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

Optimal. Leaf size=132 \[ \frac {x \left (x \left (2 a c e+b^2 (-e)+b c d\right )+a (2 c d-b e)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\left (2 a c (2 c d-3 b e)+b^3 e\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {e \log \left (a+b x+c x^2\right )}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {818, 634, 618, 206, 628} \[ \frac {\left (2 a c (2 c d-3 b e)+b^3 e\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {x \left (x \left (2 a c e+b^2 (-e)+b c d\right )+a (2 c d-b e)\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {e \log \left (a+b x+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a*(2*c*d - b*e) + (b*c*d - b^2*e + 2*a*c*e)*x))/(c*(b^2 - 4*a*c)*(a + b*x + c*x^2)) + ((b^3*e + 2*a*c*(2*c
*d - 3*b*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^2*(b^2 - 4*a*c)^(3/2)) + (e*Log[a + b*x + c*x^2])/(2*c
^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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(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))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 146, normalized size = 1.11 \[ \frac {-\frac {2 \left (2 a^2 c e+a \left (b^2 (-e)+b c (d+3 e x)-2 c^2 d x\right )+b^2 x (c d-b e)\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {2 \left (2 a c (2 c d-3 b e)+b^3 e\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+e \log (a+x (b+c x))}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.98, size = 813, normalized size = 6.16 \[ \left [\frac {{\left (4 \, a^{2} c^{2} d + {\left (4 \, a c^{3} d + {\left (b^{3} c - 6 \, a b c^{2}\right )} e\right )} x^{2} + {\left (a b^{3} - 6 \, a^{2} b c\right )} e + {\left (4 \, a b c^{2} d + {\left (b^{4} - 6 \, a b^{2} c\right )} e\right )} x\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 ) - 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} d + 2 \, {\left (a b^{4} - 6 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} e - 2 \, {\left ({\left (b^{4} c - 6 \, a b^{2} c^{2} + 8 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} e\right )} x + {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} e x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{2} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x\right )}}, \frac {2 \, {\left (4 \, a^{2} c^{2} d + {\left (4 \, a c^{3} d + {\left (b^{3} c - 6 \, a b c^{2}\right )} e\right )} x^{2} + {\left (a b^{3} - 6 \, a^{2} b c\right )} e + {\left (4 \, a b c^{2} d + {\left (b^{4} - 6 \, a b^{2} c\right )} e\right )} x\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 ) - 2 \, {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} d + 2 \, {\left (a b^{4} - 6 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} e - 2 \, {\left ({\left (b^{4} c - 6 \, a b^{2} c^{2} + 8 \, a^{2} c^{3}\right )} d - {\left (b^{5} - 7 \, a b^{3} c + 12 \, a^{2} b c^{2}\right )} e\right )} x + {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} e x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} e x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{2} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((4*a^2*c^2*d + (4*a*c^3*d + (b^3*c - 6*a*b*c^2)*e)*x^2 + (a*b^3 - 6*a^2*b*c)*e + (4*a*b*c^2*d + (b^4 - 6
*a*b^2*c)*e)*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*(a*b^3*c - 4*a^2*b*c^2)*d + 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e - 2*((b^4*c - 6*a*b^2*c^2
+ 8*a^2*c^3)*d - (b^5 - 7*a*b^3*c + 12*a^2*b*c^2)*e)*x + ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e*x^2 + (b^5 - 8
*a*b^3*c + 16*a^2*b*c^2)*e*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*e)*log(c*x^2 + b*x + a))/(a*b^4*c^2 - 8*a^2*
b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^2 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x), 1
/2*(2*(4*a^2*c^2*d + (4*a*c^3*d + (b^3*c - 6*a*b*c^2)*e)*x^2 + (a*b^3 - 6*a^2*b*c)*e + (4*a*b*c^2*d + (b^4 - 6
*a*b^2*c)*e)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 2*(a*b^3*c - 4*a^2*
b*c^2)*d + 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*e - 2*((b^4*c - 6*a*b^2*c^2 + 8*a^2*c^3)*d - (b^5 - 7*a*b^3*c +
12*a^2*b*c^2)*e)*x + ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e*x + (a*b^
4 - 8*a^2*b^2*c + 16*a^3*c^2)*e)*log(c*x^2 + b*x + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8*
a*b^2*c^4 + 16*a^2*c^5)*x^2 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x)]

