\(\int (d x)^{3/2} \sqrt {a+b x^2+c x^4} \, dx\) [1067]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int (d x)^{3/2} \sqrt {a+b x^2+c x^4} \, dx=\frac {2 (d x)^{5/2} \sqrt {a+b x^2+c x^4} \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},-\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 d \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(365\) vs. \(2(147)=294\).

Time = 11.39 (sec) , antiderivative size = 365, normalized size of antiderivative = 2.48 \[ \int (d x)^{3/2} \sqrt {a+b x^2+c x^4} \, dx=\frac {2 d \sqrt {d x} \left (5 \left (2 b+5 c x^2\right ) \left (a+b x^2+c x^4\right )-10 a b \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},\frac {1}{2},\frac {5}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+2 \left (-3 b^2+10 a c\right ) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},\frac {1}{2},\frac {9}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{225 c \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

(2*d*Sqrt[d*x]*(5*(2*b + 5*c*x^2)*(a + b*x^2 + c*x^4) - 10*a*b*Sqrt[(b - S 
qrt[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 + Sqrt[b^2 - 4*a*c])] + 2* 
(-3*b^2 + 10*a*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])]))/(225*c*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 147, 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.

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

\(\Big \downarrow \) 1461

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d x)^{3/2} \sqrt {a+b x^2+c x^4} \, dx=\frac {2 \sqrt {d}\, d \left (4 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, a +3 \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{2}-10 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{3}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a c +3 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{3}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b^{2}-2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{5}+b \,x^{3}+a x}d x \right ) a^{2}\right )}{27 b} \] Input:

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

Output:

(2*sqrt(d)*d*(4*sqrt(x)*sqrt(a + b*x**2 + c*x**4)*a + 3*sqrt(x)*sqrt(a + b 
*x**2 + c*x**4)*b*x**2 - 10*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4)*x**3)/( 
a + b*x**2 + c*x**4),x)*a*c + 3*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4)*x** 
3)/(a + b*x**2 + c*x**4),x)*b**2 - 2*int((sqrt(x)*sqrt(a + b*x**2 + c*x**4 
))/(a*x + b*x**3 + c*x**5),x)*a**2))/(27*b)