### 3.77 $$\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x}) x^4 (d+e x)^2} \, dx$$

Optimal. Leaf size=291 $\frac{\left (-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )+2 a^3 c d^2+2 a b^2 e (b d+2 c e)+b^4 \left (-e^2\right )\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{(a d-b e) \left (a b d+2 a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{e^3}{d^2 (d+e x) \left (a d^2-e (b d-c e)\right )}+\frac{e^3 \log (d+e x) \left (4 a d^2-e (3 b d-2 c e)\right )}{d^3 \left (a d^2-e (b d-c e)\right )^2}-\frac{\log (x) (b d+2 c e)}{c^2 d^3}-\frac{1}{c d^2 x}$

[Out]

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

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Rubi [A]  time = 0.563207, antiderivative size = 291, 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, 893, 634, 618, 206, 628} $\frac{\left (-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )+2 a^3 c d^2+2 a b^2 e (b d+2 c e)+b^4 \left (-e^2\right )\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{(a d-b e) \left (a b d+2 a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{e^3}{d^2 (d+e x) \left (a d^2-e (b d-c e)\right )}+\frac{e^3 \log (d+e x) \left (4 a d^2-e (3 b d-2 c e)\right )}{d^3 \left (a d^2-e (b d-c e)\right )^2}-\frac{\log (x) (b d+2 c e)}{c^2 d^3}-\frac{1}{c d^2 x}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + c/x^2 + b/x)*x^4*(d + e*x)^2),x]

[Out]

-(1/(c*d^2*x)) - e^3/(d^2*(a*d^2 - e*(b*d - c*e))*(d + e*x)) + ((2*a^3*c*d^2 - b^4*e^2 + 2*a*b^2*e*(b*d + 2*c*
e) - a^2*(b^2*d^2 + 6*b*c*d*e + 2*c^2*e^2))*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]*(a*
d^2 - e*(b*d - c*e))^2) - ((b*d + 2*c*e)*Log[x])/(c^2*d^3) + (e^3*(4*a*d^2 - e*(3*b*d - 2*c*e))*Log[d + e*x])/
(d^3*(a*d^2 - e*(b*d - c*e))^2) + ((a*d - b*e)*(a*b*d - b^2*e + 2*a*c*e)*Log[c + b*x + a*x^2])/(2*c^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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 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{1}{\left (a+\frac{c}{x^2}+\frac{b}{x}\right ) x^4 (d+e x)^2} \, dx &=\int \frac{1}{x^2 (d+e x)^2 \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac{1}{c d^2 x^2}+\frac{-b d-2 c e}{c^2 d^3 x}+\frac{e^4}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)^2}+\frac{e^4 \left (4 a d^2-e (3 b d-2 c e)\right )}{d^3 \left (a d^2-e (b d-c e)\right )^2 (d+e x)}+\frac{-a^3 c d^2+b^4 e^2-a b^2 e (2 b d+3 c e)+a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a (a d-b e) \left (a b d-b^2 e+2 a c e\right ) x}{c^2 \left (a d^2-e (b d-c e)\right )^2 \left (c+b x+a x^2\right )}\right ) \, dx\\ &=-\frac{1}{c d^2 x}-\frac{e^3}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{(b d+2 c e) \log (x)}{c^2 d^3}+\frac{e^3 \left (4 a d^2-e (3 b d-2 c e)\right ) \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\int \frac{-a^3 c d^2+b^4 e^2-a b^2 e (2 b d+3 c e)+a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a (a d-b e) \left (a b d-b^2 e+2 a c e\right ) x}{c+b x+a x^2} \, dx}{c^2 \left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac{1}{c d^2 x}-\frac{e^3}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{(b d+2 c e) \log (x)}{c^2 d^3}+\frac{e^3 \left (4 a d^2-e (3 b d-2 c e)\right ) \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left ((a d-b e) \left (a b d-b^2 e+2 a c e\right )\right ) \int \frac{b+2 a x}{c+b x+a x^2} \, dx}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{\left (2 a^3 c d^2-b^4 e^2+2 a b^2 e (b d+2 c e)-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )\right ) \int \frac{1}{c+b x+a x^2} \, dx}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac{1}{c d^2 x}-\frac{e^3}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{(b d+2 c e) \log (x)}{c^2 d^3}+\frac{e^3 \left (4 a d^2-e (3 b d-2 c e)\right ) \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{(a d-b e) \left (a b d-b^2 e+2 a c e\right ) \log \left (c+b x+a x^2\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (2 a^3 c d^2-b^4 e^2+2 a b^2 e (b d+2 c e)-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 a x\right )}{c^2 \left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac{1}{c d^2 x}-\frac{e^3}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac{\left (2 a^3 c d^2-b^4 e^2+2 a b^2 e (b d+2 c e)-a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 a x}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d+2 c e) \log (x)}{c^2 d^3}+\frac{e^3 \left (4 a d^2-e (3 b d-2 c e)\right ) \log (d+e x)}{d^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{(a d-b e) \left (a b d-b^2 e+2 a c e\right ) \log \left (c+b x+a x^2\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.380898, size = 287, normalized size = 0.99 $\frac{\left (a^2 \left (b^2 d^2+6 b c d e+2 c^2 e^2\right )-2 a^3 c d^2-2 a b^2 e (b d+2 c e)+b^4 e^2\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{c^2 \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )^2}+\frac{(a d-b e) \left (a b d+2 a c e+b^2 (-e)\right ) \log (x (a x+b)+c)}{2 c^2 \left (a d^2+e (c e-b d)\right )^2}-\frac{e^3}{d^2 (d+e x) \left (a d^2+e (c e-b d)\right )}+\frac{e^3 \log (d+e x) \left (4 a d^2+e (2 c e-3 b d)\right )}{d^3 \left (a d^2+e (c e-b d)\right )^2}-\frac{\log (x) (b d+2 c e)}{c^2 d^3}-\frac{1}{c d^2 x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + c/x^2 + b/x)*x^4*(d + e*x)^2),x]

