\(\int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx\) [191]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=-\frac {3 (a-b x)^{2/3}}{22 a b (a+b x)^{11/3}}-\frac {27 (a-b x)^{2/3}}{352 a^2 b (a+b x)^{8/3}}-\frac {81 (a-b x)^{2/3}}{1760 a^3 b (a+b x)^{5/3}}-\frac {243 (a-b x)^{2/3}}{7040 a^4 b (a+b x)^{2/3}} \] Output:

-3/22*(-b*x+a)^(2/3)/a/b/(b*x+a)^(11/3)-27/352*(-b*x+a)^(2/3)/a^2/b/(b*x+a 
)^(8/3)-81/1760*(-b*x+a)^(2/3)/a^3/b/(b*x+a)^(5/3)-243/7040*(-b*x+a)^(2/3) 
/a^4/b/(b*x+a)^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=-\frac {3 (a-b x)^{2/3} \left (689 a^3+639 a^2 b x+351 a b^2 x^2+81 b^3 x^3\right )}{7040 a^4 b (a+b x)^{11/3}} \] Input:

Integrate[1/((a - b*x)^(1/3)*(a + b*x)^(14/3)),x]
 

Output:

(-3*(a - b*x)^(2/3)*(689*a^3 + 639*a^2*b*x + 351*a*b^2*x^2 + 81*b^3*x^3))/ 
(7040*a^4*b*(a + b*x)^(11/3))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {55, 55, 55, 48}

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.

\(\displaystyle \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {9 \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{11/3}}dx}{22 a}-\frac {3 (a-b x)^{2/3}}{22 a b (a+b x)^{11/3}}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {9 \left (\frac {3 \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{8/3}}dx}{8 a}-\frac {3 (a-b x)^{2/3}}{16 a b (a+b x)^{8/3}}\right )}{22 a}-\frac {3 (a-b x)^{2/3}}{22 a b (a+b x)^{11/3}}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {9 \left (\frac {3 \left (\frac {3 \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{5/3}}dx}{10 a}-\frac {3 (a-b x)^{2/3}}{10 a b (a+b x)^{5/3}}\right )}{8 a}-\frac {3 (a-b x)^{2/3}}{16 a b (a+b x)^{8/3}}\right )}{22 a}-\frac {3 (a-b x)^{2/3}}{22 a b (a+b x)^{11/3}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {9 \left (\frac {3 \left (-\frac {9 (a-b x)^{2/3}}{40 a^2 b (a+b x)^{2/3}}-\frac {3 (a-b x)^{2/3}}{10 a b (a+b x)^{5/3}}\right )}{8 a}-\frac {3 (a-b x)^{2/3}}{16 a b (a+b x)^{8/3}}\right )}{22 a}-\frac {3 (a-b x)^{2/3}}{22 a b (a+b x)^{11/3}}\)

Input:

Int[1/((a - b*x)^(1/3)*(a + b*x)^(14/3)),x]
 

Output:

(-3*(a - b*x)^(2/3))/(22*a*b*(a + b*x)^(11/3)) + (9*((-3*(a - b*x)^(2/3))/ 
(16*a*b*(a + b*x)^(8/3)) + (3*((-3*(a - b*x)^(2/3))/(10*a*b*(a + b*x)^(5/3 
)) - (9*(a - b*x)^(2/3))/(40*a^2*b*(a + b*x)^(2/3))))/(8*a)))/(22*a)
 

Defintions of rubi rules used

rule 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46

method	result	size
gosper	\(-\frac {3 \left (-b x +a \right )^{\frac {2}{3}} \left (81 b^{3} x^{3}+351 a \,b^{2} x^{2}+639 a^{2} b x +689 a^{3}\right )}{7040 \left (b x +a \right )^{\frac {11}{3}} a^{4} b}\)	\(54\)
orering	\(-\frac {3 \left (-b x +a \right )^{\frac {2}{3}} \left (81 b^{3} x^{3}+351 a \,b^{2} x^{2}+639 a^{2} b x +689 a^{3}\right )}{7040 \left (b x +a \right )^{\frac {11}{3}} a^{4} b}\)	\(54\)

Input:

int(1/(-b*x+a)^(1/3)/(b*x+a)^(14/3),x,method=_RETURNVERBOSE)
 

Output:

-3/7040*(-b*x+a)^(2/3)*(81*b^3*x^3+351*a*b^2*x^2+639*a^2*b*x+689*a^3)/(b*x 
+a)^(11/3)/a^4/b
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=-\frac {3 \, {\left (81 \, b^{3} x^{3} + 351 \, a b^{2} x^{2} + 639 \, a^{2} b x + 689 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (-b x + a\right )}^{\frac {2}{3}}}{7040 \, {\left (a^{4} b^{5} x^{4} + 4 \, a^{5} b^{4} x^{3} + 6 \, a^{6} b^{3} x^{2} + 4 \, a^{7} b^{2} x + a^{8} b\right )}} \] Input:

integrate(1/(-b*x+a)^(1/3)/(b*x+a)^(14/3),x, algorithm="fricas")
 

Output:

-3/7040*(81*b^3*x^3 + 351*a*b^2*x^2 + 639*a^2*b*x + 689*a^3)*(b*x + a)^(1/ 
3)*(-b*x + a)^(2/3)/(a^4*b^5*x^4 + 4*a^5*b^4*x^3 + 6*a^6*b^3*x^2 + 4*a^7*b 
^2*x + a^8*b)
 

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=\int \frac {1}{\sqrt [3]{a - b x} \left (a + b x\right )^{\frac {14}{3}}}\, dx \] Input:

integrate(1/(-b*x+a)**(1/3)/(b*x+a)**(14/3),x)
 

Output:

Integral(1/((a - b*x)**(1/3)*(a + b*x)**(14/3)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {14}{3}} {\left (-b x + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(-b*x+a)^(1/3)/(b*x+a)^(14/3),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(14/3)*(-b*x + a)^(1/3)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {14}{3}} {\left (-b x + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(-b*x+a)^(1/3)/(b*x+a)^(14/3),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(14/3)*(-b*x + a)^(1/3)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{14/3}\,{\left (a-b\,x\right )}^{1/3}} \,d x \] Input:

int(1/((a + b*x)^(14/3)*(a - b*x)^(1/3)),x)
 

Output:

int(1/((a + b*x)^(14/3)*(a - b*x)^(1/3)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [3]{a-b x} (a+b x)^{14/3}} \, dx=\int \frac {1}{\left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} a^{4}+4 \left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} a^{3} b x +6 \left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} a^{2} b^{2} x^{2}+4 \left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} a \,b^{3} x^{3}+\left (b x +a \right )^{\frac {2}{3}} \left (-b x +a \right )^{\frac {1}{3}} b^{4} x^{4}}d x \] Input:

int(1/(-b*x+a)^(1/3)/(b*x+a)^(14/3),x)
 

Output:

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