\(\int \frac {x^5}{a+b x^3+c x^6} \, dx\) [139]

Optimal result

Integrand size = 18, antiderivative size = 63 \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^3+c x^6\right )}{6 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1371, 648, 632, 212, 642} \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^3+c x^6\right )}{6 c} \]

[In]

Int[x^5/(a + b*x^3 + c*x^6),x]

[Out]

(b*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*c*Sqrt[b^2 - 4*a*c]) + Log[a + b*x^3 + c*x^6]/(6*c)

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 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {x}{a+b x+c x^2} \, dx,x,x^3\right ) \\ & = \frac {\text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^3\right )}{6 c}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^3\right )}{6 c} \\ & = \frac {\log \left (a+b x^3+c x^6\right )}{6 c}+\frac {b \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^3\right )}{3 c} \\ & = \frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^3+c x^6\right )}{6 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\frac {-\frac {2 b \arctan \left (\frac {b+2 c x^3}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\log \left (a+b x^3+c x^6\right )}{6 c} \]

[In]

Integrate[x^5/(a + b*x^3 + c*x^6),x]

[Out]

((-2*b*ArcTan[(b + 2*c*x^3)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + Log[a + b*x^3 + c*x^6])/(6*c)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\ln \left (c \,x^{6}+b \,x^{3}+a \right )}{6 c}-\frac {b \arctan \left (\frac {2 c \,x^{3}+b}{\sqrt {4 a c -b^{2}}}\right )}{3 c \sqrt {4 a c -b^{2}}}\)	\(60\)
risch	\(\frac {2 \ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{3}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) a}{3 \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{3}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2}}{6 c \left (4 a c -b^{2}\right )}+\frac {\ln \left (\left (-4 a b c +b^{3}+\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{3}+2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right )}}{6 c \left (4 a c -b^{2}\right )}+\frac {2 \ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{3}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) a}{3 \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{3}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) b^{2}}{6 c \left (4 a c -b^{2}\right )}-\frac {\ln \left (\left (-4 a b c +b^{3}-\sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, b \right ) x^{3}-2 \sqrt {-b^{2} \left (4 a c -b^{2}\right )}\, a \right ) \sqrt {-b^{2} \left (4 a c -b^{2}\right )}}{6 c \left (4 a c -b^{2}\right )}\)	\(467\)

[In]

int(x^5/(c*x^6+b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.13 \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{6} + 2 \, b c x^{3} + b^{2} - 2 \, a c + {\left (2 \, c x^{3} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{6} + b x^{3} + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{3} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \]

[In]

integrate(x^5/(c*x^6+b*x^3+a),x, algorithm="fricas")

[Out]

[1/6*(sqrt(b^2 - 4*a*c)*b*log((2*c^2*x^6 + 2*b*c*x^3 + b^2 - 2*a*c + (2*c*x^3 + b)*sqrt(b^2 - 4*a*c))/(c*x^6 +
b*x^3 + a)) + (b^2 - 4*a*c)*log(c*x^6 + b*x^3 + a))/(b^2*c - 4*a*c^2), 1/6*(2*sqrt(-b^2 + 4*a*c)*b*arctan(-(2
*c*x^3 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + (b^2 - 4*a*c)*log(c*x^6 + b*x^3 + a))/(b^2*c - 4*a*c^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (54) = 108\).

Time = 0.84 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.54 \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\left (- \frac {b \sqrt {- 4 a c + b^{2}}}{6 c \left (4 a c - b^{2}\right )} + \frac {1}{6 c}\right ) \log {\left (x^{3} + \frac {- 12 a c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{6 c \left (4 a c - b^{2}\right )} + \frac {1}{6 c}\right ) + 2 a + 3 b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{6 c \left (4 a c - b^{2}\right )} + \frac {1}{6 c}\right )}{b} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{6 c \left (4 a c - b^{2}\right )} + \frac {1}{6 c}\right ) \log {\left (x^{3} + \frac {- 12 a c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{6 c \left (4 a c - b^{2}\right )} + \frac {1}{6 c}\right ) + 2 a + 3 b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{6 c \left (4 a c - b^{2}\right )} + \frac {1}{6 c}\right )}{b} \right )} \]

[In]

integrate(x**5/(c*x**6+b*x**3+a),x)

[Out]

