Integrand size = 24, antiderivative size = 125 \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {2 a^2 \sqrt {c+d x^3}}{3 b^3}-\frac {2 (b c+a d) \left (c+d x^3\right )^{3/2}}{9 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b d^2}-\frac {2 a^2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2}} \] Output:
2/3*a^2*(d*x^3+c)^(1/2)/b^3-2/9*(a*d+b*c)*(d*x^3+c)^(3/2)/b^2/d^2+2/15*(d* x^3+c)^(5/2)/b/d^2-2/3*a^2*(-a*d+b*c)^(1/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2 )/(-a*d+b*c)^(1/2))/b^(7/2)
Time = 0.38 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.97 \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {2 \sqrt {c+d x^3} \left (15 a^2 d^2-5 a b d \left (c+d x^3\right )+b^2 \left (-2 c^2+c d x^3+3 d^2 x^6\right )\right )}{45 b^3 d^2}-\frac {2 a^2 \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2}} \] Input:
Integrate[(x^8*Sqrt[c + d*x^3])/(a + b*x^3),x]
Output:
(2*Sqrt[c + d*x^3]*(15*a^2*d^2 - 5*a*b*d*(c + d*x^3) + b^2*(-2*c^2 + c*d*x ^3 + 3*d^2*x^6)))/(45*b^3*d^2) - (2*a^2*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[b] *Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(3*b^(7/2))
Time = 0.50 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 99, 2009}
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.
| \(\displaystyle \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \sqrt {d x^3+c}}{b x^3+a}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {\sqrt {d x^3+c} a^2}{b^2 \left (b x^3+a\right )}+\frac {\left (d x^3+c\right )^{3/2}}{b d}+\frac {(-b c-a d) \sqrt {d x^3+c}}{b^2 d}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\frac {2 a^2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {2 a^2 \sqrt {c+d x^3}}{b^3}-\frac {2 \left (c+d x^3\right )^{3/2} (a d+b c)}{3 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{5/2}}{5 b d^2}\right )\) |
Input:
Int[(x^8*Sqrt[c + d*x^3])/(a + b*x^3),x]
Output:
((2*a^2*Sqrt[c + d*x^3])/b^3 - (2*(b*c + a*d)*(c + d*x^3)^(3/2))/(3*b^2*d^ 2) + (2*(c + d*x^3)^(5/2))/(5*b*d^2) - (2*a^2*Sqrt[b*c - a*d]*ArcTanh[(Sqr t[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(7/2))/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]]
Time = 4.76 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99
| method | result | size |
| risch | \(\frac {2 \left (3 b^{2} d^{2} x^{6}-5 x^{3} a b \,d^{2}+x^{3} b^{2} c d +15 a^{2} d^{2}-5 a b c d -2 b^{2} c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d^{2} b^{3}}-\frac {2 a^{2} \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{3} \sqrt {\left (a d -b c \right ) b}}\) | \(124\) |
| pseudoelliptic | \(-\frac {2 \left (-\left (-\frac {2 \left (-\frac {3 d \,x^{3}}{2}+c \right ) \left (d \,x^{3}+c \right ) b^{2}}{15}-\frac {\left (d \,x^{3}+c \right ) a b d}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{3}+c}+a^{2} d^{2} \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{3 \sqrt {\left (a d -b c \right ) b}\, d^{2} b^{3}}\) | \(124\) |
| default | \(\frac {\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15}+\frac {2 c \,x^{3} \sqrt {d \,x^{3}+c}}{45 d}-\frac {4 c^{2} \sqrt {d \,x^{3}+c}}{45 d^{2}}}{b}+\frac {2 a^{2} \left (\sqrt {d \,x^{3}+c}-\frac {\left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{3}}-\frac {2 a \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9 b^{2} d}\) | \(138\) |
| elliptic | \(\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15 b}+\frac {2 \left (-\frac {a d -b c}{b^{2}}-\frac {4 c}{5 b}\right ) x^{3} \sqrt {d \,x^{3}+c}}{9 d}+\frac {2 \left (\frac {\left (a d -b c \right ) a}{b^{3}}-\frac {2 \left (-\frac {a d -b c}{b^{2}}-\frac {4 c}{5 b}\right ) c}{3 d}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i a^{2} \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {b \left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d \right )}{2 d \left (a d -b c \right )}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{3 b^{3} d^{2}}\) | \(531\) |
Input:
int(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
2/45*(3*b^2*d^2*x^6-5*a*b*d^2*x^3+b^2*c*d*x^3+15*a^2*d^2-5*a*b*c*d-2*b^2*c ^2)*(d*x^3+c)^(1/2)/d^2/b^3-2/3*a^2*(a*d-b*c)/b^3/((a*d-b*c)*b)^(1/2)*arct an(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2))
Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.24 \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\left [\frac {15 \, a^{2} d^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{6} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} + {\left (b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{45 \, b^{3} d^{2}}, -\frac {2 \, {\left (15 \, a^{2} d^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (3 \, b^{2} d^{2} x^{6} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} + {\left (b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, b^{3} d^{2}}\right ] \] Input:
integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="fricas")
Output:
[1/45*(15*a^2*d^2*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt( d*x^3 + c)*b*sqrt((b*c - a*d)/b))/(b*x^3 + a)) + 2*(3*b^2*d^2*x^6 - 2*b^2* c^2 - 5*a*b*c*d + 15*a^2*d^2 + (b^2*c*d - 5*a*b*d^2)*x^3)*sqrt(d*x^3 + c)) /(b^3*d^2), -2/45*(15*a^2*d^2*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c) *b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (3*b^2*d^2*x^6 - 2*b^2*c^2 - 5*a*b* c*d + 15*a^2*d^2 + (b^2*c*d - 5*a*b*d^2)*x^3)*sqrt(d*x^3 + c))/(b^3*d^2)]
Time = 11.47 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.38 \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\begin {cases} \frac {2 \left (\frac {a^{2} d^{3} \sqrt {c + d x^{3}}}{3 b^{3}} - \frac {a^{2} d^{3} \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{4} \sqrt {\frac {a d - b c}{b}}} + \frac {d \left (c + d x^{3}\right )^{\frac {5}{2}}}{15 b} + \frac {\left (c + d x^{3}\right )^{\frac {3}{2}} \left (- a d^{2} - b c d\right )}{9 b^{2}}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a^{2} \left (\begin {cases} \frac {x^{3}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{3} \right )}}{b} & \text {otherwise} \end {cases}\right )}{3 b^{2}} - \frac {a x^{3}}{3 b^{2}} + \frac {x^{6}}{6 b}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(d*x**3+c)**(1/2)/(b*x**3+a),x)
Output:
Piecewise((2*(a**2*d**3*sqrt(c + d*x**3)/(3*b**3) - a**2*d**3*(a*d - b*c)* atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b**4*sqrt((a*d - b*c)/b)) + d*(c + d*x**3)**(5/2)/(15*b) + (c + d*x**3)**(3/2)*(-a*d**2 - b*c*d)/(9*b* *2))/d**3, Ne(d, 0)), (sqrt(c)*(a**2*Piecewise((x**3/a, Eq(b, 0)), (log(a + b*x**3)/b, True))/(3*b**2) - a*x**3/(3*b**2) + x**6/(6*b)), True))
Exception generated. \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),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
Time = 0.12 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.11 \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {2 \, {\left (a^{2} b c - a^{3} d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{4} d^{8} - 5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{4} c d^{8} - 5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a b^{3} d^{9} + 15 \, \sqrt {d x^{3} + c} a^{2} b^{2} d^{10}\right )}}{45 \, b^{5} d^{10}} \] Input:
integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="giac")
Output:
2/3*(a^2*b*c - a^3*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt (-b^2*c + a*b*d)*b^3) + 2/45*(3*(d*x^3 + c)^(5/2)*b^4*d^8 - 5*(d*x^3 + c)^ (3/2)*b^4*c*d^8 - 5*(d*x^3 + c)^(3/2)*a*b^3*d^9 + 15*sqrt(d*x^3 + c)*a^2*b ^2*d^10)/(b^5*d^10)
Time = 4.74 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.41 \[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {2\,a^2\,\sqrt {d\,x^3+c}}{3\,b^3}+\frac {2\,{\left (d\,x^3+c\right )}^{5/2}}{15\,b\,d^2}-\frac {2\,a\,{\left (d\,x^3+c\right )}^{3/2}}{9\,b^2\,d}-\frac {2\,c\,{\left (d\,x^3+c\right )}^{3/2}}{9\,b\,d^2}+\frac {a^2\,\ln \left (\frac {a^2\,d^2\,1{}\mathrm {i}+b^2\,c^2\,2{}\mathrm {i}-2\,\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}-a\,b\,d^2\,x^3\,1{}\mathrm {i}+b^2\,c\,d\,x^3\,1{}\mathrm {i}-a\,b\,c\,d\,3{}\mathrm {i}}{2\,b\,x^3+2\,a}\right )\,\sqrt {a\,d-b\,c}\,1{}\mathrm {i}}{3\,b^{7/2}} \] Input:
int((x^8*(c + d*x^3)^(1/2))/(a + b*x^3),x)
Output:
(2*a^2*(c + d*x^3)^(1/2))/(3*b^3) + (2*(c + d*x^3)^(5/2))/(15*b*d^2) - (2* a*(c + d*x^3)^(3/2))/(9*b^2*d) - (2*c*(c + d*x^3)^(3/2))/(9*b*d^2) + (a^2* log((a^2*d^2*1i + b^2*c^2*2i - 2*b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(3/ 2) - a*b*d^2*x^3*1i + b^2*c*d*x^3*1i - a*b*c*d*3i)/(2*a + 2*b*x^3))*(a*d - b*c)^(1/2)*1i)/(3*b^(7/2))
\[ \int \frac {x^8 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {20 \sqrt {d \,x^{3}+c}\, a c d -10 \sqrt {d \,x^{3}+c}\, a \,d^{2} x^{3}-4 \sqrt {d \,x^{3}+c}\, b \,c^{2}+2 \sqrt {d \,x^{3}+c}\, b c d \,x^{3}+6 \sqrt {d \,x^{3}+c}\, b \,d^{2} x^{6}+45 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} d^{3}-45 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b c \,d^{2}}{45 b^{2} d^{2}} \] Input:
int(x^8*(d*x^3+c)^(1/2)/(b*x^3+a),x)
Output:
(20*sqrt(c + d*x**3)*a*c*d - 10*sqrt(c + d*x**3)*a*d**2*x**3 - 4*sqrt(c + d*x**3)*b*c**2 + 2*sqrt(c + d*x**3)*b*c*d*x**3 + 6*sqrt(c + d*x**3)*b*d**2 *x**6 + 45*int((sqrt(c + d*x**3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x* *6),x)*a**2*d**3 - 45*int((sqrt(c + d*x**3)*x**5)/(a*c + a*d*x**3 + b*c*x* *3 + b*d*x**6),x)*a*b*c*d**2)/(45*b**2*d**2)