3.1542 \(\int \frac{(b+2 c x) (d+e x)^2}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac{e (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{(d+e x)^2}{2 \left (a+b x+c x^2\right )^2} \]

[Out]

-(d + e*x)^2/(2*(a + b*x + c*x^2)^2) - (e*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)) +
(2*e*(2*c*d - b*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 = 0.0615039, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {768, 638, 618, 206} \[ -\frac{e (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{2 e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{(d+e x)^2}{2 \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d + e*x)^2/(2*(a + b*x + c*x^2)^2) - (e*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)) +
(2*e*(2*c*d - b*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 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{(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac{(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}+e \int \frac{d+e x}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac{(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}-\frac{e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac{(e (2 c d-b e)) \int \frac{1}{a+b x+c x^2} \, dx}{b^2-4 a c}\\ &=-\frac{(d+e x)^2}{2 \left (a+b x+c x^2\right )^2}-\frac{e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{(2 e (2 c d-b 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}{2 \left (a+b x+c x^2\right )^2}-\frac{e (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{2 e (2 c d-b 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 = 0.211893, size = 143, normalized size = 1.28 \[ \frac{1}{2} \left (\frac{e \left (4 c (c d x-2 a e)+b^2 e+2 b c (d-e x)\right )}{c \left (4 a c-b^2\right ) (a+x (b+c x))}-\frac{4 e (b e-2 c d) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+\frac{e^2 (a+b x)-c d (d+2 e x)}{c (a+x (b+c x))^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*(b^2*e + 4*c*(-2*a*e + c*d*x) + 2*b*c*(d - e*x)))/(c*(-b^2 + 4*a*c)*(a + x*(b + c*x))) + (e^2*(a + b*x) -
c*d*(d + 2*e*x))/(c*(a + x*(b + c*x))^2) - (4*e*(-2*c*d + b*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 [B]  time = 0.013, size = 229, normalized size = 2. \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ( -{\frac{ce \left ( be-2\,cd \right ){x}^{3}}{4\,ac-{b}^{2}}}-{\frac{e \left ( 8\,ace+{b}^{2}e-6\,bcd \right ){x}^{2}}{8\,ac-2\,{b}^{2}}}-{\frac{e \left ( 3\,abe+2\,acd-2\,{b}^{2}d \right ) x}{4\,ac-{b}^{2}}}-{\frac{4\,{a}^{2}{e}^{2}-2\,abde+4\,ac{d}^{2}-{b}^{2}{d}^{2}}{8\,ac-2\,{b}^{2}}} \right ) }-2\,{\frac{b{e}^{2}}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+4\,{\frac{cde}{ \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)^2/(c*x^2+b*x+a)^3,x)

[Out]

(-c*e*(b*e-2*c*d)/(4*a*c-b^2)*x^3-1/2*e*(8*a*c*e+b^2*e-6*b*c*d)/(4*a*c-b^2)*x^2-e*(3*a*b*e+2*a*c*d-2*b^2*d)/(4
*a*c-b^2)*x-1/2*(4*a^2*e^2-2*a*b*d*e+4*a*c*d^2-b^2*d^2)/(4*a*c-b^2))/(c*x^2+b*x+a)^2-2*e^2/(4*a*c-b^2)^(3/2)*a
rctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b+4*e/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c*d

