3.26 \(\int \frac {1}{b x+c x^2+d x^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1594, 705, 29, 634, 618, 206, 628} \[ \frac {c \tanh ^{-1}\left (\frac {c+2 d x}{\sqrt {c^2-4 b d}}\right )}{b \sqrt {c^2-4 b d}}-\frac {\log \left (b+c x+d x^2\right )}{2 b}+\frac {\log (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2 + d*x^3)^(-1),x]

[Out]

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

Rule 29

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

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 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 61, normalized size = 0.98 \[ -\frac {\frac {2 c \tan ^{-1}\left (\frac {c+2 d x}{\sqrt {4 b d-c^2}}\right )}{\sqrt {4 b d-c^2}}+\log (b+x (c+d x))-2 \log (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 211, normalized size = 3.40 \[ \left [\frac {\sqrt {c^{2} - 4 \, b d} c \log \left (\frac {2 \, d^{2} x^{2} + 2 \, c d x + c^{2} - 2 \, b d + \sqrt {c^{2} - 4 \, b d} {\left (2 \, d x + c\right )}}{d x^{2} + c x + b}\right ) - {\left (c^{2} - 4 \, b d\right )} \log \left (d x^{2} + c x + b\right ) + 2 \, {\left (c^{2} - 4 \, b d\right )} \log \relax (x)}{2 \, {\left (b c^{2} - 4 \, b^{2} d\right )}}, \frac {2 \, \sqrt {-c^{2} + 4 \, b d} c \arctan \left (-\frac {\sqrt {-c^{2} + 4 \, b d} {\left (2 \, d x + c\right )}}{c^{2} - 4 \, b d}\right ) - {\left (c^{2} - 4 \, b d\right )} \log \left (d x^{2} + c x + b\right ) + 2 \, {\left (c^{2} - 4 \, b d\right )} \log \relax (x)}{2 \, {\left (b c^{2} - 4 \, b^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 62, normalized size = 1.00 \[ -\frac {c \arctan \left (\frac {2 \, d x + c}{\sqrt {-c^{2} + 4 \, b d}}\right )}{\sqrt {-c^{2} + 4 \, b d} b} - \frac {\log \left (d x^{2} + c x + b\right )}{2 \, b} + \frac {\log \left ({\left | x \right |}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 62, normalized size = 1.00 \[ -\frac {c \arctan \left (\frac {2 d x +c}{\sqrt {4 b d -c^{2}}}\right )}{\sqrt {4 b d -c^{2}}\, b}+\frac {\ln \relax (x )}{b}-\frac {\ln \left (d \,x^{2}+c x +b \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(d*x^3+c*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*b*d-c^2>0)', see `assume?` f
or more details)Is 4*b*d-c^2 positive or negative?

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mupad [B]  time = 0.47, size = 213, normalized size = 3.44 \[ \frac {\ln \relax (x)}{b}-\ln \left (\left (x\,\left (6\,b\,d^2-2\,c^2\,d\right )-b\,c\,d\right )\,\left (\frac {1}{2\,b}-\frac {c\,\sqrt {c^2-4\,b\,d}}{2\,\left (b\,c^2-4\,b^2\,d\right )}\right )-c\,d-3\,d^2\,x\right )\,\left (\frac {1}{2\,b}-\frac {c\,\sqrt {c^2-4\,b\,d}}{2\,\left (b\,c^2-4\,b^2\,d\right )}\right )-\ln \left (\left (x\,\left (6\,b\,d^2-2\,c^2\,d\right )-b\,c\,d\right )\,\left (\frac {1}{2\,b}+\frac {c\,\sqrt {c^2-4\,b\,d}}{2\,\left (b\,c^2-4\,b^2\,d\right )}\right )-c\,d-3\,d^2\,x\right )\,\left (\frac {1}{2\,b}+\frac {c\,\sqrt {c^2-4\,b\,d}}{2\,\left (b\,c^2-4\,b^2\,d\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2 + d*x^3),x)

[Out]

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

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sympy [B]  time = 4.19, size = 564, normalized size = 9.10 \[ \left (- \frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right ) \log {\left (x + \frac {24 b^{4} d^{2} \left (- \frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right )^{2} - 14 b^{3} c^{2} d \left (- \frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right )^{2} - 12 b^{3} d^{2} \left (- \frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right ) + 2 b^{2} c^{4} \left (- \frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right )^{2} + 3 b^{2} c^{2} d \left (- \frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right ) - 12 b^{2} d^{2} + 11 b c^{2} d - 2 c^{4}}{9 b c d^{2} - 2 c^{3} d} \right )} + \left (\frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right ) \log {\left (x + \frac {24 b^{4} d^{2} \left (\frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right )^{2} - 14 b^{3} c^{2} d \left (\frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right )^{2} - 12 b^{3} d^{2} \left (\frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right ) + 2 b^{2} c^{4} \left (\frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right )^{2} + 3 b^{2} c^{2} d \left (\frac {c \sqrt {- 4 b d + c^{2}}}{2 b \left (4 b d - c^{2}\right )} - \frac {1}{2 b}\right ) - 12 b^{2} d^{2} + 11 b c^{2} d - 2 c^{4}}{9 b c d^{2} - 2 c^{3} d} \right )} + \frac {\log {\relax (x )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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