 \, a c e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{5}{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]

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