3.195 \(\int \frac{\text{b1}+\text{c1} x}{(a+2 b x+c x^2)^2} \, dx\)

Optimal. Leaf size=89 \[ \frac{(\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{2 \left (b^2-a c\right )^{3/2}}-\frac{-a \text{c1}+x (\text{b1} c-b \text{c1})+b \text{b1}}{2 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )} \]

[Out]

-(b*b1 - a*c1 + (b1*c - b*c1)*x)/(2*(b^2 - a*c)*(a + 2*b*x + c*x^2)) + ((b1*c - b*c1)*ArcTanh[(b + c*x)/Sqrt[b
^2 - a*c]])/(2*(b^2 - a*c)^(3/2))

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Rubi [A]  time = 0.0447332, antiderivative size = 89, normalized size of antiderivative = 1., 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} \[ \frac{(\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{2 \left (b^2-a c\right )^{3/2}}-\frac{-a \text{c1}+x (\text{b1} c-b \text{c1})+b \text{b1}}{2 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(b1 + c1*x)/(a + 2*b*x + c*x^2)^2,x]

[Out]

-(b*b1 - a*c1 + (b1*c - b*c1)*x)/(2*(b^2 - a*c)*(a + 2*b*x + c*x^2)) + ((b1*c - b*c1)*ArcTanh[(b + c*x)/Sqrt[b
^2 - a*c]])/(2*(b^2 - a*c)^(3/2))

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]

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{\text{b1}+\text{c1} x}{\left (a+2 b x+c x^2\right )^2} \, dx &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{2 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )}-\frac{(\text{b1} c-b \text{c1}) \int \frac{1}{a+2 b x+c x^2} \, dx}{2 \left (b^2-a c\right )}\\ &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{2 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )}+\frac{(\text{b1} c-b \text{c1}) \operatorname{Subst}\left (\int \frac{1}{4 \left (b^2-a c\right )-x^2} \, dx,x,2 b+2 c x\right )}{b^2-a c}\\ &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{2 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )}+\frac{(\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{2 \left (b^2-a c\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0800663, size = 88, normalized size = 0.99 \[ \frac{\frac{(b \text{c1}-\text{b1} c) \tan ^{-1}\left (\frac{b+c x}{\sqrt{a c-b^2}}\right )}{\sqrt{a c-b^2}}+\frac{a \text{c1}-b \text{b1}+b \text{c1} x-\text{b1} c x}{a+x (2 b+c x)}}{2 \left (b^2-a c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(b1 + c1*x)/(a + 2*b*x + c*x^2)^2,x]

[Out]

((-(b*b1) + a*c1 - b1*c*x + b*c1*x)/(a + x*(2*b + c*x)) + ((-(b1*c) + b*c1)*ArcTan[(b + c*x)/Sqrt[-b^2 + a*c]]
)/Sqrt[-b^2 + a*c])/(2*(b^2 - a*c))

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Maple [A]  time = 0.003, size = 146, normalized size = 1.6 \begin{align*}{\frac{ \left ( -2\,b{\it c1}+2\,{\it b1}\,c \right ) x+2\,b{\it b1}-2\,a{\it c1}}{ \left ( 4\,ac-4\,{b}^{2} \right ) \left ( c{x}^{2}+2\,bx+a \right ) }}-2\,{\frac{b{\it c1}}{ \left ( 4\,ac-4\,{b}^{2} \right ) \sqrt{ac-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,cx+2\,b}{\sqrt{ac-{b}^{2}}}} \right ) }+2\,{\frac{{\it b1}\,c}{ \left ( 4\,ac-4\,{b}^{2} \right ) \sqrt{ac-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,cx+2\,b}{\sqrt{ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c1*x+b1)/(c*x^2+2*b*x+a)^2,x)

[Out]

((-2*b*c1+2*b1*c)*x+2*b*b1-2*a*c1)/(4*a*c-4*b^2)/(c*x^2+2*b*x+a)-2/(4*a*c-4*b^2)/(a*c-b^2)^(1/2)*arctan(1/2*(2
*c*x+2*b)/(a*c-b^2)^(1/2))*b*c1+2/(4*a*c-4*b^2)/(a*c-b^2)^(1/2)*arctan(1/2*(2*c*x+2*b)/(a*c-b^2)^(1/2))*b1*c

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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((c1*x+b1)/(c*x^2+2*b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.84848, size = 964, normalized size = 10.83 \begin{align*} \left [-\frac{2 \, b^{3} b_{1} - 2 \, a b b_{1} c -{\left (a b_{1} c - a b c_{1} +{\left (b_{1} c^{2} - b c c_{1}\right )} x^{2} + 2 \,{\left (b b_{1} c - b^{2} c_{1}\right )} x\right )} \sqrt{b^{2} - a c} \log \left (\frac{c^{2} x^{2} + 2 \, b c x + 2 \, b^{2} - a c + 2 \, \sqrt{b^{2} - a c}{\left (c x + b\right )}}{c x^{2} + 2 \, b x + a}\right ) - 2 \,{\left (a b^{2} - a^{2} c\right )} c_{1} + 2 \,{\left (b^{2} b_{1} c - a b_{1} c^{2} -{\left (b^{3} - a b c\right )} c_{1}\right )} x}{4 \,{\left (a b^{4} - 2 \, a^{2} b^{2} c + a^{3} c^{2} +{\left (b^{4} c - 2 \, a b^{2} c^{2} + a^{2} c^{3}\right )} x^{2} + 2 \,{\left (b^{5} - 2 \, a b^{3} c + a^{2} b c^{2}\right )} x\right )}}, -\frac{b^{3} b_{1} - a b b_{1} c -{\left (a b_{1} c - a b c_{1} +{\left (b_{1} c^{2} - b c c_{1}\right )} x^{2} + 2 \,{\left (b b_{1} c - b^{2} c_{1}\right )} x\right )} \sqrt{-b^{2} + a c} \arctan \left (-\frac{\sqrt{-b^{2} + a c}{\left (c x + b\right )}}{b^{2} - a c}\right ) -{\left (a b^{2} - a^{2} c\right )} c_{1} +{\left (b^{2} b_{1} c - a b_{1} c^{2} -{\left (b^{3} - a b c\right )} c_{1}\right )} x}{2 \,{\left (a b^{4} - 2 \, a^{2} b^{2} c + a^{3} c^{2} +{\left (b^{4} c - 2 \, a b^{2} c^{2} + a^{2} c^{3}\right )} x^{2} + 2 \,{\left (b^{5} - 2 \, a b^{3} c + a^{2} b c^{2}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x^2+2*b*x+a)^2,x, algorithm="fricas")

