3.2374 \(\int \frac{(d+e x) (f+g x)}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=221 \[ \frac{(b+2 c x) \left (c (2 a e g-3 b (d g+e f))+b^2 e g+6 c^2 d f\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{-x \left (-c (2 a e g+b d g+b e f)+b^2 e g+2 c^2 d f\right )-b (a e g+c d f)+2 a c (d g+e f)}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \left (c (2 a e g-3 b (d g+e f))+b^2 e g+6 c^2 d f\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c*(b*e*f + b*d*g + 2*a*e*g))*x)/(2*c*(b^2 - 4*
a*c)*(a + b*x + c*x^2)^2) + ((6*c^2*d*f + b^2*e*g + c*(2*a*e*g - 3*b*(e*f + d*g)))*(b + 2*c*x))/(2*c*(b^2 - 4*
a*c)^2*(a + b*x + c*x^2)) - (2*(6*c^2*d*f + b^2*e*g + c*(2*a*e*g - 3*b*(e*f + d*g)))*ArcTanh[(b + 2*c*x)/Sqrt[
b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

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Rubi [A]  time = 0.209247, antiderivative size = 219, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {777, 614, 618, 206} \[ \frac{(b+2 c x) \left (2 a c e g+b^2 e g-3 b c (d g+e f)+6 c^2 d f\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{-x \left (-c (2 a e g+b d g+b e f)+b^2 e g+2 c^2 d f\right )-b (a e g+c d f)+2 a c (d g+e f)}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{2 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ) \left (2 a c e g+b^2 e g-3 b c (d g+e f)+6 c^2 d f\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c*(b*e*f + b*d*g + 2*a*e*g))*x)/(2*c*(b^2 - 4*
a*c)*(a + b*x + c*x^2)^2) + ((6*c^2*d*f + b^2*e*g + 2*a*c*e*g - 3*b*c*(e*f + d*g))*(b + 2*c*x))/(2*c*(b^2 - 4*
a*c)^2*(a + b*x + c*x^2)) - (2*(6*c^2*d*f + b^2*e*g + 2*a*c*e*g - 3*b*c*(e*f + d*g))*ArcTanh[(b + 2*c*x)/Sqrt[
b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

Rule 777

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

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{(d+e x) (f+g x)}{\left (a+b x+c x^2\right )^3} \, dx &=\frac{2 a c (e f+d g)-b (c d f+a e g)-\left (2 c^2 d f+b^2 e g-c (b e f+b d g+2 a e g)\right ) x}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac{\left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right ) \int \frac{1}{\left (a+b x+c x^2\right )^2} \, dx}{2 c \left (b^2-4 a c\right )}\\ &=\frac{2 a c (e f+d g)-b (c d f+a e g)-\left (2 c^2 d f+b^2 e g-c (b e f+b d g+2 a e g)\right ) x}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{\left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right ) (b+2 c x)}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac{\left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right ) \int \frac{1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=\frac{2 a c (e f+d g)-b (c d f+a e g)-\left (2 c^2 d f+b^2 e g-c (b e f+b d g+2 a e g)\right ) x}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{\left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right ) (b+2 c x)}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{\left (2 \left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right )^2}\\ &=\frac{2 a c (e f+d g)-b (c d f+a e g)-\left (2 c^2 d f+b^2 e g-c (b e f+b d g+2 a e g)\right ) x}{2 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac{\left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right ) (b+2 c x)}{2 c \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{2 \left (6 c^2 d f+b^2 e g+2 a c e g-3 b c (e f+d g)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.484153, size = 216, normalized size = 0.98 \[ \frac{1}{2} \left (\frac{(b+2 c x) \left (2 a c e g+b^2 e g-3 b c (d g+e f)+6 c^2 d f\right )}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac{a b e g-2 a c (d g+e (f+g x))+b^2 e g x+b c (d (f-g x)-e f x)+2 c^2 d f x}{c \left (4 a c-b^2\right ) (a+x (b+c x))^2}+\frac{4 \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right ) \left (2 a c e g+b^2 e g-3 b c (d g+e f)+6 c^2 d f\right )}{\left (4 a c-b^2\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((6*c^2*d*f + b^2*e*g + 2*a*c*e*g - 3*b*c*(e*f + d*g))*(b + 2*c*x))/(c*(b^2 - 4*a*c)^2*(a + x*(b + c*x))) + (
a*b*e*g + 2*c^2*d*f*x + b^2*e*g*x + b*c*(-(e*f*x) + d*(f - g*x)) - 2*a*c*(d*g + e*(f + g*x)))/(c*(-b^2 + 4*a*c
)*(a + x*(b + c*x))^2) + (4*(6*c^2*d*f + b^2*e*g + 2*a*c*e*g - 3*b*c*(e*f + d*g))*ArcTan[(b + 2*c*x)/Sqrt[-b^2
+ 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/2

