\(\int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx\) [7]

Optimal result

Integrand size = 22, antiderivative size = 263 \[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=-\frac {2 c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {2 c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {c e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {c e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}} \] Output:

-2*c*d*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2- 
4*a*c-b*(-4*a*c+b^2)^(1/2))-2*c*d*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b 
+(-4*a*c+b^2)^(1/2)))/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)-c*e*x^2*hypergeom([ 
1, 2/n],[(2+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^ 
2)^(1/2))-c*e*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1 
/2)))/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)
 

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 525, normalized size of antiderivative = 2.00 \[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=c x \left (-e x \left (\frac {1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-2/n} \operatorname {Hypergeometric2F1}\left (-\frac {2}{n},-\frac {2}{n},\frac {-2+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {1-4^{-1/n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-2/n} \operatorname {Hypergeometric2F1}\left (-\frac {2}{n},-\frac {2}{n},\frac {-2+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right )-2 d \left (\frac {1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {1-2^{-1/n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right )\right ) \] Input:

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

Output:

c*x*(-(e*x*((1 - Hypergeometric2F1[-2/n, -2/n, (-2 + n)/n, (b - Sqrt[b^2 - 
 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4* 
a*c])/c + x^n))^(2/n))/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) + (1 - Hypergeo 
metric2F1[-2/n, -2/n, (-2 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 
4*a*c] + 2*c*x^n)]/(4^n^(-1)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^( 
2/n)))/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])))) - 2*d*((1 - Hypergeom 
etric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b 
^2 - 4*a*c] + 2*c*x^n)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^n^(- 
1))/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) + (1 - Hypergeometric2F1[-n^(-1), 
-n^(-1), (-1 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c* 
x^n)]/(2^n^(-1)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^n^(-1)))/(Sqrt 
[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c]))))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2325, 2432, 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.

\(\displaystyle \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx\)

\(\Big \downarrow \) 2325

\(\displaystyle \frac {2 c \int \frac {d+e x}{2 c x^n+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {d+e x}{2 c x^n+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {2 c \int \left (-\frac {d}{-2 c x^n-b+\sqrt {b^2-4 a c}}-\frac {e x}{-2 c x^n-b+\sqrt {b^2-4 a c}}\right )dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \left (\frac {d}{2 c x^n+b+\sqrt {b^2-4 a c}}+\frac {e x}{2 c x^n+b+\sqrt {b^2-4 a c}}\right )dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 c \left (\frac {d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 \left (b-\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b}+\frac {e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \left (\sqrt {b^2-4 a c}+b\right )}\right )}{\sqrt {b^2-4 a c}}\)

Input:

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

Output:

(2*c*((d*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b 
^2 - 4*a*c])])/(b - Sqrt[b^2 - 4*a*c]) + (e*x^2*Hypergeometric2F1[1, 2/n, 
(2 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(2*(b - Sqrt[b^2 - 4*a*c]) 
)))/Sqrt[b^2 - 4*a*c] - (2*c*((d*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1) 
, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b + Sqrt[b^2 - 4*a*c]) + (e*x^2*Hy 
pergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2 
*(b + Sqrt[b^2 - 4*a*c]))))/Sqrt[b^2 - 4*a*c]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2325

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int[Pq/(b - q + 2*c*x^n), x], x] - 
 Simp[2*(c/q)   Int[Pq/(b + q + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, n}, x] 
 && EqQ[n2, 2*n] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0]
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

Maple [F]

\[\int \frac {e x +d}{a +b \,x^{n}+c \,x^{2 n}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {e x + d}{c x^{2 \, n} + b x^{n} + a} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=\int \frac {d + e x}{a + b x^{n} + c x^{2 n}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {e x + d}{c x^{2 \, n} + b x^{n} + a} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {e x + d}{c x^{2 \, n} + b x^{n} + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=\int \frac {d+e\,x}{a+b\,x^n+c\,x^{2\,n}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {d+e x}{a+b x^n+c x^{2 n}} \, dx=\left (\int \frac {x}{x^{2 n} c +x^{n} b +a}d x \right ) e +\left (\int \frac {1}{x^{2 n} c +x^{n} b +a}d x \right ) d \] Input:

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

Output:

int(x/(x**(2*n)*c + x**n*b + a),x)*e + int(1/(x**(2*n)*c + x**n*b + a),x)* 
d