\(\int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx\) [558]

Optimal result

Integrand size = 19, antiderivative size = 98 \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\frac {3}{(b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac {18 b (c+d x)^{2/3}}{5 (b c-a d)^2 (a+b x)^{5/3}}+\frac {27 b d (c+d x)^{2/3}}{5 (b c-a d)^3 (a+b x)^{2/3}} \] Output:

3/(-a*d+b*c)/(b*x+a)^(5/3)/(d*x+c)^(1/3)-18/5*b*(d*x+c)^(2/3)/(-a*d+b*c)^2 
/(b*x+a)^(5/3)+27/5*b*d*(d*x+c)^(2/3)/(-a*d+b*c)^3/(b*x+a)^(2/3)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\frac {3 \left (5 a^2 d^2+5 a b d (c+3 d x)+b^2 \left (-c^2+3 c d x+9 d^2 x^2\right )\right )}{5 (b c-a d)^3 (a+b x)^{5/3} \sqrt [3]{c+d x}} \] Input:

Integrate[1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.

\(\displaystyle \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx\)

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

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

Input:

Int[1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x]
 

Output:

-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))
 

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 = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07

method	result	size
gosper	\(-\frac {3 \left (9 d^{2} x^{2} b^{2}+15 x a b \,d^{2}+3 x \,b^{2} c d +5 a^{2} d^{2}+5 a b c d -b^{2} c^{2}\right )}{5 \left (b x +a \right )^{\frac {5}{3}} \left (x d +c \right )^{\frac {1}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)
orering	\(-\frac {3 \left (9 d^{2} x^{2} b^{2}+15 x a b \,d^{2}+3 x \,b^{2} c d +5 a^{2} d^{2}+5 a b c d -b^{2} c^{2}\right )}{5 \left (b x +a \right )^{\frac {5}{3}} \left (x d +c \right )^{\frac {1}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)

Input:

int(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x,method=_RETURNVERBOSE)
 

Output:

-3/5*(9*b^2*d^2*x^2+15*a*b*d^2*x+3*b^2*c*d*x+5*a^2*d^2+5*a*b*c*d-b^2*c^2)/ 
(b*x+a)^(5/3)/(d*x+c)^(1/3)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (82) = 164\).

Time = 0.09 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} - b^{2} c^{2} + 5 \, a b c d + 5 \, a^{2} d^{2} + 3 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{5 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \] Input:

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="fricas")
 

Output:

3/5*(9*b^2*d^2*x^2 - b^2*c^2 + 5*a*b*c*d + 5*a^2*d^2 + 3*(b^2*c*d + 5*a*b* 
d^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3 
*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^ 
3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3* 
b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2* 
c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x)
 

Sympy [F]

\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {8}{3}} \left (c + d x\right )^{\frac {4}{3}}}\, dx \] Input:

integrate(1/(b*x+a)**(8/3)/(d*x+c)**(4/3),x)
 

Output:

Integral(1/((a + b*x)**(8/3)*(c + d*x)**(4/3)), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {8}{3}} {\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(8/3)*(d*x + c)^(4/3)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {8}{3}} {\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(8/3)*(d*x + c)^(4/3)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{8/3}\,{\left (c+d\,x\right )}^{4/3}} \,d x \] Input:

int(1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x)
 

Output:

int(1/((a + b*x)^(8/3)*(c + d*x)^(4/3)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} a^{2} c +\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} a^{2} d x +2 \left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} a b c x +2 \left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} a b d \,x^{2}+\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} b^{2} c \,x^{2}+\left (d x +c \right )^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}} b^{2} d \,x^{3}}d x \] Input:

int(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x)
 

Output:

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