\(\int (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}) \, dx\) [23]

Optimal result

Integrand size = 33, antiderivative size = 118 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}\right ) \, dx=-\frac {A \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b}}+\frac {A \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}-\frac {A \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {266, 1885, 12, 206, 31, 648, 631, 210, 642} \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}\right ) \, dx=-\frac {A \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b}}-\frac {A \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b}}+\frac {A \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}} \]

[In]

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

[Out]

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

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 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 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 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 {A+C x^2}{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 {A}{a+b x^3} \, dx \\ & = A \int \frac {1}{a+b x^3} \, dx \\ & = \frac {A \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{2/3}}+\frac {A \int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 a^{2/3}} \\ & = \frac {A \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}+\frac {A \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 \sqrt [3]{a}}-\frac {A \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 a^{2/3} \sqrt [3]{b}} \\ & = \frac {A \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}-\frac {A \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b}}+\frac {A \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{a^{2/3} \sqrt [3]{b}} \\ & = -\frac {A \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b}}+\frac {A \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}-\frac {A \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

-1/6*(A*(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]))/(a^(2/3)*b^(1/3))

Maple [C] (verified)

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

Time = 1.62 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42

method	result	size
risch	\(-\frac {C \ln \left (b \,x^{3}+a \right )}{3 b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (C \,\textit {\_R}^{2}+A \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 b}\)	\(49\)
default	\(\frac {A \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {A \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {A \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\)	\(94\)

[In]

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

[Out]

-1/3*C/b*ln(b*x^3+a)+1/3/b*sum((C*_R^2+A)/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.58 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}\right ) \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} A a b \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) - \left (a^{2} b\right )^{\frac {2}{3}} A \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, \left (a^{2} b\right )^{\frac {2}{3}} A \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{6 \, a^{2} b}, \frac {6 \, \sqrt {\frac {1}{3}} A a b \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - \left (a^{2} b\right )^{\frac {2}{3}} A \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, \left (a^{2} b\right )^{\frac {2}{3}} A \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{6 \, a^{2} b}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.19 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}\right ) \, dx=A \operatorname {RootSum} {\left (27 t^{3} a^{2} b - 1, \left ( t \mapsto t \log {\left (3 t a + x \right )} \right )\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}\right ) \, dx=-\frac {A \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} A \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} A \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \left (-\frac {C x^2}{a+b x^3}+\frac {A+C x^2}{a+b x^3}\right ) \, dx=\frac {A\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{3\,a^{2/3}\,b^{1/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (A-\sqrt {3}\,A\,1{}\mathrm {i}\right )}{6\,a^{2/3}\,b^{1/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (A+\sqrt {3}\,A\,1{}\mathrm {i}\right )}{6\,a^{2/3}\,b^{1/3}} \]

[In]

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

[Out]

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