Integrand size = 20, antiderivative size = 88 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=-\frac {a (b c-a d) \left (a+b x^3\right )^{1+p}}{3 b^3 (1+p)}+\frac {(b c-2 a d) \left (a+b x^3\right )^{2+p}}{3 b^3 (2+p)}+\frac {d \left (a+b x^3\right )^{3+p}}{3 b^3 (3+p)} \] Output:
-1/3*a*(-a*d+b*c)*(b*x^3+a)^(p+1)/b^3/(p+1)+1/3*(-2*a*d+b*c)*(b*x^3+a)^(2+ p)/b^3/(2+p)+1/3*d*(b*x^3+a)^(3+p)/b^3/(3+p)
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {1}{3} \left (-\frac {a (b c-a d) \left (a+b x^3\right )^{1+p}}{b^3 (1+p)}+\frac {(b c-2 a d) \left (a+b x^3\right )^{2+p}}{b^3 (2+p)}+\frac {d \left (a+b x^3\right )^{3+p}}{b^3 (3+p)}\right ) \] Input:
Integrate[x^5*(a + b*x^3)^p*(c + d*x^3),x]
Output:
(-((a*(b*c - a*d)*(a + b*x^3)^(1 + p))/(b^3*(1 + p))) + ((b*c - 2*a*d)*(a + b*x^3)^(2 + p))/(b^3*(2 + p)) + (d*(a + b*x^3)^(3 + p))/(b^3*(3 + p)))/3
Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {948, 86, 2009}
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.
\(\displaystyle \int x^5 \left (c+d x^3\right ) \left (a+b x^3\right )^p \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^3 \left (b x^3+a\right )^p \left (d x^3+c\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {a (a d-b c) \left (b x^3+a\right )^p}{b^2}+\frac {(b c-2 a d) \left (b x^3+a\right )^{p+1}}{b^2}+\frac {d \left (b x^3+a\right )^{p+2}}{b^2}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {a (b c-a d) \left (a+b x^3\right )^{p+1}}{b^3 (p+1)}+\frac {(b c-2 a d) \left (a+b x^3\right )^{p+2}}{b^3 (p+2)}+\frac {d \left (a+b x^3\right )^{p+3}}{b^3 (p+3)}\right )\) |
Input:
Int[x^5*(a + b*x^3)^p*(c + d*x^3),x]
Output:
(-((a*(b*c - a*d)*(a + b*x^3)^(1 + p))/(b^3*(1 + p))) + ((b*c - 2*a*d)*(a + b*x^3)^(2 + p))/(b^3*(2 + p)) + (d*(a + b*x^3)^(3 + p))/(b^3*(3 + p)))/3
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.89 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44
method | result | size |
gosper | \(\frac {\left (b \,x^{3}+a \right )^{p +1} \left (b^{2} d \,p^{2} x^{6}+3 b^{2} d p \,x^{6}+2 x^{6} b^{2} d +b^{2} c \,p^{2} x^{3}-2 a b d p \,x^{3}+4 b^{2} c p \,x^{3}-2 a b d \,x^{3}+3 b^{2} c \,x^{3}-a b c p +2 d \,a^{2}-3 a b c \right )}{3 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(127\) |
orering | \(\frac {\left (b \,x^{3}+a \right ) \left (b^{2} d \,p^{2} x^{6}+3 b^{2} d p \,x^{6}+2 x^{6} b^{2} d +b^{2} c \,p^{2} x^{3}-2 a b d p \,x^{3}+4 b^{2} c p \,x^{3}-2 a b d \,x^{3}+3 b^{2} c \,x^{3}-a b c p +2 d \,a^{2}-3 a b c \right ) \left (b \,x^{3}+a \right )^{p}}{3 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(132\) |
risch | \(\frac {\left (b^{3} d \,p^{2} x^{9}+3 b^{3} d p \,x^{9}+2 b^{3} d \,x^{9}+a \,b^{2} d \,p^{2} x^{6}+b^{3} c \,p^{2} x^{6}+a \,b^{2} d p \,x^{6}+4 b^{3} c p \,x^{6}+3 b^{3} c \,x^{6}+a \,b^{2} c \,p^{2} x^{3}-2 a^{2} b d p \,x^{3}+3 a \,b^{2} c p \,x^{3}-a^{2} b c p +2 a^{3} d -3 a^{2} b c \right ) \left (b \,x^{3}+a \right )^{p}}{3 \left (2+p \right ) \left (3+p \right ) \left (p +1\right ) b^{3}}\) | \(168\) |
