\(\int \frac {a+b x^3+c x^6}{(d+e x^3)^2} \, dx\) [7]

Optimal result

Integrand size = 22, antiderivative size = 213 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx=\frac {c x}{e^2}+\frac {\left (c d^2-b d e+a e^2\right ) x}{3 d e^2 \left (d+e x^3\right )}+\frac {\left (4 c d^2-e (b d+2 a e)\right ) \arctan \left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right )}{3 \sqrt {3} d^{5/3} e^{7/3}}-\frac {\left (4 c d^2-e (b d+2 a e)\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{9 d^{5/3} e^{7/3}}+\frac {\left (4 c d^2-e (b d+2 a e)\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{18 d^{5/3} e^{7/3}} \] Output:

c*x/e^2+1/3*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)+1/9*(4*c*d^2-e*(2*a*e+b* 
d))*arctan(1/3*(d^(1/3)-2*e^(1/3)*x)*3^(1/2)/d^(1/3))*3^(1/2)/d^(5/3)/e^(7 
/3)-1/9*(4*c*d^2-e*(2*a*e+b*d))*ln(d^(1/3)+e^(1/3)*x)/d^(5/3)/e^(7/3)+1/18 
*(4*c*d^2-e*(2*a*e+b*d))*ln(d^(2/3)-d^(1/3)*e^(1/3)*x+e^(2/3)*x^2)/d^(5/3) 
/e^(7/3)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx=\frac {18 c \sqrt [3]{e} x+\frac {6 \sqrt [3]{e} \left (c d^2+e (-b d+a e)\right ) x}{d \left (d+e x^3\right )}+\frac {2 \sqrt {3} \left (4 c d^2-e (b d+2 a e)\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )}{d^{5/3}}-\frac {2 \left (4 c d^2-e (b d+2 a e)\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{d^{5/3}}+\frac {\left (4 c d^2-e (b d+2 a e)\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{d^{5/3}}}{18 e^{7/3}} \] Input:

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3)^2,x]
 

Output:

(18*c*e^(1/3)*x + (6*e^(1/3)*(c*d^2 + e*(-(b*d) + a*e))*x)/(d*(d + e*x^3)) 
 + (2*Sqrt[3]*(4*c*d^2 - e*(b*d + 2*a*e))*ArcTan[(1 - (2*e^(1/3)*x)/d^(1/3 
))/Sqrt[3]])/d^(5/3) - (2*(4*c*d^2 - e*(b*d + 2*a*e))*Log[d^(1/3) + e^(1/3 
)*x])/d^(5/3) + ((4*c*d^2 - e*(b*d + 2*a*e))*Log[d^(2/3) - d^(1/3)*e^(1/3) 
*x + e^(2/3)*x^2])/d^(5/3))/(18*e^(7/3))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.86, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {1739, 913, 750, 16, 1142, 25, 27, 1082, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx\)

\(\Big \downarrow \) 1739

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\int \frac {-3 c d e x^3+c d^2-e (b d+2 a e)}{e x^3+d}dx}{3 d e^2}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \int \frac {1}{e x^3+d}dx-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\int \frac {2 \sqrt [3]{d}-\sqrt [3]{e} x}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx}{3 d^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{e} x+\sqrt [3]{d}}dx}{3 d^{2/3}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\int \frac {2 \sqrt [3]{d}-\sqrt [3]{e} x}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\frac {3}{2} \sqrt [3]{d} \int \frac {1}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx-\frac {\int -\frac {\sqrt [3]{e} \left (\sqrt [3]{d}-2 \sqrt [3]{e} x\right )}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx}{2 \sqrt [3]{e}}}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\frac {3}{2} \sqrt [3]{d} \int \frac {1}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx+\frac {\int \frac {\sqrt [3]{e} \left (\sqrt [3]{d}-2 \sqrt [3]{e} x\right )}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx}{2 \sqrt [3]{e}}}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\frac {3}{2} \sqrt [3]{d} \int \frac {1}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}\right )}{\sqrt [3]{e}}}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{e^{2/3} x^2-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )}{\sqrt [3]{e}}}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{3 d e^2 \left (d+e x^3\right )}-\frac {\left (4 c d^2-e (2 a e+b d)\right ) \left (\frac {-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )}{\sqrt [3]{e}}-\frac {\log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{2 \sqrt [3]{e}}}{3 d^{2/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} \sqrt [3]{e}}\right )-3 c d x}{3 d e^2}\)

Input:

