\(\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx\) [334]
Optimal result
Integrand size = 20, antiderivative size = 206 \[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}+\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c+\sqrt [3]{a} d\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3}}-\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3}}+\frac {c^2 \log \left (a+b x^3\right )}{3 b}
\]
[Out]
2*c*d*x/b+1/2*d^2*x^2/b-1/3*a^(1/3)*d*(2*b^(1/3)*c-a^(1/3)*d)*ln(a^(1/3)+b^(1/3)*x)/b^(5/3)+1/6*a^(1/3)*d*(2*b
^(1/3)*c-a^(1/3)*d)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(5/3)+1/3*c^2*ln(b*x^3+a)/b+1/3*a^(1/3)*d*(2*b
^(1/3)*c+a^(1/3)*d)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/b^(5/3)*3^(1/2)
Rubi [A] (verified)
Time = 0.19 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of
steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1901, 1885, 1874, 31, 648,
631, 210, 642, 266} \[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3}}+\frac {\sqrt [3]{a} d \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (\sqrt [3]{a} d+2 \sqrt [3]{b} c\right )}{\sqrt {3} b^{5/3}}-\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {c^2 \log \left (a+b x^3\right )}{3 b}+\frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}
\]
[In]
Int[(x^2*(c + d*x)^2)/(a + b*x^3),x]
[Out]
(2*c*d*x)/b + (d^2*x^2)/(2*b) + (a^(1/3)*d*(2*b^(1/3)*c + a^(1/3)*d)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a
^(1/3))])/(Sqrt[3]*b^(5/3)) - (a^(1/3)*d*(2*b^(1/3)*c - a^(1/3)*d)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(5/3)) + (a^
(1/3)*d*(2*b^(1/3)*c - a^(1/3)*d)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(5/3)) + (c^2*Log[a + b
*x^3])/(3*b)
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 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 1874
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 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]
Rule 1901
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]
Rubi steps \begin{align*}
\text {integral}& = \int \left (\frac {2 c d}{b}+\frac {d^2 x}{b}-\frac {2 a c d+a d^2 x-b c^2 x^2}{b \left (a+b x^3\right )}\right ) \, dx \\ & = \frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {\int \frac {2 a c d+a d^2 x-b c^2 x^2}{a+b x^3} \, dx}{b} \\ & = \frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {\int \frac {2 a c d+a d^2 x}{a+b x^3} \, dx}{b}+c^2 \int \frac {x^2}{a+b x^3} \, dx \\ & = \frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}+\frac {c^2 \log \left (a+b x^3\right )}{3 b}-\frac {\int \frac {\sqrt [3]{a} \left (4 a \sqrt [3]{b} c d+a^{4/3} d^2\right )+\sqrt [3]{b} \left (-2 a \sqrt [3]{b} c d+a^{4/3} d^2\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 a^{2/3} b^{4/3}}-\frac {\left (\sqrt [3]{a} d \left (2 c-\frac {\sqrt [3]{a} d}{\sqrt [3]{b}}\right )\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b} \\ & = \frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {c^2 \log \left (a+b x^3\right )}{3 b}+\frac {\left (\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right )\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 b^{5/3}}-\frac {\left (a^{2/3} d \left (2 \sqrt [3]{b} c+\sqrt [3]{a} d\right )\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 b^{4/3}} \\ & = \frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}-\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3}}+\frac {c^2 \log \left (a+b x^3\right )}{3 b}-\frac {\left (\sqrt [3]{a} d \left (2 \sqrt [3]{b} c+\sqrt [3]{a} d\right )\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{b^{5/3}} \\ & = \frac {2 c d x}{b}+\frac {d^2 x^2}{2 b}+\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c+\sqrt [3]{a} d\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3}}-\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {\sqrt [3]{a} d \left (2 \sqrt [3]{b} c-\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3}}+\frac {c^2 \log \left (a+b x^3\right )}{3 b} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.07 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.94
