\(\int \frac {x^2 (d+e x^n)^q}{a+b x^n+c x^{2 n}} \, dx\) [146]

Optimal result

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

[Out]

-2/3*c*x^3*(d+e*x^n)^q*AppellF1(3/n,1,-q,(3+n)/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-e*x^n/d)/((1+e*x^n/d)^q)/(b^
2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2/3*c*x^3*(d+e*x^n)^q*AppellF1(3/n,1,-q,(3+n)/n,-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)),
-e*x^n/d)/((1+e*x^n/d)^q)/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1570, 525, 524} \[ \int \frac {x^2 \left (d+e x^n\right )^q}{a+b x^n+c x^{2 n}} \, dx=-\frac {2 c x^3 \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{n},1,-q,\frac {n+3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {e x^n}{d}\right )}{3 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {2 c x^3 \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {3}{n},1,-q,\frac {n+3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},-\frac {e x^n}{d}\right )}{3 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \]

[In]

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

[Out]

(-2*c*x^3*(d + e*x^n)^q*AppellF1[3/n, 1, -q, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), -((e*x^n)/d)])/(3*
(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 + (e*x^n)/d)^q) - (2*c*x^3*(d + e*x^n)^q*AppellF1[3/n, 1, -q, (3 + n)/n
, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), -((e*x^n)/d)])/(3*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*(1 + (e*x^n)/d)^q)

Rule 524

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] /; Free
Q[{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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 1570

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]}, Dist[2*(c/r), Int[(f*x)^m*((d + e*x^n)^q/(b - r + 2*c*x^n)), x], x] - Dist[
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] && Eq
Q[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(2 c) \int \frac {x^2 \left (d+e x^n\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {x^2 \left (d+e x^n\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}} \\ & = \frac {\left (2 c \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q}\right ) \int \frac {x^2 \left (1+\frac {e x^n}{d}\right )^q}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}}-\frac {\left (2 c \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q}\right ) \int \frac {x^2 \left (1+\frac {e x^n}{d}\right )^q}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{\sqrt {b^2-4 a c}} \\ & = -\frac {2 c x^3 \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} F_1\left (\frac {3}{n};1,-q;\frac {3+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {e x^n}{d}\right )}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c x^3 \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} F_1\left (\frac {3}{n};1,-q;\frac {3+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},-\frac {e x^n}{d}\right )}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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