Int[(a + b*x^3 + c*x^6)/(d + e*x^3)^2,x]
 

Output:

((c*d^2 - b*d*e + a*e^2)*x)/(3*d*e^2*(d + e*x^3)) - (-3*c*d*x + (4*c*d^2 - 
 e*(b*d + 2*a*e))*(Log[d^(1/3) + e^(1/3)*x]/(3*d^(2/3)*e^(1/3)) + (-((Sqrt 
[3]*ArcTan[(1 - (2*e^(1/3)*x)/d^(1/3))/Sqrt[3]])/e^(1/3)) - Log[d^(2/3) - 
d^(1/3)*e^(1/3)*x + e^(2/3)*x^2]/(2*e^(1/3)))/(3*d^(2/3))))/(3*d*e^2)
 

Defintions of rubi rules used

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1739

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
)), x_Symbol] :> Simp[(-(c*d^2 - b*d*e + a*e^2))*x*((d + e*x^n)^(q + 1)/(d* 
e^2*n*(q + 1))), x] + Simp[1/(n*(q + 1)*d*e^2)   Int[(d + e*x^n)^(q + 1)*Si 
mp[c*d^2 - b*d*e + a*e^2*(n*(q + 1) + 1) + c*d*e*n*(q + 1)*x^n, x], x], x] 
/; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && N 
eQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[q, -1]
 

Maple [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.41

method	result	size
risch	\(\frac {c x}{e^{2}}+\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) x}{3 d \,e^{2} \left (e \,x^{3}+d \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{3}+d \right )}{\sum }\frac {\left (2 a \,e^{2}+b d e -4 c \,d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 e^{3} d}\)	\(88\)
default	\(\frac {c x}{e^{2}}+\frac {\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) x}{3 d \left (e \,x^{3}+d \right )}+\frac {\left (2 a \,e^{2}+b d e -4 c \,d^{2}\right ) \left (\frac {\ln \left (x +\left (\frac {d}{e}\right )^{\frac {1}{3}}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {d}{e}\right )^{\frac {1}{3}} x +\left (\frac {d}{e}\right )^{\frac {2}{3}}\right )}{6 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{e}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}\right )}{3 d}}{e^{2}}\)	\(156\)

Input:

int((c*x^6+b*x^3+a)/(e*x^3+d)^2,x,method=_RETURNVERBOSE)
 

Output:

c*x/e^2+1/3*(a*e^2-b*d*e+c*d^2)*x/d/e^2/(e*x^3+d)+1/9/e^3/d*sum((2*a*e^2+b 
*d*e-4*c*d^2)/_R^2*ln(x-_R),_R=RootOf(_Z^3*e+d))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 697, normalized size of antiderivative = 3.27 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^2,x, algorithm="fricas")
 

Output:

[1/18*(18*c*d^3*e^2*x^4 - 3*sqrt(1/3)*(4*c*d^4*e - b*d^3*e^2 - 2*a*d^2*e^3 
 + (4*c*d^3*e^2 - b*d^2*e^3 - 2*a*d*e^4)*x^3)*sqrt(-(d^2*e)^(1/3)/e)*log(( 
2*d*e*x^3 - 3*(d^2*e)^(1/3)*d*x - d^2 + 3*sqrt(1/3)*(2*d*e*x^2 + (d^2*e)^( 
2/3)*x - (d^2*e)^(1/3)*d)*sqrt(-(d^2*e)^(1/3)/e))/(e*x^3 + d)) + (4*c*d^3 
- b*d^2*e - 2*a*d*e^2 + (4*c*d^2*e - b*d*e^2 - 2*a*e^3)*x^3)*(d^2*e)^(2/3) 
*log(d*e*x^2 - (d^2*e)^(2/3)*x + (d^2*e)^(1/3)*d) - 2*(4*c*d^3 - b*d^2*e - 
 2*a*d*e^2 + (4*c*d^2*e - b*d*e^2 - 2*a*e^3)*x^3)*(d^2*e)^(2/3)*log(d*e*x 
+ (d^2*e)^(2/3)) + 6*(4*c*d^4*e - b*d^3*e^2 + a*d^2*e^3)*x)/(d^3*e^4*x^3 + 
 d^4*e^3), 1/18*(18*c*d^3*e^2*x^4 - 6*sqrt(1/3)*(4*c*d^4*e - b*d^3*e^2 - 2 
*a*d^2*e^3 + (4*c*d^3*e^2 - b*d^2*e^3 - 2*a*d*e^4)*x^3)*sqrt((d^2*e)^(1/3) 
/e)*arctan(sqrt(1/3)*(2*(d^2*e)^(2/3)*x - (d^2*e)^(1/3)*d)*sqrt((d^2*e)^(1 
/3)/e)/d^2) + (4*c*d^3 - b*d^2*e - 2*a*d*e^2 + (4*c*d^2*e - b*d*e^2 - 2*a* 
e^3)*x^3)*(d^2*e)^(2/3)*log(d*e*x^2 - (d^2*e)^(2/3)*x + (d^2*e)^(1/3)*d) - 
 2*(4*c*d^3 - b*d^2*e - 2*a*d*e^2 + (4*c*d^2*e - b*d*e^2 - 2*a*e^3)*x^3)*( 
d^2*e)^(2/3)*log(d*e*x + (d^2*e)^(2/3)) + 6*(4*c*d^4*e - b*d^3*e^2 + a*d^2 
*e^3)*x)/(d^3*e^4*x^3 + d^4*e^3)]
 

