Integrand size = 19, antiderivative size = 171 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=-\frac {a^3 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p}}{b^4 (1+p)}+\frac {3 a^2 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{2+p}}{b^4 (2+p)}-\frac {3 a x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{3+p}}{b^4 (3+p)}+\frac {x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{4+p}}{b^4 (4+p)} \]
-a^3*x^4*(a+b*(c*x^n)^(1/n))^(p+1)/b^4/(p+1)/((c*x^n)^(4/n))+3*a^2*x^4*(a+ b*(c*x^n)^(1/n))^(2+p)/b^4/(2+p)/((c*x^n)^(4/n))-3*a*x^4*(a+b*(c*x^n)^(1/n ))^(3+p)/b^4/(3+p)/((c*x^n)^(4/n))+x^4*(a+b*(c*x^n)^(1/n))^(4+p)/b^4/(4+p) /((c*x^n)^(4/n))
Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p} \left (-\frac {a^3}{1+p}+\frac {3 a^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{2+p}-\frac {3 a \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2}{3+p}+\frac {\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^3}{4+p}\right )}{b^4} \]
(x^4*(a + b*(c*x^n)^n^(-1))^(1 + p)*(-(a^3/(1 + p)) + (3*a^2*(a + b*(c*x^n )^n^(-1)))/(2 + p) - (3*a*(a + b*(c*x^n)^n^(-1))^2)/(3 + p) + (a + b*(c*x^ n)^n^(-1))^3/(4 + p)))/(b^4*(c*x^n)^(4/n))
Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {892, 53, 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.
\(\displaystyle \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx\) |
\(\Big \downarrow \) 892 |
\(\displaystyle x^4 \left (c x^n\right )^{-4/n} \int \left (c x^n\right )^{3/n} \left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )^pd\left (c x^n\right )^{\frac {1}{n}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle x^4 \left (c x^n\right )^{-4/n} \int \left (-\frac {a^3 \left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )^p}{b^3}+\frac {3 a^2 \left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )^{p+1}}{b^3}-\frac {3 a \left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )^{p+2}}{b^3}+\frac {\left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )^{p+3}}{b^3}\right )d\left (c x^n\right )^{\frac {1}{n}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^4 \left (c x^n\right )^{-4/n} \left (-\frac {a^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+1}}{b^4 (p+1)}+\frac {3 a^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+2}}{b^4 (p+2)}-\frac {3 a \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+3}}{b^4 (p+3)}+\frac {\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{p+4}}{b^4 (p+4)}\right )\) |
(x^4*(-((a^3*(a + b*(c*x^n)^n^(-1))^(1 + p))/(b^4*(1 + p))) + (3*a^2*(a + b*(c*x^n)^n^(-1))^(2 + p))/(b^4*(2 + p)) - (3*a*(a + b*(c*x^n)^n^(-1))^(3 + p))/(b^4*(3 + p)) + (a + b*(c*x^n)^n^(-1))^(4 + p)/(b^4*(4 + p))))/(c*x^ n)^(4/n)
3.31.22.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1)) Subst[Int[x^m*(a + b *x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x ] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 5.93 (sec) , antiderivative size = 1127, normalized size of antiderivative = 6.59
(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x ^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)^(1+p)/(c^(1/n))*x^4/((x^n)^(1/n))*exp (-1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c* x^n))/n)/b/(1+p)-3/b/(1+p)*(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^ n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)^(1+p)/(4+p )/((x^n)^(1/n))*x^4/(c^(1/n))*exp(-1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+cs gn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)-3/b^2/(1+p)*(b*(x^n)^(1/n)*c^(1/ n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn (I*c*x^n))/n)+a)^(1+p)/(4+p)/(3+p)*a*x^4/((x^n)^(1/n))^2/(c^(1/n))^2*exp(- I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/ n)-3/b^2/(1+p)*(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I* x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)^(1+p)/(4+p)/(3+p)*a*x^ 4/((x^n)^(1/n))^2/(c^(1/n))^2*exp(-I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I *c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)*p-6/b^4/(1+p)*(b*(x^n)^(1/n)*c^(1/n) *exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I *c*x^n))/n)+a)^(1+p)/(4+p)*a^3/(2+p)*x^4/((x^n)^(1/n))^4/(c^(1/n))^4/(3+p) *exp(-2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I* c*x^n))/n)+6/b^3/(1+p)*(b*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*( -csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)^(1+p)/(4+p)/(2 +p)/(3+p)*a^2*x^4/((x^n)^(1/n))^3/(c^(1/n))^3*exp(-3/2*I*Pi*csgn(I*c*x^...
