\(\int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx\) [534]

Optimal result

Integrand size = 19, antiderivative size = 101 \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {9 d (c+d x)^{2/3}}{20 (b c-a d)^2 (a+b x)^{5/3}}-\frac {27 d^2 (c+d x)^{2/3}}{40 (b c-a d)^3 (a+b x)^{2/3}} \] Output:

-3/8*(d*x+c)^(2/3)/(-a*d+b*c)/(b*x+a)^(8/3)+9/20*d*(d*x+c)^(2/3)/(-a*d+b*c 
)^2/(b*x+a)^(5/3)-27/40*d^2*(d*x+c)^(2/3)/(-a*d+b*c)^3/(b*x+a)^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 (c+d x)^{2/3} \left (20 a^2 d^2+8 a b d (-2 c+3 d x)+b^2 \left (5 c^2-6 c d x+9 d^2 x^2\right )\right )}{40 (b c-a d)^3 (a+b x)^{8/3}} \] Input:

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

Output:

(-3*(c + d*x)^(2/3)*(20*a^2*d^2 + 8*a*b*d*(-2*c + 3*d*x) + b^2*(5*c^2 - 6* 
c*d*x + 9*d^2*x^2)))/(40*(b*c - a*d)^3*(a + b*x)^(8/3))
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13, 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.

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

\(\displaystyle -\frac {3 d \left (\frac {9 d (c+d x)^{2/3}}{10 (a+b x)^{2/3} (b c-a d)^2}-\frac {3 (c+d x)^{2/3}}{5 (a+b x)^{5/3} (b c-a d)}\right )}{4 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)}\)

Input:

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

Output:

(-3*(c + d*x)^(2/3))/(8*(b*c - a*d)*(a + b*x)^(8/3)) - (3*d*((-3*(c + d*x) 
^(2/3))/(5*(b*c - a*d)*(a + b*x)^(5/3)) + (9*d*(c + d*x)^(2/3))/(10*(b*c - 
 a*d)^2*(a + b*x)^(2/3))))/(4*(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 = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.04

method	result	size
gosper	\(\frac {3 \left (x d +c \right )^{\frac {2}{3}} \left (9 d^{2} x^{2} b^{2}+24 x a b \,d^{2}-6 x \,b^{2} c d +20 a^{2} d^{2}-16 a b c d +5 b^{2} c^{2}\right )}{40 \left (b x +a \right )^{\frac {8}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)
orering	\(\frac {3 \left (x d +c \right )^{\frac {2}{3}} \left (9 d^{2} x^{2} b^{2}+24 x a b \,d^{2}-6 x \,b^{2} c d +20 a^{2} d^{2}-16 a b c d +5 b^{2} c^{2}\right )}{40 \left (b x +a \right )^{\frac {8}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)

Input:

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

Output:

3/40*(d*x+c)^(2/3)*(9*b^2*d^2*x^2+24*a*b*d^2*x-6*b^2*c*d*x+20*a^2*d^2-16*a 
*b*c*d+5*b^2*c^2)/(b*x+a)^(8/3)/(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. 251 vs. \(2 (83) = 166\).

Time = 0.10 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.49 \[ \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 16 \, a b c d + 20 \, a^{2} d^{2} - 6 \, {\left (b^{2} c d - 4 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{40 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \] Input:

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

Output:

-3/40*(9*b^2*d^2*x^2 + 5*b^2*c^2 - 16*a*b*c*d + 20*a^2*d^2 - 6*(b^2*c*d - 
4*a*b*d^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2 
*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 
- a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^ 
4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5* 
b*d^3)*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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