\(\int \frac {(d x)^{3/2}}{(a+b x^2+c x^4)^{3/2}} \, dx\) [1079]

Optimal result

Integrand size = 24, antiderivative size = 150 \[ \int \frac {(d x)^{3/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {2 (d x)^{5/2} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},\frac {3}{2},\frac {9}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 a d \sqrt {a+b x^2+c x^4}} \] Output:

2/5*(d*x)^(5/2)*(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)*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)))/a/d/(c*x^4+b*x^2+a)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(348\) vs. \(2(150)=300\).

Time = 11.32 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.32 \[ \int \frac {(d x)^{3/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\frac {d \sqrt {d x} \left (-5 \left (b+2 c x^2\right )+5 b \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+2 c x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{5 \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

(d*Sqrt[d*x]*(-5*(b + 2*c*x^2) + 5*b*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 + Sqr 
t[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 + Sqrt[b^2 - 4*a*c])] + 2*c*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*(b^2 - 4 
*a*c)*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1461, 394}

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.

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

\(\Big \downarrow \) 1461

\(\displaystyle \frac {\sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \int \frac {(d x)^{3/2}}{\left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}}dx}{a \sqrt {a+b x^2+c x^4}}\)

\(\Big \downarrow \) 394

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

Input:

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

Output:

(2*(d*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*d*Sqrt[a + b 
*x^2 + c*x^4])
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1461

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^2 + c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + R 
t[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2])))^F 
racPart[p]))   Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2 
*c*(x^2/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(c*x^4 + b*x^2 + a)*sqrt(d*x)*d*x/(c^2*x^8 + 2*b*c*x^6 + (b^2 
 + 2*a*c)*x^4 + 2*a*b*x^2 + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d x)^{3/2}}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx=\sqrt {d}\, \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x}{c^{2} x^{8}+2 b c \,x^{6}+2 a c \,x^{4}+b^{2} x^{4}+2 a b \,x^{2}+a^{2}}d x \right ) d \] Input:

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

Output:

sqrt(d)*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4)*x)/(a**2 + 2*a*b*x**2 + 2*a 
*c*x**4 + b**2*x**4 + 2*b*c*x**6 + c**2*x**8),x)*d