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giac [A]  time = 0.20, size = 169, normalized size = 1.28 \[ -\frac {{\left (4 \, a c^{2} d + b^{3} e - 6 \, a b c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {e \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} - \frac {a b c d - a b^{2} e + 2 \, a^{2} c e + {\left (b^{2} c d - 2 \, a c^{2} d - b^{3} e + 3 \, a b c e\right )} x}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 270, normalized size = 2.05 \[ -\frac {6 a b e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} c}+\frac {4 a d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {b^{3} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} c^{2}}+\frac {2 a e \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right ) c}-\frac {b^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 \left (4 a c -b^{2}\right ) c^{2}}+\frac {\frac {\left (2 a c e -e \,b^{2}+b c d \right ) a}{\left (4 a c -b^{2}\right ) c^{2}}+\frac {\left (3 a b c e -2 a \,c^{2} d -b^{3} e +b^{2} c d \right ) x}{\left (4 a c -b^{2}\right ) c^{2}}}{c \,x^{2}+b x +a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x+d)/(c*x^2+b*x+a)^2,x)

[Out]

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

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

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mupad [B]  time = 1.54, size = 895, normalized size = 6.78 \[ \frac {2\,a^2\,c\,e}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}-\frac {a\,b^2\,e}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}-\frac {b^3\,e\,x}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}+\frac {4\,a\,d\,\mathrm {atan}\left (\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}-\frac {b^3}{{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {b^6\,e\,\ln \left (c\,x^2+b\,x+a\right )}{2\,\left (64\,a^3\,c^5-48\,a^2\,b^2\,c^4+12\,a\,b^4\,c^3-b^6\,c^2\right )}-\frac {2\,a\,c^2\,d\,x}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}+\frac {b^2\,c\,d\,x}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}+\frac {a\,b\,c\,d}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}+\frac {32\,a^3\,c^3\,e\,\ln \left (c\,x^2+b\,x+a\right )}{64\,a^3\,c^5-48\,a^2\,b^2\,c^4+12\,a\,b^4\,c^3-b^6\,c^2}+\frac {b^3\,e\,\mathrm {atan}\left (\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}-\frac {b^3}{{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{c^2\,{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {3\,a\,b\,c\,e\,x}{4\,a^2\,c^3-a\,b^2\,c^2+4\,a\,b\,c^3\,x+4\,a\,c^4\,x^2-b^3\,c^2\,x-b^2\,c^3\,x^2}-\frac {24\,a^2\,b^2\,c^2\,e\,\ln \left (c\,x^2+b\,x+a\right )}{64\,a^3\,c^5-48\,a^2\,b^2\,c^4+12\,a\,b^4\,c^3-b^6\,c^2}+\frac {6\,a\,b^4\,c\,e\,\ln \left (c\,x^2+b\,x+a\right )}{64\,a^3\,c^5-48\,a^2\,b^2\,c^4+12\,a\,b^4\,c^3-b^6\,c^2}-\frac {6\,a\,b\,e\,\mathrm {atan}\left (\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}-\frac {b^3}{{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )}{c\,{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(d + e*x))/(a + b*x + c*x^2)^2,x)

[Out]

(2*a^2*c*e)/(4*a^2*c^3 - a*b^2*c^2 + 4*a*c^4*x^2 - b^3*c^2*x - b^2*c^3*x^2 + 4*a*b*c^3*x) - (a*b^2*e)/(4*a^2*c
^3 - a*b^2*c^2 + 4*a*c^4*x^2 - b^3*c^2*x - b^2*c^3*x^2 + 4*a*b*c^3*x) - (b^3*e*x)/(4*a^2*c^3 - a*b^2*c^2 + 4*a
*c^4*x^2 - b^3*c^2*x - b^2*c^3*x^2 + 4*a*b*c^3*x) + (4*a*d*atan((2*c*x)/(4*a*c - b^2)^(1/2) - b^3/(4*a*c - b^2
)^(3/2) + (4*a*b*c)/(4*a*c - b^2)^(3/2)))/(4*a*c - b^2)^(3/2) - (b^6*e*log(a + b*x + c*x^2))/(2*(64*a^3*c^5 -
b^6*c^2 + 12*a*b^4*c^3 - 48*a^2*b^2*c^4)) - (2*a*c^2*d*x)/(4*a^2*c^3 - a*b^2*c^2 + 4*a*c^4*x^2 - b^3*c^2*x - b
^2*c^3*x^2 + 4*a*b*c^3*x) + (b^2*c*d*x)/(4*a^2*c^3 - a*b^2*c^2 + 4*a*c^4*x^2 - b^3*c^2*x - b^2*c^3*x^2 + 4*a*b
*c^3*x) + (a*b*c*d)/(4*a^2*c^3 - a*b^2*c^2 + 4*a*c^4*x^2 - b^3*c^2*x - b^2*c^3*x^2 + 4*a*b*c^3*x) + (32*a^3*c^
3*e*log(a + b*x + c*x^2))/(64*a^3*c^5 - b^6*c^2 + 12*a*b^4*c^3 - 48*a^2*b^2*c^4) + (b^3*e*atan((2*c*x)/(4*a*c
- b^2)^(1/2) - b^3/(4*a*c - b^2)^(3/2) + (4*a*b*c)/(4*a*c - b^2)^(3/2)))/(c^2*(4*a*c - b^2)^(3/2)) + (3*a*b*c*
e*x)/(4*a^2*c^3 - a*b^2*c^2 + 4*a*c^4*x^2 - b^3*c^2*x - b^2*c^3*x^2 + 4*a*b*c^3*x) - (24*a^2*b^2*c^2*e*log(a +
b*x + c*x^2))/(64*a^3*c^5 - b^6*c^2 + 12*a*b^4*c^3 - 48*a^2*b^2*c^4) + (6*a*b^4*c*e*log(a + b*x + c*x^2))/(64
*a^3*c^5 - b^6*c^2 + 12*a*b^4*c^3 - 48*a^2*b^2*c^4) - (6*a*b*e*atan((2*c*x)/(4*a*c - b^2)^(1/2) - b^3/(4*a*c -
b^2)^(3/2) + (4*a*b*c)/(4*a*c - b^2)^(3/2)))/(c*(4*a*c - b^2)^(3/2))