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Maple [B]  time = 0.01, size = 591, normalized size = 2.7 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+bx+a \right ) ^{2}} \left ({\frac{c \left ( 2\,aceg+{b}^{2}eg-3\,bcdg-3\,bcef+6\,{c}^{2}df \right ){x}^{3}}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}+{\frac{3\,b \left ( 2\,aceg+{b}^{2}eg-3\,bcdg-3\,bcef+6\,{c}^{2}df \right ){x}^{2}}{32\,{a}^{2}{c}^{2}-16\,a{b}^{2}c+2\,{b}^{4}}}-{\frac{ \left ( 2\,{a}^{2}ceg-5\,a{b}^{2}eg+5\,abcdg+5\,abcef-10\,a{c}^{2}df+{b}^{3}dg+{b}^{3}ef-2\,{b}^{2}cdf \right ) x}{16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4}}}+{\frac{6\,{a}^{2}beg-8\,{a}^{2}cdg-8\,{a}^{2}cef-a{b}^{2}dg-a{b}^{2}ef+10\,abcdf-{b}^{3}df}{32\,{a}^{2}{c}^{2}-16\,a{b}^{2}c+2\,{b}^{4}}} \right ) }+4\,{\frac{aceg}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{{b}^{2}eg}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-6\,{\frac{bcdg}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-6\,{\frac{bcef}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+12\,{\frac{{c}^{2}df}{ \left ( 16\,{a}^{2}{c}^{2}-8\,a{b}^{2}c+{b}^{4} \right ) \sqrt{4\,ac-{b}^{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((e*x+d)*(g*x+f)/(c*x^2+b*x+a)^3,x)

[Out]

(c*(2*a*c*e*g+b^2*e*g-3*b*c*d*g-3*b*c*e*f+6*c^2*d*f)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+3/2*b*(2*a*c*e*g+b^2*e*g-3
*b*c*d*g-3*b*c*e*f+6*c^2*d*f)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2-(2*a^2*c*e*g-5*a*b^2*e*g+5*a*b*c*d*g+5*a*b*c*e*f-
10*a*c^2*d*f+b^3*d*g+b^3*e*f-2*b^2*c*d*f)/(16*a^2*c^2-8*a*b^2*c+b^4)*x+1/2*(6*a^2*b*e*g-8*a^2*c*d*g-8*a^2*c*e*
f-a*b^2*d*g-a*b^2*e*f+10*a*b*c*d*f-b^3*d*f)/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2+4/(16*a^2*c^2-8*a*b^2*
c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*c*e*g+2/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^
(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2*e*g-6/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x
+b)/(4*a*c-b^2)^(1/2))*b*c*d*g-6/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/
2))*b*c*e*f+12/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c^2*d*f