parallelrisch | \(\frac {x^{9} \left (b \,x^{3}+a \right )^{p} b^{3} d \,p^{2}+3 x^{9} \left (b \,x^{3}+a \right )^{p} b^{3} d p +2 x^{9} \left (b \,x^{3}+a \right )^{p} b^{3} d +x^{6} \left (b \,x^{3}+a \right )^{p} a \,b^{2} d \,p^{2}+x^{6} \left (b \,x^{3}+a \right )^{p} b^{3} c \,p^{2}+x^{6} \left (b \,x^{3}+a \right )^{p} a \,b^{2} d p +4 x^{6} \left (b \,x^{3}+a \right )^{p} b^{3} c p +3 x^{6} \left (b \,x^{3}+a \right )^{p} b^{3} c +x^{3} \left (b \,x^{3}+a \right )^{p} a \,b^{2} c \,p^{2}-2 x^{3} \left (b \,x^{3}+a \right )^{p} a^{2} b d p +3 x^{3} \left (b \,x^{3}+a \right )^{p} a \,b^{2} c p -\left (b \,x^{3}+a \right )^{p} a^{2} b c p +2 \left (b \,x^{3}+a \right )^{p} a^{3} d -3 \left (b \,x^{3}+a \right )^{p} a^{2} b c}{3 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\) | \(285\) |
Input:
int(x^5*(b*x^3+a)^p*(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/3/b^3*(b*x^3+a)^(p+1)/(p^3+6*p^2+11*p+6)*(b^2*d*p^2*x^6+3*b^2*d*p*x^6+2* b^2*d*x^6+b^2*c*p^2*x^3-2*a*b*d*p*x^3+4*b^2*c*p*x^3-2*a*b*d*x^3+3*b^2*c*x^ 3-a*b*c*p+2*a^2*d-3*a*b*c)
Time = 0.14 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.83 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {{\left ({\left (b^{3} d p^{2} + 3 \, b^{3} d p + 2 \, b^{3} d\right )} x^{9} + {\left (3 \, b^{3} c + {\left (b^{3} c + a b^{2} d\right )} p^{2} + {\left (4 \, b^{3} c + a b^{2} d\right )} p\right )} x^{6} - a^{2} b c p - 3 \, a^{2} b c + 2 \, a^{3} d + {\left (a b^{2} c p^{2} + {\left (3 \, a b^{2} c - 2 \, a^{2} b d\right )} p\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \] Input:
integrate(x^5*(b*x^3+a)^p*(d*x^3+c),x, algorithm="fricas")
Output:
1/3*((b^3*d*p^2 + 3*b^3*d*p + 2*b^3*d)*x^9 + (3*b^3*c + (b^3*c + a*b^2*d)* p^2 + (4*b^3*c + a*b^2*d)*p)*x^6 - a^2*b*c*p - 3*a^2*b*c + 2*a^3*d + (a*b^ 2*c*p^2 + (3*a*b^2*c - 2*a^2*b*d)*p)*x^3)*(b*x^3 + a)^p/(b^3*p^3 + 6*b^3*p ^2 + 11*b^3*p + 6*b^3)
Timed out. \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\text {Timed out} \] Input:
integrate(x**5*(b*x**3+a)**p*(d*x**3+c),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {{\left (b^{2} {\left (p + 1\right )} x^{6} + a b p x^{3} - a^{2}\right )} {\left (b x^{3} + a\right )}^{p} c}{3 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2}} + \frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{9} + {\left (p^{2} + p\right )} a b^{2} x^{6} - 2 \, a^{2} b p x^{3} + 2 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{p} d}{3 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \] Input:
integrate(x^5*(b*x^3+a)^p*(d*x^3+c),x, algorithm="maxima")
Output:
1/3*(b^2*(p + 1)*x^6 + a*b*p*x^3 - a^2)*(b*x^3 + a)^p*c/((p^2 + 3*p + 2)*b ^2) + 1/3*((p^2 + 3*p + 2)*b^3*x^9 + (p^2 + p)*a*b^2*x^6 - 2*a^2*b*p*x^3 + 2*a^3)*(b*x^3 + a)^p*d/((p^3 + 6*p^2 + 11*p + 6)*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (82) = 164\).
