3.149 \(\int \frac {B+A x}{(c+2 b x+a x^2)^2} \, dx\)

Optimal. Leaf size=90 \[ -\frac {-x (A b-a B)-A c+b B}{2 \left (b^2-a c\right ) \left (a x^2+2 b x+c\right )}-\frac {(A b-a B) \tanh ^{-1}\left (\frac {a x+b}{\sqrt {b^2-a c}}\right )}{2 \left (b^2-a c\right )^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {638, 618, 206} \[ -\frac {-x (A b-a B)-A c+b B}{2 \left (b^2-a c\right ) \left (a x^2+2 b x+c\right )}-\frac {(A b-a B) \tanh ^{-1}\left (\frac {a x+b}{\sqrt {b^2-a c}}\right )}{2 \left (b^2-a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*B - A*c - (A*b - a*B)*x)/(2*(b^2 - a*c)*(c + 2*b*x + a*x^2)) - ((A*b - a*B)*ArcTanh[(b + a*x)/Sqrt[b^2 - a
*c]])/(2*(b^2 - a*c)^(3/2))

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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin {align*} \int \frac {B+A x}{\left (c+2 b x+a x^2\right )^2} \, dx &=-\frac {b B-A c-(A b-a B) x}{2 \left (b^2-a c\right ) \left (c+2 b x+a x^2\right )}+\frac {(A b-a B) \int \frac {1}{c+2 b x+a x^2} \, dx}{2 \left (b^2-a c\right )}\\ &=-\frac {b B-A c-(A b-a B) x}{2 \left (b^2-a c\right ) \left (c+2 b x+a x^2\right )}-\frac {(A b-a B) \operatorname {Subst}\left (\int \frac {1}{4 \left (b^2-a c\right )-x^2} \, dx,x,2 b+2 a x\right )}{b^2-a c}\\ &=-\frac {b B-A c-(A b-a B) x}{2 \left (b^2-a c\right ) \left (c+2 b x+a x^2\right )}-\frac {(A b-a B) \tanh ^{-1}\left (\frac {b+a x}{\sqrt {b^2-a c}}\right )}{2 \left (b^2-a c\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 88, normalized size = 0.98 \[ \frac {\frac {(A b-a B) \tan ^{-1}\left (\frac {a x+b}{\sqrt {a c-b^2}}\right )}{\sqrt {a c-b^2}}+\frac {-a B x+A b x+A c-b B}{x (a x+2 b)+c}}{2 \left (b^2-a c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 113, normalized size = 1.26 \[ \frac {-a B x+A b x+A c-b B}{2 \left (b^2-a c\right ) \left (a x^2+2 b x+c\right )}+\frac {(A b-a B) \tan ^{-1}\left (\frac {a x}{\sqrt {a c-b^2}}+\frac {b}{\sqrt {a c-b^2}}\right )}{2 \left (b^2-a c\right ) \sqrt {a c-b^2}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(B + A*x)/(c + 2*b*x + a*x^2)^2,x]

[Out]

