\(\int \frac {c+d x^3}{(a+b x^3)^{16/3}} \, dx\) [115]

Optimal result

Integrand size = 19, antiderivative size = 151 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac {27 (12 b c+a d) x}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac {81 (12 b c+a d) x}{1820 a^5 b \sqrt [3]{a+b x^3}} \] Output:

1/13*(-a*d+b*c)*x/a/b/(b*x^3+a)^(13/3)+1/130*(a*d+12*b*c)*x/a^2/b/(b*x^3+a 
)^(10/3)+9/910*(a*d+12*b*c)*x/a^3/b/(b*x^3+a)^(7/3)+27/1820*(a*d+12*b*c)*x 
/a^4/b/(b*x^3+a)^(4/3)+81/1820*(a*d+12*b*c)*x/a^5/b/(b*x^3+a)^(1/3)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {x \left (972 b^4 c x^{12}+455 a^4 \left (4 c+d x^3\right )+351 a^2 b^2 x^6 \left (20 c+d x^3\right )+81 a b^3 x^9 \left (52 c+d x^3\right )+195 a^3 b x^3 \left (28 c+3 d x^3\right )\right )}{1820 a^5 \left (a+b x^3\right )^{13/3}} \] Input:

Integrate[(c + d*x^3)/(a + b*x^3)^(16/3),x]
 

Output:

(x*(972*b^4*c*x^12 + 455*a^4*(4*c + d*x^3) + 351*a^2*b^2*x^6*(20*c + d*x^3 
) + 81*a*b^3*x^9*(52*c + d*x^3) + 195*a^3*b*x^3*(28*c + 3*d*x^3)))/(1820*a 
^5*(a + b*x^3)^(13/3))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {910, 749, 749, 749, 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.

\(\displaystyle \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx\)

\(\Big \downarrow \) 910

\(\displaystyle \frac {(a d+12 b c) \int \frac {1}{\left (b x^3+a\right )^{13/3}}dx}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {(a d+12 b c) \left (\frac {9 \int \frac {1}{\left (b x^3+a\right )^{10/3}}dx}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right )}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {(a d+12 b c) \left (\frac {9 \left (\frac {6 \int \frac {1}{\left (b x^3+a\right )^{7/3}}dx}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right )}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {(a d+12 b c) \left (\frac {9 \left (\frac {6 \left (\frac {3 \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{4 a}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right )}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 746

\(\displaystyle \frac {\left (\frac {9 \left (\frac {6 \left (\frac {3 x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right ) (a d+12 b c)}{13 a b}+\frac {x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}}\)

Input:

Int[(c + d*x^3)/(a + b*x^3)^(16/3),x]
 

Output:

((b*c - a*d)*x)/(13*a*b*(a + b*x^3)^(13/3)) + ((12*b*c + a*d)*(x/(10*a*(a 
+ b*x^3)^(10/3)) + (9*(x/(7*a*(a + b*x^3)^(7/3)) + (6*(x/(4*a*(a + b*x^3)^ 
(4/3)) + (3*x)/(4*a^2*(a + b*x^3)^(1/3))))/(7*a)))/(10*a)))/(13*a*b)
 

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 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

rule 910

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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(\frac {\left (\left (\frac {d \,x^{3}}{4}+c \right ) a^{4}+3 b \,x^{3} \left (\frac {3 d \,x^{3}}{28}+c \right ) a^{3}+\frac {27 b^{2} x^{6} \left (\frac {d \,x^{3}}{20}+c \right ) a^{2}}{7}+\frac {81 \left (\frac {d \,x^{3}}{52}+c \right ) b^{3} x^{9} a}{35}+\frac {243 b^{4} c \,x^{12}}{455}\right ) x}{\left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(90\)
gosper	\(\frac {x \left (81 a \,b^{3} d \,x^{12}+972 b^{4} c \,x^{12}+351 a^{2} b^{2} d \,x^{9}+4212 a \,b^{3} c \,x^{9}+585 a^{3} b d \,x^{6}+7020 a^{2} b^{2} c \,x^{6}+455 a^{4} d \,x^{3}+5460 a^{3} b c \,x^{3}+1820 c \,a^{4}\right )}{1820 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(105\)
trager	\(\frac {x \left (81 a \,b^{3} d \,x^{12}+972 b^{4} c \,x^{12}+351 a^{2} b^{2} d \,x^{9}+4212 a \,b^{3} c \,x^{9}+585 a^{3} b d \,x^{6}+7020 a^{2} b^{2} c \,x^{6}+455 a^{4} d \,x^{3}+5460 a^{3} b c \,x^{3}+1820 c \,a^{4}\right )}{1820 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(105\)
orering	\(\frac {x \left (81 a \,b^{3} d \,x^{12}+972 b^{4} c \,x^{12}+351 a^{2} b^{2} d \,x^{9}+4212 a \,b^{3} c \,x^{9}+585 a^{3} b d \,x^{6}+7020 a^{2} b^{2} c \,x^{6}+455 a^{4} d \,x^{3}+5460 a^{3} b c \,x^{3}+1820 c \,a^{4}\right )}{1820 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(105\)

Input:

int((d*x^3+c)/(b*x^3+a)^(16/3),x,method=_RETURNVERBOSE)
 

Output:

