### 3.78 $$\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x}) x^5 (d+e x)^2} \, dx$$

Optimal. Leaf size=372 $\frac{\left (-a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a^3 c d^2+a b^2 e (2 b d+3 c e)+b^4 \left (-e^2\right )\right ) \log \left (a x^2+b x+c\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )-a^3 c d (3 b d+4 c e)-a b^3 e (2 b d+5 c e)+b^5 e^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{\log (x) \left (-c \left (a d^2-3 c e^2\right )+b^2 d^2+2 b c d e\right )}{c^3 d^4}+\frac{e^4}{d^3 (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{e^4 \log (d+e x) \left (5 a d^2-e (4 b d-3 c e)\right )}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{b d+2 c e}{c^2 d^3 x}-\frac{1}{2 c d^2 x^2}$

[Out]

-1/(2*c*d^2*x^2) + (b*d + 2*c*e)/(c^2*d^3*x) + e^4/(d^3*(a*d^2 - e*(b*d - c*e))*(d + e*x)) + ((b^5*e^2 - a^3*c
*d*(3*b*d + 4*c*e) - a*b^3*e*(2*b*d + 5*c*e) + a^2*b*(b^2*d^2 + 8*b*c*d*e + 5*c^2*e^2))*ArcTanh[(b + 2*a*x)/Sq
rt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))^2) + ((b^2*d^2 + 2*b*c*d*e - c*(a*d^2 - 3*c*e
^2))*Log[x])/(c^3*d^4) - (e^4*(5*a*d^2 - e*(4*b*d - 3*c*e))*Log[d + e*x])/(d^4*(a*d^2 - e*(b*d - c*e))^2) + ((
a^3*c*d^2 - b^4*e^2 + a*b^2*e*(2*b*d + 3*c*e) - a^2*(b^2*d^2 + 4*b*c*d*e + c^2*e^2))*Log[c + b*x + a*x^2])/(2*
c^3*(a*d^2 - e*(b*d - c*e))^2)

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Rubi [A]  time = 0.85104, antiderivative size = 372, 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} $\frac{\left (-a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )+a^3 c d^2+a b^2 e (2 b d+3 c e)+b^4 \left (-e^2\right )\right ) \log \left (a x^2+b x+c\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )-a^3 c d (3 b d+4 c e)-a b^3 e (2 b d+5 c e)+b^5 e^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{\log (x) \left (-c \left (a d^2-3 c e^2\right )+b^2 d^2+2 b c d e\right )}{c^3 d^4}+\frac{e^4}{d^3 (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{e^4 \log (d+e x) \left (5 a d^2-e (4 b d-3 c e)\right )}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{b d+2 c e}{c^2 d^3 x}-\frac{1}{2 c d^2 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + c/x^2 + b/x)*x^5*(d + e*x)^2),x]

[Out]