Sympy [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx=\frac {c x}{e^{2}} + \frac {x \left (a e^{2} - b d e + c d^{2}\right )}{3 d^{2} e^{2} + 3 d e^{3} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} d^{5} e^{7} - 8 a^{3} e^{6} - 12 a^{2} b d e^{5} + 48 a^{2} c d^{2} e^{4} - 6 a b^{2} d^{2} e^{4} + 48 a b c d^{3} e^{3} - 96 a c^{2} d^{4} e^{2} - b^{3} d^{3} e^{3} + 12 b^{2} c d^{4} e^{2} - 48 b c^{2} d^{5} e + 64 c^{3} d^{6}, \left ( t \mapsto t \log {\left (\frac {9 t d^{2} e^{2}}{2 a e^{2} + b d e - 4 c d^{2}} + x \right )} \right )\right )} \] Input:

integrate((c*x**6+b*x**3+a)/(e*x**3+d)**2,x)
 

Output:

c*x/e**2 + x*(a*e**2 - b*d*e + c*d**2)/(3*d**2*e**2 + 3*d*e**3*x**3) + Roo 
tSum(729*_t**3*d**5*e**7 - 8*a**3*e**6 - 12*a**2*b*d*e**5 + 48*a**2*c*d**2 
*e**4 - 6*a*b**2*d**2*e**4 + 48*a*b*c*d**3*e**3 - 96*a*c**2*d**4*e**2 - b* 
*3*d**3*e**3 + 12*b**2*c*d**4*e**2 - 48*b*c**2*d**5*e + 64*c**3*d**6, Lamb 
da(_t, _t*log(9*_t*d**2*e**2/(2*a*e**2 + b*d*e - 4*c*d**2) + x)))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^2,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.99 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx=\frac {c x}{e^{2}} + \frac {\sqrt {3} {\left (4 \, c d^{2} - b d e - 2 \, a e^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {d}{e}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {d}{e}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-d e^{2}\right )^{\frac {2}{3}} d e} + \frac {{\left (4 \, c d^{2} - b d e - 2 \, a e^{2}\right )} \log \left (x^{2} + x \left (-\frac {d}{e}\right )^{\frac {1}{3}} + \left (-\frac {d}{e}\right )^{\frac {2}{3}}\right )}{18 \, \left (-d e^{2}\right )^{\frac {2}{3}} d e} + \frac {{\left (4 \, c d^{2} - b d e - 2 \, a e^{2}\right )} \left (-\frac {d}{e}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {d}{e}\right )^{\frac {1}{3}} \right |}\right )}{9 \, d^{2} e^{2}} + \frac {c d^{2} x - b d e x + a e^{2} x}{3 \, {\left (e x^{3} + d\right )} d e^{2}} \] Input:

integrate((c*x^6+b*x^3+a)/(e*x^3+d)^2,x, algorithm="giac")
 

Output:

c*x/e^2 + 1/9*sqrt(3)*(4*c*d^2 - b*d*e - 2*a*e^2)*arctan(1/3*sqrt(3)*(2*x 
+ (-d/e)^(1/3))/(-d/e)^(1/3))/((-d*e^2)^(2/3)*d*e) + 1/18*(4*c*d^2 - b*d*e 
 - 2*a*e^2)*log(x^2 + x*(-d/e)^(1/3) + (-d/e)^(2/3))/((-d*e^2)^(2/3)*d*e) 
+ 1/9*(4*c*d^2 - b*d*e - 2*a*e^2)*(-d/e)^(1/3)*log(abs(x - (-d/e)^(1/3)))/ 
(d^2*e^2) + 1/3*(c*d^2*x - b*d*e*x + a*e^2*x)/((e*x^3 + d)*d*e^2)
 

