Integrand size = 17, antiderivative size = 124 \[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\frac {c \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac {c \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}} \] Output:
1/3*c*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(1/3)+ 1/2*d*x^2*(1+b*x^3/a)^(1/3)*hypergeom([1/3, 2/3],[5/3],-b*x^3/a)/(b*x^3+a) ^(1/3)-1/2*c*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(1/3)
Time = 10.17 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.31 \[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\frac {1}{6} \left (\frac {3 d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}+\frac {c \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )-2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+\log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )\right )}{\sqrt [3]{b}}\right ) \] Input:
Integrate[(c + d*x)/(a + b*x^3)^(1/3),x]
Output:
((3*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3) /a)])/(a + b*x^3)^(1/3) + (c*(2*Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x ^3)^(1/3))/Sqrt[3]] - 2*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)] + Log[1 + ( b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]))/b^(1/3)) /6
Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2432, 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.
\(\displaystyle \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 2432 |
\(\displaystyle \int \left (\frac {c}{\sqrt [3]{a+b x^3}}+\frac {d x}{\sqrt [3]{a+b x^3}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {c \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac {d x^2 \sqrt [3]{\frac {b x^3}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}\) |
Input:
Int[(c + d*x)/(a + b*x^3)^(1/3),x]
Output:
(c*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3) ) + (d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3 )/a)])/(2*(a + b*x^3)^(1/3)) - (c*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/( 2*b^(1/3))
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly Q[Pq, x^n])
\[\int \frac {d x +c}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x\]
Input:
int((d*x+c)/(b*x^3+a)^(1/3),x)
Output:
int((d*x+c)/(b*x^3+a)^(1/3),x)
\[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {d x + c}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((d*x+c)/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
integral((d*x + c)/(b*x^3 + a)^(1/3), x)
Result contains complex when optimal does not.
Time = 1.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63 \[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} \] Input:
integrate((d*x+c)/(b*x**3+a)**(1/3),x)
Output:
c*x*gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**( 1/3)*gamma(4/3)) + d*x**2*gamma(2/3)*hyper((1/3, 2/3), (5/3,), b*x**3*exp_ polar(I*pi)/a)/(3*a**(1/3)*gamma(5/3))
\[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {d x + c}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((d*x+c)/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/ 3))/b^(1/3) - log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3 )/x^2)/b^(1/3) + 2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3))*c + d*inte grate(x/(b*x^3 + a)^(1/3), x)
\[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {d x + c}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((d*x+c)/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate((d*x + c)/(b*x^3 + a)^(1/3), x)
Timed out. \[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {c+d\,x}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \] Input:
int((c + d*x)/(a + b*x^3)^(1/3),x)
Output:
int((c + d*x)/(a + b*x^3)^(1/3), x)
\[ \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) c \] Input:
int((d*x+c)/(b*x^3+a)^(1/3),x)
Output:
int(x/(a + b*x**3)**(1/3),x)*d + int(1/(a + b*x**3)**(1/3),x)*c