Time = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.32 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {{\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} b c p + {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} d p - 2 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a d p + 3 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} b c + 2 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} d - 6 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a d}{3 \, {\left (b^{3} p^{2} + 5 \, b^{3} p + 6 \, b^{3}\right )}} - \frac {\frac {{\left (b x^{3} + a\right )}^{p + 1} a b c}{p + 1} - \frac {{\left (b x^{3} + a\right )}^{p + 1} a^{2} d}{p + 1}}{3 \, b^{3}} \] Input:
integrate(x^5*(b*x^3+a)^p*(d*x^3+c),x, algorithm="giac")
Output:
1/3*((b*x^3 + a)^2*(b*x^3 + a)^p*b*c*p + (b*x^3 + a)^3*(b*x^3 + a)^p*d*p - 2*(b*x^3 + a)^2*(b*x^3 + a)^p*a*d*p + 3*(b*x^3 + a)^2*(b*x^3 + a)^p*b*c + 2*(b*x^3 + a)^3*(b*x^3 + a)^p*d - 6*(b*x^3 + a)^2*(b*x^3 + a)^p*a*d)/(b^3 *p^2 + 5*b^3*p + 6*b^3) - 1/3*((b*x^3 + a)^(p + 1)*a*b*c/(p + 1) - (b*x^3 + a)^(p + 1)*a^2*d/(p + 1))/b^3
Time = 0.88 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.75 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx={\left (b\,x^3+a\right )}^p\,\left (\frac {d\,x^9\,\left (p^2+3\,p+2\right )}{3\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {a^2\,\left (3\,b\,c-2\,a\,d+b\,c\,p\right )}{3\,b^3\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {x^6\,\left (p+1\right )\,\left (3\,b\,c+a\,d\,p+b\,c\,p\right )}{3\,b\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,p\,x^3\,\left (3\,b\,c-2\,a\,d+b\,c\,p\right )}{3\,b^2\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \] Input:
int(x^5*(a + b*x^3)^p*(c + d*x^3),x)
Output:
(a + b*x^3)^p*((d*x^9*(3*p + p^2 + 2))/(3*(11*p + 6*p^2 + p^3 + 6)) - (a^2 *(3*b*c - 2*a*d + b*c*p))/(3*b^3*(11*p + 6*p^2 + p^3 + 6)) + (x^6*(p + 1)* (3*b*c + a*d*p + b*c*p))/(3*b*(11*p + 6*p^2 + p^3 + 6)) + (a*p*x^3*(3*b*c - 2*a*d + b*c*p))/(3*b^2*(11*p + 6*p^2 + p^3 + 6)))
Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.90 \[ \int x^5 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {\left (b \,x^{3}+a \right )^{p} \left (b^{3} d \,p^{2} x^{9}+3 b^{3} d p \,x^{9}+2 b^{3} d \,x^{9}+a \,b^{2} d \,p^{2} x^{6}+b^{3} c \,p^{2} x^{6}+a \,b^{2} d p \,x^{6}+4 b^{3} c p \,x^{6}+3 b^{3} c \,x^{6}+a \,b^{2} c \,p^{2} x^{3}-2 a^{2} b d p \,x^{3}+3 a \,b^{2} c p \,x^{3}-a^{2} b c p +2 a^{3} d -3 a^{2} b c \right )}{3 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int(x^5*(b*x^3+a)^p*(d*x^3+c),x)
Output:
((a + b*x**3)**p*(2*a**3*d - a**2*b*c*p - 3*a**2*b*c - 2*a**2*b*d*p*x**3 + a*b**2*c*p**2*x**3 + 3*a*b**2*c*p*x**3 + a*b**2*d*p**2*x**6 + a*b**2*d*p* x**6 + b**3*c*p**2*x**6 + 4*b**3*c*p*x**6 + 3*b**3*c*x**6 + b**3*d*p**2*x* *9 + 3*b**3*d*p*x**9 + 2*b**3*d*x**9))/(3*b**3*(p**3 + 6*p**2 + 11*p + 6))