Integrand size = 20, antiderivative size = 308 \[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{a-b x} \sqrt [3]{a+b x}}{2 b}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} a^3 \left (1-\frac {b^2 x^2}{a^2}\right )^{2/3} \left (1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right ) \sqrt {\frac {1+\sqrt [3]{1-\frac {b^2 x^2}{a^2}}+\left (1-\frac {b^2 x^2}{a^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}\right ),-7+4 \sqrt {3}\right )}{b^2 x (a-b x)^{2/3} (a+b x)^{2/3} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}}} \] Output:
-3/2*(-b*x+a)^(1/3)*(b*x+a)^(1/3)/b+3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a^3* (1-b^2*x^2/a^2)^(2/3)*(1-(1-b^2*x^2/a^2)^(1/3))*((1+(1-b^2*x^2/a^2)^(1/3)+ (1-b^2*x^2/a^2)^(2/3))/(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3))^2)^(1/2)*Elliptic F((1+3^(1/2)-(1-b^2*x^2/a^2)^(1/3))/(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3)),2*I- I*3^(1/2))/b^2/x/(-b*x+a)^(2/3)/(b*x+a)^(2/3)/(-(1-(1-b^2*x^2/a^2)^(1/3))/ (1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.21 \[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{2} \sqrt [3]{a-b x} \sqrt [3]{a+b x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},\frac {a-b x}{2 a}\right )}{b \sqrt [3]{\frac {a+b x}{a}}} \] Input:
Integrate[(a + b*x)^(1/3)/(a - b*x)^(2/3),x]
Output:
(-3*2^(1/3)*(a - b*x)^(1/3)*(a + b*x)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 4 /3, (a - b*x)/(2*a)])/(b*((a + b*x)/a)^(1/3))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.21, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {80, 27, 79}
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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {\sqrt [3]{2} \sqrt [3]{a+b x} \int \frac {\sqrt [3]{\frac {b x}{a}+1}}{\sqrt [3]{2} (a-b x)^{2/3}}dx}{\sqrt [3]{\frac {a+b x}{a}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt [3]{a+b x} \int \frac {\sqrt [3]{\frac {b x}{a}+1}}{(a-b x)^{2/3}}dx}{\sqrt [3]{\frac {a+b x}{a}}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {3 \sqrt [3]{2} \sqrt [3]{a-b x} \sqrt [3]{a+b x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},\frac {a-b x}{2 a}\right )}{b \sqrt [3]{\frac {a+b x}{a}}}\) |
Input:
Int[(a + b*x)^(1/3)/(a - b*x)^(2/3),x]
Output:
(-3*2^(1/3)*(a - b*x)^(1/3)*(a + b*x)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 4 /3, (a - b*x)/(2*a)])/(b*((a + b*x)/a)^(1/3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
\[\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (-b x +a \right )^{\frac {2}{3}}}d x\]
Input:
int((b*x+a)^(1/3)/(-b*x+a)^(2/3),x)
Output:
int((b*x+a)^(1/3)/(-b*x+a)^(2/3),x)
\[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (-b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((b*x+a)^(1/3)/(-b*x+a)^(2/3),x, algorithm="fricas")
Output:
integral(-(b*x + a)^(1/3)*(-b*x + a)^(1/3)/(b*x - a), x)
\[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=\int \frac {\sqrt [3]{a + b x}}{\left (a - b x\right )^{\frac {2}{3}}}\, dx \] Input:
integrate((b*x+a)**(1/3)/(-b*x+a)**(2/3),x)
Output:
Integral((a + b*x)**(1/3)/(a - b*x)**(2/3), x)
\[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (-b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((b*x+a)^(1/3)/(-b*x+a)^(2/3),x, algorithm="maxima")
Output:
integrate((b*x + a)^(1/3)/(-b*x + a)^(2/3), x)
\[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (-b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((b*x+a)^(1/3)/(-b*x+a)^(2/3),x, algorithm="giac")
Output:
integrate((b*x + a)^(1/3)/(-b*x + a)^(2/3), x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/3}}{{\left (a-b\,x\right )}^{2/3}} \,d x \] Input:
int((a + b*x)^(1/3)/(a - b*x)^(2/3),x)
Output:
int((a + b*x)^(1/3)/(a - b*x)^(2/3), x)
\[ \int \frac {\sqrt [3]{a+b x}}{(a-b x)^{2/3}} \, dx=\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (-b x +a \right )^{\frac {2}{3}}}d x \] Input:
int((b*x+a)^(1/3)/(-b*x+a)^(2/3),x)
Output:
int((a + b*x)**(1/3)/(a - b*x)**(2/3),x)