3.3.18 \(\int \frac {d+e x^2}{\sqrt {f x} (a+b x^2+c x^4)^{3/2}} \, dx\) [218]

3.3.18.1 Optimal result

Integrand size = 31, antiderivative size = 301 \[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {2 d \sqrt {f x} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{2},\frac {3}{2},\frac {5}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{5/2} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},\frac {3}{2},\frac {9}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 a f^3 \sqrt {a+b x^2+c x^4}} \]

output

2/5*e*(f*x)^(5/2)*AppellF1(5/4,3/2,3/2,9/4,-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)/a/f^3/(c*x^4+b*x^2+a)^(1/2)+2*d* 
AppellF1(1/4,3/2,3/2,5/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a 
*c+b^2)^(1/2)))*(f*x)^(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)/a/f/(c*x^4+b*x^2+a)^(1/2)
 

3.3.18.2 Mathematica [A] (verified)

Time = 11.57 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.31 \[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {x \left (-5 b^2 d+5 b \left (a e-c d x^2\right )+10 a c \left (d+e x^2\right )-5 \left (b^2 d-6 a c d+a b e\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+c (b d-2 a e) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{5 a \left (-b^2+4 a c\right ) \sqrt {f x} \sqrt {a+b x^2+c x^4}} \]

input

Integrate[(d + e*x^2)/(Sqrt[f*x]*(a + b*x^2 + c*x^4)^(3/2)),x]
 

output

(x*(-5*b^2*d + 5*b*(a*e - c*d*x^2) + 10*a*c*(d + e*x^2) - 5*(b^2*d - 6*a*c 
*d + a*b*e)*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])]*AppellF1 
[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + S 
qrt[b^2 - 4*a*c])] + c*(b*d - 2*a*e)*x^2*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])]*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b + Sqrt[b^2 
 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(5*a*(-b^2 + 4*a*c)*Sqrt 
[f*x]*Sqrt[a + b*x^2 + c*x^4])
 

3.3.18.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, 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.

\(\displaystyle \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1674

\(\displaystyle \int \left (\frac {d}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}}+\frac {e (f x)^{3/2}}{f^2 \left (a+b x^2+c x^4\right )^{3/2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 d \sqrt {f x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{2},\frac {3}{2},\frac {5}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{a f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{5/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},\frac {3}{2},\frac {9}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 a f^3 \sqrt {a+b x^2+c x^4}}\)

input

Int[(d + e*x^2)/(Sqrt[f*x]*(a + b*x^2 + c*x^4)^(3/2)),x]
 

output

(2*d*Sqrt[f*x]*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[1/4, 3/2, 3/2, 5/4, (-2*c*x^2)/(b - 
Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(a*f*Sqrt[a + b*x 
^2 + c*x^4]) + (2*e*(f*x)^(5/2)*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[5/4, 3/2, 3/2, 9/4, 
 (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/ 
(5*a*f^3*Sqrt[a + b*x^2 + c*x^4])
 

3.3.18.3.1 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]
 

3.3.18.4 Maple [F]

\[\int \frac {e \,x^{2}+d}{\sqrt {f x}\, \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}d x\]

input

int((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x)
 

output

int((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x)
 

3.3.18.5 Fricas [F]

\[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {f x}} \,d x } \]

input

integrate((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas 
")
 

output

integral(sqrt(c*x^4 + b*x^2 + a)*(e*x^2 + d)*sqrt(f*x)/(c^2*f*x^9 + 2*b*c* 
f*x^7 + (b^2 + 2*a*c)*f*x^5 + 2*a*b*f*x^3 + a^2*f*x), x)
 

3.3.18.6 Sympy [F]

\[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {d + e x^{2}}{\sqrt {f x} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]

input

integrate((e*x**2+d)/(f*x)**(1/2)/(c*x**4+b*x**2+a)**(3/2),x)
 

output

Integral((d + e*x**2)/(sqrt(f*x)*(a + b*x**2 + c*x**4)**(3/2)), x)
 

3.3.18.7 Maxima [F]

\[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {f x}} \,d x } \]

input

integrate((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima 
")
 

output

integrate((e*x^2 + d)/((c*x^4 + b*x^2 + a)^(3/2)*sqrt(f*x)), x)
 

3.3.18.8 Giac [F]

\[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {f x}} \,d x } \]

input

integrate((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")
 

output

integrate((e*x^2 + d)/((c*x^4 + b*x^2 + a)^(3/2)*sqrt(f*x)), x)
 

3.3.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x^2}{\sqrt {f x} \left (a+b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {e\,x^2+d}{\sqrt {f\,x}\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \]

input

int((d + e*x^2)/((f*x)^(1/2)*(a + b*x^2 + c*x^4)^(3/2)),x)
 

output

int((d + e*x^2)/((f*x)^(1/2)*(a + b*x^2 + c*x^4)^(3/2)), x)