3.3.27 \(\int \frac {(f x)^m (d+e x^2)}{\sqrt {a+b x^2+c x^4}} \, dx\) [227]

3.3.27.1 Optimal result

Integrand size = 29, antiderivative size = 317 \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {d (f x)^{1+m} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},\frac {1}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (1+m) \sqrt {a+b x^2+c x^4}}+\frac {e (f x)^{3+m} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},\frac {1}{2},\frac {5+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (3+m) \sqrt {a+b x^2+c x^4}} \]

output

d*(f*x)^(1+m)*AppellF1(1/2+1/2*m,1/2,1/2,3/2+1/2*m,-2*c*x^2/(b-(-4*a*c+b^2 
)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2) 
))^(1/2)*(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/f/(1+m)/(c*x^4+b*x^2+a)^ 
(1/2)+e*(f*x)^(3+m)*AppellF1(3/2+1/2*m,1/2,1/2,5/2+1/2*m,-2*c*x^2/(b-(-4*a 
*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^2/(b-(-4*a*c+b^2) 
^(1/2)))^(1/2)*(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/f^3/(3+m)/(c*x^4+b 
*x^2+a)^(1/2)
 

3.3.27.2 Mathematica [A] (verified)

Time = 2.83 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.84 \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {x (f x)^m \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \left (d (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},\frac {1}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+e (1+m) x^2 \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},\frac {1}{2},\frac {5+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{(1+m) (3+m) \sqrt {a+b x^2+c x^4}} \]

input

Integrate[((f*x)^m*(d + e*x^2))/Sqrt[a + b*x^2 + c*x^4],x]
 

output

(x*(f*x)^m*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])] 
*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*(d*(3 + m 
)*AppellF1[(1 + m)/2, 1/2, 1/2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a* 
c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^2*AppellF1[(3 + m)/ 
2, 1/2, 1/2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b 
+ Sqrt[b^2 - 4*a*c])]))/((1 + m)*(3 + m)*Sqrt[a + b*x^2 + c*x^4])
 

3.3.27.3 Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1674, 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.

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

\(\Big \downarrow \) 1674

\(\displaystyle \int \left (\frac {d (f x)^m}{\sqrt {a+b x^2+c x^4}}+\frac {e (f x)^{m+2}}{f^2 \sqrt {a+b x^2+c x^4}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d (f x)^{m+1} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {m+1}{2},\frac {1}{2},\frac {1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \sqrt {a+b x^2+c x^4}}+\frac {e (f x)^{m+3} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {m+3}{2},\frac {1}{2},\frac {1}{2},\frac {m+5}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (m+3) \sqrt {a+b x^2+c x^4}}\)

input

Int[((f*x)^m*(d + e*x^2))/Sqrt[a + b*x^2 + c*x^4],x]
 

output

(d*(f*x)^(1 + m)*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c 
*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[(1 + m)/2, 1/2, 1/2, (3 + m)/2, (- 
2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f* 
(1 + m)*Sqrt[a + b*x^2 + c*x^4]) + (e*(f*x)^(3 + m)*Sqrt[1 + (2*c*x^2)/(b 
- Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1 
[(3 + m)/2, 1/2, 1/2, (5 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c 
*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f^3*(3 + m)*Sqrt[a + b*x^2 + c*x^4])
 

3.3.27.3.1 Defintions of rubi rules used

rule 1674

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && N 
eQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

3.3.27.4 Maple [F]

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

input

int((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x)
 

output

int((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x)
 

3.3.27.5 Fricas [F]

\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \]

input

integrate((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
 

output

integral((e*x^2 + d)*(f*x)^m/sqrt(c*x^4 + b*x^2 + a), x)
 

3.3.27.6 Sympy [F]

\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {\left (f x\right )^{m} \left (d + e x^{2}\right )}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \]

input

integrate((f*x)**m*(e*x**2+d)/(c*x**4+b*x**2+a)**(1/2),x)
 

output

Integral((f*x)**m*(d + e*x**2)/sqrt(a + b*x**2 + c*x**4), x)
 

3.3.27.7 Maxima [F]

\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \]

input

integrate((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")
 

output

integrate((e*x^2 + d)*(f*x)^m/sqrt(c*x^4 + b*x^2 + a), x)
 

3.3.27.8 Giac [F]

\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \]

input

integrate((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
 

output

integrate((e*x^2 + d)*(f*x)^m/sqrt(c*x^4 + b*x^2 + a), x)
 

3.3.27.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (e\,x^2+d\right )}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

input

int(((f*x)^m*(d + e*x^2))/(a + b*x^2 + c*x^4)^(1/2),x)
 

output

int(((f*x)^m*(d + e*x^2))/(a + b*x^2 + c*x^4)^(1/2), x)