\[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\frac {12 b^{2/3} c d x+3 b^{2/3} d^2 x^2+2 \sqrt {3} \sqrt [3]{a} d \left (2 \sqrt [3]{b} c+\sqrt [3]{a} d\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 \sqrt [3]{a} d \left (-2 \sqrt [3]{b} c+\sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-\sqrt [3]{a} d \left (-2 \sqrt [3]{b} c+\sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 b^{2/3} c^2 \log \left (a+b x^3\right )}{6 b^{5/3}}
\]
[In]
Integrate[(x^2*(c + d*x)^2)/(a + b*x^3),x]
[Out]
(12*b^(2/3)*c*d*x + 3*b^(2/3)*d^2*x^2 + 2*Sqrt[3]*a^(1/3)*d*(2*b^(1/3)*c + a^(1/3)*d)*ArcTan[(1 - (2*b^(1/3)*x
)/a^(1/3))/Sqrt[3]] + 2*a^(1/3)*d*(-2*b^(1/3)*c + a^(1/3)*d)*Log[a^(1/3) + b^(1/3)*x] - a^(1/3)*d*(-2*b^(1/3)*
c + a^(1/3)*d)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 2*b^(2/3)*c^2*Log[a + b*x^3])/(6*b^(5/3))
Maple [C] (verified)
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.78 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.33
| | |
method | result | size |
| | |
risch |
\(\frac {d^{2} x^{2}}{2 b}+\frac {2 c d x}{b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{2} b \,c^{2}-\textit {\_R} a \,d^{2}-2 a c d \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 b^{2}}\) |
\(68\) |
default |
\(\frac {d \left (\frac {1}{2} d \,x^{2}+2 c x \right )}{b}+\frac {-2 a c d \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )-a \,d^{2} \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {c^{2} \ln \left (b \,x^{3}+a \right )}{3}}{b}\) |
\(227\) |
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[In]
int(x^2*(d*x+c)^2/(b*x^3+a),x,method=_RETURNVERBOSE)
[Out]
1/2*d^2*x^2/b+2*c*d*x/b+1/3/b^2*sum((_R^2*b*c^2-_R*a*d^2-2*a*c*d)/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
Fricas [C] (verification not implemented)
Result contains complex when optimal does not.
Time = 0.94 (sec) , antiderivative size = 4545, normalized size of antiderivative = 22.06
\[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\text {Too large to display}
\]
[In]
integrate(x^2*(d*x+c)^2/(b*x^3+a),x, algorithm="fricas")
[Out]
1/12*(6*d^2*x^2 + 24*c*d*x - 2*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3
+ (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)
+ (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3
*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b*log(1/4*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/
b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*
a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^
3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)^2*b^3 + 3*(2*(1/2)^(2/3
)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2
*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3
)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) -
2*c^2/b)*b^2*c^2 + 5*b*c^4 + 4*a*c*d^3 + (8*b*c^3*d + a*d^4)*x) + ((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d
^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6
- 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*
c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b + 6*c^2 + 3*sqrt(
1/3)*b*sqrt(-((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^
3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2
*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b
^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)^2*b^3 + 4*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3)
+ 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2
*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2
*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b^2*c^2 + 4*b*c^4 + 32*a*c*d^3)/b^3))*lo
g(-1/4*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^
3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^
3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/
3)*(I*sqrt(3) + 1) - 2*c^2/b)^2*b^3 - 3*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2
