\(\int (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}) \, dx\) [22]

Optimal result

Integrand size = 35, antiderivative size = 118 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=-\frac {B \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3}}-\frac {B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}+\frac {B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {266, 1607, 1885, 12, 298, 31, 648, 631, 210, 642} \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=\frac {B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3}}-\frac {B \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3}}-\frac {B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}} \]

[In]

Int[-((C*x^2)/(a + b*x^3)) + (B*x + C*x^2)/(a + b*x^3),x]

[Out]

-((B*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(1/3)*b^(2/3))) - (B*Log[a^(1/3) + b^(1/3)*
x])/(3*a^(1/3)*b^(2/3)) + (B*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(1/3)*b^(2/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1607

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

Rule 1885

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rubi steps \begin{align*} \text {integral}& = -\left (C \int \frac {x^2}{a+b x^3} \, dx\right )+\int \frac {B x+C x^2}{a+b x^3} \, dx \\ & = -\frac {C \log \left (a+b x^3\right )}{3 b}+\int \frac {x (B+C x)}{a+b x^3} \, dx \\ & = -\frac {C \log \left (a+b x^3\right )}{3 b}+C \int \frac {x^2}{a+b x^3} \, dx+\int \frac {B x}{a+b x^3} \, dx \\ & = B \int \frac {x}{a+b x^3} \, dx \\ & = -\frac {B \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 \sqrt [3]{a} \sqrt [3]{b}}+\frac {B \int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 \sqrt [3]{a} \sqrt [3]{b}} \\ & = -\frac {B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}+\frac {B \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 \sqrt [3]{a} b^{2/3}}+\frac {B \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 \sqrt [3]{b}} \\ & = -\frac {B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}+\frac {B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3}}+\frac {B \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{a} b^{2/3}} \\ & = -\frac {B \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3}}-\frac {B \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}+\frac {B \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=\frac {B \left (-2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )\right )}{6 \sqrt [3]{a} b^{2/3}} \]

[In]

Integrate[-((C*x^2)/(a + b*x^3)) + (B*x + C*x^2)/(a + b*x^3),x]

[Out]

(B*(-2*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 2*Log[a^(1/3) + b^(1/3)*x] + Log[a^(2/3) - a^(1/3
)*b^(1/3)*x + b^(2/3)*x^2]))/(6*a^(1/3)*b^(2/3))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.57 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.40

method	result	size
risch	\(-\frac {C \ln \left (b \,x^{3}+a \right )}{3 b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (C \textit {\_R} +B \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{3 b}\)	\(47\)
default	\(-\frac {B \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {B \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {B \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\)	\(94\)

[In]

int(-C*x^2/(b*x^3+a)+(C*x^2+B*x)/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

-1/3*C/b*ln(b*x^3+a)+1/3/b*sum(1/_R*(C*_R+B)*ln(x-_R),_R=RootOf(_Z^3*b+a))

Fricas [A] (verification not implemented)

none

Time = 0.91 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.63 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} B a b \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) + \left (-a b^{2}\right )^{\frac {2}{3}} B \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} B \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a b^{2}}, \frac {6 \, \sqrt {\frac {1}{3}} B a b \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + \left (-a b^{2}\right )^{\frac {2}{3}} B \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} B \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a b^{2}}\right ] \]

[In]

integrate(-C*x^2/(b*x^3+a)+(C*x^2+B*x)/(b*x^3+a),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.22 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=B \operatorname {RootSum} {\left (27 t^{3} a b^{2} + 1, \left ( t \mapsto t \log {\left (9 t^{2} a b + x \right )} \right )\right )} \]

[In]

integrate(-C*x**2/(b*x**3+a)+(C*x**2+B*x)/(b*x**3+a),x)

[Out]

B*RootSum(27*_t**3*a*b**2 + 1, Lambda(_t, _t*log(9*_t**2*a*b + x)))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.35 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=-\frac {C \log \left (b x^{3} + a\right )}{3 \, b} + \frac {{\left (2 \, C \left (\frac {a}{b}\right )^{\frac {1}{3}} + B\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (C \left (\frac {a}{b}\right )^{\frac {1}{3}} - B\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {\sqrt {3} {\left (2 \, C a - {\left (3 \, B \left (\frac {a}{b}\right )^{\frac {2}{3}} + \frac {2 \, C a}{b}\right )} b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b} \]

[In]

integrate(-C*x^2/(b*x^3+a)+(C*x^2+B*x)/(b*x^3+a),x, algorithm="maxima")

[Out]

-1/3*C*log(b*x^3 + a)/b + 1/6*(2*C*(a/b)^(1/3) + B)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b*(a/b)^(1/3)) + 1
/3*(C*(a/b)^(1/3) - B)*log(x + (a/b)^(1/3))/(b*(a/b)^(1/3)) - 1/9*sqrt(3)*(2*C*a - (3*B*(a/b)^(2/3) + 2*C*a/b)
*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=\frac {\sqrt {3} B \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {1}{3}}} - \frac {B \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {1}{3}}} - \frac {B \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a} \]

[In]

integrate(-C*x^2/(b*x^3+a)+(C*x^2+B*x)/(b*x^3+a),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.83 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {B x+C x^2}{a+b x^3}\right ) \, dx=-\frac {B\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{3\,a^{1/3}\,b^{2/3}}+\frac {\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}-\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (B-\sqrt {3}\,B\,1{}\mathrm {i}\right )}{6\,a^{1/3}\,b^{2/3}}+\frac {\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}+\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (B+\sqrt {3}\,B\,1{}\mathrm {i}\right )}{6\,a^{1/3}\,b^{2/3}} \]

[In]

int((B*x + C*x^2)/(a + b*x^3) - (C*x^2)/(a + b*x^3),x)

[Out]

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