3.2184 \(\int \frac {(d+e x)^2}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}+\frac {e^2 x}{c} \]

[Out]

e^2*x/c+1/2*e*(-b*e+2*c*d)*ln(c*x^2+b*x+a)/c^2-(2*c^2*d^2+b^2*e^2-2*c*e*(a*e+b*d))*arctanh((2*c*x+b)/(-4*a*c+b
^2)^(1/2))/c^2/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {701, 634, 618, 206, 628} \[ -\frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}+\frac {e^2 x}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + b*x + c*x^2),x]

[Out]

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

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])

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, 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] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{a+b x+c x^2} \, dx &=\int \left (\frac {e^2}{c}+\frac {c d^2-a e^2+e (2 c d-b e) x}{c \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {e^2 x}{c}+\frac {\int \frac {c d^2-a e^2+e (2 c d-b e) x}{a+b x+c x^2} \, dx}{c}\\ &=\frac {e^2 x}{c}+\frac {(e (2 c d-b e)) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c^2}+\frac {\left (-b e (2 c d-b e)+2 c \left (c d^2-a e^2\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c^2}\\ &=\frac {e^2 x}{c}+\frac {e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {\left (-b e (2 c d-b e)+2 c \left (c d^2-a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^2}\\ &=\frac {e^2 x}{c}-\frac {\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {e (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 101, normalized size = 1.00 \[ \frac {\frac {2 \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+e (2 c d-b e) \log (a+x (b+c x))+2 c e^2 x}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.00, size = 324, normalized size = 3.21 \[ \left [\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x - {\left (2 \, c^{2} d^{2} - 2 \, b c d e + {\left (b^{2} - 2 \, a c\right )} e^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e - {\left (b^{3} - 4 \, a b c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x - 2 \, {\left (2 \, c^{2} d^{2} - 2 \, b c d e + {\left (b^{2} - 2 \, a c\right )} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e - {\left (b^{3} - 4 \, a b c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(b^2*c - 4*a*c^2)*e^2*x - (2*c^2*d^2 - 2*b*c*d*e + (b^2 - 2*a*c)*e^2)*sqrt(b^2 - 4*a*c)*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)) + (2*(b^2*c - 4*a*c^2)*d*e - (b^3
- 4*a*b*c)*e^2)*log(c*x^2 + b*x + a))/(b^2*c^2 - 4*a*c^3), 1/2*(2*(b^2*c - 4*a*c^2)*e^2*x - 2*(2*c^2*d^2 - 2*
b*c*d*e + (b^2 - 2*a*c)*e^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + (2*(b^
2*c - 4*a*c^2)*d*e - (b^3 - 4*a*b*c)*e^2)*log(c*x^2 + b*x + a))/(b^2*c^2 - 4*a*c^3)]

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giac [A]  time = 0.16, size = 100, normalized size = 0.99 \[ \frac {x e^{2}}{c} + \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} + \frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^2/c + 1/2*(2*c*d*e - b*e^2)*log(c*x^2 + b*x + a)/c^2 + (2*c^2*d^2 - 2*b*c*d*e + b^2*e^2 - 2*a*c*e^2)*arcta
n((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^2)

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maple [B]  time = 0.05, size = 207, normalized size = 2.05 \[ -\frac {2 a \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {b^{2} e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {2 b d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {2 d^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {b \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {d e \ln \left (c \,x^{2}+b x +a \right )}{c}+\frac {e^{2} x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*e^2*x-1/2/c^2*ln(c*x^2+b*x+a)*b*e^2+1/c*ln(c*x^2+b*x+a)*d*e-2/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-
b^2)^(1/2))*a*e^2+2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*d^2+1/c^2/(4*a*c-b^2)^(1/2)*arctan((
2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2*e^2-2/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*d*e

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x+a),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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 0.27, size = 147, normalized size = 1.46 \[ \frac {e^2\,x}{c}+\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (b^3\,e^2-2\,d\,b^2\,c\,e-4\,a\,b\,c\,e^2+8\,a\,d\,c^2\,e\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )}+\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (b^2\,e^2-2\,b\,c\,d\,e+2\,c^2\,d^2-2\,a\,c\,e^2\right )}{c^2\,\sqrt {4\,a\,c-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + b*x + c*x^2),x)

[Out]

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

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sympy [B]  time = 2.06, size = 588, normalized size = 5.82 \[ \left (- \frac {e \left (b e - 2 c d\right )}{2 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- a b e^{2} - 4 a c^{2} \left (- \frac {e \left (b e - 2 c d\right )}{2 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + 4 a c d e + b^{2} c \left (- \frac {e \left (b e - 2 c d\right )}{2 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) - b c d^{2}}{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}} \right )} + \left (- \frac {e \left (b e - 2 c d\right )}{2 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- a b e^{2} - 4 a c^{2} \left (- \frac {e \left (b e - 2 c d\right )}{2 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + 4 a c d e + b^{2} c \left (- \frac {e \left (b e - 2 c d\right )}{2 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) - b c d^{2}}{2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}} \right )} + \frac {e^{2} x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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