Integrand size = 24, antiderivative size = 107 \[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 c^2}{3 d^2 (b c-a d) \sqrt {c+d x^3}}+\frac {2 \sqrt {c+d x^3}}{3 b d^2}-\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}} \] Output:
2/3*c^2/d^2/(-a*d+b*c)/(d*x^3+c)^(1/2)+2/3*(d*x^3+c)^(1/2)/b/d^2-2/3*a^2*a rctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(3/2)/(-a*d+b*c)^(3/2)
Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04 \[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \left (\frac {\sqrt {b} \left (a d \left (c+d x^3\right )-b c \left (2 c+d x^3\right )\right )}{d^2 (-b c+a d) \sqrt {c+d x^3}}-\frac {a^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{3/2}}\right )}{3 b^{3/2}} \] Input:
Integrate[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]
Output:
(2*((Sqrt[b]*(a*d*(c + d*x^3) - b*c*(2*c + d*x^3)))/(d^2*(-(b*c) + a*d)*Sq rt[c + d*x^3]) - (a^2*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]] )/(-(b*c) + a*d)^(3/2)))/(3*b^(3/2))
Time = 0.49 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 98, 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.
\(\displaystyle \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6}{\left (b x^3+a\right ) \left (d x^3+c\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 98 |
\(\displaystyle \frac {1}{3} \int \left (\frac {a^2}{b (b c-a d) \left (b x^3+a\right ) \sqrt {d x^3+c}}+\frac {1}{b d \sqrt {d x^3+c}}+\frac {c^2}{d (a d-b c) \left (d x^3+c\right )^{3/2}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{3/2}}+\frac {2 c^2}{d^2 \sqrt {c+d x^3} (b c-a d)}+\frac {2 \sqrt {c+d x^3}}{b d^2}\right )\) |
Input:
Int[x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x]
Output:
((2*c^2)/(d^2*(b*c - a*d)*Sqrt[c + d*x^3]) + (2*Sqrt[c + d*x^3])/(b*d^2) - (2*a^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(b^(3/2)*(b*c - a*d)^(3/2)))/3
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x _)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
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 = 1.52 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 b \,d^{2}}-\frac {2 a^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}-\frac {2 c^{2}}{3 d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\) | \(100\) |
pseudoelliptic | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 b \,d^{2}}-\frac {2 a^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}-\frac {2 c^{2}}{3 d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\) | \(100\) |
default | \(\frac {\frac {2 c}{3 d^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}}{b}+\frac {2 a^{2} \left (-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {1}{\sqrt {d \,x^{3}+c}}\right )}{3 b^{2} \left (a d -b c \right )}+\frac {2 a}{3 b^{2} d \sqrt {d \,x^{3}+c}}\) | \(127\) |
elliptic | \(-\frac {2 c^{2}}{3 d^{2} \left (a d -b c \right ) \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {2 \sqrt {d \,x^{3}+c}}{3 b \,d^{2}}+\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 \left (a d -b c \right )^{2} \sqrt {d \,x^{3}+c}}\right )}{3 b \,d^{2}}\) | \(481\) |
Input:
int(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/3*(d*x^3+c)^(1/2)/b/d^2-2/3/b*a^2/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b *(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2))-2/3/d^2*c^2/(a*d-b*c)/(d*x^3+c)^(1/2 )
Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (87) = 174\).
Time = 0.11 (sec) , antiderivative size = 440, normalized size of antiderivative = 4.11 \[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [-\frac {{\left (a^{2} d^{3} x^{3} + a^{2} c d^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \, {\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d + a^{2} b c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left (b^{4} c^{3} d^{2} - 2 \, a b^{3} c^{2} d^{3} + a^{2} b^{2} c d^{4} + {\left (b^{4} c^{2} d^{3} - 2 \, a b^{3} c d^{4} + a^{2} b^{2} d^{5}\right )} x^{3}\right )}}, \frac {2 \, {\left ({\left (a^{2} d^{3} x^{3} + a^{2} c d^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) + {\left (2 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d + a^{2} b c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{3 \, {\left (b^{4} c^{3} d^{2} - 2 \, a b^{3} c^{2} d^{3} + a^{2} b^{2} c d^{4} + {\left (b^{4} c^{2} d^{3} - 2 \, a b^{3} c d^{4} + a^{2} b^{2} d^{5}\right )} x^{3}\right )}}\right ] \] Input:
integrate(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
[-1/3*((a^2*d^3*x^3 + a^2*c*d^2)*sqrt(b^2*c - a*b*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*(2*b^3*c^3 - 3*a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x ^3)*sqrt(d*x^3 + c))/(b^4*c^3*d^2 - 2*a*b^3*c^2*d^3 + a^2*b^2*c*d^4 + (b^4 *c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^3), 2/3*((a^2*d^3*x^3 + a^2*c*d^ 2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x ^3 + b*c)) + (2*b^3*c^3 - 3*a*b^2*c^2*d + a^2*b*c*d^2 + (b^3*c^2*d - 2*a*b ^2*c*d^2 + a^2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^4*c^3*d^2 - 2*a*b^3*c^2*d^3 + a^2*b^2*c*d^4 + (b^4*c^2*d^3 - 2*a*b^3*c*d^4 + a^2*b^2*d^5)*x^3)]
\[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^{8}}{\left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**8/(b*x**3+a)/(d*x**3+c)**(3/2),x)
Output:
Integral(x**8/((a + b*x**3)*(c + d*x**3)**(3/2)), x)
Exception generated. \[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),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 = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {a^{2} d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}} + \frac {c^{2}}{\sqrt {d x^{3} + c} {\left (b c - a d\right )}} + \frac {\sqrt {d x^{3} + c}}{b}\right )}}{3 \, d^{2}} \] Input:
integrate(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
2/3*(a^2*d^2*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/((b^2*c - a*b* d)*sqrt(-b^2*c + a*b*d)) + c^2/(sqrt(d*x^3 + c)*(b*c - a*d)) + sqrt(d*x^3 + c)/b)/d^2
Time = 5.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.07 \[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2\,\sqrt {d\,x^3+c}}{3\,b\,d^2}-\frac {2\,c^2}{3\,d^2\,\sqrt {d\,x^3+c}\,\left (a\,d-b\,c\right )}+\frac {a^2\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}} \] Input:
int(x^8/((a + b*x^3)*(c + d*x^3)^(3/2)),x)
Output:
(2*(c + d*x^3)^(1/2))/(3*b*d^2) - (2*c^2)/(3*d^2*(c + d*x^3)^(1/2)*(a*d - b*c)) + (a^2*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)^(3/2))
\[ \int \frac {x^8}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {4 \sqrt {d \,x^{3}+c}\, c +2 \sqrt {d \,x^{3}+c}\, d \,x^{3}-3 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) a c \,d^{2}-3 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) a \,d^{3} x^{3}}{3 b \,d^{2} \left (d \,x^{3}+c \right )} \] Input:
int(x^8/(b*x^3+a)/(d*x^3+c)^(3/2),x)
Output:
(4*sqrt(c + d*x**3)*c + 2*sqrt(c + d*x**3)*d*x**3 - 3*int((sqrt(c + d*x**3 )*x**5)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*a*c*d**2 - 3*int((sqrt(c + d*x**3)*x**5)/(a*c**2 + 2*a*c *d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*a*d** 3*x**3)/(3*b*d**2*(c + d*x**3))