\(\int \frac {x^2 (d+e x^n)^q}{a+c x^{2 n}} \, dx\) [52]

Optimal result

Integrand size = 24, antiderivative size = 152 \[ \int \frac {x^2 \left (d+e x^n\right )^q}{a+c x^{2 n}} \, dx=\frac {x^3 \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{n},-q,1,\frac {3+n}{n},-\frac {e x^n}{d},-\frac {\sqrt {c} x^n}{\sqrt {-a}}\right )}{6 a}+\frac {x^3 \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{n},-q,1,\frac {3+n}{n},-\frac {e x^n}{d},\frac {\sqrt {c} x^n}{\sqrt {-a}}\right )}{6 a} \] Output:

1/6*x^3*(d+e*x^n)^q*AppellF1(3/n,1,-q,(3+n)/n,-c^(1/2)*x^n/(-a)^(1/2),-e*x 
^n/d)/a/((1+e*x^n/d)^q)+1/6*x^3*(d+e*x^n)^q*AppellF1(3/n,1,-q,(3+n)/n,c^(1 
/2)*x^n/(-a)^(1/2),-e*x^n/d)/a/((1+e*x^n/d)^q)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1881, 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.

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

\(\Big \downarrow \) 1881

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x^3 \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{n},1,-q,\frac {n+3}{n},-\frac {\sqrt {c} x^n}{\sqrt {-a}},-\frac {e x^n}{d}\right )}{6 a}+\frac {x^3 \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{n},1,-q,\frac {n+3}{n},\frac {\sqrt {c} x^n}{\sqrt {-a}},-\frac {e x^n}{d}\right )}{6 a}\)

Input:

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

Output:

(x^3*(d + e*x^n)^q*AppellF1[3/n, 1, -q, (3 + n)/n, -((Sqrt[c]*x^n)/Sqrt[-a 
]), -((e*x^n)/d)])/(6*a*(1 + (e*x^n)/d)^q) + (x^3*(d + e*x^n)^q*AppellF1[3 
/n, 1, -q, (3 + n)/n, (Sqrt[c]*x^n)/Sqrt[-a], -((e*x^n)/d)])/(6*a*(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 1881

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^2 \left (d+e x^n\right )^q}{a+c x^{2 n}} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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