-1/(2*c*d^2*x^2) + (b*d + 2*c*e)/(c^2*d^3*x) + e^4/(d^3*(a*d^2 - e*(b*d - c*e))*(d + e*x)) + ((b^5*e^2 - a^3*c
*d*(3*b*d + 4*c*e) - a*b^3*e*(2*b*d + 5*c*e) + a^2*b*(b^2*d^2 + 8*b*c*d*e + 5*c^2*e^2))*ArcTanh[(b + 2*a*x)/Sq
rt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))^2) + ((b^2*d^2 + 2*b*c*d*e - c*(a*d^2 - 3*c*e
^2))*Log[x])/(c^3*d^4) - (e^4*(5*a*d^2 - e*(4*b*d - 3*c*e))*Log[d + e*x])/(d^4*(a*d^2 - e*(b*d - c*e))^2) + ((
a^3*c*d^2 - b^4*e^2 + a*b^2*e*(2*b*d + 3*c*e) - a^2*(b^2*d^2 + 4*b*c*d*e + c^2*e^2))*Log[c + b*x + a*x^2])/(2*
c^3*(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^5 (d+e x)^2} \, dx &=\int \frac{1}{x^3 (d+e x)^2 \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac{1}{c d^2 x^3}+\frac{-b d-2 c e}{c^2 d^3 x^2}+\frac{b^2 d^2+2 b c d e-c \left (a d^2-3 c e^2\right )}{c^3 d^4 x}+\frac{e^5}{d^3 \left (-a d^2+e (b d-c e)\right ) (d+e x)^2}+\frac{e^5 \left (-5 a d^2+e (4 b d-3 c e)\right )}{d^4 \left (a d^2-e (b d-c e)\right )^2 (d+e x)}+\frac{-\left (a b d-b^2 e+a c e\right ) \left (a b^2 d-2 a^2 c d-b^3 e+3 a b c e\right )+a \left (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 )\right ) x}{c^3 \left (a d^2-e (b d-c e)\right )^2 \left (c+b x+a x^2\right )}\right ) \, dx\\ &=-\frac{1}{2 c d^2 x^2}+\frac{b d+2 c e}{c^2 d^3 x}+\frac{e^4}{d^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac{\left (b^2 d^2+2 b c d e-c \left (a d^2-3 c e^2\right )\right ) \log (x)}{c^3 d^4}-\frac{e^4 \left (5 a d^2-e (4 b d-3 c e)\right ) \log (d+e x)}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{\int \frac{-\left (a b d-b^2 e+a c e\right ) \left (a b^2 d-2 a^2 c d-b^3 e+3 a b c e\right )+a \left (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 )\right ) x}{c+b x+a x^2} \, dx}{c^3 \left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac{1}{2 c d^2 x^2}+\frac{b d+2 c e}{c^2 d^3 x}+\frac{e^4}{d^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac{\left (b^2 d^2+2 b c d e-c \left (a d^2-3 c e^2\right )\right ) \log (x)}{c^3 d^4}-\frac{e^4 \left (5 a d^2-e (4 b d-3 c e)\right ) \log (d+e x)}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (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 )\right ) \int \frac{b+2 a x}{c+b x+a x^2} \, dx}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}-\frac{\left (b^5 e^2-a^3 c d (3 b d+4 c e)-a b^3 e (2 b d+5 c e)+a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )\right ) \int \frac{1}{c+b x+a x^2} \, dx}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac{1}{2 c d^2 x^2}+\frac{b d+2 c e}{c^2 d^3 x}+\frac{e^4}{d^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac{\left (b^2 d^2+2 b c d e-c \left (a d^2-3 c e^2\right )\right ) \log (x)}{c^3 d^4}-\frac{e^4 \left (5 a d^2-e (4 b d-3 c e)\right ) \log (d+e x)}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (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 )\right ) \log \left (c+b x+a x^2\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (b^5 e^2-a^3 c d (3 b d+4 c e)-a b^3 e (2 b d+5 c e)+a^2 b \left (b^2 d^2+8 b c d e+5 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^3 \left (a d^2-e (b d-c e)\right )^2}\\ &=-\frac{1}{2 c d^2 x^2}+\frac{b d+2 c e}{c^2 d^3 x}+\frac{e^4}{d^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac{\left (b^5 e^2-a^3 c d (3 b d+4 c e)-a b^3 e (2 b d+5 c e)+a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 a x}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (b^2 d^2+2 b c d e-c \left (a d^2-3 c e^2\right )\right ) \log (x)}{c^3 d^4}-\frac{e^4 \left (5 a d^2-e (4 b d-3 c e)\right ) \log (d+e x)}{d^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (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 )\right ) \log \left (c+b x+a x^2\right )}{2 c^3 \left (a d^2-e (b d-c e)\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.467986, size = 370, normalized size = 0.99 $-\frac{\left (a^2 \left (b^2 d^2+4 b c d e+c^2 e^2\right )-a^3 c d^2-a b^2 e (2 b d+3 c e)+b^4 e^2\right ) \log (x (a x+b)+c)}{2 c^3 \left (a d^2+e (c e-b d)\right )^2}+\frac{\left (-a^2 b \left (b^2 d^2+8 b c d e+5 c^2 e^2\right )+a^3 c d (3 b d+4 c e)+a b^3 e (2 b d+5 c e)+b^5 \left (-e^2\right )\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{c^3 \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )^2}+\frac{\log (x) \left (c \left (3 c e^2-a d^2\right )+b^2 d^2+2 b c d e\right )}{c^3 d^4}+\frac{e^4}{d^3 (d+e x) \left (a d^2+e (c e-b d)\right )}-\frac{e^4 \log (d+e x) \left (5 a d^2+e (3 c e-4 b d)\right )}{d^4 \left (a d^2+e (c e-b d)\right )^2}+\frac{b d+2 c e}{c^2 d^3 x}-\frac{1}{2 c d^2 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + c/x^2 + b/x)*x^5*(d + e*x)^2),x]

