Integrand size = 26, antiderivative size = 571 \[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=-\frac {e^2 (b e (1+n (2+p))-3 c d (1+n (3+2 p))) x \left (a+b x^n+c x^{2 n}\right )^{1+p}}{c^2 (1+2 n (1+p)) (1+n (3+2 p))}+\frac {e^3 x^{1+n} \left (a+b x^n+c x^{2 n}\right )^{1+p}}{c (1+3 n+2 n p)}+\frac {e \left (b e (1+n+n p) (b e (1+n (2+p))-3 c d (1+n (3+2 p)))-c (1+2 n (1+p)) \left (a e^2 (1+n)-3 c d^2 (1+n (3+2 p))\right )\right ) x^{1+n} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (1+\frac {1}{n},-p,-p,2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c^2 (1+n) (1+2 n (1+p)) (1+n (3+2 p))}+\frac {\left (c^2 d^3 (1+2 n (1+p)) (1+n (3+2 p))+a e^2 (b e (1+n (2+p))-3 c d (1+n (3+2 p)))\right ) x \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1}{n},-p,-p,1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c^2 (1+2 n (1+p)) (1+n (3+2 p))} \] Output:
-e^2*(b*e*(1+n*(2+p))-3*c*d*(1+n*(3+2*p)))*x*(a+b*x^n+c*x^(2*n))^(p+1)/c^2 /(1+2*n*(p+1))/(1+n*(3+2*p))+e^3*x^(1+n)*(a+b*x^n+c*x^(2*n))^(p+1)/c/(2*n* p+3*n+1)+e*(b*e*(n*p+n+1)*(b*e*(1+n*(2+p))-3*c*d*(1+n*(3+2*p)))-c*(1+2*n*( p+1))*(a*e^2*(1+n)-3*c*d^2*(1+n*(3+2*p))))*x^(1+n)*(a+b*x^n+c*x^(2*n))^p*A ppellF1(1+1/n,-p,-p,2+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^n/(b+(-4* a*c+b^2)^(1/2)))/c^2/(1+n)/(1+2*n*(p+1))/(1+n*(3+2*p))/((1+2*c*x^n/(b-(-4* a*c+b^2)^(1/2)))^p)/((1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^p)+(c^2*d^3*(1+2*n *(p+1))*(1+n*(3+2*p))+a*e^2*(b*e*(1+n*(2+p))-3*c*d*(1+n*(3+2*p))))*x*(a+b* x^n+c*x^(2*n))^p*AppellF1(1/n,-p,-p,1+1/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)), -2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/c^2/(1+2*n*(p+1))/(1+n*(3+2*p))/((1+2*c*x ^n/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^p)
Time = 1.41 (sec) , antiderivative size = 438, normalized size of antiderivative = 0.77 \[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {x \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+x^n \left (b+c x^n\right )\right )^p \left (3 d^2 e \left (1+5 n+6 n^2\right ) x^n \operatorname {AppellF1}\left (1+\frac {1}{n},-p,-p,2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+(1+n) \left (3 d e^2 (1+3 n) x^{2 n} \operatorname {AppellF1}\left (2+\frac {1}{n},-p,-p,3+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+(1+2 n) \left (e^3 x^{3 n} \operatorname {AppellF1}\left (3+\frac {1}{n},-p,-p,4+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+d^3 (1+3 n) \operatorname {AppellF1}\left (\frac {1}{n},-p,-p,1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )\right )}{(1+n) (1+2 n) (1+3 n)} \] Input:
Integrate[(d + e*x^n)^3*(a + b*x^n + c*x^(2*n))^p,x]
Output:
(x*(a + x^n*(b + c*x^n))^p*(3*d^2*e*(1 + 5*n + 6*n^2)*x^n*AppellF1[1 + n^( -1), -p, -p, 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + (1 + n)*(3*d*e^2*(1 + 3*n)*x^(2*n)*AppellF1[2 + n ^(-1), -p, -p, 3 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/( -b + Sqrt[b^2 - 4*a*c])] + (1 + 2*n)*(e^3*x^(3*n)*AppellF1[3 + n^(-1), -p, -p, 4 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[ b^2 - 4*a*c])] + d^3*(1 + 3*n)*AppellF1[n^(-1), -p, -p, 1 + n^(-1), (-2*c* x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]))))/((1 + n)*(1 + 2*n)*(1 + 3*n)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p )
Time = 0.87 (sec) , antiderivative size = 606, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1766, 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.
