Integrand size = 30, antiderivative size = 368 \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {2 B c (d x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^2 (2+m)}-\frac {2 B c (d x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^2 (2+m)} \]
(d*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2*c*x^2/(b-(-4*a*c+b^2)^ (1/2)))*(C+(2*A*c-C*b)/(-4*a*c+b^2)^(1/2))/d/(1+m)/(b-(-4*a*c+b^2)^(1/2))+ 2*B*c*(d*x)^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-2*c*x^2/(b-(-4*a*c+b^2 )^(1/2)))/d^2/(2+m)/(b-(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)+(d*x)^(1+m)* hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(C+( -2*A*c+C*b)/(-4*a*c+b^2)^(1/2))/d/(1+m)/(b+(-4*a*c+b^2)^(1/2))-2*B*c*(d*x) ^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))/d ^2/(2+m)/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 2.26 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.19 \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {(d x)^m \left (A \left (2+3 m+m^2\right ) \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+B (2+m) \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x+\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}+m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+C \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x^2+m^2 x^2+2 m x \text {$\#$1}+m^2 x \text {$\#$1}+2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+3 m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m^2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m \left (\frac {x}{\text {$\#$1}}\right )^{-m} \text {$\#$1}^2}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]\right )}{2 m (1+m) (2+m)} \]
((d*x)^m*(A*(2 + 3*m + m^2)*RootSum[a + b*#1^2 + c*#1^4 & , Hypergeometric 2F1[-m, -m, 1 - m, -(#1/(x - #1))]/((x/(x - #1))^m*(b*#1 + 2*c*#1^3)) & ] + B*(2 + m)*RootSum[a + b*#1^2 + c*#1^4 & , (m*x + (Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1)/(x/(x - #1))^m + (m*Hypergeometric2F1[-m, - m, 1 - m, -(#1/(x - #1))]*#1)/(x/(x - #1))^m)/(b*#1 + 2*c*#1^3) & ] + C*Ro otSum[a + b*#1^2 + c*#1^4 & , (m*x^2 + m^2*x^2 + 2*m*x*#1 + m^2*x*#1 + (2* Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (3 *m*Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (m^2*Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^ m + (m*#1^2)/(x/#1)^m)/(b*#1 + 2*c*#1^3) & ]))/(2*m*(1 + m)*(2 + m))
Time = 0.66 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2193, 27, 1451, 27, 278, 1608, 27, 278}
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.
| \(\displaystyle \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 2193 |
| \(\displaystyle \int \frac {(d x)^m \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {\int \frac {B (d x)^{m+1}}{c x^4+b x^2+a}dx}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(d x)^m \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {B \int \frac {(d x)^{m+1}}{c x^4+b x^2+a}dx}{d}\) |
| \(\Big \downarrow \) 1451 |
| \(\displaystyle \int \frac {(d x)^m \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {B \left (\frac {c \int \frac {2 (d x)^{m+1}}{2 c x^2+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {c \int \frac {2 (d x)^{m+1}}{2 c x^2+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(d x)^m \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {B \left (\frac {2 c \int \frac {(d x)^{m+1}}{2 c x^2+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {(d x)^{m+1}}{2 c x^2+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\right )}{d}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \int \frac {(d x)^m \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\frac {B \left (\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\right )}{d}\) |
| \(\Big \downarrow \) 1608 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 A c-b C}{\sqrt {b^2-4 a c}}+C\right ) \int \frac {2 (d x)^m}{2 c x^2+b-\sqrt {b^2-4 a c}}dx+\frac {1}{2} \left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \int \frac {2 (d x)^m}{2 c x^2+b+\sqrt {b^2-4 a c}}dx+\frac {B \left (\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \left (\frac {2 A c-b C}{\sqrt {b^2-4 a c}}+C\right ) \int \frac {(d x)^m}{2 c x^2+b-\sqrt {b^2-4 a c}}dx+\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \int \frac {(d x)^m}{2 c x^2+b+\sqrt {b^2-4 a c}}dx+\frac {B \left (\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\right )}{d}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {(d x)^{m+1} \left (\frac {2 A c-b C}{\sqrt {b^2-4 a c}}+C\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(d x)^{m+1} \left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {B \left (\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+2) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\right )}{d}\) |
((C + (2*A*c - b*C)/Sqrt[b^2 - 4*a*c])*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c])*d*(1 + m)) + ((C - (2*A*c - b*C)/Sqrt[b^2 - 4*a*c])*(d*x)^(1 + m )*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4* a*c])])/((b + Sqrt[b^2 - 4*a*c])*d*(1 + m)) + (B*((2*c*(d*x)^(2 + m)*Hyper geometric2F1[1, (2 + m)/2, (4 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]) /(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*(2 + m)) - (2*c*(d*x)^(2 + m )*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4* a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*d*(2 + m))))/d
3.1.40.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Wi th[{q = Rt[b^2 - 4*a*c, 2]}, Simp[c/q Int[(d*x)^m/(b/2 - q/2 + c*x^2), x] , x] - Simp[c/q Int[(d*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, m}, x] && NeQ[b^2 - 4*a*c, 0]
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (b_.)*(x_)^2 + (c_.) *(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0]
Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_S ymbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(d*x)^m*(a + b*x^2 + c*x^4)^p, x] + Simp[1/d Int[Sum[Coe ff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q + 1)/2}]*(d*x)^(m + 1)*(a + b*x^2 + c *x^4)^p, x], x]] /; FreeQ[{a, b, c, d, m, p}, x] && PolyQ[Pq, x] && !PolyQ [Pq, x^2]
\[\int \frac {\left (d x \right )^{m} \left (C \,x^{2}+B x +A \right )}{c \,x^{4}+b \,x^{2}+a}d x\]
\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int \frac {\left (d x\right )^{m} \left (A + B x + C x^{2}\right )}{a + b x^{2} + c x^{4}}\, dx \]
\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int \frac {{\left (d\,x\right )}^m\,\left (C\,x^2+B\,x+A\right )}{c\,x^4+b\,x^2+a} \,d x \]