[Out]

-1/(2*c*d^2*x^2) + (b*d + 2*c*e)/(c^2*d^3*x) + e^4/(d^3*(a*d^2 + e*(-(b*d) + c*e))*(d + e*x)) + ((-(b^5*e^2) +
a^3*c*d*(3*b*d + 4*c*e) + a*b^3*e*(2*b*d + 5*c*e) - a^2*b*(b^2*d^2 + 8*b*c*d*e + 5*c^2*e^2))*ArcTan[(b + 2*a*
x)/Sqrt[-b^2 + 4*a*c]])/(c^3*Sqrt[-b^2 + 4*a*c]*(a*d^2 + e*(-(b*d) + c*e))^2) + ((b^2*d^2 + 2*b*c*d*e + c*(-(a
*d^2) + 3*c*e^2))*Log[x])/(c^3*d^4) - (e^4*(5*a*d^2 + e*(-4*b*d + 3*c*e))*Log[d + e*x])/(d^4*(a*d^2 + e*(-(b*d
) + c*e))^2) - ((-(a^3*c*d^2) + b^4*e^2 - a*b^2*e*(2*b*d + 3*c*e) + a^2*(b^2*d^2 + 4*b*c*d*e + c^2*e^2))*Log[c
+ x*(b + a*x)])/(2*c^3*(a*d^2 + e*(-(b*d) + c*e))^2)

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Maple [B]  time = 0.019, size = 993, 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^5/(e*x+d)^2,x)

[Out]

-1/2/c/d^2/x^2-8/(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*e+2
/(a*d^2-b*d*e+c*e^2)^2/c^3/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*b^4*d*e-1/2/(a*d^2-b*d*e+c*
e^2)^2/c*a^2*ln(a*x^2+b*x+c)*e^2+1/2/(a*d^2-b*d*e+c*e^2)^2/c^2*a^3*ln(a*x^2+b*x+c)*d^2-5*e^4/d^2/(a*d^2-b*d*e+
c*e^2)^2*ln(e*x+d)*a-2/(a*d^2-b*d*e+c*e^2)^2/c^2*a^2*ln(a*x^2+b*x+c)*b*d*e+1/(a*d^2-b*d*e+c*e^2)^2/c^3*a*ln(a*
x^2+b*x+c)*b^3*d*e+3/(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^3*b*d^2
+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^3*d*e-1/(a*d^2-b*d*e+c*e^2)
^2/c^3/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a^2*b^3*d^2-5/(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^2*b*e^2+5/(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*e^2+2/d^3/c^2*ln(x)*b*e-1/2/(a*d^2-b*d*e+c*e^2)^2/c^3*ln(a*x^2+b*x+c)*b^4*e^2-3
*e^6/d^4/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*c+3/d^4/c*ln(x)*e^2+e^4/d^3/(a*d^2-b*d*e+c*e^2)/(e*x+d)+1/c^2/d^2/x*b
+2/c/d^3/x*e-1/d^2/c^2*ln(x)*a+1/d^2/c^3*ln(x)*b^2+3/2/(a*d^2-b*d*e+c*e^2)^2/c^2*a*ln(a*x^2+b*x+c)*b^2*e^2-1/(
a*d^2-b*d*e+c*e^2)^2/c^3/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^5*e^2-1/2/(a*d^2-b*d*e+c*e^2)
^2/c^3*a^2*ln(a*x^2+b*x+c)*b^2*d^2+4*e^5/d^3/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*b

