Integrand size = 24, antiderivative size = 120 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=-\frac {2 a (b c-a d) \sqrt {c+d x^3}}{3 b^3}-\frac {2 a \left (c+d x^3\right )^{3/2}}{9 b^2}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b d}+\frac {2 a (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2}} \] Output:
-2/3*a*(-a*d+b*c)*(d*x^3+c)^(1/2)/b^3-2/9*a*(d*x^3+c)^(3/2)/b^2+2/15*(d*x^ 3+c)^(5/2)/b/d+2/3*a*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a* d+b*c)^(1/2))/b^(7/2)
Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 \sqrt {c+d x^3} \left (15 a^2 d^2+3 b^2 \left (c+d x^3\right )^2-5 a b d \left (4 c+d x^3\right )\right )}{45 b^3 d}-\frac {2 a (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2}} \] Input:
Integrate[(x^5*(c + d*x^3)^(3/2))/(a + b*x^3),x]
Output:
(2*Sqrt[c + d*x^3]*(15*a^2*d^2 + 3*b^2*(c + d*x^3)^2 - 5*a*b*d*(4*c + d*x^ 3)))/(45*b^3*d) - (2*a*(-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3 ])/Sqrt[-(b*c) + a*d]])/(3*b^(7/2))
Time = 0.40 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {948, 90, 60, 60, 73, 221}
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.
| \(\displaystyle \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^3 \left (d x^3+c\right )^{3/2}}{b x^3+a}dx^3\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \left (c+d x^3\right )^{5/2}}{5 b d}-\frac {a \int \frac {\left (d x^3+c\right )^{3/2}}{b x^3+a}dx^3}{b}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \left (c+d x^3\right )^{5/2}}{5 b d}-\frac {a \left (\frac {(b c-a d) \int \frac {\sqrt {d x^3+c}}{b x^3+a}dx^3}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )}{b}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \left (c+d x^3\right )^{5/2}}{5 b d}-\frac {a \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )}{b}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \left (c+d x^3\right )^{5/2}}{5 b d}-\frac {a \left (\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )}{b}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 \left (c+d x^3\right )^{5/2}}{5 b d}-\frac {a \left (\frac {(b c-a d) \left (\frac {2 \sqrt {c+d x^3}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )}{b}\right )\) |
Input:
Int[(x^5*(c + d*x^3)^(3/2))/(a + b*x^3),x]
Output:
((2*(c + d*x^3)^(5/2))/(5*b*d) - (a*((2*(c + d*x^3)^(3/2))/(3*b) + ((b*c - a*d)*((2*Sqrt[c + d*x^3])/b - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(3/2)))/b))/b)/3
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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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 = 1.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.94
| method | result | size |
| default | \(\frac {2 \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 b d}+\frac {2 a \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 b^{3} \sqrt {\left (a d -b c \right ) b}}\) | \(113\) |
| pseudoelliptic | \(-\frac {2 \left (-\left (\frac {b^{2} \left (d \,x^{3}+c \right )^{2}}{5}-\frac {4 a d \left (\frac {d \,x^{3}}{4}+c \right ) b}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{3}+c}+a d \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )\right )}{3 \sqrt {\left (a d -b c \right ) b}\, d \,b^{3}}\) | \(117\) |
| risch | \(\frac {2 \left (3 b^{2} d^{2} x^{6}-5 x^{3} a b \,d^{2}+6 x^{3} b^{2} c d +15 a^{2} d^{2}-20 a b c d +3 b^{2} c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d \,b^{3}}-\frac {2 a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{3} \sqrt {\left (a d -b c \right ) b}}\) | \(136\) |
| elliptic | \(\frac {2 d \,x^{6} \sqrt {d \,x^{3}+c}}{15 b}+\frac {2 \left (-\frac {d \left (a d -2 b c \right )}{b^{2}}-\frac {4 c d}{5 b}\right ) x^{3} \sqrt {d \,x^{3}+c}}{9 d}+\frac {2 \left (\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3}}-\frac {2 \left (-\frac {d \left (a d -2 b c \right )}{b^{2}}-\frac {4 c d}{5 b}\right ) c}{3 d}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i a \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \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 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{3 b^{3} d^{2}}\) | \(577\) |
Input:
int(x^5*(d*x^3+c)^(3/2)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
2/15*(d*x^3+c)^(5/2)/b/d+2/3*a/b^3*(-(a*d-b*c)^2*arctan(b*(d*x^3+c)^(1/2)/ ((a*d-b*c)*b)^(1/2))+(1/3*(-d*x^3-4*c)*b+a*d)*(d*x^3+c)^(1/2)*((a*d-b*c)*b )^(1/2))/((a*d-b*c)*b)^(1/2)
