3.24 \(\int (-\frac {C x^2}{a+b x^3}+\frac {A+B x+C x^2}{a+b x^3}) \, dx\)

Optimal. Leaf size=161 \[ -\frac {\left (A-\frac {\sqrt [3]{a} B}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b}}+\frac {\left (A \sqrt [3]{b}-\sqrt [3]{a} B\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{2/3}}-\frac {\left (\sqrt [3]{a} B+A \sqrt [3]{b}\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {260, 1871, 1860, 31, 634, 617, 204, 628} \[ -\frac {\left (A-\frac {\sqrt [3]{a} B}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b}}+\frac {\left (A \sqrt [3]{b}-\sqrt [3]{a} B\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{2/3}}-\frac {\left (\sqrt [3]{a} B+A \sqrt [3]{b}\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 204

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

Rule 260

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 617

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 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 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 1871

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

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Mathematica [A]  time = 0.05, size = 124, normalized size = 0.77 \[ \frac {\left (A \sqrt [3]{b}-\sqrt [3]{a} B\right ) \left (2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )\right )-2 \sqrt {3} \left (\sqrt [3]{a} B+A \sqrt [3]{b}\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{6 a^{2/3} b^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[3]*(A*b^(1/3) + a^(1/3)*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + (A*b^(1/3) - a^(1/3)*B)*(2*L
og[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^(2/3)*b^(2/3))

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fricas [C]  time = 2.32, size = 1961, normalized size = 12.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3
)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))*log(1/4*((1/2)^(1/
3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(
3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))^2*B*a^2*b - 1/2*((1/2)^(1/3)*(I*s
qrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)
/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))*A^2*a*b + 2*A*B^2*a + (B^3*a + A^3*b)*x)
+ 1/12*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(
2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)) + 3*sqrt(1/3)*sq
rt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3
)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))^2*a*b + 16*A*B)/(a
*b)))*log(-1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*
(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))^2*B*a^2*
b + 1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(
2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))*A^2*a*b - 2*A*B
^2*a + 2*(B^3*a + A^3*b)*x + 3/4*sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a -
A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b
)/(a^2*b^2))^(1/3)))*B*a^2*b + 2*A^2*a*b)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^
3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a -
A^3*b)/(a^2*b^2))^(1/3)))^2*a*b + 16*A*B)/(a*b))) + 1/12*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^
2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (
B^3*a - A^3*b)/(a^2*b^2))^(1/3)) - 3*sqrt(1/3)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2)
- (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3
*a - A^3*b)/(a^2*b^2))^(1/3)))^2*a*b + 16*A*B)/(a*b)))*log(-1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/
(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b
^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))^2*B*a^2*b + 1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^
2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (
B^3*a - A^3*b)/(a^2*b^2))^(1/3)))*A^2*a*b - 2*A*B^2*a + 2*(B^3*a + A^3*b)*x - 3/4*sqrt(1/3)*(((1/2)^(1/3)*(I*s
qrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3) + 1)
/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))*B*a^2*b + 2*A^2*a*b)*sqrt(-(((1/2)^(1/3)
*(I*sqrt(3) + 1)*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3) - 2*(1/2)^(2/3)*A*B*(-I*sqrt(3)
+ 1)/(a*b*((B^3*a + A^3*b)/(a^2*b^2) - (B^3*a - A^3*b)/(a^2*b^2))^(1/3)))^2*a*b + 16*A*B)/(a*b)))

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giac [A]  time = 0.20, size = 147, normalized size = 0.91 \[ -\frac {\sqrt {3} {\left (A b - \left (-a b^{2}\right )^{\frac {1}{3}} B\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {2}{3}}} - \frac {{\left (A b + \left (-a b^{2}\right )^{\frac {1}{3}} B\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {2}{3}}} - \frac {{\left (B b \left (-\frac {a}{b}\right )^{\frac {1}{3}} + A b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 186, normalized size = 1.16 \[ \frac {\sqrt {3}\, A \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {A \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}-\frac {A \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {\sqrt {3}\, B \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}-\frac {B \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}+\frac {B \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 188, normalized size = 1.17 \[ -\frac {C \log \left (b x^{3} + a\right )}{3 \, b} - \frac {\sqrt {3} {\left (2 \, C a - {\left (3 \, B \left (\frac {a}{b}\right )^{\frac {2}{3}} + 3 \, A \left (\frac {a}{b}\right )^{\frac {1}{3}} + \frac {2 \, C a}{b}\right )} b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b} + \frac {{\left (2 \, C \left (\frac {a}{b}\right )^{\frac {2}{3}} + B \left (\frac {a}{b}\right )^{\frac {1}{3}} - A\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (C \left (\frac {a}{b}\right )^{\frac {2}{3}} - B \left (\frac {a}{b}\right )^{\frac {1}{3}} + A\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-C*x^2/(b*x^3+a)+(C*x^2+B*x+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*B*(a/b)^(2/3) + 3*A*(a/b)^(1/3) + 2*C*a/b)*b)*arctan(1/3*sqr
t(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b) + 1/6*(2*C*(a/b)^(2/3) + B*(a/b)^(1/3) - 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) - B*(a/b)^(1/3) + A)*log(x + (a/b)^(1/3))/(b*(a/b)^(2/3)
)

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mupad [B]  time = 4.92, size = 127, normalized size = 0.79 \[ \sum _{k=1}^3\ln \left (b\,\left (B^2\,x+A\,B+{\mathrm {root}\left (27\,a^2\,b^2\,z^3+9\,A\,B\,a\,b\,z+B^3\,a-A^3\,b,z,k\right )}^2\,a\,b\,9+A\,\mathrm {root}\left (27\,a^2\,b^2\,z^3+9\,A\,B\,a\,b\,z+B^3\,a-A^3\,b,z,k\right )\,b\,x\,3\right )\right )\,\mathrm {root}\left (27\,a^2\,b^2\,z^3+9\,A\,B\,a\,b\,z+B^3\,a-A^3\,b,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(b*(B^2*x + A*B + 9*root(27*a^2*b^2*z^3 + 9*A*B*a*b*z + B^3*a - A^3*b, z, k)^2*a*b + 3*A*root(27*a^2
*b^2*z^3 + 9*A*B*a*b*z + B^3*a - A^3*b, z, k)*b*x))*root(27*a^2*b^2*z^3 + 9*A*B*a*b*z + B^3*a - A^3*b, z, k),
k, 1, 3)

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sympy [A]  time = 1.28, size = 76, normalized size = 0.47 \[ \operatorname {RootSum} {\left (27 t^{3} a^{2} b^{2} + 9 t A B a b - A^{3} b + B^{3} a, \left (t \mapsto t \log {\left (x + \frac {9 t^{2} B a^{2} b + 3 t A^{2} a b + 2 A B^{2} a}{A^{3} b + B^{3} a} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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