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sympy [B]  time = 2.84, size = 901, normalized size = 6.83 \[ \left (\frac {e}{2 c^{2}} - \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right ) \log {\left (x + \frac {- 16 a^{2} c^{3} \left (\frac {e}{2 c^{2}} - \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right ) + 8 a^{2} c e + 8 a b^{2} c^{2} \left (\frac {e}{2 c^{2}} - \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right ) - a b^{2} e - 2 a b c d - b^{4} c \left (\frac {e}{2 c^{2}} - \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right )}{6 a b c e - 4 a c^{2} d - b^{3} e} \right )} + \left (\frac {e}{2 c^{2}} + \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right ) \log {\left (x + \frac {- 16 a^{2} c^{3} \left (\frac {e}{2 c^{2}} + \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right ) + 8 a^{2} c e + 8 a b^{2} c^{2} \left (\frac {e}{2 c^{2}} + \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right ) - a b^{2} e - 2 a b c d - b^{4} c \left (\frac {e}{2 c^{2}} + \frac {\sqrt {- \left (4 a c - b^{2}\right )^{3}} \left (6 a b c e - 4 a c^{2} d - b^{3} e\right )}{2 c^{2} \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}\right )}{6 a b c e - 4 a c^{2} d - b^{3} e} \right )} + \frac {2 a^{2} c e - a b^{2} e + a b c d + x \left (3 a b c e - 2 a c^{2} d - b^{3} e + b^{2} c d\right )}{4 a^{2} c^{3} - a b^{2} c^{2} + x^{2} \left (4 a c^{4} - b^{2} c^{3}\right ) + x \left (4 a b c^{3} - b^{3} c^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e/(2*c**2) - sqrt(-(4*a*c - b**2)**3)*(6*a*b*c*e - 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*
c**2 + 12*a*b**4*c - b**6)))*log(x + (-16*a**2*c**3*(e/(2*c**2) - sqrt(-(4*a*c - b**2)**3)*(6*a*b*c*e - 4*a*c*
*2*d - b**3*e)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 8*a**2*c*e + 8*a*b**2*c**2*
(e/(2*c**2) - sqrt(-(4*a*c - b**2)**3)*(6*a*b*c*e - 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*
c**2 + 12*a*b**4*c - b**6))) - a*b**2*e - 2*a*b*c*d - b**4*c*(e/(2*c**2) - sqrt(-(4*a*c - b**2)**3)*(6*a*b*c*e
- 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))))/(6*a*b*c*e - 4*a*c*
*2*d - b**3*e)) + (e/(2*c**2) + sqrt(-(4*a*c - b**2)**3)*(6*a*b*c*e - 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c*
*3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)))*log(x + (-16*a**2*c**3*(e/(2*c**2) + sqrt(-(4*a*c - b**2)**3)*(
6*a*b*c*e - 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 8*a**2*c*
e + 8*a*b**2*c**2*(e/(2*c**2) + sqrt(-(4*a*c - b**2)**3)*(6*a*b*c*e - 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c*
*3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) - a*b**2*e - 2*a*b*c*d - b**4*c*(e/(2*c**2) + sqrt(-(4*a*c - b*
*2)**3)*(6*a*b*c*e - 4*a*c**2*d - b**3*e)/(2*c**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))))/(
6*a*b*c*e - 4*a*c**2*d - b**3*e)) + (2*a**2*c*e - a*b**2*e + a*b*c*d + x*(3*a*b*c*e - 2*a*c**2*d - b**3*e + b*
*2*c*d))/(4*a**2*c**3 - a*b**2*c**2 + x**2*(4*a*c**4 - b**2*c**3) + x*(4*a*b*c**3 - b**3*c**2))

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