\(\int \frac {(d+e x^n)^q}{x (a+b x^n+c x^{2 n})} \, dx\) [177]

Optimal result

Integrand size = 29, antiderivative size = 263 \[ \int \frac {\left (d+e x^n\right )^q}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^n\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^n\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{a \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) n (1+q)}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^n\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^n\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{a \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) n (1+q)}-\frac {\left (d+e x^n\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,1+\frac {e x^n}{d}\right )}{a d n (1+q)} \] Output:

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.83 \[ \int \frac {\left (d+e x^n\right )^q}{x \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {\left (d+e x^n\right )^{1+q} \left (\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^n\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^n\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+q,2+q,1+\frac {e x^n}{d}\right )}{d}\right )}{a n (1+q)} \] Input:

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

Output:

((d + e*x^n)^(1 + q)*((c*(1 + b/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 
+ q, 2 + q, (2*c*(d + e*x^n))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(2*c* 
d + (-b + Sqrt[b^2 - 4*a*c])*e) + (c*(1 - b/Sqrt[b^2 - 4*a*c])*Hypergeomet 
ric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^n))/(2*c*d - (b + Sqrt[b^2 - 4*a*c]) 
*e)])/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e) - Hypergeometric2F1[1, 1 + q, 2 
+ q, 1 + (e*x^n)/d]/d))/(a*n*(1 + q))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1802, 1200, 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.

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

\(\Big \downarrow \) 1802

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1802

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( 
e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1 
)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, 
c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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