[Out]

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

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Maple [B]  time = 0.016, size = 791, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c/x^2+b/x)/x^4/(e*x+d)^2,x)

[Out]

-e^3/d^2/(a*d^2-b*d*e+c*e^2)/(e*x+d)+4*e^3/d/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*a-3*e^4/d^2/(a*d^2-b*d*e+c*e^2)^2
*ln(e*x+d)*b+2*e^5/d^3/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*c+1/2/(a*d^2-b*d*e+c*e^2)^2/c^2*a^2*ln(a*x^2+b*x+c)*b*d
^2+1/(a*d^2-b*d*e+c*e^2)^2/c*a^2*ln(a*x^2+b*x+c)*d*e-1/(a*d^2-b*d*e+c*e^2)^2/c^2*a*ln(a*x^2+b*x+c)*b^2*d*e-1/(
a*d^2-b*d*e+c*e^2)^2/c*a*ln(a*x^2+b*x+c)*b*e^2+1/2/(a*d^2-b*d*e+c*e^2)^2/c^2*ln(a*x^2+b*x+c)*b^3*e^2-2/(a*d^2-
b*d*e+c*e^2)^2/c/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a^3*d^2+1/(a*d^2-b*d*e+c*e^2)^2/c^2/(4*
a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a^2*b^2*d^2+6/(a*d^2-b*d*e+c*e^2)^2/c/(4*a*c-b^2)^(1/2)*arc
tan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a^2*b*d*e+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))*a^2*e^2-2/(a*d^2-b*d*e+c*e^2)^2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*b^3*d*e
-4/(a*d^2-b*d*e+c*e^2)^2/c/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*b^2*e^2+1/(a*d^2-b*d*e+c*e^
2)^2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^4*e^2-1/c/d^2/x-1/d^2/c^2*ln(x)*b-2/d^3/c*ln(
x)*e

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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(1/(a+c/x^2+b/x)/x^4/(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(1/(a+c/x^2+b/x)/x^4/(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(1/(a+c/x**2+b/x)/x**4/(e*x+d)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.10928, size = 657, normalized size = 2.26 \begin{align*} -\frac{{\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} c d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 6 \, a^{2} b c d e^{3} + b^{4} e^{4} - 4 \, a b^{2} c e^{4} + 2 \, a^{2} c^{2} 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^{2} c^{2} d^{4} - 2 \, a b c^{2} d^{3} e + b^{2} c^{2} d^{2} e^{2} + 2 \, a c^{3} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + c^{4} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (a^{2} b d^{2} - 2 \, a b^{2} d e + 2 \, a^{2} c d e + b^{3} e^{2} - 2 \, a b c 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^{2} c^{2} d^{4} - 2 \, a b c^{2} d^{3} e + b^{2} c^{2} d^{2} e^{2} + 2 \, a c^{3} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + c^{4} e^{4}\right )}} - \frac{e^{7}}{{\left (a d^{4} e^{4} - b d^{3} e^{5} + c d^{2} e^{6}\right )}{\left (x e + d\right )}} - \frac{{\left (b d e + 2 \, c e^{2}\right )} e^{\left (-1\right )} \log \left ({\left | -\frac{d}{x e + d} + 1 \right |}\right )}{c^{2} d^{3}} + \frac{e}{c d^{3}{\left (\frac{d}{x e + d} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^4/(e*x+d)^2,x, algorithm="giac")

[Out]

-(a^2*b^2*d^2*e^2 - 2*a^3*c*d^2*e^2 - 2*a*b^3*d*e^3 + 6*a^2*b*c*d*e^3 + b^4*e^4 - 4*a*b^2*c*e^4 + 2*a^2*c^2*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^2*c^2*d^4 - 2*a*b*c^2*d^3*e + b^2*c^2*d^2*e^2 + 2*a*c^3*d^2*e^2 - 2*b*c^3*d*e^3 + c^4*e^4)*sqrt(-
b^2 + 4*a*c)) + 1/2*(a^2*b*d^2 - 2*a*b^2*d*e + 2*a^2*c*d*e + b^3*e^2 - 2*a*b*c*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^2*c^2*d^4 - 2*a*b*c^2*d^3*e + b
^2*c^2*d^2*e^2 + 2*a*c^3*d^2*e^2 - 2*b*c^3*d*e^3 + c^4*e^4) - e^7/((a*d^4*e^4 - b*d^3*e^5 + c*d^2*e^6)*(x*e +
d)) - (b*d*e + 2*c*e^2)*e^(-1)*log(abs(-d/(x*e + d) + 1))/(c^2*d^3) + e/(c*d^3*(d/(x*e + d) - 1))