\(\int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx\) [547]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}-\frac {81 d^2 \sqrt [3]{c+d x}}{140 (b c-a d)^3 (a+b x)^{4/3}}+\frac {243 d^3 \sqrt [3]{c+d x}}{140 (b c-a d)^4 \sqrt [3]{a+b x}} \] Output:

-3/10*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(10/3)+27/70*d*(d*x+c)^(1/3)/(-a*d+ 
b*c)^2/(b*x+a)^(7/3)-81/140*d^2*(d*x+c)^(1/3)/(-a*d+b*c)^3/(b*x+a)^(4/3)+2 
43/140*d^3*(d*x+c)^(1/3)/(-a*d+b*c)^4/(b*x+a)^(1/3)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {3 \sqrt [3]{c+d x} \left (140 a^3 d^3-105 a^2 b d^2 (c-3 d x)+30 a b^2 d \left (2 c^2-3 c d x+9 d^2 x^2\right )+b^3 \left (-14 c^3+18 c^2 d x-27 c d^2 x^2+81 d^3 x^3\right )\right )}{140 (b c-a d)^4 (a+b x)^{10/3}} \] Input:

Integrate[1/((a + b*x)^(13/3)*(c + d*x)^(2/3)),x]
 

Output:

(3*(c + d*x)^(1/3)*(140*a^3*d^3 - 105*a^2*b*d^2*(c - 3*d*x) + 30*a*b^2*d*( 
2*c^2 - 3*c*d*x + 9*d^2*x^2) + b^3*(-14*c^3 + 18*c^2*d*x - 27*c*d^2*x^2 + 
81*d^3*x^3)))/(140*(b*c - a*d)^4*(a + b*x)^(10/3))
 

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.

\(\displaystyle \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}}dx}{10 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \left (-\frac {6 d \int \frac {1}{(a+b x)^{7/3} (c+d x)^{2/3}}dx}{7 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)}\right )}{10 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {9 d \left (-\frac {6 d \left (-\frac {3 d \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}}dx}{4 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{4 (a+b x)^{4/3} (b c-a d)}\right )}{7 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)}\right )}{10 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {9 d \left (-\frac {6 d \left (\frac {9 d \sqrt [3]{c+d x}}{4 \sqrt [3]{a+b x} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{4 (a+b x)^{4/3} (b c-a d)}\right )}{7 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)}\right )}{10 (b c-a d)}-\frac {3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)}\)

Input:

Int[1/((a + b*x)^(13/3)*(c + d*x)^(2/3)),x]
 

Output:

(-3*(c + d*x)^(1/3))/(10*(b*c - a*d)*(a + b*x)^(10/3)) - (9*d*((-3*(c + d* 
x)^(1/3))/(7*(b*c - a*d)*(a + b*x)^(7/3)) - (6*d*((-3*(c + d*x)^(1/3))/(4* 
(b*c - a*d)*(a + b*x)^(4/3)) + (9*d*(c + d*x)^(1/3))/(4*(b*c - a*d)^2*(a + 
 b*x)^(1/3))))/(7*(b*c - a*d))))/(10*(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	\(\frac {3 \left (x d +c \right )^{\frac {1}{3}} \left (81 d^{3} x^{3} b^{3}+270 x^{2} a \,b^{2} d^{3}-27 x^{2} b^{3} c \,d^{2}+315 x \,a^{2} b \,d^{3}-90 x a \,b^{2} c \,d^{2}+18 x \,b^{3} c^{2} d +140 a^{3} d^{3}-105 a^{2} b c \,d^{2}+60 a \,b^{2} c^{2} d -14 b^{3} c^{3}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)
orering	\(\frac {3 \left (x d +c \right )^{\frac {1}{3}} \left (81 d^{3} x^{3} b^{3}+270 x^{2} a \,b^{2} d^{3}-27 x^{2} b^{3} c \,d^{2}+315 x \,a^{2} b \,d^{3}-90 x a \,b^{2} c \,d^{2}+18 x \,b^{3} c^{2} d +140 a^{3} d^{3}-105 a^{2} b c \,d^{2}+60 a \,b^{2} c^{2} d -14 b^{3} c^{3}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)

