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

Optimal. Leaf size=70 \[ -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x+c x^2\right )}{2 c^2}+\frac {x}{c} \]

[Out]

x/c-1/2*b*ln(c*x^2+b*x+a)/c^2-(-2*a*c+b^2)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/c^2/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1340, 703, 634, 618, 206, 628} \[ -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x+c x^2\right )}{2 c^2}+\frac {x}{c} \]

Antiderivative was successfully verified.

[In]

Int[(c + a/x^2 + b/x)^(-1),x]

[Out]

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

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 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 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]

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 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, 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] && GtQ[m, 1]

Rule 1340

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(2*n*p)*(c + b/x^n + a/x^(2*n))^p,
x] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && LtQ[n, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 73, normalized size = 1.04 \[ \frac {\left (b^2-2 a c\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{c^2 \sqrt {4 a c-b^2}}-\frac {b \log \left (a+b x+c x^2\right )}{2 c^2}+\frac {x}{c} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + a/x^2 + b/x)^(-1),x]

[Out]

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

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fricas [A]  time = 0.79, size = 235, normalized size = 3.36 \[ \left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x + {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x + {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 67, normalized size = 0.96 \[ \frac {x}{c} - \frac {b \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 101, normalized size = 1.44 \[ -\frac {2 a \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {b^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {b \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*x-1/2*b*ln(c*x^2+b*x+a)/c^2-2/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a+1/c^2/(4*a*c-b^2)^
(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 1.42, size = 172, normalized size = 2.46 \[ \frac {x}{c}+\frac {b^3\,\ln \left (c\,x^2+b\,x+a\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )}-\frac {2\,a\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{c\,\sqrt {4\,a\,c-b^2}}+\frac {b^2\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{c^2\,\sqrt {4\,a\,c-b^2}}-\frac {2\,a\,b\,c\,\ln \left (c\,x^2+b\,x+a\right )}{4\,a\,c^3-b^2\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c + a/x^2 + b/x),x)

[Out]

x/c + (b^3*log(a + b*x + c*x^2))/(2*(4*a*c^3 - b^2*c^2)) - (2*a*atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c -
b^2)^(1/2)))/(c*(4*a*c - b^2)^(1/2)) + (b^2*atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2)))/(c^2*(4
*a*c - b^2)^(1/2)) - (2*a*b*c*log(a + b*x + c*x^2))/(4*a*c^3 - b^2*c^2)

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sympy [B]  time = 0.60, size = 306, normalized size = 4.37 \[ \left (- \frac {b}{2 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- a b - 4 a c^{2} \left (- \frac {b}{2 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + b^{2} c \left (- \frac {b}{2 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \left (- \frac {b}{2 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- a b - 4 a c^{2} \left (- \frac {b}{2 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + b^{2} c \left (- \frac {b}{2 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \frac {x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x),x)

[Out]

(-b/(2*c**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*c**2*(4*a*c - b**2)))*log(x + (-a*b - 4*a*c**2*(-b/(2*c**
2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*c**2*(4*a*c - b**2))) + b**2*c*(-b/(2*c**2) - sqrt(-4*a*c + b**2)*(
2*a*c - b**2)/(2*c**2*(4*a*c - b**2))))/(2*a*c - b**2)) + (-b/(2*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2
*c**2*(4*a*c - b**2)))*log(x + (-a*b - 4*a*c**2*(-b/(2*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*c**2*(4*a
*c - b**2))) + b**2*c*(-b/(2*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*c**2*(4*a*c - b**2))))/(2*a*c - b**
2)) + x/c

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