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.8204, size = 3875, normalized size = 17.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(3*(2*(b^2*c^3 - 4*a*c^4)*d - (b^3*c^2 - 4*a*b*c^3)*e)*f - (3*(b^3*c^2 - 4*a*b*c^3)*d - (b^4*c - 2*a*b
^2*c^2 - 8*a^2*c^3)*e)*g)*x^3 + 3*(3*(2*(b^3*c^2 - 4*a*b*c^3)*d - (b^4*c - 4*a*b^2*c^2)*e)*f - (3*(b^4*c - 4*a
*b^2*c^2)*d - (b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*e)*g)*x^2 + 2*((3*(2*c^4*d - b*c^3*e)*f - (3*b*c^3*d - (b^2*c^2
+ 2*a*c^3)*e)*g)*x^4 + 2*(3*(2*b*c^3*d - b^2*c^2*e)*f - (3*b^2*c^2*d - (b^3*c + 2*a*b*c^2)*e)*g)*x^3 + (3*(2*(
b^2*c^2 + 2*a*c^3)*d - (b^3*c + 2*a*b*c^2)*e)*f - (3*(b^3*c + 2*a*b*c^2)*d - (b^4 + 4*a*b^2*c + 4*a^2*c^2)*e)*
g)*x^2 + 3*(2*a^2*c^2*d - a^2*b*c*e)*f - (3*a^2*b*c*d - (a^2*b^2 + 2*a^3*c)*e)*g + 2*(3*(2*a*b*c^2*d - a*b^2*c
*e)*f - (3*a*b^2*c*d - (a*b^3 + 2*a^2*b*c)*e)*g)*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)) - ((b^5 - 14*a*b^3*c + 40*a^2*b*c^2)*d + (a*b^4 + 4*a^2*b^2
*c - 32*a^3*c^2)*e)*f - ((a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*d - 6*(a^2*b^3 - 4*a^3*b*c)*e)*g + 2*((2*(b^4*c +
a*b^2*c^2 - 20*a^2*c^3)*d - (b^5 + a*b^3*c - 20*a^2*b*c^2)*e)*f - ((b^5 + a*b^3*c - 20*a^2*b*c^2)*d - (5*a*b^4
- 22*a^2*b^2*c + 8*a^3*c^2)*e)*g)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*
b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b
^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^
2 - 64*a^4*b*c^3)*x), 1/2*(2*(3*(2*(b^2*c^3 - 4*a*c^4)*d - (b^3*c^2 - 4*a*b*c^3)*e)*f - (3*(b^3*c^2 - 4*a*b*c^
3)*d - (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*e)*g)*x^3 + 3*(3*(2*(b^3*c^2 - 4*a*b*c^3)*d - (b^4*c - 4*a*b^2*c^2)*e
)*f - (3*(b^4*c - 4*a*b^2*c^2)*d - (b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*e)*g)*x^2 - 4*((3*(2*c^4*d - b*c^3*e)*f - (
3*b*c^3*d - (b^2*c^2 + 2*a*c^3)*e)*g)*x^4 + 2*(3*(2*b*c^3*d - b^2*c^2*e)*f - (3*b^2*c^2*d - (b^3*c + 2*a*b*c^2
)*e)*g)*x^3 + (3*(2*(b^2*c^2 + 2*a*c^3)*d - (b^3*c + 2*a*b*c^2)*e)*f - (3*(b^3*c + 2*a*b*c^2)*d - (b^4 + 4*a*b
^2*c + 4*a^2*c^2)*e)*g)*x^2 + 3*(2*a^2*c^2*d - a^2*b*c*e)*f - (3*a^2*b*c*d - (a^2*b^2 + 2*a^3*c)*e)*g + 2*(3*(
2*a*b*c^2*d - a*b^2*c*e)*f - (3*a*b^2*c*d - (a*b^3 + 2*a^2*b*c)*e)*g)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2
+ 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - ((b^5 - 14*a*b^3*c + 40*a^2*b*c^2)*d + (a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2
)*e)*f - ((a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*d - 6*(a^2*b^3 - 4*a^3*b*c)*e)*g + 2*((2*(b^4*c + a*b^2*c^2 - 20*
a^2*c^3)*d - (b^5 + a*b^3*c - 20*a^2*b*c^2)*e)*f - ((b^5 + a*b^3*c - 20*a^2*b*c^2)*d - (5*a*b^4 - 22*a^2*b^2*c
+ 8*a^3*c^2)*e)*g)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^
2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c
+ 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^
3)*x)]

