Integrand size = 27, antiderivative size = 52 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \sqrt {c+d x^3}}{3 d^2}+\frac {16 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^2} \] Output:
-2/3*(d*x^3+c)^(1/2)/d^2+16/9*c^(1/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2)) /d^2
Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \left (3 \sqrt {c+d x^3}-8 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{9 d^2} \] Input:
Integrate[x^5/((8*c - d*x^3)*Sqrt[c + d*x^3]),x]
Output:
(-2*(3*Sqrt[c + d*x^3] - 8*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]))/ (9*d^2)
Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {948, 90, 73, 219}
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.
\(\displaystyle \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{3} \left (\frac {8 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{d}-\frac {2 \sqrt {c+d x^3}}{d^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {16 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d^2}-\frac {2 \sqrt {c+d x^3}}{d^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {16 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^2}-\frac {2 \sqrt {c+d x^3}}{d^2}\right )\) |
Input:
Int[x^5/((8*c - d*x^3)*Sqrt[c + d*x^3]),x]
Output:
((-2*Sqrt[c + d*x^3])/d^2 + (16*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c] )])/(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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[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]]
Time = 1.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.73
method | result | size |
pseudoelliptic | \(\frac {-\frac {2 \sqrt {d \,x^{3}+c}}{3}+\frac {16 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9}}{d^{2}}\) | \(38\) |
default | \(-\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}+\frac {16 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{2}}\) | \(39\) |
risch | \(-\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}+\frac {16 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{2}}\) | \(39\) |
elliptic | \(-\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}-\frac {8 i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 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}{18 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 )}{27 d^{4}}\) | \(425\) |
Input:
int(x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/9*(8*c^(1/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))-3*(d*x^3+c)^(1/2))/d^2
Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.92 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {2 \, {\left (4 \, \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}}, -\frac {2 \, {\left (8 \, \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}}\right ] \] Input:
integrate(x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
[2/9*(4*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8* c)) - 3*sqrt(d*x^3 + c))/d^2, -2/9*(8*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^ 3 + c)) + 3*sqrt(d*x^3 + c))/d^2]
Time = 6.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.15 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\begin {cases} \frac {2 \left (- \frac {8 c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{9 \sqrt {- c}} - \frac {\sqrt {c + d x^{3}}}{3}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{6}}{48 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**5/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)
Output:
Piecewise((2*(-8*c*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(9*sqrt(-c)) - sqrt (c + d*x**3)/3)/d**2, Ne(d, 0)), (x**6/(48*c**(3/2)), True))
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \, {\left (4 \, \sqrt {c} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 3 \, \sqrt {d x^{3} + c}\right )}}{9 \, d^{2}} \] Input:
integrate(x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")
Output:
-2/9*(4*sqrt(c)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqr t(c))) + 3*sqrt(d*x^3 + c))/d^2
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \, {\left (\frac {8 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} + \frac {3 \, \sqrt {d x^{3} + c}}{d}\right )}}{9 \, d} \] Input:
integrate(x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
-2/9*(8*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) + 3*sqrt(d*x^3 + c)/d)/d
Time = 1.66 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.15 \[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {8\,\sqrt {c}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{9\,d^2}-\frac {2\,\sqrt {d\,x^3+c}}{3\,d^2} \] Input:
int(x^5/((c + d*x^3)^(1/2)*(8*c - d*x^3)),x)
Output:
(8*c^(1/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)) )/(9*d^2) - (2*(c + d*x^3)^(1/2))/(3*d^2)
\[ \int \frac {x^5}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \] Input:
int(x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)
Output:
int((sqrt(c + d*x**3)*x**5)/(8*c**2 + 7*c*d*x**3 - d**2*x**6),x)