Integrand size = 31, antiderivative size = 498 \[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {d^2 (f x)^{1+m} \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{n},-p,-p,\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m)}+\frac {2 d e x^{1+n} (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m+n}{n},-p,-p,\frac {1+m+2 n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+m+n}+\frac {e^2 x^{1+2 n} (f x)^m \left (1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m+2 n}{n},-p,-p,\frac {1+m+3 n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{1+m+2 n} \]
d^2*(f*x)^(1+m)*(a+b*x^n+c*x^(2*n))^p*AppellF1((1+m)/n,-p,-p,(1+m+n)/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)))/f/(1+m)/((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)+ 2*d*e*x^(1+n)*(f*x)^m*(a+b*x^n+c*x^(2*n))^p*AppellF1((1+m+n)/n,-p,-p,(1+m+ 2*n)/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)))/(1 +m+n)/((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)+e^2*x^(1+2*n)*(f*x)^m*(a+b*x^n+c*x^(2*n))^p*AppellF1((1+m+2*n)/n ,-p,-p,(1+m+3*n)/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)))/(1+m+2*n)/((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 = 1.19 (sec) , antiderivative size = 391, normalized size of antiderivative = 0.79 \[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\frac {x (f x)^m \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+x^n \left (b+c x^n\right )\right )^p \left (d^2 \left (1+m^2+3 n+2 n^2+m (2+3 n)\right ) \operatorname {AppellF1}\left (\frac {1+m}{n},-p,-p,\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+e (1+m) x^n \left (2 d (1+m+2 n) \operatorname {AppellF1}\left (\frac {1+m+n}{n},-p,-p,\frac {1+m+2 n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+e (1+m+n) x^n \operatorname {AppellF1}\left (\frac {1+m+2 n}{n},-p,-p,\frac {1+m+3 n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{(1+m) (1+m+n) (1+m+2 n)} \]
(x*(f*x)^m*(a + x^n*(b + c*x^n))^p*(d^2*(1 + m^2 + 3*n + 2*n^2 + m*(2 + 3* n))*AppellF1[(1 + m)/n, -p, -p, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^n*(2*d*(1 + m + 2*n)*AppellF1[(1 + m + n)/n, -p, -p, (1 + m + 2*n)/n, (-2*c*x^n)/(b + Sqr t[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m + n)*x^n*A ppellF1[(1 + m + 2*n)/n, -p, -p, (1 + m + 3*n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])))/((1 + m)*(1 + m + n)*(1 + m + 2*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 = 0.81 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1884, 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.
| \(\displaystyle \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx\) |
| \(\Big \downarrow \) 1884 |
| \(\displaystyle \int \left (d^2 (f x)^m \left (a+b x^n+c x^{2 n}\right )^p+2 d e x^n (f x)^m \left (a+b x^n+c x^{2 n}\right )^p+e^2 x^{2 n} (f x)^m \left (a+b x^n+c x^{2 n}\right )^p\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 (f x)^{m+1} \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+1}{n},-p,-p,\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1)}+\frac {2 d e x^{n+1} (f x)^m \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+n+1}{n},-p,-p,\frac {m+2 n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{m+n+1}+\frac {e^2 x^{2 n+1} (f x)^m \left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^n+c x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+2 n+1}{n},-p,-p,\frac {m+3 n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{m+2 n+1}\) |
(d^2*(f*x)^(1 + m)*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m)/n, -p, -p, ( 1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(f*(1 + m)*(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) + (2*d*e*x^(1 + n)*(f*x)^m*(a + b*x^n + c*x^(2*n))^p*AppellF1[(1 + m + n)/n, -p, -p, (1 + m + 2*n)/n, (-2*c*x^n)/ (b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + m + 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^2*x^(1 + 2*n)*(f*x)^m*(a + b*x^n + c*x^(2*n))^p*AppellF1[ (1 + m + 2*n)/n, -p, -p, (1 + m + 3*n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c ]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((1 + m + 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)
3.2.52.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*( (d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && !RationalQ[n] && ( IGtQ[p, 0] || IGtQ[q, 0])
\[\int \left (f x \right )^{m} \left (d +e \,x^{n}\right )^{2} \left (a +b \,x^{n}+c \,x^{2 n}\right )^{p}d x\]
\[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )}^{2} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m} \,d x } \]
Timed out. \[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {Timed out} \]
\[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )}^{2} {\left (c x^{2 \, n} + b x^{n} + a\right )}^{p} \left (f x\right )^{m} \,d x } \]
Exception generated. \[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\text {Exception raised: TypeError} \]
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%%%{-128,[1,0,5,3,0,6,4,1,6,0,2]%%%}+%%%{512,[1,0,5,3,0,6,4,0, 7,1,1]%%%
Timed out. \[ \int (f x)^m \left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )^p \, dx=\int {\left (f\,x\right )}^m\,{\left (d+e\,x^n\right )}^2\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^p \,d x \]