Integrand size = 27, antiderivative size = 339 \[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b+\sqrt {b^2-4 a c}\right )}-\frac {b x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {e x^2}{d}\right )}{3 c} \]
-b*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/2],-e*x^2/d)/c^2/((1+e*x^2/d)^q)+1 /3*x^3*(e*x^2+d)^q*hypergeom([3/2, -q],[5/2],-e*x^2/d)/c/((1+e*x^2/d)^q)+x *(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-e*x^2/ d)*(b^2-a*c-b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c^2/((1+e*x^2/d)^q)/(b-(-4* a*c+b^2)^(1/2))+x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-2*c*x^2/(b+(-4*a*c+b^ 2)^(1/2)),-e*x^2/d)*(b^2-a*c+b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c^2/((1+e* x^2/d)^q)/(b+(-4*a*c+b^2)^(1/2))
\[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx \]
Time = 0.66 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1626, 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.
\(\displaystyle \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1626 |
\(\displaystyle \int \left (\frac {\left (x^2 \left (b^2-a c\right )+a b\right ) \left (d+e x^2\right )^q}{c^2 \left (a+b x^2+c x^4\right )}-\frac {b \left (d+e x^2\right )^q}{c^2}+\frac {x^2 \left (d+e x^2\right )^q}{c}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},-\frac {e x^2}{d}\right )}{c^2 \left (\sqrt {b^2-4 a c}+b\right )}-\frac {b x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{c^2}+\frac {x^3 \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {e x^2}{d}\right )}{3 c}\) |
((b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^q*AppellF 1[1/2, 1, -q, 3/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), -((e*x^2)/d)])/(c^2 *(b - Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q) + ((b^2 - a*c + (b*(b^2 - 3*a* c))/Sqrt[b^2 - 4*a*c])*x*(d + e*x^2)^q*AppellF1[1/2, 1, -q, 3/2, (-2*c*x^2 )/(b + Sqrt[b^2 - 4*a*c]), -((e*x^2)/d)])/(c^2*(b + Sqrt[b^2 - 4*a*c])*(1 + (e*x^2)/d)^q) - (b*x*(d + e*x^2)^q*Hypergeometric2F1[1/2, -q, 3/2, -((e* x^2)/d)])/(c^2*(1 + (e*x^2)/d)^q) + (x^3*(d + e*x^2)^q*Hypergeometric2F1[3 /2, -q, 5/2, -((e*x^2)/d)])/(3*c*(1 + (e*x^2)/d)^q)
3.5.7.3.1 Defintions of rubi rules used
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q, (f*x)^m/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[b^2 - 4*a *c, 0] && !IntegerQ[q] && IntegerQ[m]
\[\int \frac {x^{6} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}d x\]
\[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{6}}{c x^{4} + b x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
\[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{6}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{6}}{c x^{4} + b x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^6 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int \frac {x^6\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \]