Integrand size = 24, antiderivative size = 77 \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2}{3 (b c-a d) \sqrt {c+d x^3}}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 (b c-a d)^{3/2}} \] Output:
2/3/(-a*d+b*c)/(d*x^3+c)^(1/2)-2/3*b^(1/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2) /(-a*d+b*c)^(1/2))/(-a*d+b*c)^(3/2)
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2}{(3 b c-3 a d) \sqrt {c+d x^3}}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 (-b c+a d)^{3/2}} \] Input:
Integrate[x^2/((a + b*x^3)*(c + d*x^3)^(3/2)),x]
Output:
2/((3*b*c - 3*a*d)*Sqrt[c + d*x^3]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(3*(-(b*c) + a*d)^(3/2))
Time = 0.35 (sec) , antiderivative size = 77, 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 = {946, 61, 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.
\(\displaystyle \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\left (b x^3+a\right ) \left (d x^3+c\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{3} \left (\frac {b \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b c-a d}+\frac {2}{\sqrt {c+d x^3} (b c-a d)}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {2 b \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{d (b c-a d)}+\frac {2}{\sqrt {c+d x^3} (b c-a d)}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{\sqrt {c+d x^3} (b c-a d)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )\) |
Input:
Int[x^2/((a + b*x^3)*(c + d*x^3)^(3/2)),x]
Output:
(2/((b*c - a*d)*Sqrt[c + d*x^3]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[c + d* x^3])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2))/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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_)^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[(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] && EqQ[m - n + 1, 0]
Time = 1.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {-\frac {2 b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \sqrt {\left (a d -b c \right ) b}}-\frac {2}{3 \sqrt {d \,x^{3}+c}}}{a d -b c}\) | \(64\) |
pseudoelliptic | \(\frac {-\frac {2 b \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \sqrt {\left (a d -b c \right ) b}}-\frac {2}{3 \sqrt {d \,x^{3}+c}}}{a d -b c}\) | \(64\) |
elliptic | \(-\frac {2}{3 \left (a d -b c \right ) \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {i b \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 ) \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{3 d^{2}}\) | \(463\) |
Input:
int(x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/3/(a*d-b*c)*(-b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)* b)^(1/2))-1/(d*x^3+c)^(1/2))
Time = 0.10 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.77 \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [-\frac {{\left (d x^{3} + c\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) - 2 \, \sqrt {d x^{3} + c}}{3 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )}}, \frac {2 \, {\left ({\left (d x^{3} + c\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\sqrt {d x^{3} + c} \sqrt {-\frac {b}{b c - a d}}\right ) + \sqrt {d x^{3} + c}\right )}}{3 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + b c^{2} - a c d\right )}}\right ] \] Input:
integrate(x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
[-1/3*((d*x^3 + c)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt (d*x^3 + c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) - 2*sqrt(d*x^3 + c))/((b*c*d - a*d^2)*x^3 + b*c^2 - a*c*d), 2/3*((d*x^3 + c)*sqrt(-b/(b*c - a*d))*arctan(sqrt(d*x^3 + c)*sqrt(-b/(b*c - a*d))) + sqrt(d*x^3 + c))/(( b*c*d - a*d^2)*x^3 + b*c^2 - a*c*d)]
Time = 10.78 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.57 \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\begin {cases} \frac {2 \left (- \frac {d}{3 \sqrt {c + d x^{3}} \left (a d - b c\right )} - \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )}\right )}{d} & \text {for}\: d \neq 0 \\\begin {cases} \frac {x^{3}}{3 a c^{\frac {3}{2}}} & \text {for}\: b = 0 \\\tilde {\infty } x^{3} & \text {for}\: c^{\frac {3}{2}} = 0 \\\frac {\log {\left (3 a c^{\frac {3}{2}} + 3 b c^{\frac {3}{2}} x^{3} \right )}}{3 b c^{\frac {3}{2}}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \] Input:
integrate(x**2/(b*x**3+a)/(d*x**3+c)**(3/2),x)
Output:
Piecewise((2*(-d/(3*sqrt(c + d*x**3)*(a*d - b*c)) - d*atan(sqrt(c + d*x**3 )/sqrt((a*d - b*c)/b))/(3*sqrt((a*d - b*c)/b)*(a*d - b*c)))/d, Ne(d, 0)), (Piecewise((x**3/(3*a*c**(3/2)), Eq(b, 0)), (zoo*x**3, Eq(c**(3/2), 0)), ( log(3*a*c**(3/2) + 3*b*c**(3/2)*x**3)/(3*b*c**(3/2)), True)), True))
Exception generated. \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^2/(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 = 73, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \, b \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} {\left (b c - a d\right )}} + \frac {2}{3 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )}} \] Input:
integrate(x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
2/3*b*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d) *(b*c - a*d)) + 2/3/(sqrt(d*x^3 + c)*(b*c - a*d))
Time = 4.40 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16 \[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2}{3\,\sqrt {d\,x^3+c}\,\left (a\,d-b\,c\right )}+\frac {\sqrt {b}\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,{\left (a\,d-b\,c\right )}^{3/2}} \] Input:
int(x^2/((a + b*x^3)*(c + d*x^3)^(3/2)),x)
Output:
(b^(1/2)*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*(a*d - b*c)^(3/2)) - 2/(3*(c + d*x^3)^(1/2 )*(a*d - b*c))
\[ \int \frac {x^2}{\left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \] Input:
int(x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x)
Output:
int((sqrt(c + d*x**3)*x**2)/(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)