Mupad [B] (verification not implemented)

Time = 10.94 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.88 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx=\frac {c\,x}{e^2}+\frac {\ln \left (e^{1/3}\,x+d^{1/3}\right )\,\left (-4\,c\,d^2+b\,d\,e+2\,a\,e^2\right )}{9\,d^{5/3}\,e^{7/3}}+\frac {x\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{3\,d\,\left (e^3\,x^3+d\,e^2\right )}+\frac {\ln \left (2\,e^{1/3}\,x-d^{1/3}+\sqrt {3}\,d^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-4\,c\,d^2+b\,d\,e+2\,a\,e^2\right )}{9\,d^{5/3}\,e^{7/3}}-\frac {\ln \left (d^{1/3}-2\,e^{1/3}\,x+\sqrt {3}\,d^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-4\,c\,d^2+b\,d\,e+2\,a\,e^2\right )}{9\,d^{5/3}\,e^{7/3}} \] Input:

int((a + b*x^3 + c*x^6)/(d + e*x^3)^2,x)
 

Output:

(c*x)/e^2 + (log(e^(1/3)*x + d^(1/3))*(2*a*e^2 - 4*c*d^2 + b*d*e))/(9*d^(5 
/3)*e^(7/3)) + (x*(a*e^2 + c*d^2 - b*d*e))/(3*d*(d*e^2 + e^3*x^3)) + (log( 
3^(1/2)*d^(1/3)*1i + 2*e^(1/3)*x - d^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(2*a*e^ 
2 - 4*c*d^2 + b*d*e))/(9*d^(5/3)*e^(7/3)) - (log(3^(1/2)*d^(1/3)*1i - 2*e^ 
(1/3)*x + d^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(2*a*e^2 - 4*c*d^2 + b*d*e))/(9* 
d^(5/3)*e^(7/3))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 538, normalized size of antiderivative = 2.53 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^2} \, dx =\text {Too large to display} \] Input:

int((c*x^6+b*x^3+a)/(e*x^3+d)^2,x)
 

Output:

( - 4*d**(1/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))* 
a*d*e**2 - 4*d**(1/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqr 
t(3)))*a*e**3*x**3 - 2*d**(1/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3)*x)/(d* 
*(1/3)*sqrt(3)))*b*d**2*e - 2*d**(1/3)*sqrt(3)*atan((d**(1/3) - 2*e**(1/3) 
*x)/(d**(1/3)*sqrt(3)))*b*d*e**2*x**3 + 8*d**(1/3)*sqrt(3)*atan((d**(1/3) 
- 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*c*d**3 + 8*d**(1/3)*sqrt(3)*atan((d**( 
1/3) - 2*e**(1/3)*x)/(d**(1/3)*sqrt(3)))*c*d**2*e*x**3 - 2*d**(1/3)*log(d* 
*(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*a*d*e**2 - 2*d**(1/3)*log(d* 
*(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*a*e**3*x**3 - d**(1/3)*log(d 
**(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*b*d**2*e - d**(1/3)*log(d** 
(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*b*d*e**2*x**3 + 4*d**(1/3)*lo 
g(d**(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*c*d**3 + 4*d**(1/3)*log( 
d**(2/3) - e**(1/3)*d**(1/3)*x + e**(2/3)*x**2)*c*d**2*e*x**3 + 4*d**(1/3) 
*log(d**(1/3) + e**(1/3)*x)*a*d*e**2 + 4*d**(1/3)*log(d**(1/3) + e**(1/3)* 
x)*a*e**3*x**3 + 2*d**(1/3)*log(d**(1/3) + e**(1/3)*x)*b*d**2*e + 2*d**(1/ 
3)*log(d**(1/3) + e**(1/3)*x)*b*d*e**2*x**3 - 8*d**(1/3)*log(d**(1/3) + e* 
*(1/3)*x)*c*d**3 - 8*d**(1/3)*log(d**(1/3) + e**(1/3)*x)*c*d**2*e*x**3 + 6 
*e**(1/3)*a*d*e**2*x - 6*e**(1/3)*b*d**2*e*x + 24*e**(1/3)*c*d**3*x + 18*e 
**(1/3)*c*d**2*e*x**4)/(18*e**(1/3)*d**2*e**2*(d + e*x**3))