Integrand size = 23, antiderivative size = 97 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=-\frac {(a+b c) x^{-2 n} (e x)^{2 n} \left (a+b c+b d x^n\right )^{1+p}}{b^2 d^2 e n (1+p)}+\frac {x^{-2 n} (e x)^{2 n} \left (a+b c+b d x^n\right )^{2+p}}{b^2 d^2 e n (2+p)} \] Output:
-(b*c+a)*(e*x)^(2*n)*(a+b*c+b*d*x^n)^(p+1)/b^2/d^2/e/n/(p+1)/(x^(2*n))+(e* x)^(2*n)*(a+b*c+b*d*x^n)^(2+p)/b^2/d^2/e/n/(2+p)/(x^(2*n))
Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(97)=194\).
Time = 0.54 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.57 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\frac {x^{-2 n} (e x)^{2 n} \left (1+\frac {b d x^n}{a+b c}\right )^{-p} \left (a+b \left (c+d x^n\right )\right )^p \left (-a^2 \left (-1+\left (\frac {a+b c+b d x^n}{a+b c}\right )^p\right )+a b \left (d p x^n \left (\frac {a+b c+b d x^n}{a+b c}\right )^p-2 c \left (-1+\left (\frac {a+b c+b d x^n}{a+b c}\right )^p\right )\right )-b^2 \left (-c d p x^n \left (\frac {a+b c+b d x^n}{a+b c}\right )^p-d^2 (1+p) x^{2 n} \left (\frac {a+b c+b d x^n}{a+b c}\right )^p+c^2 \left (-1+\left (\frac {a+b c+b d x^n}{a+b c}\right )^p\right )\right )\right )}{b^2 d^2 e n (1+p) (2+p)} \] Input:
Integrate[(e*x)^(-1 + 2*n)*(a + b*(c + d*x^n))^p,x]
Output:
((e*x)^(2*n)*(a + b*(c + d*x^n))^p*(-(a^2*(-1 + ((a + b*c + b*d*x^n)/(a + b*c))^p)) + a*b*(d*p*x^n*((a + b*c + b*d*x^n)/(a + b*c))^p - 2*c*(-1 + ((a + b*c + b*d*x^n)/(a + b*c))^p)) - b^2*(-(c*d*p*x^n*((a + b*c + b*d*x^n)/( a + b*c))^p) - d^2*(1 + p)*x^(2*n)*((a + b*c + b*d*x^n)/(a + b*c))^p + c^2 *(-1 + ((a + b*c + b*d*x^n)/(a + b*c))^p))))/(b^2*d^2*e*n*(1 + p)*(2 + p)* x^(2*n)*(1 + (b*d*x^n)/(a + b*c))^p)
Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2073, 800, 798, 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.
\(\displaystyle \int (e x)^{2 n-1} \left (a+b \left (c+d x^n\right )\right )^p \, dx\) |
\(\Big \downarrow \) 2073 |
\(\displaystyle \int (e x)^{2 n-1} \left (a+b c+b d x^n\right )^pdx\) |
\(\Big \downarrow \) 800 |
\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int x^{2 n-1} \left (b d x^n+a+b c\right )^pdx}{e}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int x^n \left (b d x^n+a+b c\right )^pdx^n}{e n}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \left (\frac {(-a-b c) \left (b d x^n+a+b c\right )^p}{b d}+\frac {\left (b d x^n+a+b c\right )^{p+1}}{b d}\right )dx^n}{e n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \left (\frac {\left (a+b c+b d x^n\right )^{p+2}}{b^2 d^2 (p+2)}-\frac {(a+b c) \left (a+b c+b d x^n\right )^{p+1}}{b^2 d^2 (p+1)}\right )}{e n}\) |
Input:
Int[(e*x)^(-1 + 2*n)*(a + b*(c + d*x^n))^p,x]
Output:
((e*x)^(2*n)*(-(((a + b*c)*(a + b*c + b*d*x^n)^(1 + p))/(b^2*d^2*(1 + p))) + (a + b*c + b*d*x^n)^(2 + p)/(b^2*d^2*(2 + p))))/(e*n*x^(2*n))
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^Int Part[m]*((c*x)^FracPart[m]/x^FracPart[m]) Int[x^m*(a + b*x^n)^p, x], x] / ; FreeQ[{a, b, c, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_)^(p_.)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{c, m, p}, x] && BinomialQ[u, x] && !BinomialMatchQ[u, x ]
