### 3.74 $$\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x}) x (d+e x)^2} \, dx$$

Optimal. Leaf size=183 $\frac{\left (a d (b d-4 c e)+b c e^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (a d^2-c e^2\right ) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac{d}{(d+e x) \left (a d^2-b d e+c e^2\right )}-\frac{\left (a d^2-c e^2\right ) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}$

[Out]

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

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Rubi [A]  time = 0.236279, antiderivative size = 183, 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, 800, 634, 618, 206, 628} $\frac{\left (a d (b d-4 c e)+b c e^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (a d^2-c e^2\right ) \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )^2}+\frac{d}{(d+e x) \left (a d^2-b d e+c e^2\right )}-\frac{\left (a d^2-c e^2\right ) \log (d+e x)}{\left (a d^2-e (b d-c e)\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

d/((a*d^2 - b*d*e + c*e^2)*(d + e*x)) + ((b*c*e^2 + a*d*(b*d - 4*c*e))*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])
/(Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))^2) - ((a*d^2 - c*e^2)*Log[d + e*x])/(a*d^2 - e*(b*d - c*e))^2 + ((
a*d^2 - c*e^2)*Log[c + b*x + a*x^2])/(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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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

Mathematica [A]  time = 0.262249, size = 148, normalized size = 0.81 $\frac{-\frac{2 \left (a d (b d-4 c e)+b c e^2\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+\left (a d^2-c e^2\right ) \log (x (a x+b)+c)+\frac{2 d \left (a d^2+e (c e-b d)\right )}{d+e x}+\left (2 c e^2-2 a d^2\right ) \log (d+e x)}{2 \left (a d^2+e (c e-b d)\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 328, normalized size = 1.8 \begin{align*}{\frac{d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) \left ( ex+d \right ) }}-{\frac{\ln \left ( ex+d \right ) a{d}^{2}}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}+{\frac{\ln \left ( ex+d \right ){e}^{2}c}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ){d}^{2}}{2\, \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ) c{e}^{2}}{2\, \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}}-{\frac{ab{d}^{2}}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+4\,{\frac{acde}{ \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{bc{e}^{2}}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) ^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 31.5593, size = 2217, normalized size = 12.11 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.13598, size = 441, normalized size = 2.41 \begin{align*} \frac{1}{2} \,{\left (\frac{2 \,{\left (a b d^{2} e - 4 \, a c d e^{2} + b c e^{3}\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} d^{4} - 2 \, a b d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, b c d e^{3} + c^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (a d^{2} - 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 )}{a^{2} d^{4} e - 2 \, a b d^{3} e^{2} + b^{2} d^{2} e^{3} + 2 \, a c d^{2} e^{3} - 2 \, b c d e^{4} + c^{2} e^{5}} + \frac{2 \, d e}{{\left (a d^{2} e^{2} - b d e^{3} + c e^{4}\right )}{\left (x e + d\right )}}\right )} e \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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