*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b
^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 -
2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b^2*c^2 - 5*b*c^4 - 4*a*c*d^3 + 2*(8*b*c^3*d +
a*d^4)*x + 3/4*sqrt(1/3)*((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*
b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1
/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3
+ a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b^3 - 6*b^2*c^2)*sqrt(-((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*
a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2
*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 +
2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)^2*b^3 + 4*(2*(
1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(
b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^
3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(
3) + 1) - 2*c^2/b)*b^2*c^2 + 4*b*c^4 + 32*a*c*d^3)/b^3)) + ((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)
*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*
c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c
^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b + 6*c^2 - 3*sqrt(1/3)*b*s
qrt(-((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3
/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3
+ (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3
)*(I*sqrt(3) + 1) - 2*c^2/b)^2*b^3 + 4*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*
c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^
5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2
*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b^2*c^2 + 4*b*c^4 + 32*a*c*d^3)/b^3))*log(-1/4*(
2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 -
3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b
*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sq
rt(3) + 1) - 2*c^2/b)^2*b^3 - 3*(2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3
+ (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3
) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^
3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b^2*c^2 - 5*b*c^4 - 4*a*c*d^3 + 2*(8*b*c^3*d + a*d^4)*x
- 3/4*sqrt(1/3)*((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 +
a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3
)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^
6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)*b^3 - 6*b^2*c^2)*sqrt(-((2*(1/2)^(2/3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)
/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2
*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d
^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1) - 2*c^2/b)^2*b^3 + 4*(2*(1/2)^(2/
3)*(c^4/b^2 - (b*c^4 + 2*a*c*d^3)/b^3)*(-I*sqrt(3) + 1)/(2*c^6/b^3 + (8*b*c^3 + a*d^3)*a*d^3/b^5 - 3*(b*c^4 +
2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3) + (1/2)^(1/3)*(2*c^6/b^3 + (8*b*c^3 + a*d^
3)*a*d^3/b^5 - 3*(b*c^4 + 2*a*c*d^3)*c^2/b^4 + (b^2*c^6 - 2*a*b*c^3*d^3 + a^2*d^6)/b^5)^(1/3)*(I*sqrt(3) + 1)
- 2*c^2/b)*b^2*c^2 + 4*b*c^4 + 32*a*c*d^3)/b^3)))/b
Sympy [A] (verification not implemented)
Time = 0.73 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.67
\[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\operatorname {RootSum} {\left (27 t^{3} b^{5} - 27 t^{2} b^{4} c^{2} + t \left (18 a b^{2} c d^{3} + 9 b^{3} c^{4}\right ) - a^{2} d^{6} + 2 a b c^{3} d^{3} - b^{2} c^{6}, \left ( t \mapsto t \log {\left (x + \frac {9 t^{2} b^{3} - 18 t b^{2} c^{2} + 4 a c d^{3} + 5 b c^{4}}{a d^{4} + 8 b c^{3} d} \right )} \right )\right )} + \frac {2 c d x}{b} + \frac {d^{2} x^{2}}{2 b}
\]
[In]
integrate(x**2*(d*x+c)**2/(b*x**3+a),x)
[Out]
RootSum(27*_t**3*b**5 - 27*_t**2*b**4*c**2 + _t*(18*a*b**2*c*d**3 + 9*b**3*c**4) - a**2*d**6 + 2*a*b*c**3*d**3
- b**2*c**6, Lambda(_t, _t*log(x + (9*_t**2*b**3 - 18*_t*b**2*c**2 + 4*a*c*d**3 + 5*b*c**4)/(a*d**4 + 8*b*c**
3*d)))) + 2*c*d*x/b + d**2*x**2/(2*b)
Maxima [A] (verification not implemented)
