\(\int \frac {(f x)^m (d+e x^n)}{a+b x^n+c x^{2 n}} \, dx\) [141]

Optimal result

Integrand size = 29, antiderivative size = 196 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \]

[Out]

(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)))*(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(
1/2))/f/(1+m)/(b-(-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)))*(e+(b*e-2*c*d)/(-4*a*c+b^2)^(1/2))/f/(1+m)/(b+(-4*a*c+b^2)^(1/2))

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1574, 371} \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {(f x)^{m+1} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(f x)^{m+1} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (\sqrt {b^2-4 a c}+b\right )} \]

[In]

Int[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)),x]

[Out]

((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))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1574

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> 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] && (IGtQ[p, 0] || IGtQ[q, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n}\right ) \, dx \\ & = \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {(f x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx+\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {(f x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx \\ & = \frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\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} \, _2F_1\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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\frac {x (f x)^m \left (\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+\left (-b d+\sqrt {b^2-4 a c} d+2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \]

[In]

Integrate[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)),x]

[Out]

(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))

Maple [F]

\[\int \frac {\left (f x \right )^{m} \left (d +e \,x^{n}\right )}{a +b \,x^{n}+c \,x^{2 n}}d x\]

[In]

int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x)

[Out]

int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x)

Fricas [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

integral((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a), x)

Sympy [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int \frac {\left (f x\right )^{m} \left (d + e x^{n}\right )}{a + b x^{n} + c x^{2 n}}\, dx \]

[In]

integrate((f*x)**m*(d+e*x**n)/(a+b*x**n+c*x**(2*n)),x)

[Out]

Integral((f*x)**m*(d + e*x**n)/(a + b*x**n + c*x**(2*n)), x)

Maxima [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a), x)

Giac [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(f x)^m \left (d+e x^n\right )}{a+b x^n+c x^{2 n}} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (d+e\,x^n\right )}{a+b\,x^n+c\,x^{2\,n}} \,d x \]

[In]

int(((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)),x)

[Out]

int(((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n)), x)