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

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fricas [B]  time = 1.13, size = 448, normalized size = 4.98 \[ \left [-\frac {2 \, B b^{3} + 2 \, A a c^{2} - {\left ({\left (B a^{2} - A a b\right )} x^{2} + {\left (B a - A b\right )} c + 2 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {b^{2} - a c} \log \left (\frac {a^{2} x^{2} + 2 \, a b x + 2 \, b^{2} - a c + 2 \, \sqrt {b^{2} - a c} {\left (a x + b\right )}}{a x^{2} + 2 \, b x + c}\right ) - 2 \, {\left (B a b + A b^{2}\right )} c + 2 \, {\left (B a b^{2} - A b^{3} - {\left (B a^{2} - A a b\right )} c\right )} x}{4 \, {\left (b^{4} c - 2 \, a b^{2} c^{2} + a^{2} c^{3} + {\left (a b^{4} - 2 \, a^{2} b^{2} c + a^{3} c^{2}\right )} x^{2} + 2 \, {\left (b^{5} - 2 \, a b^{3} c + a^{2} b c^{2}\right )} x\right )}}, -\frac {B b^{3} + A a c^{2} - {\left ({\left (B a^{2} - A a b\right )} x^{2} + {\left (B a - A b\right )} c + 2 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {-b^{2} + a c} \arctan \left (-\frac {\sqrt {-b^{2} + a c} {\left (a x + b\right )}}{b^{2} - a c}\right ) - {\left (B a b + A b^{2}\right )} c + {\left (B a b^{2} - A b^{3} - {\left (B a^{2} - A a b\right )} c\right )} x}{2 \, {\left (b^{4} c - 2 \, a b^{2} c^{2} + a^{2} c^{3} + {\left (a b^{4} - 2 \, a^{2} b^{2} c + a^{3} c^{2}\right )} x^{2} + 2 \, {\left (b^{5} - 2 \, a b^{3} c + a^{2} b c^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*B*b^3 + 2*A*a*c^2 - ((B*a^2 - A*a*b)*x^2 + (B*a - A*b)*c + 2*(B*a*b - A*b^2)*x)*sqrt(b^2 - a*c)*log((
a^2*x^2 + 2*a*b*x + 2*b^2 - a*c + 2*sqrt(b^2 - a*c)*(a*x + b))/(a*x^2 + 2*b*x + c)) - 2*(B*a*b + A*b^2)*c + 2*
(B*a*b^2 - A*b^3 - (B*a^2 - A*a*b)*c)*x)/(b^4*c - 2*a*b^2*c^2 + a^2*c^3 + (a*b^4 - 2*a^2*b^2*c + a^3*c^2)*x^2
+ 2*(b^5 - 2*a*b^3*c + a^2*b*c^2)*x), -1/2*(B*b^3 + A*a*c^2 - ((B*a^2 - A*a*b)*x^2 + (B*a - A*b)*c + 2*(B*a*b
- A*b^2)*x)*sqrt(-b^2 + a*c)*arctan(-sqrt(-b^2 + a*c)*(a*x + b)/(b^2 - a*c)) - (B*a*b + A*b^2)*c + (B*a*b^2 -
A*b^3 - (B*a^2 - A*a*b)*c)*x)/(b^4*c - 2*a*b^2*c^2 + a^2*c^3 + (a*b^4 - 2*a^2*b^2*c + a^3*c^2)*x^2 + 2*(b^5 -
2*a*b^3*c + a^2*b*c^2)*x)]

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giac [A]  time = 0.99, size = 92, normalized size = 1.02 \[ -\frac {{\left (B a - A b\right )} \arctan \left (\frac {a x + b}{\sqrt {-b^{2} + a c}}\right )}{2 \, {\left (b^{2} - a c\right )} \sqrt {-b^{2} + a c}} - \frac {B a x - A b x + B b - A c}{2 \, {\left (a x^{2} + 2 \, b x + c\right )} {\left (b^{2} - a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*a - A*b)*arctan((a*x + b)/sqrt(-b^2 + a*c))/((b^2 - a*c)*sqrt(-b^2 + a*c)) - 1/2*(B*a*x - A*b*x + B*b
- A*c)/((a*x^2 + 2*b*x + c)*(b^2 - a*c))

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maple [A]  time = 0.59, size = 103, normalized size = 1.14

method	result	size
default	\(\frac {\left (-2 A b +2 B a \right ) x +2 b B -2 A c}{\left (4 a c -4 b^{2}\right ) \left (a \,x^{2}+2 b x +c \right )}+\frac {\left (-2 A b +2 B a \right ) \arctan \left (\frac {2 a x +2 b}{2 \sqrt {a c -b^{2}}}\right )}{\left (4 a c -4 b^{2}\right ) \sqrt {a c -b^{2}}}\)	\(103\)
risch	\(\frac {-\frac {\left (A b -B a \right ) x}{2 \left (a c -b^{2}\right )}-\frac {A c -b B}{2 \left (a c -b^{2}\right )}}{a \,x^{2}+2 b x +c}+\frac {\ln \left (\left (-a^{2} c +a \,b^{2}\right ) x -\left (-a c +b^{2}\right )^{\frac {3}{2}}-a b c +b^{3}\right ) A b}{4 \left (-a c +b^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left (\left (-a^{2} c +a \,b^{2}\right ) x -\left (-a c +b^{2}\right )^{\frac {3}{2}}-a b c +b^{3}\right ) B a}{4 \left (-a c +b^{2}\right )^{\frac {3}{2}}}-\frac {\ln \left (\left (a^{2} c -a \,b^{2}\right ) x -\left (-a c +b^{2}\right )^{\frac {3}{2}}+a b c -b^{3}\right ) A b}{4 \left (-a c +b^{2}\right )^{\frac {3}{2}}}+\frac {\ln \left (\left (a^{2} c -a \,b^{2}\right ) x -\left (-a c +b^{2}\right )^{\frac {3}{2}}+a b c -b^{3}\right ) B a}{4 \left (-a c +b^{2}\right )^{\frac {3}{2}}}\)	\(262\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A*x+B)/(a*x^2+2*b*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