| \(\displaystyle \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx\) |
| \(\Big \downarrow \) 1766 |
| \(\displaystyle \int \left (d^3 \left (a+b x^n+c x^{2 n}\right )^p+3 d^2 e x^n \left (a+b x^n+c x^{2 n}\right )^p+3 d e^2 x^{2 n} \left (a+b x^n+c x^{2 n}\right )^p+e^3 x^{3 n} \left (a+b x^n+c x^{2 n}\right )^p\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle d^3 x \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1}{n},-p,-p,1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )+\frac {3 d^2 e x^{n+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (1+\frac {1}{n},-p,-p,2+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{n+1}+\frac {3 d e^2 x^{2 n+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (2+\frac {1}{n},-p,-p,3+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 n+1}+\frac {e^3 x^{3 n+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (3+\frac {1}{n},-p,-p,4+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 n+1}\) |
Input:
Int[(d + e*x^n)^3*(a + b*x^n + c*x^(2*n))^p,x]
Output:
(3*d^2*e*x^(1 + n)*(a + b*x^n + c*x^(2*n))^p*AppellF1[1 + n^(-1), -p, -p, 2 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + n)*(1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^ n)/(b + Sqrt[b^2 - 4*a*c]))^p) + (3*d*e^2*x^(1 + 2*n)*(a + b*x^n + c*x^(2* n))^p*AppellF1[2 + n^(-1), -p, -p, 3 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + 2*n)*(1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p) + (e^3 *x^(1 + 3*n)*(a + b*x^n + c*x^(2*n))^p*AppellF1[3 + n^(-1), -p, -p, 4 + n^ (-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c ])])/((1 + 3*n)*(1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^n)/( b + Sqrt[b^2 - 4*a*c]))^p) + (d^3*x*(a + b*x^n + c*x^(2*n))^p*AppellF1[n^( -1), -p, -p, 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2 *c*x^n)/(b + Sqrt[b^2 - 4*a*c]))^p)
Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ ))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q*(a + b*x^n + c*x^(2 *n))^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ [b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ((IntegersQ[p, q] && !IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] && !IntegerQ[n]))
\[\int \left (d +e \,x^{n}\right )^{3} \left (a +b \,x^{n}+c \,x^{2 n}\right )^{p}d x\]
Input:
int((d+e*x^n)^3*(a+b*x^n+c*x^(2*n))^p,x)
Output:
int((d+e*x^n)^3*(a+b*x^n+c*x^(2*n))^p,x)
\[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )}^{3} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \,d x } \] Input:
integrate((d+e*x^n)^3*(a+b*x^n+c*x^(2*n))^p,x, algorithm="fricas")
Output:
integral((e^3*x^(3*n) + 3*d*e^2*x^(2*n) + 3*d^2*e*x^n + d^3)*(c*x^(2*n) + b*x^n + a)^p, x)
Timed out. \[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {Timed out} \] Input:
integrate((d+e*x**n)**3*(a+b*x**n+c*x**(2*n))**p,x)
Output:
Timed out
\[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )}^{3} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \,d x } \] Input:
integrate((d+e*x^n)^3*(a+b*x^n+c*x^(2*n))^p,x, algorithm="maxima")
Output:
integrate((e*x^n + d)^3*(c*x^(2*n) + b*x^n + a)^p, x)
Exception generated. \[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {Exception raised: TypeError} \] Input:
integrate((d+e*x^n)^3*(a+b*x^n+c*x^(2*n))^p,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{512,[1,0,7,4,9,5,1,8,0,3]%%%}+%%%{-3072,[1,0,7,4,9,5,0,9,1 ,2]%%%}+%
Timed out. \[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int {\left (d+e\,x^n\right )}^3\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^p \,d x \] Input:
int((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))^p,x)
Output:
int((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))^p, x)
\[ \int \left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {too large to display} \] Input:
int((d+e*x^n)^3*(a+b*x^n+c*x^(2*n))^p,x)
Output:
(4*x**(3*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*e**3*n**3*p**3*x + 6*x**(3 *n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*e**3*n**3*p**2*x + 2*x**(3*n)*(x** (2*n)*c + x**n*b + a)**p*b*c**2*e**3*n**3*p*x + 8*x**(3*n)*(x**(2*n)*c + x **n*b + a)**p*b*c**2*e**3*n**2*p**2*x + 9*x**(3*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*e**3*n**2*p*x + 2*x**(3*n)*(x**(2*n)*c + x**n*b + a)**p*b*c** 2*e**3*n**2*x + 5*x**(3*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*e**3*n*p*x + 3*x**(3*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*e**3*n*x + x**(3*n)*(x**( 2*n)*c + x**n*b + a)**p*b*c**2*e**3*x + 2*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b**2*c*e**3*n**3*p**3*x + x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b**2 *c*e**3*n**3*p**2*x + 3*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b**2*c*e**3* n**2*p**2*x + x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b**2*c*e**3*n**2*p*x + x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b**2*c*e**3*n*p*x + 12*x**(2*n)*(x* *(2*n)*c + x**n*b + a)**p*b*c**2*d*e**2*n**3*p**3*x + 24*x**(2*n)*(x**(2*n )*c + x**n*b + a)**p*b*c**2*d*e**2*n**3*p**2*x + 9*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*d*e**2*n**3*p*x + 24*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*d*e**2*n**2*p**2*x + 36*x**(2*n)*(x**(2*n)*c + x**n*b + a)** p*b*c**2*d*e**2*n**2*p*x + 9*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2* d*e**2*n**2*x + 15*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*d*e**2*n*p *x + 12*x**(2*n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*d*e**2*n*x + 3*x**(2* n)*(x**(2*n)*c + x**n*b + a)**p*b*c**2*d*e**2*x + 4*x**n*(x**(2*n)*c + ...