3.1543 \(\int \frac{(b+2 c x) (d+e x)}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=91 \[ -\frac{e (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{2 c e \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{d+e x}{2 \left (a+b x+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0423818, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {768, 614, 618, 206} \[ -\frac{e (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{2 c e \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{d+e x}{2 \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(d + e*x))/(a + b*x + c*x^2)^3,x]

[Out]

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

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rule 614

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

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

Mathematica [A]  time = 0.137787, size = 93, normalized size = 1.02 \[ \frac{1}{2} \left (-\frac{e (b+2 c x)}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac{4 c e \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}-\frac{d+e x}{(a+x (b+c x))^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(d + e*x))/(a + b*x + c*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.008, size = 146, normalized size = 1.6 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ({\frac{{c}^{2}e{x}^{3}}{4\,ac-{b}^{2}}}+{\frac{3\,ceb{x}^{2}}{8\,ac-2\,{b}^{2}}}-{\frac{e \left ( ac-{b}^{2} \right ) x}{4\,ac-{b}^{2}}}+{\frac{abe-4\,acd+{b}^{2}d}{8\,ac-2\,{b}^{2}}} \right ) }+2\,{\frac{ce}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\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*x+b)*(e*x+d)/(c*x^2+b*x+a)^3,x)

[Out]

(c^2*e/(4*a*c-b^2)*x^3+3/2*b*c*e/(4*a*c-b^2)*x^2-e*(a*c-b^2)/(4*a*c-b^2)*x+1/2*(a*b*e-4*a*c*d+b^2*d)/(4*a*c-b^
2))/(c*x^2+b*x+a)^2+2*c*e/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(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((2*c*x+b)*(e*x+d)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.79851, size = 1466, normalized size = 16.11 \begin{align*} \left [-\frac{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e x^{3} + 3 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} e x^{2} + 2 \,{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} e x + 2 \,{\left (c^{3} e x^{4} + 2 \, b c^{2} e x^{3} + 2 \, a b c e x + a^{2} c e +{\left (b^{2} c + 2 \, a c^{2}\right )} e x^{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 (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d +{\left (a b^{3} - 4 \, a^{2} b c\right )} e}{2 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}}, -\frac{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e x^{3} + 3 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} e x^{2} + 2 \,{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} e x - 4 \,{\left (c^{3} e x^{4} + 2 \, b c^{2} e x^{3} + 2 \, a b c e x + a^{2} c e +{\left (b^{2} c + 2 \, a c^{2}\right )} e x^{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 (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d +{\left (a b^{3} - 4 \, a^{2} b c\right )} e}{2 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

[-1/2*(2*(b^2*c^2 - 4*a*c^3)*e*x^3 + 3*(b^3*c - 4*a*b*c^2)*e*x^2 + 2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*e*x + 2*(c^
3*e*x^4 + 2*b*c^2*e*x^3 + 2*a*b*c*e*x + a^2*c*e + (b^2*c + 2*a*c^2)*e*x^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)) + (b^4 - 8*a*b^2*c + 16*a^2*c^2)*d +
(a*b^3 - 4*a^2*b*c)*e)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^
5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b
*c^2)*x), -1/2*(2*(b^2*c^2 - 4*a*c^3)*e*x^3 + 3*(b^3*c - 4*a*b*c^2)*e*x^2 + 2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*e*
x - 4*(c^3*e*x^4 + 2*b*c^2*e*x^3 + 2*a*b*c*e*x + a^2*c*e + (b^2*c + 2*a*c^2)*e*x^2)*sqrt(-b^2 + 4*a*c)*arctan(
-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + (b^4 - 8*a*b^2*c + 16*a^2*c^2)*d + (a*b^3 - 4*a^2*b*c)*e)/(a^
2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*
b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)]

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Sympy [B]  time = 3.3819, size = 377, normalized size = 4.14 \begin{align*} - c e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x + \frac{- 16 a^{2} c^{3} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{2} c^{2} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - b^{4} c e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b c e}{2 c^{2} e} \right )} + c e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (x + \frac{16 a^{2} c^{3} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{2} c^{2} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{4} c e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b c e}{2 c^{2} e} \right )} + \frac{a b e - 4 a c d + b^{2} d + 3 b c e x^{2} + 2 c^{2} e x^{3} + x \left (- 2 a c e + 2 b^{2} e\right )}{8 a^{3} c - 2 a^{2} b^{2} + x^{4} \left (8 a c^{3} - 2 b^{2} c^{2}\right ) + x^{3} \left (16 a b c^{2} - 4 b^{3} c\right ) + x^{2} \left (16 a^{2} c^{2} + 4 a b^{2} c - 2 b^{4}\right ) + x \left (16 a^{2} b c - 4 a b^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)/(c*x**2+b*x+a)**3,x)

[Out]

-c*e*sqrt(-1/(4*a*c - b**2)**3)*log(x + (-16*a**2*c**3*e*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**2*c**2*e*sqrt(-1/
(4*a*c - b**2)**3) - b**4*c*e*sqrt(-1/(4*a*c - b**2)**3) + b*c*e)/(2*c**2*e)) + c*e*sqrt(-1/(4*a*c - b**2)**3)
*log(x + (16*a**2*c**3*e*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**2*c**2*e*sqrt(-1/(4*a*c - b**2)**3) + b**4*c*e*sq
rt(-1/(4*a*c - b**2)**3) + b*c*e)/(2*c**2*e)) + (a*b*e - 4*a*c*d + b**2*d + 3*b*c*e*x**2 + 2*c**2*e*x**3 + x*(
-2*a*c*e + 2*b**2*e))/(8*a**3*c - 2*a**2*b**2 + x**4*(8*a*c**3 - 2*b**2*c**2) + x**3*(16*a*b*c**2 - 4*b**3*c)
+ x**2*(16*a**2*c**2 + 4*a*b**2*c - 2*b**4) + x*(16*a**2*b*c - 4*a*b**3))

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Giac [A]  time = 1.21571, size = 165, normalized size = 1.81 \begin{align*} -\frac{2 \, c \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right ) e}{{\left (b^{2} - 4 \, a c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{2 \, c^{2} x^{3} e + 3 \, b c x^{2} e + 2 \, b^{2} x e - 2 \, a c x e + b^{2} d - 4 \, a c d + a b e}{2 \,{\left (c x^{2} + b x + a\right )}^{2}{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

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