Integrand size = 29, antiderivative size = 374 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left ((b d-2 a e) (1+m-n)-\frac {4 a c d (1+m-2 n)-b^2 d (1+m-n)+2 a b e n}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) f (1+m) n}-\frac {c \left ((b d-2 a e) (1+m-n)+\frac {4 a c d (1+m-2 n)-b^2 d (1+m-n)+2 a b e n}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) f (1+m) n} \]
(f*x)^(1+m)*(b^2*d-2*a*c*d-a*b*e+c*(-2*a*e+b*d)*x^n)/a/(-4*a*c+b^2)/f/n/(a +b*x^n+c*x^(2*n))-c*(f*x)^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-2*c*x^ n/(b-(-4*a*c+b^2)^(1/2)))*((-2*a*e+b*d)*(1+m-n)+(-4*a*c*d*(1+m-2*n)+b^2*d* (1+m-n)-2*a*b*e*n)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)/f/(1+m)/n/(b-(-4*a*c +b^2)^(1/2))-c*(f*x)^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-2*c*x^n/(b+ (-4*a*c+b^2)^(1/2)))*((-2*a*e+b*d)*(1+m-n)+(4*a*c*d*(1+m-2*n)-b^2*d*(1+m-n )+2*a*b*e*n)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)/f/(1+m)/n/(b+(-4*a*c+b^2)^ (1/2))
Leaf count is larger than twice the leaf count of optimal. \(5363\) vs. \(2(374)=748\).
Time = 7.23 (sec) , antiderivative size = 5363, normalized size of antiderivative = 14.34 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \]
Time = 0.99 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1882, 25, 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.
| \(\displaystyle \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\) |
| \(\Big \downarrow \) 1882 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac {\int -\frac {(f x)^m \left (-c (b d-2 a e) (m-n+1) x^n+a b e (m+1)+2 a c d (m-2 n+1)-b^2 d (m-n+1)\right )}{b x^n+c x^{2 n}+a}dx}{a n \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(f x)^m \left (-c (b d-2 a e) (m-n+1) x^n+a b e (m+1)+2 a c d (m-2 n+1)-b^2 d (m-n+1)\right )}{b x^n+c x^{2 n}+a}dx}{a n \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\) |
| \(\Big \downarrow \) 1884 |
| \(\displaystyle \frac {\int \left (\frac {\left (-c (b d-2 a e) (m-n+1)-\frac {c \left (d b^2+d m b^2-d n b^2-2 a e n b-4 a c d-4 a c d m+8 a c d n\right )}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{2 c x^n+b-\sqrt {b^2-4 a c}}+\frac {\left (\frac {c \left (d b^2+d m b^2-d n b^2-2 a e n b-4 a c d-4 a c d m+8 a c d n\right )}{\sqrt {b^2-4 a c}}-c (b d-2 a e) (m-n+1)\right ) (f x)^m}{2 c x^n+b+\sqrt {b^2-4 a c}}\right )dx}{a n \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {c (f x)^{m+1} \left ((m-n+1) (b d-2 a e)-\frac {2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (f x)^{m+1} \left (\frac {2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt {b^2-4 a c}}+(m-n+1) (b d-2 a e)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (\sqrt {b^2-4 a c}+b\right )}}{a n \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\) |
((f*x)^(1 + m)*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^n))/(a*(b^2 - 4*a*c)*f*n*(a + b*x^n + c*x^(2*n))) + (-((c*((b*d - 2*a*e)*(1 + m - n) - ( 4*a*c*d*(1 + m - 2*n) - b^2*d*(1 + m - n) + 2*a*b*e*n)/Sqrt[b^2 - 4*a*c])* (f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c])*f*(1 + m))) - (c*((b*d - 2*a*e)*(1 + m - n) + (4*a*c*d*(1 + m - 2*n) - b^2*d*(1 + m - n) + 2*a*b*e* n)/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((b + Sqrt[b^2 - 4*a*c])*f*( 1 + m)))/(a*(b^2 - 4*a*c)*n)
3.2.42.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + b*x^n + c*x^ (2*n))^(p + 1)*((d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^n)/(a*f*n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(a*n*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^m* (a + b*x^n + c*x^(2*n))^(p + 1)*Simp[d*(b^2*(m + n*(p + 1) + 1) - 2*a*c*(m + 2*n*(p + 1) + 1)) - a*b*e*(m + 1) + (m + n*(2*p + 3) + 1)*(b*d - 2*a*e)*c *x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[n2, 2*n] && N eQ[b^2 - 4*a*c, 0] && !RationalQ[n] && ILtQ[p + 1, 0]
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])
\[\int \frac {\left (f x \right )^{m} \left (d +e \,x^{n}\right )}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
integral((e*x^n + d)*(f*x)^m/(c^2*x^(4*n) + b^2*x^(2*n) + 2*a*b*x^n + a^2 + 2*(b*c*x^n + a*c)*x^(2*n)), x)
Timed out. \[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
((b^2*d*f^m - (2*c*d*f^m + b*e*f^m)*a)*x*x^m + (b*c*d*f^m - 2*a*c*e*f^m)*x *e^(m*log(x) + n*log(x)))/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2* n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n) - integrate(((b^2*d*f^m*(m - n + 1) - (2*c*d*f^m*(m - 2*n + 1) + b*e*f^m*(m + 1))*a)*x^m + (b*c*d*f^m*(m - n + 1) - 2*a*c*e*f^m*(m - n + 1))*e^(m*log(x) + n*log(x)))/(a^2*b^2*n - 4 *a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n ), x)
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (d+e\,x^n\right )}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]