\(\int \frac {1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx\) [535]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int \frac {1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx=-\frac {3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac {27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}-\frac {81 d^2 (c+d x)^{2/3}}{220 (b c-a d)^3 (a+b x)^{5/3}}+\frac {243 d^3 (c+d x)^{2/3}}{440 (b c-a d)^4 (a+b x)^{2/3}} \] Output:

-3/11*(d*x+c)^(2/3)/(-a*d+b*c)/(b*x+a)^(11/3)+27/88*d*(d*x+c)^(2/3)/(-a*d+ 
b*c)^2/(b*x+a)^(8/3)-81/220*d^2*(d*x+c)^(2/3)/(-a*d+b*c)^3/(b*x+a)^(5/3)+2 
43/440*d^3*(d*x+c)^(2/3)/(-a*d+b*c)^4/(b*x+a)^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx=\frac {3 (c+d x)^{2/3} \left (220 a^3 d^3+132 a^2 b d^2 (-2 c+3 d x)+33 a b^2 d \left (5 c^2-6 c d x+9 d^2 x^2\right )+b^3 \left (-40 c^3+45 c^2 d x-54 c d^2 x^2+81 d^3 x^3\right )\right )}{440 (b c-a d)^4 (a+b x)^{11/3}} \] Input:

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

Output:

(3*(c + d*x)^(2/3)*(220*a^3*d^3 + 132*a^2*b*d^2*(-2*c + 3*d*x) + 33*a*b^2* 
d*(5*c^2 - 6*c*d*x + 9*d^2*x^2) + b^3*(-40*c^3 + 45*c^2*d*x - 54*c*d^2*x^2 
 + 81*d^3*x^3)))/(440*(b*c - a*d)^4*(a + b*x)^(11/3))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.19, 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.

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \left (-\frac {3 d \int \frac {1}{(a+b x)^{8/3} \sqrt [3]{c+d x}}dx}{4 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)}\right )}{11 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \left (-\frac {3 d \left (-\frac {3 d \int \frac {1}{(a+b x)^{5/3} \sqrt [3]{c+d x}}dx}{5 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{5 (a+b x)^{5/3} (b c-a d)}\right )}{4 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)}\right )}{11 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {9 d \left (-\frac {3 d \left (\frac {9 d (c+d x)^{2/3}}{10 (a+b x)^{2/3} (b c-a d)^2}-\frac {3 (c+d x)^{2/3}}{5 (a+b x)^{5/3} (b c-a d)}\right )}{4 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)}\right )}{11 (b c-a d)}-\frac {3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)}\)

Input:

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

Output:

(-3*(c + d*x)^(2/3))/(11*(b*c - a*d)*(a + b*x)^(11/3)) - (9*d*((-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))))/(11*(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 = 1.26

method	result	size
gosper	\(\frac {3 \left (x d +c \right )^{\frac {2}{3}} \left (81 d^{3} x^{3} b^{3}+297 x^{2} a \,b^{2} d^{3}-54 x^{2} b^{3} c \,d^{2}+396 x \,a^{2} b \,d^{3}-198 x a \,b^{2} c \,d^{2}+45 x \,b^{3} c^{2} d +220 a^{3} d^{3}-264 a^{2} b c \,d^{2}+165 a \,b^{2} c^{2} d -40 b^{3} c^{3}\right )}{440 \left (b x +a \right )^{\frac {11}{3}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)
orering	\(\frac {3 \left (x d +c \right )^{\frac {2}{3}} \left (81 d^{3} x^{3} b^{3}+297 x^{2} a \,b^{2} d^{3}-54 x^{2} b^{3} c \,d^{2}+396 x \,a^{2} b \,d^{3}-198 x a \,b^{2} c \,d^{2}+45 x \,b^{3} c^{2} d +220 a^{3} d^{3}-264 a^{2} b c \,d^{2}+165 a \,b^{2} c^{2} d -40 b^{3} c^{3}\right )}{440 \left (b x +a \right )^{\frac {11}{3}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)

Input:

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

Output:

3/440*(d*x+c)^(2/3)*(81*b^3*d^3*x^3+297*a*b^2*d^3*x^2-54*b^3*c*d^2*x^2+396 
*a^2*b*d^3*x-198*a*b^2*c*d^2*x+45*b^3*c^2*d*x+220*a^3*d^3-264*a^2*b*c*d^2+ 
165*a*b^2*c^2*d-40*b^3*c^3)/(b*x+a)^(11/3)/(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. 420 vs. \(2 (112) = 224\).

Time = 0.17 (sec) , antiderivative size = 420, normalized size of antiderivative = 3.09 \[ \int \frac {1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx=\frac {3 \, {\left (81 \, b^{3} d^{3} x^{3} - 40 \, b^{3} c^{3} + 165 \, a b^{2} c^{2} d - 264 \, a^{2} b c d^{2} + 220 \, a^{3} d^{3} - 27 \, {\left (2 \, b^{3} c d^{2} - 11 \, a b^{2} d^{3}\right )} x^{2} + 9 \, {\left (5 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 44 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{440 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \] Input:

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

Output:

3/440*(81*b^3*d^3*x^3 - 40*b^3*c^3 + 165*a*b^2*c^2*d - 264*a^2*b*c*d^2 + 2 
20*a^3*d^3 - 27*(2*b^3*c*d^2 - 11*a*b^2*d^3)*x^2 + 9*(5*b^3*c^2*d - 22*a*b 
^2*c*d^2 + 44*a^2*b*d^3)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^4*b^4*c^4 - 
 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 
- 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 
 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3 + a^ 
5*b^3*d^4)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4* 
a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2 + 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^ 
5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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