Integrand size = 24, antiderivative size = 74 \[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {2 \sqrt {c+d x^3}}{3 b d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} \sqrt {b c-a d}} \] Output:
2/3*(d*x^3+c)^(1/2)/b/d+2/3*a*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^( 1/2))/b^(3/2)/(-a*d+b*c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {2 \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{d}-\frac {a \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}\right )}{3 b^{3/2}} \] Input:
Integrate[x^5/((a + b*x^3)*Sqrt[c + d*x^3]),x]
Output:
(2*((Sqrt[b]*Sqrt[c + d*x^3])/d - (a*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt [-(b*c) + a*d]])/Sqrt[-(b*c) + a*d]))/(3*b^(3/2))
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {948, 90, 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.
\(\displaystyle \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \sqrt {c+d x^3}}{b d}-\frac {a \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \sqrt {c+d x^3}}{b d}-\frac {2 a \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {2 a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2} \sqrt {b c-a d}}+\frac {2 \sqrt {c+d x^3}}{b d}\right )\) |
Input:
Int[x^5/((a + b*x^3)*Sqrt[c + d*x^3]),x]
Output:
((2*Sqrt[c + d*x^3])/(b*d) + (2*a*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b *c - a*d]])/(b^(3/2)*Sqrt[b*c - a*d]))/3
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 = 1.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \sqrt {d \,x^{3}+c}}{3 d}-\frac {2 a \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \sqrt {\left (a d -b c \right ) b}}}{b}\) | \(59\) |
default | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 b d}-\frac {2 a \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b \sqrt {\left (a d -b c \right ) b}}\) | \(61\) |
risch | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 b d}-\frac {2 a \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b \sqrt {\left (a d -b c \right ) b}}\) | \(61\) |
elliptic | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 b 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 (-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 \,d^{2}}\) | \(448\) |
Input:
int(x^5/(b*x^3+a)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3/b*((d*x^3+c)^(1/2)/d-a/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x^3+c)^(1/2)/(( a*d-b*c)*b)^(1/2)))
Time = 0.12 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.77 \[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {\sqrt {b^{2} c - a b d} a d \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) + 2 \, \sqrt {d x^{3} + c} {\left (b^{2} c - a b d\right )}}{3 \, {\left (b^{3} c d - a b^{2} d^{2}\right )}}, -\frac {2 \, {\left (\sqrt {-b^{2} c + a b d} a d \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - \sqrt {d x^{3} + c} {\left (b^{2} c - a b d\right )}\right )}}{3 \, {\left (b^{3} c d - a b^{2} d^{2}\right )}}\right ] \] Input:
integrate(x^5/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
[1/3*(sqrt(b^2*c - a*b*d)*a*d*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*sqrt(b^2*c - a*b*d))/(b*x^3 + a)) + 2*sqrt(d*x^3 + c)*(b^2*c - a*b*d))/ (b^3*c*d - a*b^2*d^2), -2/3*(sqrt(-b^2*c + a*b*d)*a*d*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) - sqrt(d*x^3 + c)*(b^2*c - a*b*d) )/(b^3*c*d - a*b^2*d^2)]
\[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {x^{5}}{\left (a + b x^{3}\right ) \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(x**5/(b*x**3+a)/(d*x**3+c)**(1/2),x)
Output:
Integral(x**5/((a + b*x**3)*sqrt(c + d*x**3)), x)
Exception generated. \[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^5/(b*x^3+a)/(d*x^3+c)^(1/2),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(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \, {\left (\frac {a d \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b} - \frac {\sqrt {d x^{3} + c}}{b}\right )}}{3 \, d} \] Input:
integrate(x^5/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
-2/3*(a*d*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a* b*d)*b) - sqrt(d*x^3 + c)/b)/d
Time = 3.79 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {2\,\sqrt {d\,x^3+c}}{3\,b\,d}+\frac {a\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,b^{3/2}\,\sqrt {a\,d-b\,c}} \] Input:
int(x^5/((a + b*x^3)*(c + d*x^3)^(1/2)),x)
Output:
(2*(c + d*x^3)^(1/2))/(3*b*d) + (a*log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^ (1/2)*(a*d - b*c)^(1/2)*2i - b*d*x^3)/(a + b*x^3))*1i)/(3*b^(3/2)*(a*d - b *c)^(1/2))
\[ \int \frac {x^5}{\left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \] Input:
int(x^5/(b*x^3+a)/(d*x^3+c)^(1/2),x)
Output:
int((sqrt(c + d*x**3)*x**5)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)