### 3.71 $$\int \frac{x^2}{(a+\frac{c}{x^2}+\frac{b}{x}) (d+e x)^2} \, dx$$

Optimal. Leaf size=274 $-\frac{\left (-b^2 c \left (4 a d^2-c e^2\right )+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d-c e) \left (-2 a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{d^4}{e^3 (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{d^3 \log (d+e x) \left (2 a d^2-e (3 b d-4 c e)\right )}{e^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{x}{a e^2}$

[Out]

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

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Rubi [A]  time = 0.56307, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.24, Rules used = {1569, 1628, 634, 618, 206, 628} $-\frac{\left (-b^2 c \left (4 a d^2-c e^2\right )+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d-c e) \left (-2 a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{d^4}{e^3 (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{d^3 \log (d+e x) \left (2 a d^2-e (3 b d-4 c e)\right )}{e^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{x}{a e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1569

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbo
l] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && E
qQ[mn, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 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])

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]

Rubi steps

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

Mathematica [A]  time = 0.315942, size = 269, normalized size = 0.98 $\frac{\left (b^2 c \left (c e^2-4 a d^2\right )+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{a^2 \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )^2}+\frac{(b d-c e) \left (2 a c d+b^2 (-d)+b c e\right ) \log (x (a x+b)+c)}{2 a^2 \left (a d^2+e (c e-b d)\right )^2}-\frac{d^4}{e^3 (d+e x) \left (a d^2+e (c e-b d)\right )}-\frac{\log (d+e x) \left (2 a d^5+d^3 e (4 c e-3 b d)\right )}{e^3 \left (a d^2+e (c e-b d)\right )^2}+\frac{x}{a e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 765, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/e^3*d^4/(a*d^2-b*d*e+c*e^2)/(e*x+d)-2/e^3*d^5/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*a+3/e^2*d^4/(a*d^2-b*d*e+c*e^
2)^2*ln(e*x+d)*b-4/e*d^3/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*c+x/a/e^2+1/(a*d^2-b*d*e+c*e^2)^2/a*ln(a*x^2+b*x+c)*b
*c*d^2-1/(a*d^2-b*d*e+c*e^2)^2/a*ln(a*x^2+b*x+c)*c^2*d*e-1/2/(a*d^2-b*d*e+c*e^2)^2/a^2*ln(a*x^2+b*x+c)*b^3*d^2
+1/(a*d^2-b*d*e+c*e^2)^2/a^2*ln(a*x^2+b*x+c)*b^2*c*d*e-1/2/(a*d^2-b*d*e+c*e^2)^2/a^2*ln(a*x^2+b*x+c)*b*c^2*e^2
+2/(a*d^2-b*d*e+c*e^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*c^2*d^2-4/(a*d^2-b*d*e+c*e^2)^2
/a/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^2*c*d^2+6/(a*d^2-b*d*e+c*e^2)^2/a/(4*a*c-b^2)^(1/2)
*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b*c^2*d*e-2/(a*d^2-b*d*e+c*e^2)^2/a/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4
*a*c-b^2)^(1/2))*c^3*e^2+1/(a*d^2-b*d*e+c*e^2)^2/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^4
*d^2-2/(a*d^2-b*d*e+c*e^2)^2/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^3*c*d*e+1/(a*d^2-b*d*
e+c*e^2)^2/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^2*c^2*e^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(x^2/(a+c/x^2+b/x)/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.13066, size = 643, normalized size = 2.35 \begin{align*} -\frac{d^{4} e^{3}}{{\left (a d^{2} e^{6} - b d e^{7} + c e^{8}\right )}{\left (x e + d\right )}} - \frac{{\left (b^{4} d^{2} e^{2} - 4 \, a b^{2} c d^{2} e^{2} + 2 \, a^{2} c^{2} d^{2} e^{2} - 2 \, b^{3} c d e^{3} + 6 \, a b c^{2} d e^{3} + b^{2} c^{2} e^{4} - 2 \, a c^{3} e^{4}\right )} \arctan \left (-\frac{{\left (2 \, a d - \frac{2 \, a d^{2}}{x e + d} - b e + \frac{2 \, b d e}{x e + d} - \frac{2 \, c e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (a^{4} d^{4} - 2 \, a^{3} b d^{3} e + a^{2} b^{2} d^{2} e^{2} + 2 \, a^{3} c d^{2} e^{2} - 2 \, a^{2} b c d e^{3} + a^{2} c^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (x e + d\right )} e^{\left (-3\right )}}{a} - \frac{{\left (b^{3} d^{2} - 2 \, a b c d^{2} - 2 \, b^{2} c d e + 2 \, a c^{2} d e + b c^{2} e^{2}\right )} \log \left (-a + \frac{2 \, a d}{x e + d} - \frac{a d^{2}}{{\left (x e + d\right )}^{2}} - \frac{b e}{x e + d} + \frac{b d e}{{\left (x e + d\right )}^{2}} - \frac{c e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (a^{4} d^{4} - 2 \, a^{3} b d^{3} e + a^{2} b^{2} d^{2} e^{2} + 2 \, a^{3} c d^{2} e^{2} - 2 \, a^{2} b c d e^{3} + a^{2} c^{2} e^{4}\right )}} + \frac{{\left (2 \, a d + b e\right )} e^{\left (-3\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right )}{a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d^4*e^3/((a*d^2*e^6 - b*d*e^7 + c*e^8)*(x*e + d)) - (b^4*d^2*e^2 - 4*a*b^2*c*d^2*e^2 + 2*a^2*c^2*d^2*e^2 - 2*
b^3*c*d*e^3 + 6*a*b*c^2*d*e^3 + b^2*c^2*e^4 - 2*a*c^3*e^4)*arctan(-(2*a*d - 2*a*d^2/(x*e + d) - b*e + 2*b*d*e/
(x*e + d) - 2*c*e^2/(x*e + d))*e^(-1)/sqrt(-b^2 + 4*a*c))*e^(-2)/((a^4*d^4 - 2*a^3*b*d^3*e + a^2*b^2*d^2*e^2 +
2*a^3*c*d^2*e^2 - 2*a^2*b*c*d*e^3 + a^2*c^2*e^4)*sqrt(-b^2 + 4*a*c)) + (x*e + d)*e^(-3)/a - 1/2*(b^3*d^2 - 2*
a*b*c*d^2 - 2*b^2*c*d*e + 2*a*c^2*d*e + b*c^2*e^2)*log(-a + 2*a*d/(x*e + d) - a*d^2/(x*e + d)^2 - b*e/(x*e + d
) + b*d*e/(x*e + d)^2 - c*e^2/(x*e + d)^2)/(a^4*d^4 - 2*a^3*b*d^3*e + a^2*b^2*d^2*e^2 + 2*a^3*c*d^2*e^2 - 2*a^
2*b*c*d*e^3 + a^2*c^2*e^4) + (2*a*d + b*e)*e^(-3)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2)/a^2