(-b*sqrt(-4*a*c + b**2)/(6*c*(4*a*c - b**2)) + 1/(6*c))*log(x**3 + (-12*a*c*(-b*sqrt(-4*a*c + b**2)/(6*c*(4*a*
c - b**2)) + 1/(6*c)) + 2*a + 3*b**2*(-b*sqrt(-4*a*c + b**2)/(6*c*(4*a*c - b**2)) + 1/(6*c)))/b) + (b*sqrt(-4*
a*c + b**2)/(6*c*(4*a*c - b**2)) + 1/(6*c))*log(x**3 + (-12*a*c*(b*sqrt(-4*a*c + b**2)/(6*c*(4*a*c - b**2)) +
1/(6*c)) + 2*a + 3*b**2*(b*sqrt(-4*a*c + b**2)/(6*c*(4*a*c - b**2)) + 1/(6*c)))/b)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^5/(c*x^6+b*x^3+a),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 = 0.43 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=-\frac {b \arctan \left (\frac {2 \, c x^{3} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt {-b^{2} + 4 \, a c} c} + \frac {\log \left (c x^{6} + b x^{3} + a\right )}{6 \, c} \]

[In]

integrate(x^5/(c*x^6+b*x^3+a),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.64 (sec) , antiderivative size = 1199, normalized size of antiderivative = 19.03 \[ \int \frac {x^5}{a+b x^3+c x^6} \, dx=\text {Too large to display} \]

[In]

int(x^5/(a + b*x^3 + c*x^6),x)

[Out]

(log(a + b*x^3 + c*x^6)*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)) + (b*atan((4*x^3*((b*(b^2 - ((12*b^2*c - ((
45*b^2*c^2 - (27*b^2*c^3*(12*a*c - 3*b^2))/(36*a*c^2 - 9*b^2*c))*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)))*(
12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)) - (b*((b*(45*b^2*c^2 - (27*b^2*c^3*(12*a*c - 3*b^2))/(36*a*c^2 - 9*b
^2*c)))/(6*c*(4*a*c - b^2)^(1/2)) - (9*b^3*c^2*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)*(4*a*c - b^2)^(1/2)))
)/(6*c*(4*a*c - b^2)^(1/2)) + (3*b^4*c*(12*a*c - 3*b^2))/(4*(36*a*c^2 - 9*b^2*c)*(4*a*c - b^2))))/(4*a^2*c) +
((2*a*c - b^2)*(b^5/(4*(4*a*c - b^2)^(3/2)) + ((12*a*c - 3*b^2)*((b*(45*b^2*c^2 - (27*b^2*c^3*(12*a*c - 3*b^2)
)/(36*a*c^2 - 9*b^2*c)))/(6*c*(4*a*c - b^2)^(1/2)) - (9*b^3*c^2*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)*(4*a
*c - b^2)^(1/2))))/(2*(36*a*c^2 - 9*b^2*c)) - (b*(12*b^2*c - ((45*b^2*c^2 - (27*b^2*c^3*(12*a*c - 3*b^2))/(36*
a*c^2 - 9*b^2*c))*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c))))/(6*c*(4*a*c - b^2)^(1/2))))/(4*a^2*c*(4*a*c - b
^2)^(1/2)))*(4*a*c - b^2)^(3/2))/b^3 + ((4*a*c - b^2)^(3/2)*(a*b + ((((72*a*b*c^2 - (54*a*b*c^3*(12*a*c - 3*b^
2))/(36*a*c^2 - 9*b^2*c))*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)) - 15*a*b*c)*(12*a*c - 3*b^2))/(2*(36*a*c^
2 - 9*b^2*c)) - (b*((b*(72*a*b*c^2 - (54*a*b*c^3*(12*a*c - 3*b^2))/(36*a*c^2 - 9*b^2*c)))/(6*c*(4*a*c - b^2)^(
1/2)) - (9*a*b^2*c^2*(12*a*c - 3*b^2))/((36*a*c^2 - 9*b^2*c)*(4*a*c - b^2)^(1/2))))/(6*c*(4*a*c - b^2)^(1/2))
+ (3*a*b^3*c*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)*(4*a*c - b^2))))/(a^2*b^2*c) + ((2*a*c - b^2)*(4*a*c -
b^2)*((a*b^4)/(2*(4*a*c - b^2)^(3/2)) + (((b*(72*a*b*c^2 - (54*a*b*c^3*(12*a*c - 3*b^2))/(36*a*c^2 - 9*b^2*c))
)/(6*c*(4*a*c - b^2)^(1/2)) - (9*a*b^2*c^2*(12*a*c - 3*b^2))/((36*a*c^2 - 9*b^2*c)*(4*a*c - b^2)^(1/2)))*(12*a
*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)) + (b*(((72*a*b*c^2 - (54*a*b*c^3*(12*a*c - 3*b^2))/(36*a*c^2 - 9*b^2*c))
*(12*a*c - 3*b^2))/(2*(36*a*c^2 - 9*b^2*c)) - 15*a*b*c))/(6*c*(4*a*c - b^2)^(1/2))))/(a^2*b^3*c)))/(3*c*(4*a*c
- b^2)^(1/2))