\(\int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx\) [183]

Optimal result

Integrand size = 20, antiderivative size = 364 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=-\frac {3 (a-b x)^{4/3}}{8 b (a+b x)^{8/3}}+\frac {3 \sqrt [3]{a-b x}}{10 b (a+b x)^{5/3}}-\frac {3 \sqrt [3]{a-b x}}{40 a b (a+b x)^{2/3}}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} a \left (1-\frac {b^2 x^2}{a^2}\right )^{2/3} \left (1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right ) \sqrt {\frac {1+\sqrt [3]{1-\frac {b^2 x^2}{a^2}}+\left (1-\frac {b^2 x^2}{a^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}\right ),-7+4 \sqrt {3}\right )}{40 b^2 x (a-b x)^{2/3} (a+b x)^{2/3} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}}} \] Output:

-3/8*(-b*x+a)^(4/3)/b/(b*x+a)^(8/3)+3/10*(-b*x+a)^(1/3)/b/(b*x+a)^(5/3)-3/ 
40*(-b*x+a)^(1/3)/a/b/(b*x+a)^(2/3)+1/40*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2)) 
*a*(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)*Ellip 
ticF((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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.19 \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=-\frac {3 (a-b x)^{7/3} \left (\frac {a+b x}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {7}{3},\frac {11}{3},\frac {10}{3},\frac {a-b x}{2 a}\right )}{56\ 2^{2/3} a^3 b (a+b x)^{2/3}} \] Input:

Integrate[(a - b*x)^(4/3)/(a + b*x)^(11/3),x]
 

Output:

(-3*(a - b*x)^(7/3)*((a + b*x)/a)^(2/3)*Hypergeometric2F1[7/3, 11/3, 10/3, 
 (a - b*x)/(2*a)])/(56*2^(2/3)*a^3*b*(a + b*x)^(2/3))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.19, 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.

\(\displaystyle \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\left (\frac {a+b x}{a}\right )^{2/3} \int \frac {8\ 2^{2/3} (a-b x)^{4/3}}{\left (\frac {b x}{a}+1\right )^{11/3}}dx}{8\ 2^{2/3} a^3 (a+b x)^{2/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (\frac {a+b x}{a}\right )^{2/3} \int \frac {(a-b x)^{4/3}}{\left (\frac {b x}{a}+1\right )^{11/3}}dx}{a^3 (a+b x)^{2/3}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {3 (a-b x)^{7/3} \left (\frac {a+b x}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {7}{3},\frac {11}{3},\frac {10}{3},\frac {a-b x}{2 a}\right )}{56\ 2^{2/3} a^3 b (a+b x)^{2/3}}\)

Input:

Int[(a - b*x)^(4/3)/(a + b*x)^(11/3),x]
 

Output:

(-3*(a - b*x)^(7/3)*((a + b*x)/a)^(2/3)*Hypergeometric2F1[7/3, 11/3, 10/3, 
 (a - b*x)/(2*a)])/(56*2^(2/3)*a^3*b*(a + b*x)^(2/3))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

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]))
 

rule 80

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])
 

Maple [F]

\[\int \frac {\left (-b x +a \right )^{\frac {4}{3}}}{\left (b x +a \right )^{\frac {11}{3}}}d x\]

Input:

int((-b*x+a)^(4/3)/(b*x+a)^(11/3),x)
 

Output:

int((-b*x+a)^(4/3)/(b*x+a)^(11/3),x)
 

Fricas [F]

\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((-b*x+a)^(4/3)/(b*x+a)^(11/3),x, algorithm="fricas")
 

Output:

integral((b*x + a)^(1/3)*(-b*x + a)^(4/3)/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b 
^2*x^2 + 4*a^3*b*x + a^4), x)
 

Sympy [F]

\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=\int \frac {\left (a - b x\right )^{\frac {4}{3}}}{\left (a + b x\right )^{\frac {11}{3}}}\, dx \] Input:

integrate((-b*x+a)**(4/3)/(b*x+a)**(11/3),x)
 

Output:

Integral((a - b*x)**(4/3)/(a + b*x)**(11/3), x)
 

Maxima [F]

\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((-b*x+a)^(4/3)/(b*x+a)^(11/3),x, algorithm="maxima")
 

Output:

integrate((-b*x + a)^(4/3)/(b*x + a)^(11/3), x)
 

Giac [F]

\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=\int { \frac {{\left (-b x + a\right )}^{\frac {4}{3}}}{{\left (b x + a\right )}^{\frac {11}{3}}} \,d x } \] Input:

integrate((-b*x+a)^(4/3)/(b*x+a)^(11/3),x, algorithm="giac")
 

Output:

integrate((-b*x + a)^(4/3)/(b*x + a)^(11/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=\int \frac {{\left (a-b\,x\right )}^{4/3}}{{\left (a+b\,x\right )}^{11/3}} \,d x \] Input:

int((a - b*x)^(4/3)/(a + b*x)^(11/3),x)
 

Output:

int((a - b*x)^(4/3)/(a + b*x)^(11/3), x)
 

Reduce [F]

\[ \int \frac {(a-b x)^{4/3}}{(a+b x)^{11/3}} \, dx=\left (\int \frac {\left (-b x +a \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {2}{3}} a^{3}+3 \left (b x +a \right )^{\frac {2}{3}} a^{2} b x +3 \left (b x +a \right )^{\frac {2}{3}} a \,b^{2} x^{2}+\left (b x +a \right )^{\frac {2}{3}} b^{3} x^{3}}d x \right ) a -\left (\int \frac {\left (-b x +a \right )^{\frac {1}{3}} x}{\left (b x +a \right )^{\frac {2}{3}} a^{3}+3 \left (b x +a \right )^{\frac {2}{3}} a^{2} b x +3 \left (b x +a \right )^{\frac {2}{3}} a \,b^{2} x^{2}+\left (b x +a \right )^{\frac {2}{3}} b^{3} x^{3}}d x \right ) b \] Input:

int((-b*x+a)^(4/3)/(b*x+a)^(11/3),x)
 

Output:

int((a - b*x)**(1/3)/((a + b*x)**(2/3)*a**3 + 3*(a + b*x)**(2/3)*a**2*b*x 
+ 3*(a + b*x)**(2/3)*a*b**2*x**2 + (a + b*x)**(2/3)*b**3*x**3),x)*a - int( 
((a - b*x)**(1/3)*x)/((a + b*x)**(2/3)*a**3 + 3*(a + b*x)**(2/3)*a**2*b*x 
+ 3*(a + b*x)**(2/3)*a*b**2*x**2 + (a + b*x)**(2/3)*b**3*x**3),x)*b