none
Time = 0.32 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.97
\[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=-\frac {\sqrt {3} {\left (a d^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a c d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a b} + \frac {d^{2} x^{2} + 4 \, c d x}{2 \, b} + \frac {{\left (2 \, b c^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - a d^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a c d\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (b c^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + a d^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a c d\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}
\]
[In]
integrate(x^2*(d*x+c)^2/(b*x^3+a),x, algorithm="maxima")
[Out]
-1/3*sqrt(3)*(a*d^2*(a/b)^(2/3) + 2*a*c*d*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*
b) + 1/2*(d^2*x^2 + 4*c*d*x)/b + 1/6*(2*b*c^2*(a/b)^(2/3) - a*d^2*(a/b)^(1/3) + 2*a*c*d)*log(x^2 - x*(a/b)^(1/
3) + (a/b)^(2/3))/(b^2*(a/b)^(2/3)) + 1/3*(b*c^2*(a/b)^(2/3) + a*d^2*(a/b)^(1/3) - 2*a*c*d)*log(x + (a/b)^(1/3
))/(b^2*(a/b)^(2/3))
Giac [A] (verification not implemented)
none
Time = 0.27 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.01
\[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\frac {c^{2} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b} - \frac {\sqrt {3} {\left (2 \, \left (-a b^{2}\right )^{\frac {1}{3}} b c d - \left (-a b^{2}\right )^{\frac {2}{3}} d^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3}} + \frac {b d^{2} x^{2} + 4 \, b c d x}{2 \, b^{2}} - \frac {{\left (2 \, \left (-a b^{2}\right )^{\frac {1}{3}} b c d + \left (-a b^{2}\right )^{\frac {2}{3}} d^{2}\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3}} + \frac {{\left (a b^{4} d^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a b^{4} c d\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{5}}
\]
[In]
integrate(x^2*(d*x+c)^2/(b*x^3+a),x, algorithm="giac")
[Out]
1/3*c^2*log(abs(b*x^3 + a))/b - 1/3*sqrt(3)*(2*(-a*b^2)^(1/3)*b*c*d - (-a*b^2)^(2/3)*d^2)*arctan(1/3*sqrt(3)*(
2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^3 + 1/2*(b*d^2*x^2 + 4*b*c*d*x)/b^2 - 1/6*(2*(-a*b^2)^(1/3)*b*c*d + (-a*b^
2)^(2/3)*d^2)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^3 + 1/3*(a*b^4*d^2*(-a/b)^(1/3) + 2*a*b^4*c*d)*(-a/b)
^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^5)
Mupad [B] (verification not implemented)
Time = 9.36 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.73
\[
\int \frac {x^2 (c+d x)^2}{a+b x^3} \, dx=\left (\sum _{k=1}^3\ln \left (\frac {a\,\left (b\,c^4+{\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c^2\,z^2+18\,a\,b^2\,c\,d^3\,z+9\,b^3\,c^4\,z+2\,a\,b\,c^3\,d^3-b^2\,c^6-a^2\,d^6,z,k\right )}^2\,b^3\,9-\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c^2\,z^2+18\,a\,b^2\,c\,d^3\,z+9\,b^3\,c^4\,z+2\,a\,b\,c^3\,d^3-b^2\,c^6-a^2\,d^6,z,k\right )\,b^2\,c^2\,6+2\,a\,c\,d^3+a\,d^4\,x+2\,b\,c^3\,d\,x-\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c^2\,z^2+18\,a\,b^2\,c\,d^3\,z+9\,b^3\,c^4\,z+2\,a\,b\,c^3\,d^3-b^2\,c^6-a^2\,d^6,z,k\right )\,b^2\,c\,d\,x\,6\right )}{b}\right )\,\mathrm {root}\left (27\,b^5\,z^3-27\,b^4\,c^2\,z^2+18\,a\,b^2\,c\,d^3\,z+9\,b^3\,c^4\,z+2\,a\,b\,c^3\,d^3-b^2\,c^6-a^2\,d^6,z,k\right )\right )+\frac {d^2\,x^2}{2\,b}+\frac {2\,c\,d\,x}{b}
\]
[In]
int((x^2*(c + d*x)^2)/(a + b*x^3),x)
[Out]
symsum(log((a*(b*c^4 + 9*root(27*b^5*z^3 - 27*b^4*c^2*z^2 + 18*a*b^2*c*d^3*z + 9*b^3*c^4*z + 2*a*b*c^3*d^3 - b
^2*c^6 - a^2*d^6, z, k)^2*b^3 - 6*root(27*b^5*z^3 - 27*b^4*c^2*z^2 + 18*a*b^2*c*d^3*z + 9*b^3*c^4*z + 2*a*b*c^
3*d^3 - b^2*c^6 - a^2*d^6, z, k)*b^2*c^2 + 2*a*c*d^3 + a*d^4*x + 2*b*c^3*d*x - 6*root(27*b^5*z^3 - 27*b^4*c^2*
z^2 + 18*a*b^2*c*d^3*z + 9*b^3*c^4*z + 2*a*b*c^3*d^3 - b^2*c^6 - a^2*d^6, z, k)*b^2*c*d*x))/b)*root(27*b^5*z^3
- 27*b^4*c^2*z^2 + 18*a*b^2*c*d^3*z + 9*b^3*c^4*z + 2*a*b*c^3*d^3 - b^2*c^6 - a^2*d^6, z, k), k, 1, 3) + (d^2
*x^2)/(2*b) + (2*c*d*x)/b