((1/4*d*x^3+c)*a^4+3*b*x^3*(3/28*d*x^3+c)*a^3+27/7*b^2*x^6*(1/20*d*x^3+c)* 
a^2+81/35*(1/52*d*x^3+c)*b^3*x^9*a+243/455*b^4*c*x^12)/(b*x^3+a)^(13/3)*x/ 
a^5
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {{\left (81 \, {\left (12 \, b^{4} c + a b^{3} d\right )} x^{13} + 351 \, {\left (12 \, a b^{3} c + a^{2} b^{2} d\right )} x^{10} + 585 \, {\left (12 \, a^{2} b^{2} c + a^{3} b d\right )} x^{7} + 1820 \, a^{4} c x + 455 \, {\left (12 \, a^{3} b c + a^{4} d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{1820 \, {\left (a^{5} b^{5} x^{15} + 5 \, a^{6} b^{4} x^{12} + 10 \, a^{7} b^{3} x^{9} + 10 \, a^{8} b^{2} x^{6} + 5 \, a^{9} b x^{3} + a^{10}\right )}} \] Input:

integrate((d*x^3+c)/(b*x^3+a)^(16/3),x, algorithm="fricas")
 

Output:

1/1820*(81*(12*b^4*c + a*b^3*d)*x^13 + 351*(12*a*b^3*c + a^2*b^2*d)*x^10 + 
 585*(12*a^2*b^2*c + a^3*b*d)*x^7 + 1820*a^4*c*x + 455*(12*a^3*b*c + a^4*d 
)*x^4)*(b*x^3 + a)^(2/3)/(a^5*b^5*x^15 + 5*a^6*b^4*x^12 + 10*a^7*b^3*x^9 + 
 10*a^8*b^2*x^6 + 5*a^9*b*x^3 + a^10)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\text {Timed out} \] Input:

integrate((d*x**3+c)/(b*x**3+a)**(16/3),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=-\frac {{\left (140 \, b^{3} - \frac {546 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {780 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} d x^{13}}{1820 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{4}} + \frac {{\left (35 \, b^{4} - \frac {182 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {390 \, {\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3} b}{x^{9}} + \frac {455 \, {\left (b x^{3} + a\right )}^{4}}{x^{12}}\right )} c x^{13}}{455 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{5}} \] Input:

integrate((d*x^3+c)/(b*x^3+a)^(16/3),x, algorithm="maxima")
 

Output:

-1/1820*(140*b^3 - 546*(b*x^3 + a)*b^2/x^3 + 780*(b*x^3 + a)^2*b/x^6 - 455 
*(b*x^3 + a)^3/x^9)*d*x^13/((b*x^3 + a)^(13/3)*a^4) + 1/455*(35*b^4 - 182* 
(b*x^3 + a)*b^3/x^3 + 390*(b*x^3 + a)^2*b^2/x^6 - 455*(b*x^3 + a)^3*b/x^9 
+ 455*(b*x^3 + a)^4/x^12)*c*x^13/((b*x^3 + a)^(13/3)*a^5)
 

Giac [F]

\[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\int { \frac {d x^{3} + c}{{\left (b x^{3} + a\right )}^{\frac {16}{3}}} \,d x } \] Input:

integrate((d*x^3+c)/(b*x^3+a)^(16/3),x, algorithm="giac")
 

Output:

integrate((d*x^3 + c)/(b*x^3 + a)^(16/3), x)
 

Mupad [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.87 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {x\,\left (\frac {c}{13\,a}-\frac {d}{13\,b}\right )}{{\left (b\,x^3+a\right )}^{13/3}}+\frac {x\,\left (a\,d+12\,b\,c\right )}{130\,a^2\,b\,{\left (b\,x^3+a\right )}^{10/3}}+\frac {x\,\left (9\,a\,d+108\,b\,c\right )}{910\,a^3\,b\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (27\,a\,d+324\,b\,c\right )}{1820\,a^4\,b\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (81\,a\,d+972\,b\,c\right )}{1820\,a^5\,b\,{\left (b\,x^3+a\right )}^{1/3}} \] Input:

int((c + d*x^3)/(a + b*x^3)^(16/3),x)
 

Output:

(x*(c/(13*a) - d/(13*b)))/(a + b*x^3)^(13/3) + (x*(a*d + 12*b*c))/(130*a^2 
*b*(a + b*x^3)^(10/3)) + (x*(9*a*d + 108*b*c))/(910*a^3*b*(a + b*x^3)^(7/3 
)) + (x*(27*a*d + 324*b*c))/(1820*a^4*b*(a + b*x^3)^(4/3)) + (x*(81*a*d + 
972*b*c))/(1820*a^5*b*(a + b*x^3)^(1/3))
 

Reduce [F]

\[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b \,x^{3}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{2} x^{6}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{3} x^{9}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{4} x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{5} x^{15}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b \,x^{3}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{2} x^{6}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{3} x^{9}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{4} x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{5} x^{15}}d x \right ) c \] Input:

int((d*x^3+c)/(b*x^3+a)^(16/3),x)
 

Output:

int(x**3/((a + b*x**3)**(1/3)*a**5 + 5*(a + b*x**3)**(1/3)*a**4*b*x**3 + 1 
0*(a + b*x**3)**(1/3)*a**3*b**2*x**6 + 10*(a + b*x**3)**(1/3)*a**2*b**3*x* 
*9 + 5*(a + b*x**3)**(1/3)*a*b**4*x**12 + (a + b*x**3)**(1/3)*b**5*x**15), 
x)*d + int(1/((a + b*x**3)**(1/3)*a**5 + 5*(a + b*x**3)**(1/3)*a**4*b*x**3 
 + 10*(a + b*x**3)**(1/3)*a**3*b**2*x**6 + 10*(a + b*x**3)**(1/3)*a**2*b** 
3*x**9 + 5*(a + b*x**3)**(1/3)*a*b**4*x**12 + (a + b*x**3)**(1/3)*b**5*x** 
15),x)*c