Integrand size = 20, antiderivative size = 338 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\frac {12 a \sqrt [3]{a-b x} \sqrt [3]{a+b x}}{5 b}+\frac {3 (a-b x)^{4/3} \sqrt [3]{a+b x}}{5 b}+\frac {8\ 3^{3/4} \sqrt {2-\sqrt {3}} a^4 \left (1-\frac {b^2 x^2}{a^2}\right )^{2/3} \left (1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right ) \sqrt {\frac {1+\sqrt [3]{1-\frac {b^2 x^2}{a^2}}+\left (1-\frac {b^2 x^2}{a^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x (a-b x)^{2/3} (a+b x)^{2/3} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}}} \] Output:
12/5*a*(-b*x+a)^(1/3)*(b*x+a)^(1/3)/b+3/5*(-b*x+a)^(4/3)*(b*x+a)^(1/3)/b+8 /5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a^4*(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)*EllipticF((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 = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.20 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=-\frac {3 (a-b x)^{7/3} \left (\frac {a+b x}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {10}{3},\frac {a-b x}{2 a}\right )}{7\ 2^{2/3} b (a+b x)^{2/3}} \] Input:
Integrate[(a - b*x)^(4/3)/(a + b*x)^(2/3),x]
Output:
(-3*(a - b*x)^(7/3)*((a + b*x)/a)^(2/3)*Hypergeometric2F1[2/3, 7/3, 10/3, (a - b*x)/(2*a)])/(7*2^(2/3)*b*(a + b*x)^(2/3))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.20, 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.
\(\displaystyle \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {\left (\frac {a+b x}{a}\right )^{2/3} \int \frac {2^{2/3} (a-b x)^{4/3}}{\left (\frac {b x}{a}+1\right )^{2/3}}dx}{2^{2/3} (a+b x)^{2/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (\frac {a+b x}{a}\right )^{2/3} \int \frac {(a-b x)^{4/3}}{\left (\frac {b x}{a}+1\right )^{2/3}}dx}{(a+b x)^{2/3}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {3 (a-b x)^{7/3} \left (\frac {a+b x}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {10}{3},\frac {a-b x}{2 a}\right )}{7\ 2^{2/3} b (a+b x)^{2/3}}\) |
Input:
Int[(a - b*x)^(4/3)/(a + b*x)^(2/3),x]
Output:
(-3*(a - b*x)^(7/3)*((a + b*x)/a)^(2/3)*Hypergeometric2F1[2/3, 7/3, 10/3, (a - b*x)/(2*a)])/(7*2^(2/3)*b*(a + b*x)^(2/3))
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])
\[\int \frac {\left (-b x +a \right )^{\frac {4}{3}}}{\left (b x +a \right )^{\frac {2}{3}}}d x\]
Input:
int((-b*x+a)^(4/3)/(b*x+a)^(2/3),x)
Output:
int((-b*x+a)^(4/3)/(b*x+a)^(2/3),x)
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(4/3)/(b*x+a)^(2/3),x, algorithm="fricas")
Output:
integral((-b*x + a)^(4/3)/(b*x + a)^(2/3), x)
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\int \frac {\left (a - b x\right )^{\frac {4}{3}}}{\left (a + b x\right )^{\frac {2}{3}}}\, dx \] Input:
integrate((-b*x+a)**(4/3)/(b*x+a)**(2/3),x)
Output:
Integral((a - b*x)**(4/3)/(a + b*x)**(2/3), x)
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(4/3)/(b*x+a)^(2/3),x, algorithm="maxima")
Output:
integrate((-b*x + a)^(4/3)/(b*x + a)^(2/3), x)
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((-b*x+a)^(4/3)/(b*x+a)^(2/3),x, algorithm="giac")
Output:
integrate((-b*x + a)^(4/3)/(b*x + a)^(2/3), x)
Timed out. \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\int \frac {{\left (a-b\,x\right )}^{4/3}}{{\left (a+b\,x\right )}^{2/3}} \,d x \] Input:
int((a - b*x)^(4/3)/(a + b*x)^(2/3),x)
Output:
int((a - b*x)^(4/3)/(a + b*x)^(2/3), x)
\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{2/3}} \, dx=\left (\int \frac {\left (-b x +a \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {2}{3}}}d x \right ) a -\left (\int \frac {\left (-b x +a \right )^{\frac {1}{3}} x}{\left (b x +a \right )^{\frac {2}{3}}}d x \right ) b \] Input:
int((-b*x+a)^(4/3)/(b*x+a)^(2/3),x)
Output:
int((a - b*x)**(1/3)/(a + b*x)**(2/3),x)*a - int(((a - b*x)**(1/3)*x)/(a + b*x)**(2/3),x)*b