\(\int \frac {x^5}{a x^2+b x^3+c x^4} \, dx\) [11]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=-\frac {b x}{c^2}+\frac {x^2}{2 c}+\frac {b \left (b^2-3 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 c^3} \]

[Out]

-b*x/c^2+1/2*x^2/c+1/2*(-a*c+b^2)*ln(c*x^2+b*x+a)/c^3+b*(-3*a*c+b^2)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/c^3
/(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1599, 715, 648, 632, 212, 642} \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=\frac {b \left (b^2-3 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 c^3}-\frac {b x}{c^2}+\frac {x^2}{2 c} \]

[In]

Int[x^5/(a*x^2 + b*x^3 + c*x^4),x]

[Out]

-((b*x)/c^2) + x^2/(2*c) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c]) +
((b^2 - a*c)*Log[a + b*x + c*x^2])/(2*c^3)

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 715

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, 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] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 1599

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=\frac {c x (-2 b+c x)-\frac {2 b \left (b^2-3 a c\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\left (b^2-a c\right ) \log (a+x (b+c x))}{2 c^3} \]

[In]

Integrate[x^5/(a*x^2 + b*x^3 + c*x^4),x]

[Out]

(c*x*(-2*b + c*x) - (2*b*(b^2 - 3*a*c)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (b^2 - a*c
)*Log[a + x*(b + c*x)])/(2*c^3)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.10

method	result	size
default	\(-\frac {-\frac {1}{2} c \,x^{2}+b x}{c^{2}}+\frac {\frac {\left (-a c +b^{2}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (a b -\frac {\left (-a c +b^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{2}}\)	\(98\)
risch	\(\frac {x^{2}}{2 c}-\frac {b x}{c^{2}}-\frac {2 \ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x -\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) a^{2}}{c \left (4 a c -b^{2}\right )}+\frac {5 \ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x -\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) a \,b^{2}}{2 c^{2} \left (4 a c -b^{2}\right )}-\frac {\ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x -\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) b^{4}}{2 c^{3} \left (4 a c -b^{2}\right )}+\frac {\ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x -\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}}{2 c^{3} \left (4 a c -b^{2}\right )}-\frac {2 \ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x +\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) a^{2}}{c \left (4 a c -b^{2}\right )}+\frac {5 \ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x +\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) a \,b^{2}}{2 c^{2} \left (4 a c -b^{2}\right )}-\frac {\ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x +\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) b^{4}}{2 c^{3} \left (4 a c -b^{2}\right )}-\frac {\ln \left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, c x +\sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}\, b \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right ) \left (3 a c -b^{2}\right )^{2}}}{2 c^{3} \left (4 a c -b^{2}\right )}\)	\(915\)

[In]

int(x^5/(c*x^4+b*x^3+a*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.34 \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=\left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - {\left (b^{3} - 3 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 2 \, {\left (b^{3} - 3 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \]

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2),x, algorithm="fricas")

[Out]

[1/2*((b^2*c^2 - 4*a*c^3)*x^2 - (b^3 - 3*a*b*c)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqr
t(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - 2*(b^3*c - 4*a*b*c^2)*x + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*log(c
*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4), 1/2*((b^2*c^2 - 4*a*c^3)*x^2 + 2*(b^3 - 3*a*b*c)*sqrt(-b^2 + 4*a*c)*arct
an(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 2*(b^3*c - 4*a*b*c^2)*x + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*lo
g(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (83) = 166\).

Time = 0.48 (sec) , antiderivative size = 381, normalized size of antiderivative = 4.28 \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=- \frac {b x}{c^{2}} + \left (- \frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{2 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{2 c^{3}}\right ) \log {\left (x + \frac {2 a^{2} c - a b^{2} + 4 a c^{3} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{2 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{2 c^{3}}\right ) - b^{2} c^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{2 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{2 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{2 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{2 c^{3}}\right ) \log {\left (x + \frac {2 a^{2} c - a b^{2} + 4 a c^{3} \left (\frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{2 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{2 c^{3}}\right ) - b^{2} c^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}} \cdot \left (3 a c - b^{2}\right )}{2 c^{3} \cdot \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{2 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \frac {x^{2}}{2 c} \]

[In]

integrate(x**5/(c*x**4+b*x**3+a*x**2),x)

[Out]

-b*x/c**2 + (-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3))*log(x + (2
*a**2*c - a*b**2 + 4*a*c**3*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c
**3)) - b**2*c**2*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3)))/(3*
a*b*c - b**3)) + (b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3))*log(x
+ (2*a**2*c - a*b**2 + 4*a*c**3*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(
2*c**3)) - b**2*c**2*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3)))/(
3*a*b*c - b**3)) + x**2/(2*c)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2),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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=\frac {c x^{2} - 2 \, b x}{2 \, c^{2}} + \frac {{\left (b^{2} - a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} - \frac {{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{3}} \]

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {x^5}{a x^2+b x^3+c x^4} \, dx=\frac {x^2}{2\,c}-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (4\,a^2\,c^2-5\,a\,b^2\,c+b^4\right )}{2\,\left (4\,a\,c^4-b^2\,c^3\right )}-\frac {b\,x}{c^2}+\frac {b\,\mathrm {atan}\left (\frac {b+2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (3\,a\,c-b^2\right )}{c^3\,\sqrt {4\,a\,c-b^2}} \]

[In]

int(x^5/(a*x^2 + b*x^3 + c*x^4),x)

[Out]

x^2/(2*c) - (log(a + b*x + c*x^2)*(b^4 + 4*a^2*c^2 - 5*a*b^2*c))/(2*(4*a*c^4 - b^2*c^3)) - (b*x)/c^2 + (b*atan
((b + 2*c*x)/(4*a*c - b^2)^(1/2))*(3*a*c - b^2))/(c^3*(4*a*c - b^2)^(1/2))