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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)^2/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.00193, size = 2111, normalized size = 18.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(2*(b^2*c^2 - 4*a*c^3)*d*e - (b^3*c - 4*a*b*c^2)*e^2)*x^3 + (b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^2 + 2*(a
*b^3 - 4*a^2*b*c)*d*e - 4*(a^2*b^2 - 4*a^3*c)*e^2 + (6*(b^3*c - 4*a*b*c^2)*d*e - (b^4 + 4*a*b^2*c - 32*a^2*c^2
)*e^2)*x^2 - 2*(2*a^2*c*d*e - a^2*b*e^2 + (2*c^3*d*e - b*c^2*e^2)*x^4 + 2*(2*b*c^2*d*e - b^2*c*e^2)*x^3 + (2*(
b^2*c + 2*a*c^2)*d*e - (b^3 + 2*a*b*c)*e^2)*x^2 + 2*(2*a*b*c*d*e - a*b^2*e^2)*x)*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*(2*(b^4 - 5*a*b^2*c + 4*a^
2*c^2)*d*e - 3*(a*b^3 - 4*a^2*b*c)*e^2)*x)/(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*(2*(b^2*c^2 - 4*a*c^3)*d*e - (b^3*c - 4*a*b*c^2)*e^2)*x^3 + (b^4 - 8*a*b^
2*c + 16*a^2*c^2)*d^2 + 2*(a*b^3 - 4*a^2*b*c)*d*e - 4*(a^2*b^2 - 4*a^3*c)*e^2 + (6*(b^3*c - 4*a*b*c^2)*d*e - (
b^4 + 4*a*b^2*c - 32*a^2*c^2)*e^2)*x^2 - 4*(2*a^2*c*d*e - a^2*b*e^2 + (2*c^3*d*e - b*c^2*e^2)*x^4 + 2*(2*b*c^2
*d*e - b^2*c*e^2)*x^3 + (2*(b^2*c + 2*a*c^2)*d*e - (b^3 + 2*a*b*c)*e^2)*x^2 + 2*(2*a*b*c*d*e - a*b^2*e^2)*x)*s
qrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*d
*e - 3*(a*b^3 - 4*a^2*b*c)*e^2)*x)/(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 = 12.7692, size = 530, normalized size = 4.73 \begin{align*} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) \log{\left (x + \frac{- 16 a^{2} c^{2} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + 8 a b^{2} c e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) - b^{4} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{2} e^{2} - 2 b c d e}{2 b c e^{2} - 4 c^{2} d e} \right )} - e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) \log{\left (x + \frac{16 a^{2} c^{2} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) - 8 a b^{2} c e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{4} e \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \left (b e - 2 c d\right ) + b^{2} e^{2} - 2 b c d e}{2 b c e^{2} - 4 c^{2} d e} \right )} - \frac{4 a^{2} e^{2} - 2 a b d e + 4 a c d^{2} - b^{2} d^{2} + x^{3} \left (2 b c e^{2} - 4 c^{2} d e\right ) + x^{2} \left (8 a c e^{2} + b^{2} e^{2} - 6 b c d e\right ) + x \left (6 a b e^{2} + 4 a c d e - 4 b^{2} d 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)**2/(c*x**2+b*x+a)**3,x)

[Out]

e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d)*log(x + (-16*a**2*c**2*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) +
8*a*b**2*c*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) - b**4*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**
2*e**2 - 2*b*c*d*e)/(2*b*c*e**2 - 4*c**2*d*e)) - e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d)*log(x + (16*a**2*c
**2*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) - 8*a*b**2*c*e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**4*
e*sqrt(-1/(4*a*c - b**2)**3)*(b*e - 2*c*d) + b**2*e**2 - 2*b*c*d*e)/(2*b*c*e**2 - 4*c**2*d*e)) - (4*a**2*e**2
- 2*a*b*d*e + 4*a*c*d**2 - b**2*d**2 + x**3*(2*b*c*e**2 - 4*c**2*d*e) + x**2*(8*a*c*e**2 + b**2*e**2 - 6*b*c*d
*e) + x*(6*a*b*e**2 + 4*a*c*d*e - 4*b**2*d*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 = 1.17463, size = 247, normalized size = 2.21 \begin{align*} -\frac{2 \,{\left (2 \, c d e - b e^{2}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{4 \, c^{2} d x^{3} e - 2 \, b c x^{3} e^{2} + 6 \, b c d x^{2} e - b^{2} x^{2} e^{2} - 8 \, a c x^{2} e^{2} + 4 \, b^{2} d x e - 4 \, a c d x e + b^{2} d^{2} - 4 \, a c d^{2} - 6 \, a b x e^{2} + 2 \, a b d e - 4 \, a^{2} e^{2}}{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)^2/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-2*(2*c*d*e - b*e^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)) - 1/2*(4*c^2*d*
x^3*e - 2*b*c*x^3*e^2 + 6*b*c*d*x^2*e - b^2*x^2*e^2 - 8*a*c*x^2*e^2 + 4*b^2*d*x*e - 4*a*c*d*x*e + b^2*d^2 - 4*
a*c*d^2 - 6*a*b*x*e^2 + 2*a*b*d*e - 4*a^2*e^2)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c))