\(\int \frac {(d+e x^n)^q}{x^3 (a+b x^n+c x^{2 n})} \, dx\) [179]

Optimal result

Integrand size = 29, antiderivative size = 210 \[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {c \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {AppellF1}\left (-\frac {2}{n},-q,1,-\frac {2-n}{n},-\frac {e x^n}{d},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) x^2}+\frac {c \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {AppellF1}\left (-\frac {2}{n},-q,1,-\frac {2-n}{n},-\frac {e x^n}{d},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) x^2} \] Output:

c*(d+e*x^n)^q*AppellF1(-2/n,1,-q,-(2-n)/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)), 
-e*x^n/d)/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/x^2/((1+e*x^n/d)^q)+c*(d+e*x^n) 
^q*AppellF1(-2/n,1,-q,-(2-n)/n,-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)),-e*x^n/d)/( 
b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/x^2/((1+e*x^n/d)^q)
 

Mathematica [F]

\[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx \] Input:

Integrate[(d + e*x^n)^q/(x^3*(a + b*x^n + c*x^(2*n))),x]
 

Output:

Integrate[(d + e*x^n)^q/(x^3*(a + b*x^n + c*x^(2*n))), x]
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1880, 1013, 1012}

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.

\(\displaystyle \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx\)

\(\Big \downarrow \) 1880

\(\displaystyle \frac {2 c \int \frac {\left (e x^n+d\right )^q}{x^3 \left (2 c x^n+b-\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {\left (e x^n+d\right )^q}{x^3 \left (2 c x^n+b+\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {2 c \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \int \frac {\left (\frac {e x^n}{d}+1\right )^q}{x^3 \left (2 c x^n+b-\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \int \frac {\left (\frac {e x^n}{d}+1\right )^q}{x^3 \left (2 c x^n+b+\sqrt {b^2-4 a c}\right )}dx}{\sqrt {b^2-4 a c}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {c \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {AppellF1}\left (-\frac {2}{n},1,-q,-\frac {2-n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},-\frac {e x^n}{d}\right )}{x^2 \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {c \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {AppellF1}\left (-\frac {2}{n},1,-q,-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {e x^n}{d}\right )}{x^2 \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}\)

Input:

Int[(d + e*x^n)^q/(x^3*(a + b*x^n + c*x^(2*n))),x]
 

Output:

-((c*(d + e*x^n)^q*AppellF1[-2/n, 1, -q, -((2 - n)/n), (-2*c*x^n)/(b - Sqr 
t[b^2 - 4*a*c]), -((e*x^n)/d)])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c]) 
*x^2*(1 + (e*x^n)/d)^q)) + (c*(d + e*x^n)^q*AppellF1[-2/n, 1, -q, -((2 - n 
)/n), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), -((e*x^n)/d)])/(Sqrt[b^2 - 4*a*c 
]*(b + Sqrt[b^2 - 4*a*c])*x^2*(1 + (e*x^n)/d)^q)
 

Defintions of rubi rules used

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1880

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( 
n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[ 
2*(c/r)   Int[(f*x)^m*((d + e*x^n)^q/(b - r + 2*c*x^n)), x], x] - Simp[2*(c 
/r)   Int[(f*x)^m*((d + e*x^n)^q/(b + r + 2*c*x^n)), x], x]] /; FreeQ[{a, b 
, c, d, e, f, m, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !Rati 
onalQ[n]
 

Maple [F]

\[\int \frac {\left (d +e \,x^{n}\right )^{q}}{x^{3} \left (a +b \,x^{n}+c \,x^{2 n}\right )}d x\]

Input:

int((d+e*x^n)^q/x^3/(a+b*x^n+c*x^(2*n)),x)
 

Output:

int((d+e*x^n)^q/x^3/(a+b*x^n+c*x^(2*n)),x)
 

Fricas [F]

\[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{q}}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{3}} \,d x } \] Input:

integrate((d+e*x^n)^q/x^3/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")
 

Output:

integral((e*x^n + d)^q/(c*x^3*x^(2*n) + b*x^3*x^n + a*x^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Timed out} \] Input:

integrate((d+e*x**n)**q/x**3/(a+b*x**n+c*x**(2*n)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{q}}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{3}} \,d x } \] Input:

integrate((d+e*x^n)^q/x^3/(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^3), x)
 

Giac [F]

\[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{q}}{{\left (c x^{2 \, n} + b x^{n} + a\right )} x^{3}} \,d x } \] Input:

integrate((d+e*x^n)^q/x^3/(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^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {{\left (d+e\,x^n\right )}^q}{x^3\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \] Input:

int((d + e*x^n)^q/(x^3*(a + b*x^n + c*x^(2*n))),x)
 

Output:

int((d + e*x^n)^q/(x^3*(a + b*x^n + c*x^(2*n))), x)
 

Reduce [F]

\[ \int \frac {\left (d+e x^n\right )^q}{x^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {\left (x^{n} e +d \right )^{q}}{x^{2 n} c \,x^{3}+x^{n} b \,x^{3}+a \,x^{3}}d x \] Input:

int((d+e*x^n)^q/x^3/(a+b*x^n+c*x^(2*n)),x)
 

Output:

int((x**n*e + d)**q/(x**(2*n)*c*x**3 + x**n*b*x**3 + a*x**3),x)