### 3.1160 $$\int \frac{(b d+2 c d x)^2}{a+b x+c x^2} \, dx$$

Optimal. Leaf size=49 $2 d^2 (b+2 c x)-2 d^2 \sqrt{b^2-4 a c} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )$

[Out]

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

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Rubi [A]  time = 0.0308715, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {692, 618, 206} $2 d^2 (b+2 c x)-2 d^2 \sqrt{b^2-4 a c} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], 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{(b d+2 c d x)^2}{a+b x+c x^2} \, dx &=2 d^2 (b+2 c x)+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac{1}{a+b x+c x^2} \, dx\\ &=2 d^2 (b+2 c x)-\left (2 \left (b^2-4 a c\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )\\ &=2 d^2 (b+2 c x)-2 \sqrt{b^2-4 a c} d^2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0243931, size = 47, normalized size = 0.96 $d^2 \left (4 c x-2 \sqrt{4 a c-b^2} \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

d^2*(4*c*x - 2*Sqrt[-b^2 + 4*a*c]*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])

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Maple [A]  time = 0.148, size = 88, normalized size = 1.8 \begin{align*} 4\,c{d}^{2}x-8\,{\frac{a{d}^{2}c}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{{b}^{2}{d}^{2}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

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),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71929, size = 300, normalized size = 6.12 \begin{align*} \left [4 \, c d^{2} x + \sqrt{b^{2} - 4 \, a c} d^{2} \log \left (\frac{2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt{b^{2} - 4 \, a c}{\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ), 4 \, c d^{2} x - 2 \, \sqrt{-b^{2} + 4 \, a c} d^{2} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right )\right ] \end{align*}

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),x, algorithm="fricas")

[Out]

[4*c*d^2*x + sqrt(b^2 - 4*a*c)*d^2*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*
x^2 + b*x + a)), 4*c*d^2*x - 2*sqrt(-b^2 + 4*a*c)*d^2*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c))]

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Sympy [B]  time = 1.0984, size = 99, normalized size = 2.02 \begin{align*} 4 c d^{2} x + d^{2} \sqrt{- 4 a c + b^{2}} \log{\left (x + \frac{b d^{2} - d^{2} \sqrt{- 4 a c + b^{2}}}{2 c d^{2}} \right )} - d^{2} \sqrt{- 4 a c + b^{2}} \log{\left (x + \frac{b d^{2} + d^{2} \sqrt{- 4 a c + b^{2}}}{2 c d^{2}} \right )} \end{align*}

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),x)

[Out]

4*c*d**2*x + d**2*sqrt(-4*a*c + b**2)*log(x + (b*d**2 - d**2*sqrt(-4*a*c + b**2))/(2*c*d**2)) - d**2*sqrt(-4*a
*c + b**2)*log(x + (b*d**2 + d**2*sqrt(-4*a*c + b**2))/(2*c*d**2))

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Giac [A]  time = 1.13945, size = 77, normalized size = 1.57 \begin{align*} 4 \, c d^{2} x + \frac{2 \,{\left (b^{2} d^{2} - 4 \, a c d^{2}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c}} \end{align*}

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),x, algorithm="giac")

[Out]

4*c*d^2*x + 2*(b^2*d^2 - 4*a*c*d^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/sqrt(-b^2 + 4*a*c)