Integrand size = 24, antiderivative size = 186 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {c \sqrt [3]{a+b x^3}}{d^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3}}-\frac {c \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{7/3}}+\frac {c \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{7/3}} \]
-c*(b*x^3+a)^(1/3)/d^2+1/4*(b*x^3+a)^(4/3)/b/d-1/6*c*(-a*d+b*c)^(1/3)*ln(d *x^3+c)/d^(7/3)+1/2*c*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+ a)^(1/3))/d^(7/3)-1/3*c*(-a*d+b*c)^(1/3)*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a) ^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))/d^(7/3)*3^(1/2)
Time = 0.38 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.22 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x^3} \left (-4 b c+a d+b d x^3\right )}{b}-4 \sqrt {3} c \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+4 c \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-2 c \sqrt [3]{b c-a d} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 d^{7/3}} \]
((3*d^(1/3)*(a + b*x^3)^(1/3)*(-4*b*c + a*d + b*d*x^3))/b - 4*Sqrt[3]*c*(b *c - a*d)^(1/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3 ))/Sqrt[3]] + 4*c*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b *x^3)^(1/3)] - 2*c*(b*c - a*d)^(1/3)*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)])/(12*d^(7/3))
Time = 0.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {948, 90, 60, 70, 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.
| \(\displaystyle \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^3 \sqrt [3]{b x^3+a}}{d x^3+c}dx^3\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \int \frac {\sqrt [3]{b x^3+a}}{d x^3+c}dx^3}{d}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx^3}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 70 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}+\frac {3 \int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 \sqrt [3]{d} (b c-a d)^{2/3}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (\frac {3 \int \frac {1}{-x^6-3}d\left (1-\frac {2 \sqrt [3]{d} \sqrt [3]{b x^3+a}}{\sqrt [3]{b c-a d}}\right )}{\sqrt [3]{d} (b c-a d)^{2/3}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3}}{4 b d}-\frac {c \left (\frac {3 \sqrt [3]{a+b x^3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt [3]{d} (b c-a d)^{2/3}}-\frac {\log \left (c+d x^3\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}+\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{d} (b c-a d)^{2/3}}\right )}{d}\right )}{d}\right )\) |
((3*(a + b*x^3)^(4/3))/(4*b*d) - (c*((3*(a + b*x^3)^(1/3))/d - ((b*c - a*d )*(-((Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3)) /Sqrt[3]])/(d^(1/3)*(b*c - a*d)^(2/3))) - Log[c + d*x^3]/(2*d^(1/3)*(b*c - a*d)^(2/3)) + (3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d ^(1/3)*(b*c - a*d)^(2/3))))/d))/d)/3
3.7.60.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) , x] + (Simp[3/(2*b*q) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 /3)], x] + Simp[3/(2*b*q^2) 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]
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]
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])
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]]
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]
Time = 4.82 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10
| method | result | size |
| pseudoelliptic | \(\frac {\frac {3 d \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} \left (d \left (b \,x^{3}+a \right )-4 b c \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2}+b c \left (a d -b c \right ) \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} b \,d^{3}}\) | \(205\) |
1/6/(1/d*(a*d-b*c))^(2/3)*(3/2*d*(1/d*(a*d-b*c))^(2/3)*(d*(b*x^3+a)-4*b*c) *(b*x^3+a)^(1/3)+b*c*(a*d-b*c)*(2*arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)+(1 /d*(a*d-b*c))^(1/3))/(1/d*(a*d-b*c))^(1/3))*3^(1/2)+ln((b*x^3+a)^(2/3)+(1/ d*(a*d-b*c))^(1/3)*(b*x^3+a)^(1/3)+(1/d*(a*d-b*c))^(2/3))-2*ln((b*x^3+a)^( 1/3)-(1/d*(a*d-b*c))^(1/3))))/b/d^3
Time = 0.28 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.19 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {4 \, \sqrt {3} b c \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 2 \, b c \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 4 \, b c \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (b d x^{3} - 4 \, b c + a d\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{12 \, b d^{2}} \]
-1/12*(4*sqrt(3)*b*c*((b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*((b*c - a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 2* b*c*((b*c - a*d)/d)^(1/3)*log((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)) - 4*b*c*((b*c - a*d)/d)^(1/3)*log ((b*x^3 + a)^(1/3) + ((b*c - a*d)/d)^(1/3)) - 3*(b*d*x^3 - 4*b*c + a*d)*(b *x^3 + a)^(1/3))/(b*d^2)
\[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int \frac {x^{5} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \]
Exception generated. \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.48 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {{\left (b^{6} c^{2} d^{2} - a b^{5} c d^{3}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{6} c d^{4} - a b^{5} d^{5}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{3}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{3}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{4} c d^{2} - {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{3} d^{3}}{4 \, b^{4} d^{4}} \]
-1/3*(b^6*c^2*d^2 - a*b^5*c*d^3)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a )^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^6*c*d^4 - a*b^5*d^5) + 1/3*sqrt(3)*( -b*c*d^2 + a*d^3)^(1/3)*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))/d^3 + 1/6*(-b*c*d^2 + a*d^3)^(1/ 3)*c*log((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))/d^3 - 1/4*(4*(b*x^3 + a)^(1/3)*b^4*c*d^2 - (b*x^3 + a)^(4/3)*b^3*d^3)/(b^4*d^4)
Time = 8.54 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.60 \[ \int \frac {x^5 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {{\left (b\,x^3+a\right )}^{4/3}}{4\,b\,d}-{\left (b\,x^3+a\right )}^{1/3}\,\left (\frac {a}{b\,d}+\frac {b^2\,c-a\,b\,d}{b^2\,d^2}\right )-\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,b\,c^2-3\,a\,c\,d\right )+\frac {c\,{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{7/3}}-\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,b\,c^2-3\,a\,c\,d\right )+\frac {c\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{7/3}}+\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (3\,b\,c^2-3\,a\,c\,d\right )-\frac {c\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{7/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{7/3}} \]
(a + b*x^3)^(4/3)/(4*b*d) - (a + b*x^3)^(1/3)*(a/(b*d) + (b^2*c - a*b*d)/( b^2*d^2)) - (c*log((a + b*x^3)^(1/3)*(3*b*c^2 - 3*a*c*d) + (c*(a*d - b*c)^ (1/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*(a*d - b*c)^(1/3))/(3*d^(7/3)) - (c*log((a + b*x^3)^(1/3)*(3*b*c^2 - 3*a*c*d) + (c*((3^(1/2)*1i)/2 - 1/2)* (a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(7/3)))*((3^(1/2)*1i)/2 - 1/ 2)*(a*d - b*c)^(1/3))/(3*d^(7/3)) + (c*log((a + b*x^3)^(1/3)*(3*b*c^2 - 3* a*c*d) - (c*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(1/3)*(9*a*d^3 - 9*b*c*d^2) )/(3*d^(7/3)))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(1/3))/(3*d^(7/3))