\(\int \frac {(d+e x^2)^q}{a+c x^4} \, dx\) [477]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{2 a}+\frac {x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{2 a} \] Output:

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

Mathematica [F]

\[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx \] Input:

Integrate[(d + e*x^2)^q/(a + c*x^4),x]
 

Output:

Integrate[(d + e*x^2)^q/(a + c*x^4), x]
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1489, 27, 334, 333}

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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx\)

\(\Big \downarrow \) 1489

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},-\frac {\sqrt {c} x^2}{\sqrt {-a}},-\frac {e x^2}{d}\right )}{2 a}+\frac {x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},1,-q,\frac {3}{2},\frac {\sqrt {c} x^2}{\sqrt {-a}},-\frac {e x^2}{d}\right )}{2 a}\)

Input:

Int[(d + e*x^2)^q/(a + c*x^4),x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1489

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{r 
= Rt[(-a)*c, 2]}, Simp[-c/(2*r)   Int[(d + e*x^2)^q/(r - c*x^2), x], x] - S 
imp[c/(2*r)   Int[(d + e*x^2)^q/(r + c*x^2), x], x]] /; FreeQ[{a, c, d, e, 
q}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]
 

Maple [F]

\[\int \frac {\left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+a}d x\]

Input:

int((e*x^2+d)^q/(c*x^4+a),x)
 

Output:

int((e*x^2+d)^q/(c*x^4+a),x)
 

Fricas [F]

\[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q}}{c x^{4} + a} \,d x } \] Input:

integrate((e*x^2+d)^q/(c*x^4+a),x, algorithm="fricas")
 

Output:

integral((e*x^2 + d)^q/(c*x^4 + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**q/(c*x**4+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q}}{c x^{4} + a} \,d x } \] Input:

integrate((e*x^2+d)^q/(c*x^4+a),x, algorithm="maxima")
 

Output:

integrate((e*x^2 + d)^q/(c*x^4 + a), x)
 

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q}}{c x^{4} + a} \,d x } \] Input:

integrate((e*x^2+d)^q/(c*x^4+a),x, algorithm="giac")
 

Output:

integrate((e*x^2 + d)^q/(c*x^4 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\int \frac {{\left (e\,x^2+d\right )}^q}{c\,x^4+a} \,d x \] Input:

int((d + e*x^2)^q/(a + c*x^4),x)
 

Output:

int((d + e*x^2)^q/(a + c*x^4), x)
 

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^q}{a+c x^4} \, dx=\int \frac {\left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+a}d x \] Input:

int((e*x^2+d)^q/(c*x^4+a),x)
 

Output:

int((d + e*x**2)**q/(a + c*x**4),x)