Input:

int(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x,method=_RETURNVERBOSE)
 

Output:

3/140*(d*x+c)^(1/3)*(81*b^3*d^3*x^3+270*a*b^2*d^3*x^2-27*b^3*c*d^2*x^2+315 
*a^2*b*d^3*x-90*a*b^2*c*d^2*x+18*b^3*c^2*d*x+140*a^3*d^3-105*a^2*b*c*d^2+6 
0*a*b^2*c^2*d-14*b^3*c^3)/(b*x+a)^(10/3)/(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. 419 vs. \(2 (112) = 224\).

Time = 0.09 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.08 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {3 \, {\left (81 \, b^{3} d^{3} x^{3} - 14 \, b^{3} c^{3} + 60 \, a b^{2} c^{2} d - 105 \, a^{2} b c d^{2} + 140 \, a^{3} d^{3} - 27 \, {\left (b^{3} c d^{2} - 10 \, a b^{2} d^{3}\right )} x^{2} + 9 \, {\left (2 \, b^{3} c^{2} d - 10 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{140 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \] Input:

integrate(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x, algorithm="fricas")
 

Output:

3/140*(81*b^3*d^3*x^3 - 14*b^3*c^3 + 60*a*b^2*c^2*d - 105*a^2*b*c*d^2 + 14 
0*a^3*d^3 - 27*(b^3*c*d^2 - 10*a*b^2*d^3)*x^2 + 9*(2*b^3*c^2*d - 10*a*b^2* 
c*d^2 + 35*a^2*b*d^3)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3)/(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]

\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {13}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \] Input:

integrate(1/(b*x+a)**(13/3)/(d*x+c)**(2/3),x)
 

Output:

Integral(1/((a + b*x)**(13/3)*(c + d*x)**(2/3)), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(13/3)*(d*x + c)^(2/3)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {13}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(13/3)*(d*x + c)^(2/3)), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {243\,d^3\,x^3}{140\,{\left (a\,d-b\,c\right )}^4}+\frac {420\,a^3\,d^3-315\,a^2\,b\,c\,d^2+180\,a\,b^2\,c^2\,d-42\,b^3\,c^3}{140\,b^3\,{\left (a\,d-b\,c\right )}^4}+\frac {27\,d\,x\,\left (35\,a^2\,d^2-10\,a\,b\,c\,d+2\,b^2\,c^2\right )}{140\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {81\,d^2\,x^2\,\left (10\,a\,d-b\,c\right )}{140\,b\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,{\left (a+b\,x\right )}^{1/3}+\frac {a^3\,{\left (a+b\,x\right )}^{1/3}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/3}}{b^2}} \] Input:

int(1/((a + b*x)^(13/3)*(c + d*x)^(2/3)),x)
 

Output:

((c + d*x)^(1/3)*((243*d^3*x^3)/(140*(a*d - b*c)^4) + (420*a^3*d^3 - 42*b^ 
3*c^3 + 180*a*b^2*c^2*d - 315*a^2*b*c*d^2)/(140*b^3*(a*d - b*c)^4) + (27*d 
*x*(35*a^2*d^2 + 2*b^2*c^2 - 10*a*b*c*d))/(140*b^2*(a*d - b*c)^4) + (81*d^ 
2*x^2*(10*a*d - b*c))/(140*b*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/3) + (a^3* 
(a + b*x)^(1/3))/b^3 + (3*a*x^2*(a + b*x)^(1/3))/b + (3*a^2*x*(a + b*x)^(1 
/3))/b^2)
 

Reduce [F]

\[ \int \frac {1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a^{4}+4 \left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a^{3} b x +6 \left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a^{2} b^{2} x^{2}+4 \left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} a \,b^{3} x^{3}+\left (d x +c \right )^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} b^{4} x^{4}}d x \] Input:

int(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x)
 

Output:

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