Integrand size = 20, antiderivative size = 85 \[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {B x^2}{b \sqrt [3]{a+b x^3}}+\frac {(A b-2 a B) x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 a b \sqrt [3]{a+b x^3}} \] Output:
B*x^2/b/(b*x^3+a)^(1/3)+1/2*(A*b-2*B*a)*x^2*(1+b*x^3/a)^(1/3)*hypergeom([2 /3, 4/3],[5/3],-b*x^3/a)/a/b/(b*x^3+a)^(1/3)
Time = 10.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \left (5 A x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},-\frac {b x^3}{a}\right )+2 B x^5 \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )\right )}{10 a \sqrt [3]{a+b x^3}} \] Input:
Integrate[(x*(A + B*x^3))/(a + b*x^3)^(4/3),x]
Output:
((1 + (b*x^3)/a)^(1/3)*(5*A*x^2*Hypergeometric2F1[2/3, 4/3, 5/3, -((b*x^3) /a)] + 2*B*x^5*Hypergeometric2F1[4/3, 5/3, 8/3, -((b*x^3)/a)]))/(10*a*(a + b*x^3)^(1/3))
Time = 0.36 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {957, 889, 888}
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.
\(\displaystyle \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {x^2 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {(A b-2 a B) \int \frac {x}{\sqrt [3]{b x^3+a}}dx}{a b}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {x^2 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {\sqrt [3]{\frac {b x^3}{a}+1} (A b-2 a B) \int \frac {x}{\sqrt [3]{\frac {b x^3}{a}+1}}dx}{a b \sqrt [3]{a+b x^3}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {x^2 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {x^2 \sqrt [3]{\frac {b x^3}{a}+1} (A b-2 a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 a b \sqrt [3]{a+b x^3}}\) |
Input:
Int[(x*(A + B*x^3))/(a + b*x^3)^(4/3),x]
Output:
((A*b - a*B)*x^2)/(a*b*(a + b*x^3)^(1/3)) - ((A*b - 2*a*B)*x^2*(1 + (b*x^3 )/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*a*b*(a + b*x ^3)^(1/3))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
\[\int \frac {x \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {4}{3}}}d x\]
Input:
int(x*(B*x^3+A)/(b*x^3+a)^(4/3),x)
Output:
int(x*(B*x^3+A)/(b*x^3+a)^(4/3),x)
\[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x}{{\left (b x^{3} + a\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate(x*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="fricas")
Output:
integral((B*x^4 + A*x)*(b*x^3 + a)^(2/3)/(b^2*x^6 + 2*a*b*x^3 + a^2), x)
Result contains complex when optimal does not.
Time = 3.43 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {A x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} + \frac {B x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} \Gamma \left (\frac {8}{3}\right )} \] Input:
integrate(x*(B*x**3+A)/(b*x**3+a)**(4/3),x)
Output:
A*x**2*gamma(2/3)*hyper((2/3, 4/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a **(4/3)*gamma(5/3)) + B*x**5*gamma(5/3)*hyper((4/3, 5/3), (8/3,), b*x**3*e xp_polar(I*pi)/a)/(3*a**(4/3)*gamma(8/3))
\[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x}{{\left (b x^{3} + a\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate(x*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)*x/(b*x^3 + a)^(4/3), x)
\[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x}{{\left (b x^{3} + a\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate(x*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)*x/(b*x^3 + a)^(4/3), x)
Timed out. \[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int \frac {x\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{4/3}} \,d x \] Input:
int((x*(A + B*x^3))/(a + b*x^3)^(4/3),x)
Output:
int((x*(A + B*x^3))/(a + b*x^3)^(4/3), x)
\[ \int \frac {x \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \] Input:
int(x*(B*x^3+A)/(b*x^3+a)^(4/3),x)
Output:
int(x/(a + b*x**3)**(1/3),x)