[Out]

[-1/4*(2*b^3*b1 - 2*a*b*b1*c - (a*b1*c - a*b*c1 + (b1*c^2 - b*c*c1)*x^2 + 2*(b*b1*c - b^2*c1)*x)*sqrt(b^2 - a*
c)*log((c^2*x^2 + 2*b*c*x + 2*b^2 - a*c + 2*sqrt(b^2 - a*c)*(c*x + b))/(c*x^2 + 2*b*x + a)) - 2*(a*b^2 - a^2*c
)*c1 + 2*(b^2*b1*c - a*b1*c^2 - (b^3 - a*b*c)*c1)*x)/(a*b^4 - 2*a^2*b^2*c + a^3*c^2 + (b^4*c - 2*a*b^2*c^2 + a
^2*c^3)*x^2 + 2*(b^5 - 2*a*b^3*c + a^2*b*c^2)*x), -1/2*(b^3*b1 - a*b*b1*c - (a*b1*c - a*b*c1 + (b1*c^2 - b*c*c
1)*x^2 + 2*(b*b1*c - b^2*c1)*x)*sqrt(-b^2 + a*c)*arctan(-sqrt(-b^2 + a*c)*(c*x + b)/(b^2 - a*c)) - (a*b^2 - a^
2*c)*c1 + (b^2*b1*c - a*b1*c^2 - (b^3 - a*b*c)*c1)*x)/(a*b^4 - 2*a^2*b^2*c + a^3*c^2 + (b^4*c - 2*a*b^2*c^2 +
a^2*c^3)*x^2 + 2*(b^5 - 2*a*b^3*c + a^2*b*c^2)*x)]

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Sympy [B]  time = 1.0452, size = 323, normalized size = 3.63 \begin{align*} \frac{\sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) \log{\left (x + \frac{- a^{2} c^{2} \sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) + 2 a b^{2} c \sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) - b^{4} \sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) + b^{2} c_{1} - b b_{1} c}{b c c_{1} - b_{1} c^{2}} \right )}}{4} - \frac{\sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) \log{\left (x + \frac{a^{2} c^{2} \sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) - 2 a b^{2} c \sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) + b^{4} \sqrt{- \frac{1}{\left (a c - b^{2}\right )^{3}}} \left (b c_{1} - b_{1} c\right ) + b^{2} c_{1} - b b_{1} c}{b c c_{1} - b_{1} c^{2}} \right )}}{4} - \frac{a c_{1} - b b_{1} + x \left (b c_{1} - b_{1} c\right )}{2 a^{2} c - 2 a b^{2} + x^{2} \left (2 a c^{2} - 2 b^{2} c\right ) + x \left (4 a b c - 4 b^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x**2+2*b*x+a)**2,x)

[Out]

sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c)*log(x + (-a**2*c**2*sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c) + 2*a*b**2*c
*sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c) - b**4*sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c) + b**2*c1 - b*b1*c)/(b*c
*c1 - b1*c**2))/4 - sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c)*log(x + (a**2*c**2*sqrt(-1/(a*c - b**2)**3)*(b*c1 -
b1*c) - 2*a*b**2*c*sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c) + b**4*sqrt(-1/(a*c - b**2)**3)*(b*c1 - b1*c) + b**
2*c1 - b*b1*c)/(b*c*c1 - b1*c**2))/4 - (a*c1 - b*b1 + x*(b*c1 - b1*c))/(2*a**2*c - 2*a*b**2 + x**2*(2*a*c**2 -
2*b**2*c) + x*(4*a*b*c - 4*b**3))

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Giac [A]  time = 1.0706, size = 124, normalized size = 1.39 \begin{align*} -\frac{{\left (b_{1} c - b c_{1}\right )} \arctan \left (\frac{c x + b}{\sqrt{-b^{2} + a c}}\right )}{2 \,{\left (b^{2} - a c\right )} \sqrt{-b^{2} + a c}} - \frac{b_{1} c x - b c_{1} x + b b_{1} - a c_{1}}{2 \,{\left (c x^{2} + 2 \, b x + a\right )}{\left (b^{2} - a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x^2+2*b*x+a)^2,x, algorithm="giac")

[Out]

-1/2*(b1*c - b*c1)*arctan((c*x + b)/sqrt(-b^2 + a*c))/((b^2 - a*c)*sqrt(-b^2 + a*c)) - 1/2*(b1*c*x - b*c1*x +
b*b1 - a*c1)/((c*x^2 + 2*b*x + a)*(b^2 - a*c))