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Sympy [B]  time = 17.8789, size = 1234, normalized size = 5.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(4*a*c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f)*log(x + (-64*a**3*c**3
*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) + 48*a**2*b**2*c**2*sq
rt(-1/(4*a*c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) - 12*a*b**4*c*sqrt(-1/(4*
a*c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) + 2*a*b*c*e*g + b**6*sqrt(-1/(4*a*
c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) + b**3*e*g - 3*b**2*c*d*g - 3*b**2*c
*e*f + 6*b*c**2*d*f)/(4*a*c**2*e*g + 2*b**2*c*e*g - 6*b*c**2*d*g - 6*b*c**2*e*f + 12*c**3*d*f)) + sqrt(-1/(4*a
*c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f)*log(x + (64*a**3*c**3*sqrt(-1/(4*a*
c - b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) - 48*a**2*b**2*c**2*sqrt(-1/(4*a*c -
b**2)**5)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) + 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5
)*(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) + 2*a*b*c*e*g - b**6*sqrt(-1/(4*a*c - b**2)**5)*
(2*a*c*e*g + b**2*e*g - 3*b*c*d*g - 3*b*c*e*f + 6*c**2*d*f) + b**3*e*g - 3*b**2*c*d*g - 3*b**2*c*e*f + 6*b*c**
2*d*f)/(4*a*c**2*e*g + 2*b**2*c*e*g - 6*b*c**2*d*g - 6*b*c**2*e*f + 12*c**3*d*f)) + (6*a**2*b*e*g - 8*a**2*c*d
*g - 8*a**2*c*e*f - a*b**2*d*g - a*b**2*e*f + 10*a*b*c*d*f - b**3*d*f + x**3*(4*a*c**2*e*g + 2*b**2*c*e*g - 6*
b*c**2*d*g - 6*b*c**2*e*f + 12*c**3*d*f) + x**2*(6*a*b*c*e*g + 3*b**3*e*g - 9*b**2*c*d*g - 9*b**2*c*e*f + 18*b
*c**2*d*f) + x*(-4*a**2*c*e*g + 10*a*b**2*e*g - 10*a*b*c*d*g - 10*a*b*c*e*f + 20*a*c**2*d*f - 2*b**3*d*g - 2*b
**3*e*f + 4*b**2*c*d*f))/(32*a**4*c**2 - 16*a**3*b**2*c + 2*a**2*b**4 + x**4*(32*a**2*c**4 - 16*a*b**2*c**3 +
2*b**4*c**2) + x**3*(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5*c) + x**2*(64*a**3*c**3 - 12*a*b**4*c + 2*b**6)
+ x*(64*a**3*b*c**2 - 32*a**2*b**3*c + 4*a*b**5))

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Giac [A]  time = 1.12387, size = 498, normalized size = 2.25 \begin{align*} \frac{2 \,{\left (6 \, c^{2} d f - 3 \, b c d g - 3 \, b c f e + b^{2} g e + 2 \, a c g e\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{12 \, c^{3} d f x^{3} - 6 \, b c^{2} d g x^{3} - 6 \, b c^{2} f x^{3} e + 2 \, b^{2} c g x^{3} e + 4 \, a c^{2} g x^{3} e + 18 \, b c^{2} d f x^{2} - 9 \, b^{2} c d g x^{2} - 9 \, b^{2} c f x^{2} e + 3 \, b^{3} g x^{2} e + 6 \, a b c g x^{2} e + 4 \, b^{2} c d f x + 20 \, a c^{2} d f x - 2 \, b^{3} d g x - 10 \, a b c d g x - 2 \, b^{3} f x e - 10 \, a b c f x e + 10 \, a b^{2} g x e - 4 \, a^{2} c g x e - b^{3} d f + 10 \, a b c d f - a b^{2} d g - 8 \, a^{2} c d g - a b^{2} f e - 8 \, a^{2} c f e + 6 \, a^{2} b g e}{2 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}{\left (c x^{2} + b x + a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(6*c^2*d*f - 3*b*c*d*g - 3*b*c*f*e + b^2*g*e + 2*a*c*g*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a
*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)) + 1/2*(12*c^3*d*f*x^3 - 6*b*c^2*d*g*x^3 - 6*b*c^2*f*x^3*e + 2*b^2*c*g
*x^3*e + 4*a*c^2*g*x^3*e + 18*b*c^2*d*f*x^2 - 9*b^2*c*d*g*x^2 - 9*b^2*c*f*x^2*e + 3*b^3*g*x^2*e + 6*a*b*c*g*x^
2*e + 4*b^2*c*d*f*x + 20*a*c^2*d*f*x - 2*b^3*d*g*x - 10*a*b*c*d*g*x - 2*b^3*f*x*e - 10*a*b*c*f*x*e + 10*a*b^2*
g*x*e - 4*a^2*c*g*x*e - b^3*d*f + 10*a*b*c*d*f - a*b^2*d*g - 8*a^2*c*d*g - a*b^2*f*e - 8*a^2*c*f*e + 6*a^2*b*g
*e)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x + a)^2)