\[\int \left (e x \right )^{-1+2 n} {\left (a +b \left (c +d \,x^{n}\right )\right )}^{p}d x\]
Input:
int((e*x)^(-1+2*n)*(a+b*(c+d*x^n))^p,x)
Output:
int((e*x)^(-1+2*n)*(a+b*(c+d*x^n))^p,x)
Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.28 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\frac {{\left ({\left (b^{2} c + a b\right )} d e^{2 \, n - 1} p x^{n} + {\left (b^{2} d^{2} p + b^{2} d^{2}\right )} e^{2 \, n - 1} x^{2 \, n} - {\left (b^{2} c^{2} + 2 \, a b c + a^{2}\right )} e^{2 \, n - 1}\right )} {\left (b d x^{n} + b c + a\right )}^{p}}{b^{2} d^{2} n p^{2} + 3 \, b^{2} d^{2} n p + 2 \, b^{2} d^{2} n} \] Input:
integrate((e*x)^(-1+2*n)*(a+b*(c+d*x^n))^p,x, algorithm="fricas")
Output:
((b^2*c + a*b)*d*e^(2*n - 1)*p*x^n + (b^2*d^2*p + b^2*d^2)*e^(2*n - 1)*x^( 2*n) - (b^2*c^2 + 2*a*b*c + a^2)*e^(2*n - 1))*(b*d*x^n + b*c + a)^p/(b^2*d ^2*n*p^2 + 3*b^2*d^2*n*p + 2*b^2*d^2*n)
\[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\int \left (e x\right )^{2 n - 1} \left (a + b c + b d x^{n}\right )^{p}\, dx \] Input:
integrate((e*x)**(-1+2*n)*(a+b*(c+d*x**n))**p,x)
Output:
Integral((e*x)**(2*n - 1)*(a + b*c + b*d*x**n)**p, x)
Time = 0.05 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.21 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\frac {{\left (b^{2} d^{2} e^{2 \, n} {\left (p + 1\right )} x^{2 \, n} - b^{2} c^{2} e^{2 \, n} - 2 \, a b c e^{2 \, n} - a^{2} e^{2 \, n} + {\left (b^{2} c d e^{2 \, n} p + a b d e^{2 \, n} p\right )} x^{n}\right )} {\left (b d x^{n} + b c + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2} d^{2} e n} \] Input:
integrate((e*x)^(-1+2*n)*(a+b*(c+d*x^n))^p,x, algorithm="maxima")
Output:
(b^2*d^2*e^(2*n)*(p + 1)*x^(2*n) - b^2*c^2*e^(2*n) - 2*a*b*c*e^(2*n) - a^2 *e^(2*n) + (b^2*c*d*e^(2*n)*p + a*b*d*e^(2*n)*p)*x^n)*(b*d*x^n + b*c + a)^ p/((p^2 + 3*p + 2)*b^2*d^2*e*n)
\[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\int { {\left ({\left (d x^{n} + c\right )} b + a\right )}^{p} \left (e x\right )^{2 \, n - 1} \,d x } \] Input:
integrate((e*x)^(-1+2*n)*(a+b*(c+d*x^n))^p,x, algorithm="giac")
Output:
integrate(((d*x^n + c)*b + a)^p*(e*x)^(2*n - 1), x)
Timed out. \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\int {\left (e\,x\right )}^{2\,n-1}\,{\left (a+b\,\left (c+d\,x^n\right )\right )}^p \,d x \] Input:
int((e*x)^(2*n - 1)*(a + b*(c + d*x^n))^p,x)
Output:
int((e*x)^(2*n - 1)*(a + b*(c + d*x^n))^p, x)
Time = 0.40 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06 \[ \int (e x)^{-1+2 n} \left (a+b \left (c+d x^n\right )\right )^p \, dx=\frac {e^{2 n} \left (x^{n} b d +a +b c \right )^{p} \left (x^{2 n} b^{2} d^{2} p +x^{2 n} b^{2} d^{2}+x^{n} a b d p +x^{n} b^{2} c d p -a^{2}-2 a b c -b^{2} c^{2}\right )}{b^{2} d^{2} e n \left (p^{2}+3 p +2\right )} \] Input:
int((e*x)^(-1+2*n)*(a+b*(c+d*x^n))^p,x)
Output:
(e**(2*n)*(x**n*b*d + a + b*c)**p*(x**(2*n)*b**2*d**2*p + x**(2*n)*b**2*d* *2 + x**n*a*b*d*p + x**n*b**2*c*d*p - a**2 - 2*a*b*c - b**2*c**2))/(b**2*d **2*e*n*(p**2 + 3*p + 2))