\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{13/3}} \, dx\) [128]

Optimal result

Integrand size = 21, antiderivative size = 162 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {(b c-a d)^2 x}{10 a b^2 \left (a+b x^3\right )^{10/3}}+\frac {(b c-a d) (9 b c+11 a d) x}{70 a^2 b^2 \left (a+b x^3\right )^{7/3}}+\frac {\left (27 b^2 c^2+6 a b c d+2 a^2 d^2\right ) x}{140 a^3 b^2 \left (a+b x^3\right )^{4/3}}+\frac {3 \left (27 b^2 c^2+6 a b c d+2 a^2 d^2\right ) x}{140 a^4 b^2 \sqrt [3]{a+b x^3}} \] Output:

1/10*(-a*d+b*c)^2*x/a/b^2/(b*x^3+a)^(10/3)+1/70*(-a*d+b*c)*(11*a*d+9*b*c)* 
x/a^2/b^2/(b*x^3+a)^(7/3)+1/140*(2*a^2*d^2+6*a*b*c*d+27*b^2*c^2)*x/a^3/b^2 
/(b*x^3+a)^(4/3)+3/140*(2*a^2*d^2+6*a*b*c*d+27*b^2*c^2)*x/a^4/b^2/(b*x^3+a 
)^(1/3)
 

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x \left (81 b^3 c^2 x^9+18 a b^2 c x^6 \left (15 c+d x^3\right )+10 a^3 \left (14 c^2+7 c d x^3+2 d^2 x^6\right )+3 a^2 b x^3 \left (105 c^2+20 c d x^3+2 d^2 x^6\right )\right )}{140 a^4 \left (a+b x^3\right )^{10/3}} \] Input:

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

Output:

(x*(81*b^3*c^2*x^9 + 18*a*b^2*c*x^6*(15*c + d*x^3) + 10*a^3*(14*c^2 + 7*c* 
d*x^3 + 2*d^2*x^6) + 3*a^2*b*x^3*(105*c^2 + 20*c*d*x^3 + 2*d^2*x^6)))/(140 
*a^4*(a + b*x^3)^(10/3))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {907, 903, 903, 746}

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.

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

\(\Big \downarrow \) 907

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

\(\Big \downarrow \) 903

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

\(\Big \downarrow \) 903

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

\(\Big \downarrow \) 746

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

Input:

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

Output:

(b*x*(c + d*x^3)^3)/(10*a*(b*c - a*d)*(a + b*x^3)^(10/3)) + ((9*b*c - 10*a 
*d)*((x*(c + d*x^3)^2)/(7*a*(a + b*x^3)^(7/3)) + (6*c*((3*c*x)/(4*a^2*(a + 
 b*x^3)^(1/3)) + (x*(c + d*x^3))/(4*a*(a + b*x^3)^(4/3))))/(7*a)))/(10*a*( 
b*c - a*d))
 

Defintions of rubi rules used

rule 746

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) 
/a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
 

rule 903

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] 
 :> Simp[(-x)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*n*(p + 1))), x] - Simp[ 
c*(q/(a*(p + 1)))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /; 
FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 
 0] && GtQ[q, 0] && NeQ[p, -1]
 

rule 907

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - 
 a*d))), x] + Simp[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d)) 
  Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q} 
, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  ! 
LtQ[q, -1]) && NeQ[p, -1]
 

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.59

method	result	size
pseudoelliptic	\(\frac {\left (\left (\frac {1}{7} d^{2} x^{6}+\frac {1}{2} c d \,x^{3}+c^{2}\right ) a^{3}+\frac {9 \left (\frac {2}{105} d^{2} x^{6}+\frac {4}{21} c d \,x^{3}+c^{2}\right ) b \,x^{3} a^{2}}{4}+\frac {27 c \left (\frac {d \,x^{3}}{15}+c \right ) b^{2} x^{6} a}{14}+\frac {81 b^{3} c^{2} x^{9}}{140}\right ) x}{\left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(96\)
gosper	\(\frac {x \left (6 a^{2} b \,d^{2} x^{9}+18 a \,b^{2} c d \,x^{9}+81 b^{3} c^{2} x^{9}+20 a^{3} d^{2} x^{6}+60 a^{2} b c d \,x^{6}+270 a \,b^{2} c^{2} x^{6}+70 a^{3} c d \,x^{3}+315 a^{2} b \,c^{2} x^{3}+140 a^{3} c^{2}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(115\)
trager	\(\frac {x \left (6 a^{2} b \,d^{2} x^{9}+18 a \,b^{2} c d \,x^{9}+81 b^{3} c^{2} x^{9}+20 a^{3} d^{2} x^{6}+60 a^{2} b c d \,x^{6}+270 a \,b^{2} c^{2} x^{6}+70 a^{3} c d \,x^{3}+315 a^{2} b \,c^{2} x^{3}+140 a^{3} c^{2}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(115\)
orering	\(\frac {x \left (6 a^{2} b \,d^{2} x^{9}+18 a \,b^{2} c d \,x^{9}+81 b^{3} c^{2} x^{9}+20 a^{3} d^{2} x^{6}+60 a^{2} b c d \,x^{6}+270 a \,b^{2} c^{2} x^{6}+70 a^{3} c d \,x^{3}+315 a^{2} b \,c^{2} x^{3}+140 a^{3} c^{2}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(115\)

Input:

