\(\int \frac {1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx\) [559]

Optimal result

Integrand size = 19, antiderivative size = 134 \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx=\frac {3}{(b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}-\frac {27 b (c+d x)^{2/3}}{8 (b c-a d)^2 (a+b x)^{8/3}}+\frac {81 b d (c+d x)^{2/3}}{20 (b c-a d)^3 (a+b x)^{5/3}}-\frac {243 b d^2 (c+d x)^{2/3}}{40 (b c-a d)^4 (a+b x)^{2/3}} \] Output:

3/(-a*d+b*c)/(b*x+a)^(8/3)/(d*x+c)^(1/3)-27/8*b*(d*x+c)^(2/3)/(-a*d+b*c)^2 
/(b*x+a)^(8/3)+81/20*b*d*(d*x+c)^(2/3)/(-a*d+b*c)^3/(b*x+a)^(5/3)-243/40*b 
*d^2*(d*x+c)^(2/3)/(-a*d+b*c)^4/(b*x+a)^(2/3)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx=-\frac {3 \left (40 a^3 d^3+60 a^2 b d^2 (c+3 d x)+24 a b^2 d \left (-c^2+3 c d x+9 d^2 x^2\right )+b^3 \left (5 c^3-9 c^2 d x+27 c d^2 x^2+81 d^3 x^3\right )\right )}{40 (b c-a d)^4 (a+b x)^{8/3} \sqrt [3]{c+d x}} \] Input:

Integrate[1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x]
 

Output:

(-3*(40*a^3*d^3 + 60*a^2*b*d^2*(c + 3*d*x) + 24*a*b^2*d*(-c^2 + 3*c*d*x + 
9*d^2*x^2) + b^3*(5*c^3 - 9*c^2*d*x + 27*c*d^2*x^2 + 81*d^3*x^3)))/(40*(b* 
c - a*d)^4*(a + b*x)^(8/3)*(c + d*x)^(1/3))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.

\(\displaystyle \int \frac {1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}}dx}{8 (b c-a d)}-\frac {3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \left (-\frac {6 d \int \frac {1}{(a+b x)^{5/3} (c+d x)^{4/3}}dx}{5 (b c-a d)}-\frac {3}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)}\right )}{8 (b c-a d)}-\frac {3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \left (-\frac {6 d \left (-\frac {3 d \int \frac {1}{(a+b x)^{2/3} (c+d x)^{4/3}}dx}{2 (b c-a d)}-\frac {3}{2 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {3}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)}\right )}{8 (b c-a d)}-\frac {3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {9 d \left (-\frac {6 d \left (-\frac {9 d \sqrt [3]{a+b x}}{2 \sqrt [3]{c+d x} (b c-a d)^2}-\frac {3}{2 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)}\right )}{5 (b c-a d)}-\frac {3}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)}\right )}{8 (b c-a d)}-\frac {3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)}\)

Input:

Int[1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x]
 

Output:

-3/(8*(b*c - a*d)*(a + b*x)^(8/3)*(c + d*x)^(1/3)) - (9*d*(-3/(5*(b*c - a* 
d)*(a + b*x)^(5/3)*(c + d*x)^(1/3)) - (6*d*(-3/(2*(b*c - a*d)*(a + b*x)^(2 
/3)*(c + d*x)^(1/3)) - (9*d*(a + b*x)^(1/3))/(2*(b*c - a*d)^2*(c + d*x)^(1 
/3))))/(5*(b*c - a*d))))/(8*(b*c - a*d))
 

