Integrand size = 22, antiderivative size = 85 \[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=-\frac {A \left (a+b x^3\right )^{2/3}}{4 a x^4}+\frac {(A b-2 a B) \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {2}{3},-\frac {b x^3}{a}\right )}{2 a x \sqrt [3]{a+b x^3}} \] Output:
-1/4*A*(b*x^3+a)^(2/3)/a/x^4+1/2*(A*b-2*B*a)*(1+b*x^3/a)^(1/3)*hypergeom([ -1/3, 1/3],[2/3],-b*x^3/a)/a/x/(b*x^3+a)^(1/3)
Time = 10.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\frac {-A \left (a+b x^3\right )+2 (A b-2 a B) x^3 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {2}{3},-\frac {b x^3}{a}\right )}{4 a x^4 \sqrt [3]{a+b x^3}} \] Input:
Integrate[(A + B*x^3)/(x^5*(a + b*x^3)^(1/3)),x]
Output:
(-(A*(a + b*x^3)) + 2*(A*b - 2*a*B)*x^3*(1 + (b*x^3)/a)^(1/3)*Hypergeometr ic2F1[-1/3, 1/3, 2/3, -((b*x^3)/a)])/(4*a*x^4*(a + b*x^3)^(1/3))
Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 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.
\(\displaystyle \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(A b-2 a B) \int \frac {1}{x^2 \sqrt [3]{b x^3+a}}dx}{2 a}-\frac {A \left (a+b x^3\right )^{2/3}}{4 a x^4}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle -\frac {\sqrt [3]{\frac {b x^3}{a}+1} (A b-2 a B) \int \frac {1}{x^2 \sqrt [3]{\frac {b x^3}{a}+1}}dx}{2 a \sqrt [3]{a+b x^3}}-\frac {A \left (a+b x^3\right )^{2/3}}{4 a x^4}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {\sqrt [3]{\frac {b x^3}{a}+1} (A b-2 a B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {2}{3},-\frac {b x^3}{a}\right )}{2 a x \sqrt [3]{a+b x^3}}-\frac {A \left (a+b x^3\right )^{2/3}}{4 a x^4}\) |
Input:
Int[(A + B*x^3)/(x^5*(a + b*x^3)^(1/3)),x]
Output:
-1/4*(A*(a + b*x^3)^(2/3))/(a*x^4) + ((A*b - 2*a*B)*(1 + (b*x^3)/a)^(1/3)* Hypergeometric2F1[-1/3, 1/3, 2/3, -((b*x^3)/a)])/(2*a*x*(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[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
\[\int \frac {B \,x^{3}+A}{x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x\]
Input:
int((B*x^3+A)/x^5/(b*x^3+a)^(1/3),x)
Output:
int((B*x^3+A)/x^5/(b*x^3+a)^(1/3),x)
\[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{5}} \,d x } \] Input:
integrate((B*x^3+A)/x^5/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
integral((B*x^3 + A)*(b*x^3 + a)^(2/3)/(b*x^8 + a*x^5), x)
Result contains complex when optimal does not.
Time = 1.56 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\frac {A \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{3} \\ - \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} x^{4} \Gamma \left (- \frac {1}{3}\right )} + \frac {B \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} x \Gamma \left (\frac {2}{3}\right )} \] Input:
integrate((B*x**3+A)/x**5/(b*x**3+a)**(1/3),x)
Output:
A*gamma(-4/3)*hyper((-4/3, 1/3), (-1/3,), b*x**3*exp_polar(I*pi)/a)/(3*a** (1/3)*x**4*gamma(-1/3)) + B*gamma(-1/3)*hyper((-1/3, 1/3), (2/3,), b*x**3* exp_polar(I*pi)/a)/(3*a**(1/3)*x*gamma(2/3))
\[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{5}} \,d x } \] Input:
integrate((B*x^3+A)/x^5/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(1/3)*x^5), x)
\[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{5}} \,d x } \] Input:
integrate((B*x^3+A)/x^5/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(1/3)*x^5), x)
Timed out. \[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\int \frac {B\,x^3+A}{x^5\,{\left (b\,x^3+a\right )}^{1/3}} \,d x \] Input:
int((A + B*x^3)/(x^5*(a + b*x^3)^(1/3)),x)
Output:
int((A + B*x^3)/(x^5*(a + b*x^3)^(1/3)), x)
\[ \int \frac {A+B x^3}{x^5 \sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}d x \right ) a +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2}}d x \right ) b \] Input:
int((B*x^3+A)/x^5/(b*x^3+a)^(1/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**5),x)*a + int(1/((a + b*x**3)**(1/3)*x**2),x )*b