### 3.76 $$\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x}) x^3 (d+e x)^2} \, dx$$

Optimal. Leaf size=248 $-\frac{\left (a^2 d^2-a e (2 b d+c e)+b^2 e^2\right ) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (a^2 d (b d+4 c e)-a b e (2 b d+3 c e)+b^3 e^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{e^2}{d (d+e x) \left (a d^2-b d e+c e^2\right )}-\frac{e^2 \log (d+e x) \left (3 a d^2-e (2 b d-c e)\right )}{d^2 \left (a d^2-e (b d-c e)\right )^2}+\frac{\log (x)}{c d^2}$

[Out]

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

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Rubi [A]  time = 0.408682, antiderivative size = 249, 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} $-\frac{\left (a^2 d^2-a e (2 b d+c e)+b^2 e^2\right ) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (a^2 d (b d+4 c e)-a b e (2 b d+3 c e)+b^3 e^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{e^2}{d (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{e^2 \log (d+e x) \left (3 a d^2-e (2 b d-c e)\right )}{d^2 \left (a d^2-e (b d-c e)\right )^2}+\frac{\log (x)}{c d^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + c/x^2 + b/x)*x^3*(d + e*x)^2),x]

[Out]

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

Mathematica [A]  time = 0.284886, size = 246, normalized size = 0.99 $\frac{\left (-a^2 d^2+a e (2 b d+c e)-b^2 e^2\right ) \log (x (a x+b)+c)}{2 c \left (a d^2+e (c e-b d)\right )^2}-\frac{\left (a^2 d (b d+4 c e)-a b e (2 b d+3 c e)+b^3 e^2\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{c \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )^2}+\frac{e^2}{d (d+e x) \left (a d^2+e (c e-b d)\right )}-\frac{e^2 \log (d+e x) \left (3 a d^2+e (c e-2 b d)\right )}{\left (a d^3+d e (c e-b d)\right )^2}+\frac{\log (x)}{c d^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + c/x^2 + b/x)*x^3*(d + e*x)^2),x]

[Out]

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

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Maple [B]  time = 0.013, size = 589, normalized size = 2.4 \begin{align*}{\frac{{e}^{2}}{d \left ( a{d}^{2}-bde+{e}^{2}c \right ) \left ( ex+d \right ) }}-3\,{\frac{{e}^{2}\ln \left ( ex+d \right ) a}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}+2\,{\frac{{e}^{3}\ln \left ( ex+d \right ) b}{d \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}-{\frac{{e}^{4}\ln \left ( ex+d \right ) c}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}{d}^{2}}}-{\frac{{a}^{2}\ln \left ( a{x}^{2}+bx+c \right ){d}^{2}}{2\, \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}c}}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ) bde}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}c}}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ){e}^{2}}{2\, \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ){b}^{2}{e}^{2}}{2\, \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}c}}-{\frac{{a}^{2}b{d}^{2}}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}c}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-4\,{\frac{{a}^{2}de}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{a{b}^{2}de}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+3\,{\frac{ab{e}^{2}}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}{e}^{2}}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}c}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{\ln \left ( x \right ) }{c{d}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x)

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.13596, size = 528, normalized size = 2.13 \begin{align*} -\frac{{\left (a^{2} b d^{2} e^{2} - 2 \, a b^{2} d e^{3} + 4 \, a^{2} c d e^{3} + b^{3} e^{4} - 3 \, a b c 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 d^{4} - 2 \, a b c d^{3} e + b^{2} c d^{2} e^{2} + 2 \, a c^{2} d^{2} e^{2} - 2 \, b c^{2} d e^{3} + c^{3} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{{\left (a^{2} d^{2} - 2 \, a b d e + b^{2} e^{2} - a 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 d^{4} - 2 \, a b c d^{3} e + b^{2} c d^{2} e^{2} + 2 \, a c^{2} d^{2} e^{2} - 2 \, b c^{2} d e^{3} + c^{3} e^{4}\right )}} + \frac{e^{5}}{{\left (a d^{3} e^{3} - b d^{2} e^{4} + c d e^{5}\right )}{\left (x e + d\right )}} + \frac{\log \left ({\left | -\frac{d}{x e + d} + 1 \right |}\right )}{c d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d)^2,x, algorithm="giac")

[Out]

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