Time = 0.37 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.07 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {{\left (6 \, a^{3} b c^{\left (\frac {1}{n}\right )} p x + {\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} c^{\frac {4}{n}} x^{4} + {\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} c^{\frac {3}{n}} x^{3} - 6 \, a^{4} - 3 \, {\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} c^{\frac {2}{n}} x^{2}\right )} {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p}}{{\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )} c^{\frac {4}{n}}} \]
(6*a^3*b*c^(1/n)*p*x + (b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*c^(4/n)*x^ 4 + (a*b^3*p^3 + 3*a*b^3*p^2 + 2*a*b^3*p)*c^(3/n)*x^3 - 6*a^4 - 3*(a^2*b^2 *p^2 + a^2*b^2*p)*c^(2/n)*x^2)*(b*c^(1/n)*x + a)^p/((b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^4)*c^(4/n))
\[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\int x^{3} \left (a + b \left (c x^{n}\right )^{\frac {1}{n}}\right )^{p}\, dx \]
\[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\int { {\left (\left (c x^{n}\right )^{\left (\frac {1}{n}\right )} b + a\right )}^{p} x^{3} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (179) = 358\).
Time = 0.36 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.25 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {{\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} p^{3} x^{4} + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac {3}{n}} p^{3} x^{3} + 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} p^{2} x^{4} + 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac {3}{n}} p^{2} x^{3} + 11 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} p x^{4} - 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2} b^{2} c^{\frac {2}{n}} p^{2} x^{2} + 2 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac {3}{n}} p x^{3} + 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} x^{4} - 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2} b^{2} c^{\frac {2}{n}} p x^{2} + 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{3} b c^{\left (\frac {1}{n}\right )} p x - 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{4}}{b^{4} c^{\frac {4}{n}} p^{4} + 10 \, b^{4} c^{\frac {4}{n}} p^{3} + 35 \, b^{4} c^{\frac {4}{n}} p^{2} + 50 \, b^{4} c^{\frac {4}{n}} p + 24 \, b^{4} c^{\frac {4}{n}}} \]
((b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p^3*x^4 + (b*c^(1/n)*x + a)^p*a*b^3*c^(3/ n)*p^3*x^3 + 6*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p^2*x^4 + 3*(b*c^(1/n)*x + a)^p*a*b^3*c^(3/n)*p^2*x^3 + 11*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p*x^4 - 3* (b*c^(1/n)*x + a)^p*a^2*b^2*c^(2/n)*p^2*x^2 + 2*(b*c^(1/n)*x + a)^p*a*b^3* c^(3/n)*p*x^3 + 6*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*x^4 - 3*(b*c^(1/n)*x + a )^p*a^2*b^2*c^(2/n)*p*x^2 + 6*(b*c^(1/n)*x + a)^p*a^3*b*c^(1/n)*p*x - 6*(b *c^(1/n)*x + a)^p*a^4)/(b^4*c^(4/n)*p^4 + 10*b^4*c^(4/n)*p^3 + 35*b^4*c^(4 /n)*p^2 + 50*b^4*c^(4/n)*p + 24*b^4*c^(4/n))
Timed out. \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\int x^3\,{\left (a+b\,{\left (c\,x^n\right )}^{1/n}\right )}^p \,d x \]