Integrand size = 29, antiderivative size = 196 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \] Output:
(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*(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)))/(b-(-4*a*c+b^2)^(1/2))/f/(1+m)+( e-(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*(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)))/(b+(-4*a*c+b^2)^(1/2))/f/(1+m)
Time = 0.78 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {x (f x)^m \left (\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+\left (-b d+\sqrt {b^2-4 a c} d+2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \] Input:
Integrate[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)),x]
Output:
(x*(f*x)^m*((b*d + Sqrt[b^2 - 4*a*c]*d - 2*a*e)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + (-(b*d) + Sqrt[ b^2 - 4*a*c]*d + 2*a*e)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2 *c*x^n)/(b + Sqrt[b^2 - 4*a*c])]))/(2*a*Sqrt[b^2 - 4*a*c]*(1 + m))
Time = 0.38 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, 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.
\(\displaystyle \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx\) |
\(\Big \downarrow \) 1884 |
\(\displaystyle \int \left (\frac {(f x)^m \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right )}{-\sqrt {b^2-4 a c}+b+2 c x^n}+\frac {(f x)^m \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b+2 c x^n}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(f x)^{m+1} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(f x)^{m+1} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (\sqrt {b^2-4 a c}+b\right )}\) |
Input:
Int[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)),x]
Output:
((e + (2*c*d - b*e)/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)) + ((e - (2*c*d - b*e)/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))
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])
\[\int \frac {\left (f x \right )^{m} \left (d +e \,x^{n}\right )}{a +b \,x^{n}+c \,x^{2 n}}d x\]
Input:
int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x)
Output:
int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x)
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \] Input:
integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")
Output:
integral((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a), x)
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int \frac {\left (f x\right )^{m} \left (d + e x^{n}\right )}{a + b x^{n} + c x^{2 n}}\, dx \] Input:
integrate((f*x)**m*(d+e*x**n)/(a+b*x**n+c*x**(2*n)),x)
Output:
Integral((f*x)**m*(d + e*x**n)/(a + b*x**n + c*x**(2*n)), x)
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \] Input:
integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")
Output:
integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a), x)
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \] Input:
integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")
Output:
integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a), x)
Timed out. \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (d+e\,x^n\right )}{a+b\,x^n+c\,x^{2\,n}} \,d x \] Input:
int(((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)),x)
Output:
int(((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)), x)
\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=f^{m} \left (\left (\int \frac {x^{m +n}}{x^{2 n} c +x^{n} b +a}d x \right ) e +\left (\int \frac {x^{m}}{x^{2 n} c +x^{n} b +a}d x \right ) d \right ) \] Input:
int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x)
Output:
f**m*(int(x**(m + n)/(x**(2*n)*c + x**n*b + a),x)*e + int(x**m/(x**(2*n)*c + x**n*b + a),x)*d)