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

Output:

((1/7*d^2*x^6+1/2*c*d*x^3+c^2)*a^3+9/4*(2/105*d^2*x^6+4/21*c*d*x^3+c^2)*b* 
x^3*a^2+27/14*c*(1/15*d*x^3+c)*b^2*x^6*a+81/140*b^3*c^2*x^9)/(b*x^3+a)^(10 
/3)*x/a^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {{\left (3 \, {\left (27 \, b^{3} c^{2} + 6 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{10} + 10 \, {\left (27 \, a b^{2} c^{2} + 6 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x^{7} + 140 \, a^{3} c^{2} x + 35 \, {\left (9 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{140 \, {\left (a^{4} b^{4} x^{12} + 4 \, a^{5} b^{3} x^{9} + 6 \, a^{6} b^{2} x^{6} + 4 \, a^{7} b x^{3} + a^{8}\right )}} \] Input:

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

Output:

1/140*(3*(27*b^3*c^2 + 6*a*b^2*c*d + 2*a^2*b*d^2)*x^10 + 10*(27*a*b^2*c^2 
+ 6*a^2*b*c*d + 2*a^3*d^2)*x^7 + 140*a^3*c^2*x + 35*(9*a^2*b*c^2 + 2*a^3*c 
*d)*x^4)*(b*x^3 + a)^(2/3)/(a^4*b^4*x^12 + 4*a^5*b^3*x^9 + 6*a^6*b^2*x^6 + 
 4*a^7*b*x^3 + a^8)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=-\frac {{\left (7 \, b - \frac {10 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} d^{2} x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} + \frac {{\left (14 \, b^{2} - \frac {40 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c d x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{3}} - \frac {{\left (14 \, b^{3} - \frac {60 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {105 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {140 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} c^{2} x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{4}} \] Input:

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

Output:

-1/70*(7*b - 10*(b*x^3 + a)/x^3)*d^2*x^10/((b*x^3 + a)^(10/3)*a^2) + 1/70* 
(14*b^2 - 40*(b*x^3 + a)*b/x^3 + 35*(b*x^3 + a)^2/x^6)*c*d*x^10/((b*x^3 + 
a)^(10/3)*a^3) - 1/140*(14*b^3 - 60*(b*x^3 + a)*b^2/x^3 + 105*(b*x^3 + a)^ 
2*b/x^6 - 140*(b*x^3 + a)^3/x^9)*c^2*x^10/((b*x^3 + a)^(10/3)*a^4)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x\,\left (\frac {c^2}{10\,a}+\frac {a\,\left (\frac {d^2}{10\,b}-\frac {c\,d}{5\,a}\right )}{b}\right )}{{\left (b\,x^3+a\right )}^{10/3}}-\frac {x\,\left (\frac {d^2}{7\,b^2}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+9\,b^2\,c^2}{70\,a^2\,b^2}\right )}{{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (2\,a^2\,d^2+6\,a\,b\,c\,d+27\,b^2\,c^2\right )}{140\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (6\,a^2\,d^2+18\,a\,b\,c\,d+81\,b^2\,c^2\right )}{140\,a^4\,b^2\,{\left (b\,x^3+a\right )}^{1/3}} \] Input:

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

Output:

(x*(c^2/(10*a) + (a*(d^2/(10*b) - (c*d)/(5*a)))/b))/(a + b*x^3)^(10/3) - ( 
x*(d^2/(7*b^2) - (9*b^2*c^2 - a^2*d^2 + 2*a*b*c*d)/(70*a^2*b^2)))/(a + b*x 
^3)^(7/3) + (x*(2*a^2*d^2 + 27*b^2*c^2 + 6*a*b*c*d))/(140*a^3*b^2*(a + b*x 
^3)^(4/3)) + (x*(6*a^2*d^2 + 81*b^2*c^2 + 18*a*b*c*d))/(140*a^4*b^2*(a + b 
*x^3)^(1/3))
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b \,x^{3}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{2} x^{6}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{3} x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4} x^{12}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b \,x^{3}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{2} x^{6}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{3} x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4} x^{12}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b \,x^{3}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{2} x^{6}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{3} x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4} x^{12}}d x \right ) c^{2} \] Input:

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

Output:

int(x**6/((a + b*x**3)**(1/3)*a**4 + 4*(a + b*x**3)**(1/3)*a**3*b*x**3 + 6 
*(a + b*x**3)**(1/3)*a**2*b**2*x**6 + 4*(a + b*x**3)**(1/3)*a*b**3*x**9 + 
(a + b*x**3)**(1/3)*b**4*x**12),x)*d**2 + 2*int(x**3/((a + b*x**3)**(1/3)* 
a**4 + 4*(a + b*x**3)**(1/3)*a**3*b*x**3 + 6*(a + b*x**3)**(1/3)*a**2*b**2 
*x**6 + 4*(a + b*x**3)**(1/3)*a*b**3*x**9 + (a + b*x**3)**(1/3)*b**4*x**12 
),x)*c*d + int(1/((a + b*x**3)**(1/3)*a**4 + 4*(a + b*x**3)**(1/3)*a**3*b* 
x**3 + 6*(a + b*x**3)**(1/3)*a**2*b**2*x**6 + 4*(a + b*x**3)**(1/3)*a*b**3 
*x**9 + (a + b*x**3)**(1/3)*b**4*x**12),x)*c**2