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 = 0.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.28

method	result	size
gosper	\(-\frac {3 \left (81 d^{3} x^{3} b^{3}+216 x^{2} a \,b^{2} d^{3}+27 x^{2} b^{3} c \,d^{2}+180 x \,a^{2} b \,d^{3}+72 x a \,b^{2} c \,d^{2}-9 x \,b^{3} c^{2} d +40 a^{3} d^{3}+60 a^{2} b c \,d^{2}-24 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right )}{40 \left (b x +a \right )^{\frac {8}{3}} \left (x d +c \right )^{\frac {1}{3}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)
orering	\(-\frac {3 \left (81 d^{3} x^{3} b^{3}+216 x^{2} a \,b^{2} d^{3}+27 x^{2} b^{3} c \,d^{2}+180 x \,a^{2} b \,d^{3}+72 x a \,b^{2} c \,d^{2}-9 x \,b^{3} c^{2} d +40 a^{3} d^{3}+60 a^{2} b c \,d^{2}-24 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right )}{40 \left (b x +a \right )^{\frac {8}{3}} \left (x d +c \right )^{\frac {1}{3}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)

Input:

int(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x,method=_RETURNVERBOSE)
 

Output:

-3/40*(81*b^3*d^3*x^3+216*a*b^2*d^3*x^2+27*b^3*c*d^2*x^2+180*a^2*b*d^3*x+7 
2*a*b^2*c*d^2*x-9*b^3*c^2*d*x+40*a^3*d^3+60*a^2*b*c*d^2-24*a*b^2*c^2*d+5*b 
^3*c^3)/(b*x+a)^(8/3)/(d*x+c)^(1/3)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d 
^2-4*a*b^3*c^3*d+b^4*c^4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (112) = 224\).

Time = 0.11 (sec) , antiderivative size = 456, normalized size of antiderivative = 3.40 \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx=-\frac {3 \, {\left (81 \, b^{3} d^{3} x^{3} + 5 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 60 \, a^{2} b c d^{2} + 40 \, a^{3} d^{3} + 27 \, {\left (b^{3} c d^{2} + 8 \, a b^{2} d^{3}\right )} x^{2} - 9 \, {\left (b^{3} c^{2} d - 8 \, a b^{2} c d^{2} - 20 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{40 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}} \] Input:

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x, algorithm="fricas")
 

Output:

-3/40*(81*b^3*d^3*x^3 + 5*b^3*c^3 - 24*a*b^2*c^2*d + 60*a^2*b*c*d^2 + 40*a 
^3*d^3 + 27*(b^3*c*d^2 + 8*a*b^2*d^3)*x^2 - 9*(b^3*c^2*d - 8*a*b^2*c*d^2 - 
 20*a^2*b*d^3)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^3*b^4*c^5 - 4*a^4*b^3 
*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4* 
a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + ( 
b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a^4*b^ 
3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4* 
c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^ 
5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 - a^6*b* 
c*d^4 + a^7*d^5)*x)
 

Sympy [F]

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

integrate(1/(b*x+a)**(11/3)/(d*x+c)**(4/3),x)
 

Output:

Integral(1/((a + b*x)**(11/3)*(c + d*x)**(4/3)), x)
 

Maxima [F]

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

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(4/3)), x)
 

Giac [F]

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

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(4/3)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int(1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x)
 

Output:

int(1/((a + b*x)^(11/3)*(c + d*x)^(4/3)), x)
 

Reduce [F]

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

int(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x)
 

Output:

int(1/((c + d*x)**(1/3)*(a + b*x)**(2/3)*a**3*c + (c + d*x)**(1/3)*(a + b* 
x)**(2/3)*a**3*d*x + 3*(c + d*x)**(1/3)*(a + b*x)**(2/3)*a**2*b*c*x + 3*(c 
 + d*x)**(1/3)*(a + b*x)**(2/3)*a**2*b*d*x**2 + 3*(c + d*x)**(1/3)*(a + b* 
x)**(2/3)*a*b**2*c*x**2 + 3*(c + d*x)**(1/3)*(a + b*x)**(2/3)*a*b**2*d*x** 
3 + (c + d*x)**(1/3)*(a + b*x)**(2/3)*b**3*c*x**3 + (c + d*x)**(1/3)*(a + 
b*x)**(2/3)*b**3*d*x**4),x)