((-2*A*b+2*B*a)*x+2*b*B-2*A*c)/(4*a*c-4*b^2)/(a*x^2+2*b*x+c)+(-2*A*b+2*B*a)/(4*a*c-4*b^2)/(a*c-b^2)^(1/2)*arct
an(1/2*(2*a*x+2*b)/(a*c-b^2)^(1/2))

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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((A*x+B)/(a*x^2+2*b*x+c)^2,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*b^2-4*a*c>0)', see `assume?`
for more details)Is 4*b^2-4*a*c positive or negative?

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mupad [B]  time = 0.29, size = 159, normalized size = 1.77 \[ \frac {\mathrm {atan}\left (\frac {2\,\left (a\,c-b^2\right )\,\left (\frac {\left (4\,b^3-4\,a\,b\,c\right )\,\left (A\,b-B\,a\right )}{8\,{\left (a\,c-b^2\right )}^{5/2}}-\frac {a\,x\,\left (A\,b-B\,a\right )}{2\,{\left (a\,c-b^2\right )}^{3/2}}\right )}{A\,b-B\,a}\right )\,\left (A\,b-B\,a\right )}{2\,{\left (a\,c-b^2\right )}^{3/2}}-\frac {\frac {A\,c-B\,b}{2\,\left (a\,c-b^2\right )}+\frac {x\,\left (A\,b-B\,a\right )}{2\,\left (a\,c-b^2\right )}}{a\,x^2+2\,b\,x+c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B + A*x)/(c + 2*b*x + a*x^2)^2,x)

[Out]

(atan((2*(a*c - b^2)*(((4*b^3 - 4*a*b*c)*(A*b - B*a))/(8*(a*c - b^2)^(5/2)) - (a*x*(A*b - B*a))/(2*(a*c - b^2)
^(3/2))))/(A*b - B*a))*(A*b - B*a))/(2*(a*c - b^2)^(3/2)) - ((A*c - B*b)/(2*(a*c - b^2)) + (x*(A*b - B*a))/(2*
(a*c - b^2)))/(c + 2*b*x + a*x^2)

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sympy [B]  time = 1.01, size = 323, normalized size = 3.59 \[ - \frac {\sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right ) \log {\left (x + \frac {- A b^{2} + B a b - a^{2} c^{2} \sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right ) + 2 a b^{2} c \sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right ) - b^{4} \sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right )}{- A a b + B a^{2}} \right )}}{4} + \frac {\sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right ) \log {\left (x + \frac {- A b^{2} + B a b + a^{2} c^{2} \sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right ) - 2 a b^{2} c \sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right ) + b^{4} \sqrt {- \frac {1}{\left (a c - b^{2}\right )^{3}}} \left (- A b + B a\right )}{- A a b + B a^{2}} \right )}}{4} + \frac {- A c + B b + x \left (- A b + B a\right )}{2 a c^{2} - 2 b^{2} c + x^{2} \left (2 a^{2} c - 2 a b^{2}\right ) + x \left (4 a b c - 4 b^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A*x+B)/(a*x**2+2*b*x+c)**2,x)

[Out]

-sqrt(-1/(a*c - b**2)**3)*(-A*b + B*a)*log(x + (-A*b**2 + B*a*b - a**2*c**2*sqrt(-1/(a*c - b**2)**3)*(-A*b + B
*a) + 2*a*b**2*c*sqrt(-1/(a*c - b**2)**3)*(-A*b + B*a) - b**4*sqrt(-1/(a*c - b**2)**3)*(-A*b + B*a))/(-A*a*b +
B*a**2))/4 + sqrt(-1/(a*c - b**2)**3)*(-A*b + B*a)*log(x + (-A*b**2 + B*a*b + a**2*c**2*sqrt(-1/(a*c - b**2)*
*3)*(-A*b + B*a) - 2*a*b**2*c*sqrt(-1/(a*c - b**2)**3)*(-A*b + B*a) + b**4*sqrt(-1/(a*c - b**2)**3)*(-A*b + B*
a))/(-A*a*b + B*a**2))/4 + (-A*c + B*b + x*(-A*b + B*a))/(2*a*c**2 - 2*b**2*c + x**2*(2*a**2*c - 2*a*b**2) + x
*(4*a*b*c - 4*b**3))

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