Time = 0.13 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.48 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\left [-\frac {15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{6} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{45 \, b^{3} d}, \frac {2 \, {\left (15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (3 \, b^{2} d^{2} x^{6} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, b^{3} d}\right ] \] Input:
integrate(x^5*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="fricas")
Output:
[-1/45*(15*(a*b*c*d - 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 + 3*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^3)* sqrt(d*x^3 + c))/(b^3*d), 2/45*(15*(a*b*c*d - 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 + 3*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^3)* sqrt(d*x^3 + c))/(b^3*d)]
Time = 20.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.28 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\begin {cases} \frac {2 \left (- \frac {a d \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b^{2}} - \frac {a d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{4} \sqrt {\frac {a d - b c}{b}}} + \frac {\left (c + d x^{3}\right )^{\frac {5}{2}}}{15 b} + \frac {\sqrt {c + d x^{3}} \left (a^{2} d^{2} - a b c d\right )}{3 b^{3}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (- \frac {a \left (\begin {cases} \frac {x^{3}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{3} \right )}}{b} & \text {otherwise} \end {cases}\right )}{3 b} + \frac {x^{3}}{3 b}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(d*x**3+c)**(3/2)/(b*x**3+a),x)
Output:
Piecewise((2*(-a*d*(c + d*x**3)**(3/2)/(9*b**2) - a*d*(a*d - b*c)**2*atan( sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b**4*sqrt((a*d - b*c)/b)) + (c + d*x**3)**(5/2)/(15*b) + sqrt(c + d*x**3)*(a**2*d**2 - a*b*c*d)/(3*b**3))/d , Ne(d, 0)), (c**(3/2)*(-a*Piecewise((x**3/a, Eq(b, 0)), (log(a + b*x**3)/ b, True))/(3*b) + x**3/(3*b)), True))
Exception generated. \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^5*(d*x^3+c)^(3/2)/(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 = 151, normalized size of antiderivative = 1.26 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=-\frac {2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{4} d^{4} - 5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a b^{3} d^{5} - 15 \, \sqrt {d x^{3} + c} a b^{3} c d^{5} + 15 \, \sqrt {d x^{3} + c} a^{2} b^{2} d^{6}\right )}}{45 \, b^{5} d^{5}} \] Input:
integrate(x^5*(d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="giac")
Output:
-2/3*(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*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 ^4 - 5*(d*x^3 + c)^(3/2)*a*b^3*d^5 - 15*sqrt(d*x^3 + c)*a*b^3*c*d^5 + 15*s qrt(d*x^3 + c)*a^2*b^2*d^6)/(b^5*d^5)
Time = 4.57 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.79 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,c^2}{b}+\frac {2\,a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}+\frac {2\,c\,\left (\frac {2\,a\,d^2}{b^2}-\frac {12\,c\,d}{5\,b}\right )}{3\,d}\right )}{3\,d}+\frac {2\,d\,x^6\,\sqrt {d\,x^3+c}}{15\,b}-\frac {x^3\,\sqrt {d\,x^3+c}\,\left (\frac {2\,a\,d^2}{b^2}-\frac {12\,c\,d}{5\,b}\right )}{9\,d}+\frac {a\,\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d+\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,b^{7/2}} \] Input:
int((x^5*(c + d*x^3)^(3/2))/(a + b*x^3),x)
Output:
((c + d*x^3)^(1/2)*((2*c^2)/b + (2*a*((a*d^2)/b^2 - (2*c*d)/b))/b + (2*c*( (2*a*d^2)/b^2 - (12*c*d)/(5*b)))/(3*d)))/(3*d) + (2*d*x^6*(c + d*x^3)^(1/2 ))/(15*b) - (x^3*(c + d*x^3)^(1/2)*((2*a*d^2)/b^2 - (12*c*d)/(5*b)))/(9*d) + (a*log((a^2*d^2 + 2*b^2*c^2 + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(3/ 2)*2i - a*b*d^2*x^3 + b^2*c*d*x^3 - 3*a*b*c*d)/(a + b*x^3))*(a*d - b*c)^(3 /2)*1i)/(3*b^(7/2))
\[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {20 \sqrt {d \,x^{3}+c}\, a c d -10 \sqrt {d \,x^{3}+c}\, a \,d^{2} x^{3}-24 \sqrt {d \,x^{3}+c}\, b \,c^{2}+12 \sqrt {d \,x^{3}+c}\, b c d \,x^{3}+6 \sqrt {d \,x^{3}+c}\, b \,d^{2} x^{6}+45 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} d^{3}-90 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b c \,d^{2}+45 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{2} c^{2} d}{45 b^{2} d} \] Input:
int(x^5*(d*x^3+c)^(3/2)/(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 - 24*sqrt(c + d*x**3)*b*c**2 + 12*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 - 90*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*int((sqrt(c + d*x**3)*x**5)/(a*c + a*d *x**3 + b*c*x**3 + b*d*x**6),x)*b**2*c**2*d)/(45*b**2*d)