Integrand size = 26, antiderivative size = 59 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {2 \sqrt {c+d x^3}}{3 d^2}-\frac {8 \sqrt {c} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^2} \] Output:
2/3*(d*x^3+c)^(1/2)/d^2-8/9*c^(1/2)*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^( 1/2))*3^(1/2)/d^2
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {6 \sqrt {c+d x^3}-8 \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{9 d^2} \] Input:
Integrate[x^5/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]
Output:
(6*Sqrt[c + d*x^3] - 8*Sqrt[3]*Sqrt[c]*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqr t[c])])/(9*d^2)
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {948, 90, 73, 216}
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.
\(\displaystyle \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}dx^3\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \sqrt {c+d x^3}}{d^2}-\frac {4 c \int \frac {1}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}dx^3}{d}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \sqrt {c+d x^3}}{d^2}-\frac {8 c \int \frac {1}{x^6+3 c}d\sqrt {d x^3+c}}{d^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \sqrt {c+d x^3}}{d^2}-\frac {8 \sqrt {c} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^2}\right )\) |
Input:
Int[x^5/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]
Output:
((2*Sqrt[c + d*x^3])/d^2 - (8*Sqrt[c]*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt [c])])/(Sqrt[3]*d^2))/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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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]]
Time = 1.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(\frac {-8 \sqrt {c}\, \sqrt {3}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right )+6 \sqrt {d \,x^{3}+c}}{9 d^{2}}\) | \(44\) |
default | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}-\frac {8 \sqrt {c}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{9 d^{2}}\) | \(45\) |
risch | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}-\frac {8 \sqrt {c}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{9 d^{2}}\) | \(45\) |
elliptic | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}+\frac {4 i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+4 c \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 {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}{6 d c}, \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 )}{9 d^{4}}\) | \(425\) |
Input:
int(x^5/(d*x^3+c)^(1/2)/(d*x^3+4*c),x,method=_RETURNVERBOSE)
Output:
1/9*(-8*c^(1/2)*3^(1/2)*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))+6*(d*x ^3+c)^(1/2))/d^2
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.83 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\left [\frac {2 \, {\left (2 \, \sqrt {\frac {1}{3}} \sqrt {-c} \log \left (\frac {d x^{3} - 6 \, \sqrt {\frac {1}{3}} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + \sqrt {d x^{3} + c}\right )}}{3 \, d^{2}}, \frac {2 \, {\left (4 \, \sqrt {\frac {1}{3}} \sqrt {c} \arctan \left (\frac {3 \, \sqrt {\frac {1}{3}} \sqrt {c}}{\sqrt {d x^{3} + c}}\right ) + \sqrt {d x^{3} + c}\right )}}{3 \, d^{2}}\right ] \] Input:
integrate(x^5/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
Output:
[2/3*(2*sqrt(1/3)*sqrt(-c)*log((d*x^3 - 6*sqrt(1/3)*sqrt(d*x^3 + c)*sqrt(- c) - 2*c)/(d*x^3 + 4*c)) + sqrt(d*x^3 + c))/d^2, 2/3*(4*sqrt(1/3)*sqrt(c)* arctan(3*sqrt(1/3)*sqrt(c)/sqrt(d*x^3 + c)) + sqrt(d*x^3 + c))/d^2]
Time = 4.88 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {4 \sqrt {3} \sqrt {c} \operatorname {atan}{\left (\frac {\sqrt {3} \sqrt {c + d x^{3}}}{3 \sqrt {c}} \right )}}{9} + \frac {\sqrt {c + d x^{3}}}{3}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{6}}{24 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**5/(d*x**3+c)**(1/2)/(d*x**3+4*c),x)
Output:
Piecewise((2*(-4*sqrt(3)*sqrt(c)*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c)) )/9 + sqrt(c + d*x**3)/3)/d**2, Ne(d, 0)), (x**6/(24*c**(3/2)), True))
Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {2 \, {\left (4 \, \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) - 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}} \] Input:
integrate(x^5/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
Output:
-2/9*(4*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) - 3*sq rt(d*x^3 + c))/d^2
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {2 \, {\left (\frac {4 \, \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{d} - \frac {3 \, \sqrt {d x^{3} + c}}{d}\right )}}{9 \, d} \] Input:
integrate(x^5/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
Output:
-2/9*(4*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/d - 3* sqrt(d*x^3 + c)/d)/d
Time = 3.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {2\,\sqrt {d\,x^3+c}}{3\,d^2}+\frac {\sqrt {3}\,\sqrt {c}\,\ln \left (\frac {2\,\sqrt {3}\,c-\sqrt {3}\,d\,x^3+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,4{}\mathrm {i}}{9\,d^2} \] Input:
int(x^5/((c + d*x^3)^(1/2)*(4*c + d*x^3)),x)
Output:
(2*(c + d*x^3)^(1/2))/(3*d^2) + (3^(1/2)*c^(1/2)*log((2*3^(1/2)*c + c^(1/2 )*(c + d*x^3)^(1/2)*6i - 3^(1/2)*d*x^3)/(4*c + d*x^3))*4i)/(9*d^2)
\[ \int \frac {x^5}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \] Input:
int(x^5/(d*x^3+c)^(1/2)/(d*x^3+4*c),x)
Output:
int((sqrt(c + d*x**3)*x**5)/(4*c**2 + 5*c*d*x**3 + d**2*x**6),x)