3.5.18 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x}) x^3} \, dx\) [418]

Optimal. Leaf size=62 \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}+\frac {\log (x)}{a}-\frac {\log \left (a+b x+c x^2\right )}{2 a} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1368, 719, 29, 648, 632, 212, 642} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {\log \left (a+b x+c x^2\right )}{2 a}+\frac {\log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)*x^3),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 648

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 719

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 1368

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

Rubi steps

\begin {align*} \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^3} \, dx &=\int \frac {1}{x \left (a+b x+c x^2\right )} \, dx\\ &=\frac {\int \frac {1}{x} \, dx}{a}+\frac {\int \frac {-b-c x}{a+b x+c x^2} \, dx}{a}\\ &=\frac {\log (x)}{a}-\frac {\int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 a}-\frac {b \int \frac {1}{a+b x+c x^2} \, dx}{2 a}\\ &=\frac {\log (x)}{a}-\frac {\log \left (a+b x+c x^2\right )}{2 a}+\frac {b \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{a}\\ &=\frac {b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}+\frac {\log (x)}{a}-\frac {\log \left (a+b x+c x^2\right )}{2 a}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 61, normalized size = 0.98 \begin {gather*} -\frac {\frac {2 b \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 \log (x)+\log (a+x (b+c x))}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*b*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] - 2*Log[x] + Log[a + x*(b + c*x)])/a

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Maple [A]
time = 0.03, size = 61, normalized size = 0.98

method	result	size
default	\(\frac {-\frac {\ln \left (c \,x^{2}+b x +a \right )}{2}-\frac {b \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a}+\frac {\ln \left (x \right )}{a}\)	\(61\)
risch	\(\frac {\ln \left (x \right )}{a}+\left (\munderset {\textit {\_R} =\RootOf \left (\left (4 a^{2} c -a \,b^{2}\right ) \textit {\_Z}^{2}+\left (4 a c -b^{2}\right ) \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (6 a c -2 b^{2}\right ) \textit {\_R} +3 c \right ) x -a b \textit {\_R} +b \right )\right )\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)/x^3,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 deta

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Fricas [A]
time = 0.35, size = 211, normalized size = 3.40 \begin {gather*} \left [\frac {\sqrt {b^{2} - 4 \, a c} b \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 ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (54) = 108\).
time = 4.61, size = 564, normalized size = 9.10 \begin {gather*} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right ) \log {\left (x + \frac {24 a^{4} c^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right )^{2} - 14 a^{3} b^{2} c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right )^{2} - 12 a^{3} c^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right ) + 2 a^{2} b^{4} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right )^{2} + 3 a^{2} b^{2} c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right ) - 12 a^{2} c^{2} + 11 a b^{2} c - 2 b^{4}}{9 a b c^{2} - 2 b^{3} c} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right ) \log {\left (x + \frac {24 a^{4} c^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right )^{2} - 14 a^{3} b^{2} c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right )^{2} - 12 a^{3} c^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right ) + 2 a^{2} b^{4} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right )^{2} + 3 a^{2} b^{2} c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 a \left (4 a c - b^{2}\right )} - \frac {1}{2 a}\right ) - 12 a^{2} c^{2} + 11 a b^{2} c - 2 b^{4}}{9 a b c^{2} - 2 b^{3} c} \right )} + \frac {\log {\left (x \right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)/x**3,x)

[Out]

(-b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a))*log(x + (24*a**4*c**2*(-b*sqrt(-4*a*c + b**2)/(2*a*(4*
a*c - b**2)) - 1/(2*a))**2 - 14*a**3*b**2*c*(-b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a))**2 - 12*a*
*3*c**2*(-b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a)) + 2*a**2*b**4*(-b*sqrt(-4*a*c + b**2)/(2*a*(4*
a*c - b**2)) - 1/(2*a))**2 + 3*a**2*b**2*c*(-b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a)) - 12*a**2*c
**2 + 11*a*b**2*c - 2*b**4)/(9*a*b*c**2 - 2*b**3*c)) + (b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a))*
log(x + (24*a**4*c**2*(b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a))**2 - 14*a**3*b**2*c*(b*sqrt(-4*a*
c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a))**2 - 12*a**3*c**2*(b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2
*a)) + 2*a**2*b**4*(b*sqrt(-4*a*c + b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a))**2 + 3*a**2*b**2*c*(b*sqrt(-4*a*c +
b**2)/(2*a*(4*a*c - b**2)) - 1/(2*a)) - 12*a**2*c**2 + 11*a*b**2*c - 2*b**4)/(9*a*b*c**2 - 2*b**3*c)) + log(x)
/a

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Giac [A]
time = 4.11, size = 62, normalized size = 1.00 \begin {gather*} -\frac {b \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a} - \frac {\log \left (c x^{2} + b x + a\right )}{2 \, a} + \frac {\log \left ({\left | x \right |}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.72, size = 213, normalized size = 3.44 \begin {gather*} \frac {\ln \left (x\right )}{a}-\ln \left (b\,c-\left (x\,\left (6\,a\,c^2-2\,b^2\,c\right )-a\,b\,c\right )\,\left (\frac {1}{2\,a}-\frac {b\,\sqrt {b^2-4\,a\,c}}{2\,\left (a\,b^2-4\,a^2\,c\right )}\right )+3\,c^2\,x\right )\,\left (\frac {1}{2\,a}-\frac {b\,\sqrt {b^2-4\,a\,c}}{2\,\left (a\,b^2-4\,a^2\,c\right )}\right )-\ln \left (\left (x\,\left (6\,a\,c^2-2\,b^2\,c\right )-a\,b\,c\right )\,\left (\frac {1}{2\,a}+\frac {b\,\sqrt {b^2-4\,a\,c}}{2\,\left (a\,b^2-4\,a^2\,c\right )}\right )-b\,c-3\,c^2\,x\right )\,\left (\frac {1}{2\,a}+\frac {b\,\sqrt {b^2-4\,a\,c}}{2\,\left (a\,b^2-4\,a^2\,c\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(c + a/x^2 + b/x)),x)

[Out]

log(x)/a - log(b*c - (x*(6*a*c^2 - 2*b^2*c) - a*b*c)*(1/(2*a) - (b*(b^2 - 4*a*c)^(1/2))/(2*(a*b^2 - 4*a^2*c)))
+ 3*c^2*x)*(1/(2*a) - (b*(b^2 - 4*a*c)^(1/2))/(2*(a*b^2 - 4*a^2*c))) - log((x*(6*a*c^2 - 2*b^2*c) - a*b*c)*(1
/(2*a) + (b*(b^2 - 4*a*c)^(1/2))/(2*(a*b^2 - 4*a^2*c))) - b*c - 3*c^2*x)*(1/(2*a) + (b*(b^2 - 4*a*c)^(1/2))/(2
*(a*b^2 - 4*a^2*c)))

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