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

Optimal. Leaf size=123 \[ -\frac{(2 a d-b e) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac{e \log \left (a x^2+b x+c\right )}{2 \left (a d^2-b d e+c e^2\right )}+\frac{e \log (d+e x)}{a d^2-b d e+c e^2} \]

[Out]

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

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Rubi [A]  time = 0.107172, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {1569, 705, 31, 634, 618, 206, 628} \[ -\frac{(2 a d-b e) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac{e \log \left (a x^2+b x+c\right )}{2 \left (a d^2-b d e+c e^2\right )}+\frac{e \log (d+e x)}{a d^2-b d e+c e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*a*d - b*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)))) + (e*Log
[d + e*x])/(a*d^2 - b*d*e + c*e^2) - (e*Log[c + b*x + a*x^2])/(2*(a*d^2 - b*d*e + 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 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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

Mathematica [A]  time = 0.077105, size = 105, normalized size = 0.85 \[ \frac{e \sqrt{4 a c-b^2} (\log (x (a x+b)+c)-2 \log (d+e x))+(2 b e-4 a d) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{2 \sqrt{4 a c-b^2} \left (e (b d-c e)-a d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 168, normalized size = 1.4 \begin{align*}{\frac{e\ln \left ( ex+d \right ) }{a{d}^{2}-bde+{e}^{2}c}}-{\frac{e\ln \left ( a{x}^{2}+bx+c \right ) }{2\,a{d}^{2}-2\,bde+2\,{e}^{2}c}}+2\,{\frac{ad}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{be}{a{d}^{2}-bde+{e}^{2}c}\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^2/(e*x+d),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.29332, size = 697, normalized size = 5.67 \begin{align*} \left [-\frac{{\left (b^{2} - 4 \, a c\right )} e \log \left (a x^{2} + b x + c\right ) - 2 \,{\left (b^{2} - 4 \, a c\right )} e \log \left (e x + d\right ) + \sqrt{b^{2} - 4 \, a c}{\left (2 \, a d - b e\right )} \log \left (\frac{2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c + \sqrt{b^{2} - 4 \, a c}{\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right )}{2 \,{\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} -{\left (b^{3} - 4 \, a b c\right )} d e +{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )}}, -\frac{{\left (b^{2} - 4 \, a c\right )} e \log \left (a x^{2} + b x + c\right ) - 2 \,{\left (b^{2} - 4 \, a c\right )} e \log \left (e x + d\right ) + 2 \, \sqrt{-b^{2} + 4 \, a c}{\left (2 \, a d - b e\right )} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right )}{2 \,{\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} -{\left (b^{3} - 4 \, a b c\right )} d e +{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.09606, size = 170, normalized size = 1.38 \begin{align*} -\frac{e \log \left (a x^{2} + b x + c\right )}{2 \,{\left (a d^{2} - b d e + c e^{2}\right )}} + \frac{e^{2} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e - b d e^{2} + c e^{3}} + \frac{{\left (2 \, a d - b e\right )} \arctan \left (\frac{2 \, a x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a d^{2} - b d e + c e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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