\(\int \frac {x^5}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\) [732]

Optimal result

Integrand size = 24, antiderivative size = 168 \[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\left (a+b x^3\right )^{2/3}}{2 b d}+\frac {c \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{5/3} \sqrt [3]{b c-a d}}-\frac {c \log \left (c+d x^3\right )}{6 d^{5/3} \sqrt [3]{b c-a d}}+\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{5/3} \sqrt [3]{b c-a d}} \] Output:

1/2*(b*x^3+a)^(2/3)/b/d+1/3*c*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a* 
d+b*c)^(1/3))*3^(1/2))*3^(1/2)/d^(5/3)/(-a*d+b*c)^(1/3)-1/6*c*ln(d*x^3+c)/ 
d^(5/3)/(-a*d+b*c)^(1/3)+1/2*c*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3) 
)/d^(5/3)/(-a*d+b*c)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.21 \[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {3 d^{2/3} \sqrt [3]{b c-a d} \left (a+b x^3\right )^{2/3}+2 \sqrt {3} b c \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+2 b c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-b c \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 b d^{5/3} \sqrt [3]{b c-a d}} \] Input:

Integrate[x^5/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {948, 90, 68, 16, 1082, 217}

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.

\(\displaystyle \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 68

\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 b d}-\frac {c \left (-\frac {3 \int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}+\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 b d}-\frac {c \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{2/3}}{2 b d}-\frac {c \left (\frac {3 \int \frac {1}{-x^6-3}d\left (1-\frac {2 \sqrt [3]{d} \sqrt [3]{b x^3+a}}{\sqrt [3]{b c-a d}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )\)

\(\Big \downarrow \) 217

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

Input:

Int[x^5/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
 

Output:

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

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 68

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 948

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

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]
 

Maple [A] (verified)

Time = 1.61 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11

method	result	size
pseudoelliptic	\(\frac {3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} d \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}\, b c -2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right ) b c +\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right ) b c}{6 b \,d^{2} \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\)	\(186\)

Input:

int(x^5/(b*x^3+a)^(1/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (134) = 268\).

Time = 0.11 (sec) , antiderivative size = 667, normalized size of antiderivative = 3.97 \[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

integrate(x^5/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{5}}{\sqrt [3]{a + b x^{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**5/(b*x**3+a)**(1/3)/(d*x**3+c),x)
 

Output:

Integral(x**5/((a + b*x**3)**(1/3)*(c + d*x**3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5/(b*x^3+a)^(1/3)/(d*x^3+c),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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.53 \[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\frac {2 \, b c d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{b c d^{2} - a d^{3}} + \frac {6 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{3} - \sqrt {3} a d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{b c d^{3} - a d^{4}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{d}}{6 \, b} \] Input:

integrate(x^5/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

1/6*(2*b*c*d*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a 
*d)/d)^(1/3)))/(b*c*d^2 - a*d^3) + 6*(-b*c*d^2 + a*d^3)^(2/3)*b*c*arctan(1 
/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d) 
^(1/3))/(sqrt(3)*b*c*d^3 - sqrt(3)*a*d^4) - (-b*c*d^2 + a*d^3)^(2/3)*b*c*l 
og((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - 
 a*d)/d)^(2/3))/(b*c*d^3 - a*d^4) + 3*(b*x^3 + a)^(2/3)/d)/b
 

Mupad [B] (verification not implemented)

Time = 3.69 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.30 \[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {{\left (b\,x^3+a\right )}^{2/3}}{2\,b\,d}+\frac {\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {c^2\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{4/3}}\right )\,\left (c-\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}}{d}-\frac {c^2\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{4/3}}\right )\,\left (c+\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {c\,\ln \left (\frac {c^2\,{\left (b\,x^3+a\right )}^{1/3}}{d}+\frac {b\,c^3-a\,c^2\,d}{d^{4/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{5/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \] Input:

int(x^5/((a + b*x^3)^(1/3)*(c + d*x^3)),x)
 

Output:

(a + b*x^3)^(2/3)/(2*b*d) + (log((c^2*(a + b*x^3)^(1/3))/d - (c^2*(3^(1/2) 
*1i - 1)^2*(a*d - b*c)^(1/3))/(4*d^(4/3)))*(c - 3^(1/2)*c*1i))/(6*d^(5/3)* 
(a*d - b*c)^(1/3)) + (log((c^2*(a + b*x^3)^(1/3))/d - (c^2*(3^(1/2)*1i + 1 
)^2*(a*d - b*c)^(1/3))/(4*d^(4/3)))*(c + 3^(1/2)*c*1i))/(6*d^(5/3)*(a*d - 
b*c)^(1/3)) - (c*log((c^2*(a + b*x^3)^(1/3))/d + (b*c^3 - a*c^2*d)/(d^(4/3 
)*(a*d - b*c)^(2/3))))/(3*d^(5/3)*(a*d - b*c)^(1/3))
 

Reduce [F]

\[ \int \frac {x^5}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{5}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} d \,x^{3}}d x \] Input:

int(x^5/(b*x^3+a)^(1/3)/(d*x^3+c),x)
 

Output:

int(x**5/((a + b*x**3)**(1/3)*c + (a + b*x**3)**(1/3)*d*x**3),x)