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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^5/(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^5/(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**5/(e*x+d)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.11945, size = 792, normalized size = 2.13 \begin{align*} \frac{{\left (a^{2} b^{3} d^{2} e^{2} - 3 \, a^{3} b c d^{2} e^{2} - 2 \, a b^{4} d e^{3} + 8 \, a^{2} b^{2} c d e^{3} - 4 \, a^{3} c^{2} d e^{3} + b^{5} e^{4} - 5 \, a b^{3} c e^{4} + 5 \, a^{2} b 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^{3} d^{4} - 2 \, a b c^{3} d^{3} e + b^{2} c^{3} d^{2} e^{2} + 2 \, a c^{4} d^{2} e^{2} - 2 \, b c^{4} d e^{3} + c^{5} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{{\left (a^{2} b^{2} d^{2} - a^{3} c d^{2} - 2 \, a b^{3} d e + 4 \, a^{2} b c d e + b^{4} e^{2} - 3 \, a b^{2} c e^{2} + a^{2} 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^{2} c^{3} d^{4} - 2 \, a b c^{3} d^{3} e + b^{2} c^{3} d^{2} e^{2} + 2 \, a c^{4} d^{2} e^{2} - 2 \, b c^{4} d e^{3} + c^{5} e^{4}\right )}} + \frac{e^{9}}{{\left (a d^{5} e^{5} - b d^{4} e^{6} + c d^{3} e^{7}\right )}{\left (x e + d\right )}} + \frac{{\left (b^{2} d^{2} e - a c d^{2} e + 2 \, b c d e^{2} + 3 \, c^{2} e^{3}\right )} e^{\left (-1\right )} \log \left ({\left | -\frac{d}{x e + d} + 1 \right |}\right )}{c^{3} d^{4}} + \frac{2 \, b c d e + 5 \, c^{2} e^{2} - \frac{2 \,{\left (b c d^{2} e^{2} + 3 \, c^{2} d e^{3}\right )} e^{\left (-1\right )}}{x e + d}}{2 \, c^{3} d^{4}{\left (\frac{d}{x e + d} - 1\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^5/(e*x+d)^2,x, algorithm="giac")

[Out]

(a^2*b^3*d^2*e^2 - 3*a^3*b*c*d^2*e^2 - 2*a*b^4*d*e^3 + 8*a^2*b^2*c*d*e^3 - 4*a^3*c^2*d*e^3 + b^5*e^4 - 5*a*b^3
*c*e^4 + 5*a^2*b*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^3*d^4 - 2*a*b*c^3*d^3*e + b^2*c^3*d^2*e^2 + 2*a*c^4*d^2*e^2 - 2*b*c^4*
d*e^3 + c^5*e^4)*sqrt(-b^2 + 4*a*c)) - 1/2*(a^2*b^2*d^2 - a^3*c*d^2 - 2*a*b^3*d*e + 4*a^2*b*c*d*e + b^4*e^2 -
3*a*b^2*c*e^2 + a^2*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^2*c^3*d^4 - 2*a*b*c^3*d^3*e + b^2*c^3*d^2*e^2 + 2*a*c^4*d^2*e^2 - 2*b*c^4*d*e^3 + c^5*
e^4) + e^9/((a*d^5*e^5 - b*d^4*e^6 + c*d^3*e^7)*(x*e + d)) + (b^2*d^2*e - a*c*d^2*e + 2*b*c*d*e^2 + 3*c^2*e^3)
*e^(-1)*log(abs(-d/(x*e + d) + 1))/(c^3*d^4) + 1/2*(2*b*c*d*e + 5*c^2*e^2 - 2*(b*c*d^2*e^2 + 3*c^2*d*e^3)*e^(-
1)/(x*e + d))/(c^3*d^4*(d/(x*e + d) - 1)^2)